SeriesChap2 PDF

2.
I NFINITE S ERIES
2.1. A P RE -R EQUISITE : S EQUENCES

We concluded the last section by asking what we would get if we considered the “Taylor
polynomial of degree for the function ex centered at 0”,
x2 x3
1 x
2! 3!
As we said at the time, we have a lot of groundwork to consider first, such as the funda-
mental question of what it even means to add an infinite list of numbers together. As we
will see in the next section, this is a delicate question. In order to put our explorations on
solid ground, we begin by studying sequences.
A sequence is just an ordered list of objects. Our sequences are (almost) always lists
of real numbers, so another definition for us would be that a sequence is a real-valued
function whose domain is the positive integers. The sequence whose nth term is an is
denoted an , or if there might be confusion otherwise, an n 1 , which indicates that the
sequence starts when n 1 and continues forever.
Sequences are specified in several different ways. Perhaps the simplest way is to spec-
ify the first few terms, for example
an 2, 4, 6, 8, 10, 12, 14, . . . ,
is a perfectly clear definition of the sequence of positive even integers. This method is
slightly less clear when
an 2, 3, 5, 7, 11, 13, 17, . . . ,
although with a bit of imagination, one can deduce that an is the nth prime number (for
technical reasons, 1 is not considered to be a prime number). Of course, this method com-
pletely breaks down when the sequence has no discernible pattern, such as
an 0, 4, 3, 2, 11, 29, 54, 59, 35, 41, 46, . . . .
To repeat: the above is not a good definition of a series. Indeed, this sequence is the number
of home runs that Babe Ruth hit from 1914 onward; if the Red Sox had been able to predict
the pattern, would they have sold his contract to the Yankees after the 1919 season?
19
20 C HAPTER 2 I NFINITE S ERIES
Another method of specifying a sequence is by giving a formula for the general (nth)
term. For example,
an 2n
is another definition of the positive even integers, while an n2 defines the sequence
an 1, 4, 9, 16, 25, . . . of squares.
Example 1. Guess a formula for the general term of the sequence
3 5 7 9 11
, , , , ,... .
1 4 9 16 25
Solution. It is good to tackle this problem one piece at a time. First, notice that the
sequence alternates in sign. Since the sequence begins with a positive term, this shows that
we should have a factor of 1 n 1 in the formula for an . Next, the numerators of these
fractions list all the odd numbers starting with 3, so we guess 2n 1 for the numerators.
The denominators seem to be the squares, so we guess n2 for these. Putting this together
we have an 1 n 1 2n 1 n2 .
A sequence grows geometrically (or exponentially) if each term is obtained by multiplying

the previous term by a fixed common ratio, typically denoted by r. Therefore, letting a
denote the first term of a geometric sequence, the sequence can be defined as ar n n 0 ; note
here that the subscripted n 0 indicates that the sequence starts with n 0. Geometric
sequences are also called “geometric progressions”.
There is a legendary (but probably fictitious) myth involving a geometric sequence. It
is said that when the inventor of chess (an ancient Indian mathematician, in most accounts)
showed his invention to his ruler, the ruler was so pleased that he gave the inventor the
right to name his prize. The inventor asked for 1 grain of wheat for the first square of the
board, 2 grains for the second square, 4 grains for the third square, 8 grains for the fourth
square, and so on. The ruler, although initially offended that the inventor asked for so
little, agreed to the offer. Days later, the ruler asked his treasurer why it was taking so long
to pay the inventor, and the treasurer pointed out that to pay for the 64th square alone, it
would take
264 18, 446, 744, 073, 709, 551, 616
grains of wheat. To put this in perspective, a single grain of wheat contains about 1 5 of a
calorie, so 264 grains of wheat contain approximately 3, 719, 465, 121, 917, 178, 880 calories.
Assuming a 2000 calorie per day diet, the amount of wheat just for the 64th square of the
chessboard would feed 6 billion people for almost 850 years (although they should proba-
bly supplement their diet with Vitamin C to prevent scurvy). The legend concludes with
Wolfram Alpha (http://www.wolframalpha.com/) is an incredible resource for performing calculations

such as this; simply type “2ˆ64 grains of wheat” into its search box and wonder at the results.
S ECTION 2.1 A P RE -R EQUISITE : S EQUENCES 21
the ruler insisting that the inventor participate in the grain counting, in order to make sure
that it is “accurate,” an offer which the inventor presumably refused.
While sequences are interesting in their own right, we are mostly interested in apply-
ing tools for sequences to our study of infinite sums. Therefore, the two most important
questions about sequences for us are:
Does the sequence converge or diverge?

If the sequence converges, what does it converge to?
Intuitively, the notion of convergence is often quite clear. For example, the sequence
0.3, 0.33, 0.333, 0.3333, . . .
converges to 1 3, while the sequence
1n 1 1, 1 2, 1 3, 1 4, . . . ,
converges to 0. Slightly more formally, the sequence an converges to the number L if

by taking n large enough, we can make the terms of the sequence as close to L as we
like. By being a bit more specific in this description, we arrive at the formal definition of
convergence below.
Converges to L. The sequence an is said to converge to L if for

every number ! 0, there is some number N so that an L ! for
all n N .
To indicate that the sequence an converges to L, then we may write
lim an L,
n
or simply
an L as n .
Before moving on, we check one of our previous observations using the formal defini-
tion of convergence.
Example 2. Show, using the definition, that the sequence 1n converges to 0.
Solution. Let ! 0 be any positive number. We want to show that
an 0 1n !
for sufficiently large n. Solving the above inequality for n, we see that an 0 ! for all
n 1 !, proving that 1 n 0 as n .
In practice, we rarely use the formal definition of convergence for examples such as
this. After all, the numerator of 1 n is constant and the denominator increases without
bound, so it is clear that the limit is 0. Many types of limits can be computed with this
reasoning and a few techniques, as we show in the next two examples.
7n 3
Example 3. Compute lim .
n 5n n
Solution. A common technique with limits is to divide both the numerator and denom-
inator by the fastest growing function of n involved in the expression. In this case the
fastest growing function of n involved is n (choosing 7n would work just as well), so we
divide by n:
7n 3 n 7 3n 7 3n
lim lim lim .
n 5n n n n 5 nn n 5 1 n
As n ,3n 0 and 1 n 0, so the limit of this sequence is 7 5.
Example 4. Determine the limit as n of the sequence n 2 n .
Solution. Here we use another frequent technique: when dealing with square roots, it is
often helpful to multiply by the “conjugate”:
n 2 n 2
n 2 n n 2 n .
n 2 n n 2 n
Now we can analyze this fraction instead. The numerator is constant and the denominator
grows without bound, so the sequence converges to 0.
You should convince yourself that a given sequence can converge to at most one num-
ber. If the sequence an does not converge to any number, then we say that it diverges.
There are two different types of divergence, and it is important to distinguish them. First
we have the type of divergence exhibited by the sequence n! :
Diverges to Infinity. The sequence an is said to diverge to infinity

if for every number " 0, there is some number N so that an "
for all n N .
If the sequence an diverges to infinity, then we write lim an . Note that the
n
sequence 1n 1, 1, 1, 1, . . . demonstrates a different type of divergence, which
is sometimes referred to as “oscillatory divergence.” Both types of divergence show up in
our next example.
Example 5. Showing, using only the definition of convergence, that:

(a) lim r n if r 1,
n
(b) lim r n 1 if r 1,
n
(b) lim r n 0 if 1 r 1 (i.e., if r 1),

n
(c) lim r n does not exist if r 1.

n
Solution. Beginning with (a), assume that r 1 and fix a number M 1. By taking
logarithms, we see that r n M if and only if ln r n ln M , or equivalently, if and only if
n ln r ln M.
Since both M and r are greater than 1, ln r, ln M 0. This shows that r n ln M for all
n ln M ln r, proving that ln r n when r 1.
Part (b) is obvious, as 1n 1 for all n.
For part (c), let us assume that r 1. We want to prove that lim r n 0, which means
n
(according to our definition above) that for every ! 0, there is some number N so that
rn 0 rn r n
!
for all n N . Again taking logarithms, this holds if n ln r ln !. Since r 1, its

logarithm is negative, so when we divide both sides of this inequality by ln r we flip the
inequality. This r n ! for all n ln ! ln r , proving that lim r n 0 when r 1.
n
This leaves us with (d). We have already observed that the sequence 1 n diverges,
so suppose that r 1. In this case the sequence r n can be viewed as a “shuffle” of two
sequences, the negative sequence r, r 3 , r 5 , . . . and the positive sequence r 2 , r 4 , r 6 , . . . .
By part (a), r 2n r2 n as n , while r 2n 1 r r 2n as n , so in this
n
case the sequence r does not have a limit.
Another way to specify a sequence is with initial conditions and a recurrence. For
example, the factorials can be specified by the recurrence
an n an 1 for n 2,
and the initial condition a1 1.

If the inventor of chess had been extremely greedy, he would have asked for 1! grains of wheat for the first
square, 2! grains of wheat for the second square, 3! grains of wheat for the third square, and so on, because
64! 126, 886, 932, 185, 884, 164, 103, 433, 389, 335, 161, 480, 802, 865, 516,
174, 545, 192, 198, 801, 894, 375, 214, 704, 230, 400, 000, 000, 000, 000,
almost 127 octovigintillion, and about a billion times the estimated number of atoms in the universe.
A more complicated recurrence relation is provided by the Fibonacci numbers fn n 0,

defined by
fn fn 1 fn 2 for n 2,
and the initial conditions f0 f1 1. This sequence begins
fn n 0 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, . . . .
We now make an important definition. In this definition, note that we twist the notion
of “increasing” a bit; what we call increasing sequences should really be called “nonde-
creasing sequences”, but this awkward term is rarely used.
Monotone Sequences. The sequence an is (weakly) increasing if

an an 1 for all n, and (weakly) decreasing if an an 1 for all n. A
sequence is monotone if it is either increasing or decreasing.
It is good to practice identifying monotone sequences. Here are a few examples to

practice on:
• 1n is decreasing,
• 1 1n is increasing,
n
• is decreasing.
n2 1
You should also verify that the geometric sequence ar n is decreasing for 0 r 1,
increasing for r 1, and not monotone if r is negative.
It is frequently helpful to know that a given sequence converges, even if we do not
know its limit. We have a powerful tool to establish this for monotone sequences. First,
we need another definition.
Bounded Sequences. The sequence an is bounded if there are

numbers " and u such that " an u for all n.
For monotone sequences, boundedness implies convergence:
The Monotone Convergence Theorem. Every bounded mono-

tone sequence converges.
We delay the proof of this theorem to Exercise 92. Knowing that a limit exists can
sometimes be enough to solve for its true value, as our next example demonstrates.
Example 6. Prove that the sequence defined recursively by a1 2 and

2an
an 1
1 an
for n 1 is decreasing, and use the Monotone Convergence Theorem to compute its limit.
Solution. First we ask: for which values of n is an an 1? Substituting the definition of

an 1 , this inequality becomes
2an
an ,
1 an
which simplifies to a2n an . So we have our answer: an an 1 whenever an 1. We are
given that a1 2, so a1 a2 . For the other values of n, notice that if an 1, then
2an an an 1 an
an 1 ,
1 an 1 an 1 an
so since a1 1, we see that a2 1, which implies that a3 1, and so on. In the end, we
may conclude that an 1 for all values of n. By our work above, this shows that an an 1
for all values of n, so the sequence is decreasing.
When dealing with monotone sequences, a common technique is to first prove that the
sequence has a limit, and then use this fact to find the limit. In order to prove that an has
a limit, we need to show that it is bounded. From our previous work, we know that an 1
for all n, and an upper bound is also easy:
2an 2an
an 1 2.
1 an an
Therefore, 1 an 2 for all n, so the sequence an has a limit.
Let L denote this limit. Then we have
2an
an 1 L
1 an
as n . But as n , an as well, so we must have
2L
L,
1 L
which simplifies to L2 L. There are two possible solutions, L 0 and L 1, but we
can rule out L 0 because it lies outside of our bounds, so it must be the case that this
sequence converges to 1.
For the rest of the section, we study some other methods for computing limits. One
technique to find the limit of the sequence an is to “sandwich” it between a lower bound
"n and an upper bound un .
The Sandwich Theorem. Suppose the sequences an , "n , and

un satisfy "n an un for all large n and "n L and un L
then an L as well.
The Sandwich Theorem hopefully seems intuitively obvious. We ask the reader to
give a formal proof in Exercise 94. Examples 7 and 8 illustrate its use.
sin n 1
n n
Solution. For all n we have
1 sin n 1
,
n n n
sin n
so since 1n 0 and 1 n 0, 0 by the Sand-
n
wich Theorem.
n! 1
Example 8. Show that lim 0.
n nn
Solution. Evaluating this limit merely requires find-
ing the right bound:
1 2 3 4 5 6 7 8 9 10
n! n n 1 2 1 1
,
nn n n n n n
so n! nn is sandwiched between 0 and 1 n. Since 1 n
0 and 0 is 0, we see by the Sandwich Theorem that n! nn 0.
The following useful result allows us to switch from limits of sequences to limits of
functions.
Theorem. If an is a function satisfying an f n and lim f x

x
exists, then lim an lim f n .
n x
Proof. If lim f x L then for each ! 0 there is a number N such that f x L ! whenever
n
x N . Of course, this means that an L ! whenever n N , proving the theorem.
In 1942, Arther Stone and John Tukey proved a theorem called the Ham Sandwich Theorem, which states
that given any sandwich composed of bread, ham, and cheese, there is some plane (i.e., straight cut) that slices
the sandwich into two pieces, each containing the same amount of bread, the same amount of ham, and the
same amount of cheese.
Why might we want to switch from sequences to functions? Generally we do so in

order to invoke l’Hôpital’s Rule :
l’Hôpital’s Rule. Suppose that c is a real number or c , and

that f and g are differentiable and g x 0 near c (except possibly
at x c). If either
lim f x lim g x 0
x c x c
or
lim f x lim g x
x c x c
then
f x f x
lim lim ,
x c g x x c g x
provided that this limit exists.
ln n 1
n n
Solution. We know that
ln n ln x
lim lim ,
n n x x 1 2 3 4 5 6 7 8 9 10
and we can use l’Hôpital’s Rule to evaluate this sec-
ond limit since both ln x and x tend to as x :
d 1
ln x ln x 1
lim lim dx lim x lim 0.
x x x d x 1 x x
x
dx
ln n
This shows that lim 0.
n n
Another very useful result allows us to “move” limits inside continuous functions:
The Continuous Function Limits Theorem. Suppose the se-

quence an converges to L and that f is continuous at L and
defined for all values an . Then the sequence f an converges to
f L.
l’Hôpital’s Rule is one of the many misnomers in mathematics. It is named after Guillaume de l’Hôpital
(1661–1704) because it appeared in a calculus book he authored (the first calculus book ever written, in fact),
but l’Hôpital’s Rule was actually discovered by his mathematics tutor, Johann Bernoulli (1667–1748). The two
had a contract entitling l’Hôpital to use Bernoulli’s discoveries however he wished.
We conclude the section with two examples using this theorem.
π 1
Example 10. Compute lim sin .
n 2 n
Solution. Since 1 n 0 as n ,
π 1 π
lim .
n 2 n 2
Therefore,
π 1 π
lim sin sin 1,
n 2 n 2
by the Continuous Function Limits Theorem.
Our final example is less straight-forward, but more important.

n
1
Example 11. Compute lim 1 .
n n
Solution. As with most problems which have a variable in the exponent, it is a good idea
to rewrite the limit using e and ln:
n
1 1 n 1
lim 1 lim eln 1 n lim en ln 1 n
n n n n
1
Consider only the exponent for now, n ln 1 n . This is an 0 indeterminate form, so
we rewrite it to give a 00 form:
1
1 ln 1 n
n ln 1 1 .
n n
Applying l’Hôpital’s Rule, we see that

1 1
ln 1 n n2
lim 1 lim 1 1.
n n
n n2
Thus the exponents converge to 1. Because ex is a continuous function, we can now apply
the Continuous Function Limits Theorem to see that
1 1
n ln 1 lim n ln 1
lim e n n
e n e1 e,
n
so the solution to the example is e.

A wonderful resource for integer sequences is the On-Line Encyclopedia of Integer Se-
quences, maintained by Neil Sloane (1939–), available at
http://www.research.att.com/~njas/sequences/
The encyclopedia currently contains more than 165,000 sequences. Each of these sequences
is numbered, and so the sequence also contains a sequence an (number 91,967) in which
an is the nth term of the nth sequence. What is a91967 ?
E XERCISES FOR S ECTION 2.1

Write down a formula for the general term, an , of Determine if the sequences in Exercises 15–22 con-
each of the sequences in Exercises 1–8. verge or diverge. If they converge, find their limits.
1. 3, 5, 7, 9, 11, . . . 7n 5
15. an
4n 2
2. 4, 16, 64, 256, 1024, . . .
7n 5
16. an
3. 2, 4, 8, 16, 32, . . . 4n 2
4. 5 2, 7 5, 9 10, 11 17, 13 26, . . . 7n3 5

17. an
4n3 2
5. 3, 1, 3, 1, 3, . . .
74n 5
18. an
6. 4 3, 7 9, 10 27, 13 81, 16 243, . . . 44n 2
7. 1 2, 2 4, 6 8, 24 16, 120 32, . . . 19. an 2n 2 2n
8. 2, 4 2, 6 4 2, 8 6 4 2, 10 8 6 4 2, . . . (Try to 20. an 2n 2 2n
find something simpler than 2n 2n 2 4 2).
21. an 2 arctan 3n2
22. an 2 arctan n 10
Exercises 9–12 give initial terms and recurrence re-
lations for sequences. Use these to compute the first
5 terms and try to write a formula for the general
term, an . Compute the limits in Exercises 23–30.
9. a1 2, an an 1 3 7n 5
23. lim
n 4n 2
10. a1 2, an nan 1
n 2!
24. lim 2
11. a1 1, an an 1 n n n! 3 5n
5n
12. a1 1, an n3 an 1 25. lim
n ln 2 3en
13. Suppose that an is a geometric sequence. If n 2!
a2 6 and a5 162, what are the possibilities for 26. lim
n n2 n!
a1 ?
ln n
14. Suppose that an is a geometric sequence. If 27. lim
n ln 3n
a2 2 and a4 6, what are the possibilities for a1 ?
28. lim e1 n
n
n
29. lim 2n 43. Arrange the functions
n
2
30. lim 7n 3 5 n n, nn , ln n, 3n , n ln n, 2n , n6 1
n
in increasing order, so that (for large n) each func-
tion is very much larger than the one that it follows.
Prove the limit laws states in Exercises 31–34.

44. Where does n! fit in the list you made for Exer-
31. If an and bn are convergent sequences,
cise 43?
lim an bn lim an lim bn . 45. Define the sequence an recursively by
n n n
2an
an 1 .
32. If an and bn are convergent sequences, 1 an
Show that if a1 1 2 then an is increasing. (Com-
lim an bn lim an lim bn . pare this to the result of Example 6).
n n n
46. Define the sequence an recursively by

33. If an and bn are convergent sequences, 3 3an
an 1 .
3 an
lim an bn lim an lim bn .
n n n Show that if a1 1 then an is increasing, while if
a1 2 then an is decreasing.
34. If an is a convergent sequence, then
lim can c lim an

n n In Exercises 47–50, compute the integral to give
a simplified formula for an and then determine
for any number c.
lim an .
n
n
1
47. an dx
1 x
Compute the limits in Exercises 35–42. All of these n
1
limits are either 0 or infinity, explanation is re- 48. an dx
quired, but you need not apply l’Hôpital’s Rule. 1 x2
n
nn 1
35. lim 49. an dx
n n! 1 x ln x
n
n! 1
36. lim n 50. an 2
dx
n 4 1 x ln x
n
n1 4 2
37. lim 51. What is lim 1 ? More generally, what
n ln n 4 n n
x n
2n 1 10 is lim 1 ?
38. lim n n
n en 52. Let sn denote the sequence given by 1 2, 1 2
n 1 4, 1 2 14 1 8, 1 2 14 18 1 16, . . . . Conjecture
39. lim
n n 3 ln n a formula for sn . What does this mean for lim sn ?
n
n3 2n
40. lim
n n7 2n6
n 2
41. lim Recall that the function f x is periodic with period
n ln n 10 p if f x p f x for all values of x. Similarly,
n
3n2 n the sequence an n 1 is periodic with period p if
42. lim an p an for all n 1 and p is the least number
n 4n2 1
with this property. Determine the periods of the se-

quences in Exercises 53–56.
53. sin nπ n 1
nπ
54. cos
m n 1 65. Compute a4 and a5 . Based on the first 5 terms
nπ nπ of the sequence, conjecture a formula for the general
55. cos sin
m m n 1 term an .
2 nπ
m
56. sin
m n 1
Compute the sequence f 0 , f 0 , f 0 , f 3 0 , . . .

of derivatives for the functions f x listed in Exer-
66. Compute a6 . Does this match your conjecture?
cises 57–63.
57. f x ex
58. f x sin x
59. f x cos x
60. f x sin 2x
61. f x xex
3
62. f x 1 x
63. f x 1 x 67. (Due to Solomon Golomb (1932–)) There is a
64. Consider the sequence of figures below. unique sequence an of positive integers which is
nondecreasing and contains exactly an occurrences
of the number n for each n. Compute the first 10
terms of this sequence.
68. Define the sequence an recursively by
an 2 if an is even,
an 1
3an 1 if an is odd.
Compute this sequence when a1 13 and a1 24.

The 3n 1 Conjecture, first posed by Lothar Collatz
(1910–1990) in 1937 but still unproved, states that if
Let an denote the number of non-overlapping small
you start with any positive integer as a1 then this
squares in these figures, so a1 1, a2 5, a3 13,
sequence will eventually reach 1, where it will end
and a4 25. Write a formula for an . Hint: it may
in the infinite periodic sequence 1, 4, 2, 1, 4, 2, . . . .
be helpful to consider the squares that are missing
About this conjecture, the prolific Hungarian math-
from the figures.
ematician Paul Erdős (1913–1996) said “mathemat-
ics is not yet ready for such problems.”
69. Compute the first 10 terms of the sequence

Exercises 65 and 66 concern the following sequence. an n2 n 41. What do these numbers have
Choose n points on a circle, and join each point to in common?
all the others. This divides the circle into a number,
say an , of regions. For example, a1 1 (by defini- 70. Do all terms in the sequence of Exercise 69
tion), a2 2, and a3 4: share this property?
1 5
Amazingly, an is the nearest integer to ϕ2 ϕ
nϕ 1
where ϕ denotes the golden ratio.
2
71. It has been known since Euclid (see also Ex- greatest element, ai1 . Now consider the sequence
ercises 32–34 in Section 2.3) that there are infinitely ai1 1 , ai1 2 , . . . .
many primes , but how far apart can they be? Prove
79. Prove that every sequence a1 , a2 , . . . , an2 1
that for any positive integer n, the sequence
of length n2 1 has a monotone subsequence
n! 2, n! 3, . . . , n! n ai1 , ai2 , . . . , ain 1 (with i1 i2 in 1 ) of
length at least n 1. Hint: let di denote the longest
contains no prime numbers. weakly increasing subsequence beginning with ai ,
i.e., the largest m so that you can find ai ai1
72. Let an denote the sum of the integers 1 up to ai2 aim for i i1 i2 im . If di n
n, so a4 1 2 3 4 10. Compute the first 6 for all i 1, 2, . . . , n2 1, how many terms of the
terms of an . Can you give a formula for an ? sequence must share the same di ? What does this
73. Say that a number is polite if it can be writ- mean for that subsequence? This result is known as
ten as the sum of two or more consecutive pos- the Erdős-Szekeres Theorem, after Paul Erdős and
itive integers. For example, 14 is polite because George Szekeres (1911–2005), who proved it in 1935.
14 2 3 4 5. Let an denote the nth polite
number. Compute the first 6 terms of an . Do you 80. For each positive integer n, let an denote
spot a pattern? the greatest number which can be expressed as the
product of positive integers with sum n. For exam-
74. A composition of the integer n is a way of writing
ple, a6 9 because 3 3 is the greatest product of
n as a sum of positive integers, in which the order
positive integers with sum 6. Find a formula for an .
of the integers does matter. For example, there are
eight partitions of 4: 4, 3 1, 1 3, 2 2, 2 1 1,
1 2 1, 1 1 2, and 1 1 1 1. Let an denote
the number of compositions of n. Compute the first
6 terms of an . Can you conjecture a formula?
Describe the sequences in Exercises 81–84. Warning:
75. Let an denote the number of compositions of n some of these have very creative definitions. You
into 1s and 2s. Relate an to a sequence from this should use all tools at your disposal, including the
section. Internet.
81. 1, 2, 4, 6, 10, 12, 16, 18, . . .
76. For n 2, let an denote the number of compo-
sitions of n into parts greater than 1. Relate an to 82. 1896, 1900, 1904, 1906, 1908, 1912, 1920, 1924, . . .
a sequence from this section.
77. A partition of the integer n is a way of writing 83. 3, 3, 5, 4, 4, 3, 5, 5, . . .

n as a sum of positive integers, in which the order
84. 1, 2, 3, 2, 1, 2, 3, 4, . . .
of the integers does not matter. For example, there
are five partitions of 4: 4, 3 1, 2 2, 2 1 1, and 85. How many 0s are there in the decimal ex-
1 1 1 1. Let an denote the number of parti- pansion of 10!? What about 50! and 100!? Hint: the
tions of n. Compute the first 6 terms of an . Can you number of 0s is equal to the number of 10s which
conjecture a formula? divide these numbers. These 10s can come by mul-
tiplying a number divisible by 10 itself, or by multi-
78. Prove that every infinite sequence an has
plying a number divisible by 5 by a number divisi-
an infinite monotone subsequence ai1 , ai2 , ai3 , . . .
ble by 2.
(with i1 i2 ). Hint: a sequence without
a greatest element must clearly have an infinite in- 86. How many ways are there to order a deck of
creasing subsequence. So suppose that an has a cards so that each of the suits is together?
The distinguished Israeli mathematician Noga Alon recounts:

“I was interviewed in the Israeli Radio for five minutes and I said that more than 2000 years ago,
Euclid proved that there are infinitely many primes. Immediately the host interuppted me and
asked: ‘Are there still infinitely many primes?”’
87. How many ways are there to order a deck of In order to show that this holds for all n 1, since
cards so that all of the spades are together? (But the f2n counts tilings of a 2n 1 board, we need only
cards of the other suits may be in any order.) show that there are fn2 fn2 1 such tilings. Take a
particular tiling of a 2n 1 board and chop it into
88. Use the figure below to conjecture and prove
two n 1 boards. There are two possibilities. First,
a simplified formula for the sum n 2
k 0 fk , where fk
we might chop the board into two tilings of an n 1
denotes the kth Fibonacci number.
board, as shown below with an 8 1 board:
1 1
3
2
8 Secondly, we might chop through a 2 1 domino,

thereby getting two n 1 1 tilings:
5
The number of n 1 tilings is fn , so there are fn2

ways in which our chop could break up the 2n 1
tiling in the first manner. Similarly, the number of
In addition to their recurrence, Fibonacci numbers n 1 1 tilings is fn 1 , so there are fn2 1 in which
can also be described in a more concrete way: the our chop could break up the 2n 1 tiling in the sec-
nth Fibonacci number fn counts the number of dif- ond manner. Therefore, since we have accounted
ferent ways to tile a board of size n 1 using for all 2n 1 tilings, we see that f2n fn2 fn2 1 , as
“squares” of size 1 1 and “dominos” of size 2 1. desired.
For example, f4 5 because there are four ways to Give similar arguments to verify the identities
tile a 4 1 board with these pieces: in Exercises 90 and 91.
n
90. Prove that fk fn 2 1.
k 0
89. Verify that the nth Fibonacci number counts 91. Prove that f3n 2 fn3 1 3fn2 1 fn fn3 .
the different ways to tile an n 1 board using 1 1
squares and 2 1 dominos. 92. Prove the Monotone Convergence Theorem
in the case where the sequence is monotonically in-
creasing.
This interpretation can be quite useful in proving 93. Prove that the converse to the Monotone Con-
identities involving the Fibonacci numbers. Con- vergence Theorem is also true, i.e., that every mono-
sider, for example, the identity tone sequence that is not bounded diverges.
f2n fn2 fn2 1. 94. Prove the Sandwich Theorem.

A NSWERS TO S ELECTED E XERCISES , S ECTION 2.1

1. an 2n 1
n 1 n
3. an 1 2
3 if n is odd,
5. an
1 if n is even.
n!
7. an
2n
9. 2, 5, 8, 11, 14, . . . , an 3n 1
11. 1, 3, 6, 10, 15, . . . , an n n 1 2
13. The common ratio must be 3, so a1 2.

15. Converges to 7 4.
19. Converges to 0.
21. Converges to π.
S ECTION 2.2 A N I NTRODUCTION TO S ERIES 35
2.2. A N I NTRODUCTION TO S ERIES

In everyday language, the words series and sequence mean the same thing. However,
in mathematics, it is vital to recognize the difference. A series is the result of adding a
sequence of numbers together. While you may never have thought of it this way, we deal
with series all the time when we write expressions like
1
0.333 . . . ,
3
since this means that
1 3 3 3
3 10 100 1000
In general we are concerned with infinite series such as
an a1 a2 a3
n 1
for various sequences an . First though, we need to decide what it means to add an
infinite sequence of numbers together. Clearly we can’t just add the numbers together
until we reach the end (like we do with finite sums), because we won’t ever get to the end.
Instead, we adopt the following limit-based definition.
Convergence and Divergence of Series. If the sequence sn of

partial sums defined by
sn a1 a2 an
has a limit as n then we say that
an lim sn ,
n
n 1
and in this case we say that an converges. Otherwise, an di-

n 1 n 1
verges.
We begin with a particularly simple example.
1
Example 1 (Powers of 2). The series n
converges to 1.
n 1
2
Solution. We begin by computing a few partial sums:
s1 12 12 1 12
s2 12 14 34 1 14
s3 12 14 18 78 1 18
s4 12 14 18 1 16 15 16 1 1 16
These partial sums suggest that sn 1 1 2n . Once we have guessed this pattern, it is easy
to prove. If sn 1 1 2n , then sn 1 1 1 2n 1 2n 1 1 1 2n 1 , so the formula is correct
for all values of n (this technique of proof is known as mathematical induction). With this
formula, we see that lim sn 1, so 1 2n 1.

n
n 1
There is an alternative, more geometrical, way to ..

.
see that this series converges to 1. Divide the unit 18
square in half, giving two squares of area 1 2. Now di-
1 16
vide one of these squares in half, giving two squares
of area 1 4. Now divide one of these in half, giving 12
two squares of area 1 16. If we continue forever, we

will subdivide the unit square (which has area 1) into
14
squares of area 1 2, 1 4, 1 8, . . . , verifying that
1
1.
n 1
2n
Note that although we were able to find an explicit

formula for the partial sums sn in Example 1, this is not possible in general.
Our next example shows a series an which diverges even though the sequence an
gets arbitrarily small, thereby demonstrating that the difference between convergent and
divergent series is quite subtle.
1
Example 2 (The Harmonic Series). The series 3 sn
n 1
n
2
diverges.
1
an
Solution. The proof we give is due to the French 1 2 3 4 5 6 7 8 9 10
philosopher Nicolas Oresme (1323–1382), and stands
as one of the pinacles of medieval mathematical achieve-
ment. We simply group the terms together so that each group sums to at least 1 2:
1 1 1 1
1
n 1
n 2 3 4
1 1
2 4 2
1 1 1 1
5 6 7 8
1 1
4 8 2
1 1 1 1 1 1 1 1
9 10 11 12 13 14 15 16
1 1
8 16 2
1 1 1 1
1 ,
2 2 2 2
and therefore the series diverges.
The name of this series is due to Pythagoras, and his first experiments with music.
Pythagoras noticed that striking a glass half-full of water would produce a note one oc-
tave higher than striking a glass full of water. A glass one-third full of water similarly
produces a note at a “perfect fifth” of a whole glass, while a glass one-quarter full pro-
duces a note two octaves higher, and a glass one-fifth full produces a “major third.” These
higher frequencies are referred to as harmonics, and all musical instruments produce har-
monics in addition to the fundamental frequency which they are playing (the instrument’s
“timbre” describes the amounts in which these different harmonics occur). This is what
led Pythagoras to call the series 1 1 2 1 3 the harmonic series.
Example 3. Suppose that scientists measure the total yearly precipitation at a certain point
for 100 years. On average, how many of those years will have record high precipitation?
Solution. Suppose that the data is uncorrelated from year to year (in particular, that the
amount of precipitation one year has no effect on the precipitation the next), and that the
data shows no long-term trends (such as might be suggested if the climate were changing).
In other words, suppose that the precipitations by year are independent identically distributed
random variables.
Letting an denote the amount of precipitation in the nth year, the question asks: for
how many values n can we expect an to be the maximum of a1 , . . . , an ? By definition, a1
is a maximum. The second year precipitation, a2 , then has a 1 2 of being the maximum
of a1 , a2 , while in general there is a 1 n chance that an is the maximum of a1 , . . . , an . This
shows that the expected number of record high years of precipitation is
1 1 1
1 .
2 3 100
The value of this sum is approximately 5.18738, so this is the expected number of record
high years of precipitation.
Example 2 provides a bit of intuition as to why the harmonic series diverges. Suppose
that the precipitation data is collected forever. Then the expected number of record years
is 1 n. On the other hand, it seems natural that we should expect to see new record
years no matter how long the data has been collected, although the record years will occur
increasingly rarely. Therefore we should expect the harmonic series to diverge.
By making half the terms of the harmonic series negative, we obtain a convergent se-
ries:
Example 4 (The Alternating Harmonic Series). The series
1n 1 1 1 1 1 1 1 1
1
n 1
n 2 3 4 5 6 7 8
converges.
1
Solution. Let sn denote the nth partial sum of this sn
series. If we group the first 2n terms in pairs, we have
an
1 1 1 1 1
s2n 1 .
2 3 4 2n 1 2n
Since 1 1 2, 1 3 1 4, and so on, each of these groups

is positive. Therefore s2n 2 s2n , so the even partial
sums are monotonically increasing.
Moreover, by grouping the terms in a different way,
1 1 1 1 1 1 1
s2n 1 ,
2 3 4 5 2n 2 2n 1 2n
we see that s2n 1 for all n. This shows that the sequence s2n is bounded and monotone.
The Monotone Convergence Theorem (from the previous section) therefore implies that
the sequence s2n has a limit; suppose that lim s2n L.
n
Now we consider the odd partial sums: s2n 1 s2n 1 2n 1, so
1
lim s2n 1 lim s2n lim L.
n n n 2n 1
Since both the even and odd partial sums converge to

the same value, the sum of the series exists and is at
most 1.
This leaves open a natural question: what is the sum of the alternating harmonic series?
Our proof shows that the sum of this series is sandwiched between its even partial sums
(which are under-estimates) and its odd partial sums (which are over-estimates), so
1 1 1n 1 1 1 5
1 1 ,
2 2 n 1
n 2 3 6
and we could get better estimates by including more terms. Back in Euler’s time it would
be difficult to guess what this series converges to, but with computers and the web, it is
quite easy. In only a few seconds, a computer can compute that the 5 millionth partial
sum of the alternating harmonic series is approximately 0.693147, and the Inverse Symbolic
Calculator at
http://oldweb.cecm.sfu.ca/projects/ISC/
(which attempts to find “nice” expressions for decimal numbers) lists ln 2 as its best guess
for 0.693147. Exercises 46 and 47 in Section 2.4 and Exercise 24 in Section 3.1 prove that
this is indeed the sum.
We next consider the result of adding the same terms in a different order.
Example 5 (A Troublesome Inequality). The series
1 1 1 1 1 1 1 1 1 1 1
1
3 2 5 7 4 9 11 6 n 1
4n 3 4n 1 2n
does not equal 1 n 1 n, despite having the same terms.
Solution. Simplifying the inside of this series,

2
1 sn
1 1 1 8n 3
4n 3 4n 1 2n 4n 3 4n 1 2n an
1 2 3 4 5 6 7 8 9 10
shows that it contains only positive terms. Therefore
its value is at least its first two terms added together,
1 1 1 1 1 1 1 1 389 9
1 .
n 1
4n 3 4n 1 2n 3 2 5 7 4 420 10
However, if we remove the parentheses then it is clear that the fractions we are adding in
this series are precisely the terms of the alternating harmonic series, whose value is strictly
less than 9 10. To repeat,
1 1 1 1n 1
n 1
4n 3 4n 1 2n n 1
n
even though both sides contain the same fractions with the same signs!
We will explore this phenomenon more in Section 2.8. Until then, just remember that
the order in which you sum a series which has both negative and positive terms might affect the
answer. In other words, when you are adding infinitely many numbers, some of which are
positive and some of which are negative, addition is not necessarily commutative.
It is because of examples such as this that we need to be extremely careful when dealing
with series. This is why we will make sure to prove every tool we use, even when those
tools are “obvious.”
Next we consider a series which is sometimes referred to as Grandi’s series, after the
Italian mathematician, philosopher, and priest Guido Grandi (1671–1742), who studied the
series in a 1703 work.
n 1
Example 6 (An Oscillating Sum). The series 1 diverges.
n 1
Solution. While the harmonic series diverges to infinity, the series 1 1 1 1

diverges because its partial sums oscillate between 0 and 1:
1 if n is odd,
sn
0 if n is even,
so lim sn does not exist.
n
This is an example of a series which can be shown to diverge by the first general test in
our toolbox:
The Test for Divergence. If lim an 0 then an diverges.

n
Proof. It is easier to prove the contrapositive: ifan converges, then an 0 as n . Since we

are assuming that an converges, lim sn exists. Suppose that lim sn L. Then
n n
an sn sn 1 L L 0
as n 0, proving the theorem.
It is important to remember that the converse to the Test for Divergence is false, i.e.,
even if the terms of a series tend to 0, the series may still diverge. Indeed, the harmonic
series is just such a series: 1 n 0 as n , but 1 n diverges.
Before concluding the section, we make one more general observation. The conver-
gence of a series depends only on how small its “tail” is. Thus it does not matter from the
point of view of convergence/divergence if we ignore the first 10 (or the first 10 octovigin-
tillion) terms of a series:
Tail Observation. The series an converges if and only if its

n 1
“tail”
an
n N
converges for some value of N .
While our techniques in this section have mostly been ad hoc, our goal in this chapter
is to develop several tests which we can apply to a wide range of series. Our list of tests
will grow to include :
• The Test for Divergence
• The Integral Test
• The p-Series Test
• The Comparison Test
• The Ratio Test
• The Absolute Convergence Theorem
• The Alternating Series Test
It is important to realize that each test has distinct strengths and weaknesses, so if one test
is inconclusive, you need to push onward and try more tests until you find one that can
handle the series in question.
This is in addition to the numerous more specialized tests developed in the exercises.

n
In all of the following problems, sn denotes the nth 1
14. 1
partial sum of an , that is, n 1
n
n
sn a1 a2 an . 1
15.
n 1
n
1. If an 3 then what are lim an and 16. 3 5 7 11 13 17 , the sum of the

n
n 1 primes.
lim sn ?
n
17. 1 3 1 5 1 7 1 11 1 13 1 17 , the sum
2. If lim sn 4 then what are lim an and of the reciprocals of the primes.
n n
an ?
n 1
In Exercises 18–21, reindex the series so that they
2 begin at n 1.
3. Compute the 4th partial sum of .
n 1
n 2
n2
18.
2 n 2
2n
4. Compute the 4th partial sum of .
n 1
n2 2
n2 n
19.
n 4
n 53
In Exercises 5–10, write down a formula for an and n2 n

20.
sum the series if it converges. n 4
n 53
3n 2
5. sn n sin n
n 4 21.
n 23
3 n 0
6. sn
n 4
n
7. sn 1
Determine if the series in Exercises 22–25 converge
n n 1
8. sn 2 at x 2 and at x 2.
9. sn sin n xn
22.
n 1
2n
10. sn arctan n
xn
23.
n 1
n2n
Determine if the series in Exercises 11–17 diverge by 1 n xn

24.
the Test for Divergence. (Note that if they do not di- n 1
2n
verge by the Test for Divergence, then we don’t yet
know if they converge or not.) 1 n xn
25.
n2n
1 n n2 n 1
11.
n 1
3n2 2n 1
sin n
12. Use the rules of limits described in Exercises 31–
n 1
n
34 of Section 2.1 to prove the statements in Exer-
cises 26–29.
13. cos 1 n2
n 1
26. If an and bn are both convergent series,
prove that partial sums are bounded but such that an does
not converge.
an bn an bn .
n 1 n 1 n 1
33. Suppose that an 0. Prove that an con-
n 1
27. If an and bn are both convergent series, verges if and only if a2n a2n 1 converges.
n 1
prove that
34. Give an example showing that the hypothesis
an 0 in Exercise 33 is necessary.
an bn an bn .
n 1 n 1 n 1
28. If an is a convergent series, prove that Sometimes a series can be rewritten in a such a way
that nearly every term cancels with a succeeding or
preceding term, for example,
can c an
n 1 n 1 1 1 1
nn 1 n n 1
for any number c. n 1 n 1
1 1 1
29. If an is a divergent series, prove that 1
2 2 3
1.
can
n 1
These series are called telescoping. In Exercises 35–
36, express the series as telescoping series to com-
diverges for any number c. pute their sums.
30. Archimedes (circa 287 BC–212 BC) was one of 2n 1
35. .
the first mathematicians to consider infinite series. n 1
n2 n 12
In his treatise The Quadrature of the Parabola, he uses
the figure shown below to prove that 1
36. , where fn is the nth Fibonacci
n 2
fn 1 fn 1
14 1 42 1 43 1 3. number.
37. Explain why the series
1 1 1 1
1 32 1
n 1
n2 2 2 3 3 4 4
is smaller than
1 16
1 1 1
1 ,
1 2 2 3 3 4
and then show that this second series 1 nn 1
is a convergent telescoping series.
14
Exercises 38–41 concern the situation described in

Example 2.
38. How many record years should we expect in 10
Explain his proof in words.
years of observations?
31. Prove that if an converges, then its partial 39. What is the probability that in 100 years of ob-
sums sn are bounded. servations, the first year is the only record?
32. Give an example showing that the converse to 40. What is the probability that in 100 years of ob-
Exercise 31 is false, i.e., give a sequence an whose servations, there are only two record years?
41. What is the probability that in 100 years of ob- 48. The expected (or average) number of steps that
servations, there are at most three record years? this procedure takes is
n Pr the procedure takes precisely n steps .

n 1
When a musical instrument produces a sound, it
also produces harmonics of that sound to some de- Verify that another way to write this is
gree. For example, if you play a middle A on an
instrument, you are playing a note at 440 Hz, but Pr the procedure takes at least n steps ,
the instrument also produces sounds at 880 Hz, n 1
1320 Hz, 1760 Hz, and so on. In a remarkable ef-
fect known as restoration of the fundamental, if the and use this to compute the expected number of
sound at 440 Hz is artificially removed, most listen- steps that this procedure takes.
ers’ brains will “fill it in,” and perceive the collec-
tion of sounds as a middle A nonetheless. Indeed,
your brain will perform the same feat even if several
Cesàro summability, named for the Italian math-
of the first few frequencies are removed. In Exer-
ematician Ernesto Cesàro (1859–1906), is a differ-
cises 42–45, determine what fundamental frequency
ent notion of sums, given by averaging the partial
your brain will perceive the given collection of fre-
sums. Define
quencies as.
s1 s2 sn
42. 646 Hz, 969 Hz, 1292 Hz, 1615 Hz, . . . σn .
n
43. 784 Hz, 1176 Hz, 1568 Hz, 1960 Hz, . . . We say that the series an is Cesàro summable to
44. 789 Hz, 1052 Hz, 1315 Hz, 1578 Hz, . . . L if lim σn L. Exercise 52 shows that if an
n
converges to L then an is Cesàro summable to L,
45. Discuss how restoration of the fundamental can however, the converse does not hold. We explore
be used to play music on a speaker which can’t pro- Cesàro summability in Exercises 49 and 50. In both
duce low notes. of these exercises, you should use the formulas for
sn found in the text. Exercises 52 and 53 explore the
more theoretical aspects of Cesàro summability.
Exercises 46–48 concern the following procedure, 49. Show that the Cesàro sum of 2 n
is 1.
which was brought to my attention by Professor n 1
Pete Winkler. Start with a1 2. At stage n, choose
an integer m from 1 to an uniformly at random (i.e., 50. Show that the Cesàro sum of 1 n 1
is 1 2.
each number has a 1 an chance of occurring). If n 1
m 1 then stop. Otherwise, set an 1 an 1
and repeat. For example, this procedure has a 1 2 51. Show that 1 n 1
n is not Cesàro
probability of stopping after only one step, while n 1
otherwise it goes on to the second step, with a2 3. summable. (Compare this with Exercise 22 in Sec-
tion 3.2.)
46. Compute the probability that this procedure 52. Prove that if an L then an is Cesàro
continues for at least 2 steps, at least 3 steps, and at summable to L.
least n steps.
53. Prove the result, due to Alfred Tauber (1866–
47. Compute the probability that this procedure 1942), that if an is a positive sequence and is
never terminates. Cesàro summable to L, then an L.

1. lim an 0, lim sn 3
n n
3. 23 24 25 26
3n 2 3n 1 2
5. an and the sum of the series is 3.
n 4 n 5
n n 1
7. an 1 1 and the series diverges.
9. an sin n sin n 1 and the series diverges.
11. Diverges by the Test for Divergence
13. Diverges by the Test for Divergence
15. The terms do go to 0, so the Test for Divergence does not apply.
17. The terms do go to 0, so the Test for Divergence does not apply.
n 32 n 3
19.
n 1
n 3 53
n sin n
21. The n 0 term of this series is already 0, so one answer is simply . Another correct answer
n 1
n 23
n 1 sin n 1
is .
n 1
n 1 23
23. Converges at x 2 (alternating harmonic series) and diverges at x 2 (harmonic series).
25. Diverges at x 2 (harmonic series) and converges at x 2 (alternating harmonic series).
2.3. G EOMETRIC S ERIES

One of the most important types of infinite series are geometric series. A geometric series is
simply the sum of a geometric sequence,
ar n .
n 0
Fortunately, geometric series are also the easiest type of series to analyze. We dealt a little
bit with geometric series in the last section; Example 1 showed that
1
1,
n 1
2n
while Exercise 26 presented Archimedes’ computation that
1 1
n
.
n 1
4 3
(Note that in this section we will sometimes begin our series at n 0 and sometimes begin
them at n 1.)
Geometric series are some of the only series for which we can not only determine con-
vergence and divergence easily, but also find their sums, if they converge:
Geometric Series. The geometric series
a ar ar 2 ar n
n 0
converges to
a
1 r
if r 1, and diverges otherwise.
Proof. If r 1, then the geometric series diverges by the Test for Divergence, so let us suppose
that r 1. Let sn denote the sum of the first n terms,
sn a ar ar2 arn 1
,
so
rsn ar ar2 arn 1
arn .
S ECTION 2.3 G EOMETRIC S ERIES 47
Subtracting these two, we find that
sn a ar ar2 arn 1
rsn ar ar2 arn 1

arn
1 r sn a arn .
This allows us to solve for the partial sums sn ,
a arn a arn
sn .
1 r 1 r 1 r
Now we know (see Example 5 of Section 2.1) that for r 1, rn 0 as n , so
arn
0
1 r
as n as well, and thus
a arn a
lim sn lim ,
n n 1 r 1 r 1 r
proving the result.
An easy way to remember this theorem is
first term
geometric series .
1 ratio between terms
We begin with two basic examples.
4 4 4
Example 1. Compute 12 4 3 9 27 .
Solution. The first term is 12 and the ratio between terms is 1 3, so
4 4 4 first term 12
12 4 1 18,
3 9 27 1 ratio between terms 1 3
solving the problem.
n 3
n2
Example 2. Compute 1 .
n 6
3n
Solution. This series is geometric with common ratio

n 4
1 n 12
an 1 3n 1 2
r ,
an 2 3
n 3
1n n
3
and so it converges because 23 1. Its sum is
n 3 29 36
n2 first term
1 ,
n 6
3n 1 ratio between terms 1 23
which simplifies to 512 1215.
The use of the geometric series formula is of course not limited to single geometric
series, as our next example demonstrates.
2n 1 9n 2
Example 3. Compute .
n 1
5n
Solution. We break this series into two:
2n 1 9n 2 2n 1 9n 2
n 1
5n n 1
5n n 1
5n
(if both series converge). The first of these series has common ratio 2 5, so it converges. To
n n
analyze the second series, note that 9n 2 91 2 9 3n , so this series has common
ratio 3 5. Since both series converge, we may proceed with the addition:
2n 1 9n 2 2n 1 3n 22 5 35 4 3
.
n 1
5n n 1
5n n 1
5n 1 25 1 35 3 2
This answer simplifies to 17 6.
If a geometric series involves a variable x, then it may only converge for certain values
of x. Where the series does converge, it defines a function of x, which we can compute
from the summation formula. Our next example illustrates these points.
Example 4. For which values of x does the series 1 n x2n converge? For those values
n 0
of x, which function does this series define?
Solution. The common ratio of this series is
an 1 1 n 1 x2n 2
r x2 .
an 1 n x2n
Since geometric series converge if and only if r 1, we need x2 1 for this series to
converge. This expression simplifies to x 1.
Where this series does converge (i.e., for 1 x 1), its sum can be found by the
geometric series formula:
first term 1 1
1 n x2n .
n 0
1 ratio between terms 1 x2 1 x2
Although note that this only holds for x 1.
The geometric series formula may also be used to convert repeating decimals into frac-
tions, as we show next.
Example 5. Express the number 4.342342342 . . . as a fraction in the form p q where p and q
have no common factors.
Solution. Our first step is to express this number as the sum of a geometric series. Since
the decimal seems to repeat every 3 digits, we can write this as
342 342 342

4.342342342 4 4 .
1000 10002 n 1
1000n
This series is geometric, so we can use the formula to evaluate it:
342 first term 342 1000 342 1000 342

n
.
n 1
1000 1 ratio between terms 1 1 1000 1000 999 999
Our initial answer is therefore

342 3996 342 4338
4 .
999 999 999
Since we were asked for a fraction with no common factors between the numerator and
denominator, we now have to factor out the 9 which divides both 4338 and 999, leaving
482
4.342342342 .
111
To be sure that 482 and 111 have no common factors, we need to verify that 111 3 37,
and the prime 37 does not divide 482.
Our next example in some sense goes in the other direction. Here we are given a frac-
tion and asked to use geometric series to approximate its decimal expansion.
Example 6. Use geometric series to approximate the decimal expansion of 1 48.
Solution. First we find a number near 1 48 with a simple decimal expansion; 1 50 will work
nicely. Now we express 1 48 as 1 50 times a fraction of the form 1 1 r :
1 1 1 1
48 50 2 2 .
50 1 50
Now we can expand the fraction on the righthand side as a geometric series,
2 3
1 1 2 2 2
1 .
48 50 50 50 50
Using the first two terms of this series, we obtain the approximation 1 48 0.02 1 0.02
0.0204.
Example 7. Suppose that you draw a 2” by 2” square, then you join the midpoints of its
sides to draw another square, then you join the midpoints of that square’s sides to draw
another square, and so on, as shown below.
Would you need infinitely many pencils to continue this process forever?
Solution. If we look just at the upper left-hand corner of this figure, we see a triangle with
two sides of length 1” and a hypotenuse of length 2”:
1”
1”
2”
So the sequence of side lengths of these rectangles is geometric with ratio 2 2: 2”, 2”,
1 2”, . . . . Since 2 2 1, the sum of the side lengths of all these (infinitely many) squares
therefore converges to
2 2
22 2.
2 2 2
1
2 2
As the perimeter of a square is four times its side length, the total perimeter of this infinite
construction is 8 2 2 ”, so we would only need finitely many pencils to draw the figure
forever.
Our last two examples are significantly more advanced that the previous examples. On
the other hand, they are also more interesting.
Example 8 (A Game of Chance). A gambler offers you a proposition. He carries a fair

coin, with two different sides, heads and tails (“fair” here means that it is equally likely to
land heads or tails), which he will toss. If it comes up heads, he will pay you $1. If it comes
up tails, he will toss the coin again. If, on the second toss, it comes up heads, he will pay
you $2, and if it comes up tails again he will toss it again. On the third toss, if it comes up
heads, he will pay you $3, and if it comes up tails again, he will toss it again... How much
should you be willing to pay to play this game?
Solution. Would you pay the gambler $1 to play this game? Of course. You’ve got to win
at least a dollar. Would you pay the gamble $2 to play this game? Here things get more
complicated, since you have a 50% chance of losing money if you pay $2 to play, but you
also have a 25% chance to get your $2 back, and a 25% chance to win money. Would you
pay the gambler one million dollars to play the game? The gambler asserts that there are
infinitely many positive integers greater than one million. Thus (according to the gambler,
at least) you have infinitely many chances of winning more than one million dollars!
To figure out how much you should pay to play this game, we compute its average
payoff (its expectation). In the chart below, we write H for heads and T for tails.
Outcome Probability Payoff Expected Winnings
H 12 $1 1 2 $1 $0.50
TH 14 $2 1 4 $2 $0.50
TTH 18 $3 1 8 $3 $0.375
TTTH 1 16 $4 1 16 $4 $0.25
.. .. .. ..
. . . .
TT TH 1 2 n $n 1 2 n $n $n 2n
n 1
.. .. .. ..
. . . .
There are several observations we can make about this table of payoffs. First, the sum of
the probabilities of the various outcomes is
1
n
1,
n 1
2
which indicates that we have indeed listed all the (non-negligible) outcomes. But what if
the coin never lands heads? The probability of this happening is lim 1 2 n 0, and so it is
n
safely ignored. Therefore, we ignore this possibility.
To figure out how much you should pay to play this game, it seems reasonable to
compute the sum of the expected winnings, over all possible outcomes. This sum is
1 2 3 4 n
.
2 22 23 24 n 1
2n
We demonstrate two methods to compute this sum. The first is elementary but uses a
clever trick, while the second uses calculus. sn
2
Applying the geometric series formula, we can write:
1
1 12 1 22 1 23 1 24
12 1 22 1 23 1 24 an
1 22 1 23 1 24 1 2 3 4 5 6 7 8 9 10
1 23 1 24
.. ..
. .
If we now add this array vertically, we obtain an equa-

tion whose left-hand side is 2 (it is the sum of a geo-
metric series itself), and whose right-hand side, 1 2
2 22 3 23 , is the sum of the expected winnings, so
n
2.
n 1
2n
Because the expected winnings are $2, you should be willing to pay anything less than
$2 to play the game, because in the long-run, you will make money.
Now we present a method using calculus. Since we know that
1
1 x x2 x3 x4
1 x
we can differentiate this formula to get
1
2
0 1 2x 3x2 4x3 .
1 x
Now multiply both sides by x:

x
2
x 2x2 3x3 4x4 .
1 x
Finally, if we set x 12 we obtain the sum we are looking for:

2 3 4
1 1 1 1 n
2 2 3 4 n
.
2 2 2 2 n 1
2
One might (quite rightly) complain that we don’t know that we can take the derivative of
an infinite series in this way. We consider these issues in Section 3.2.
Example 9 (The St. Petersburg Paradox ). Impressed with the calculation of the expected
winnings, suppose that the gambler offers you a different wager. This time, he will pay $1
for the outcome H, $2 for the outcome T H, $4 for the outcome T T H, $8 for the outcome
T T T H, and in general, $2n if the coin lands tails n times before landing heads. Computing
the expected winnings as before, we obtain the following chart.
Outcome Probability Payoff Expected Winnings

H 12 $1 $0.50
TH 14 $2 $0.50
TTH 18 $4 $0.50
TTTH 1 16 $8 $0.50
.. .. .. ..
. . . .
TT TH 1 2n $2n $0.50
n
.. .. .. ..
. . . .
The gambler now suggests that you should be willing to pay any amount of money to play
this game since the expected winnings, $0.50 $0.50 $0.50 , are infinite!
Solution. That is why this game is referred to as a paradox. Seemingly, there is no “fair”
price to pay to play this game. Bernoulli, who introduced this paradox, attempted to
resolve it by asserting that money has “declining marginal utility.” This is not in dispute;
imagine what your life would be like with $1 billion, and then imagine what it would be
like with $2 billion. Clearly the first $1 billion makes a much bigger difference than the
second $1 billion, so it has higher “utility.” However, the gambler can simply adjust his
n 1
payoffs, for example, he could offer you $22 if the coin lands tails n 1 times before
landing heads, so even taking into account the declining marginal utility of money, there
is always some payoff function which results in a paradox.
A practical way out of this paradox is to note that the gambler can’t keep his promises.
If we assume generously that the gambler has $1 billion, then the gambler cannot pay you
the full amount if the coin lands tails 30 times before landing heads (229 536, 870, 912,
but 230 1, 073, 741, 824). The payouts in this case will be as above up to n 29, but then
beginning at n 30, all you can win is $1 billion. This gives an expected winnings of
n 29 $1,000,000,000
$1,000,000,000 230
$0.50 $14.50 $16.36.
n 1 n 30
2n 1 1 12
This problem is known as the St. Petersburg Paradox because it was introduced in a 1738 paper of Daniel
Bernoulli (1700–1782) published in the St. Petersburg Academy Proceedings.
So by making this rather innocuous assumption, the game goes from being “priceless” to
being worth the same as a new T-shirt.
Another valid point is that you don’t have unlimited money, so there is a high proba-
bility that by repeatedly playing this game, you will go broke before you hit the rare but
gigantic jackpot which makes you rich.
One amusing but nevertheless accurate way to summarize the St. Petersburg Paradox
is therefore: If both you and the gambler had an infinite amount of time and money, you
could earn another infinite amount of money playing this game forever. But wouldn’t you
have better things to do with an infinite supply of time and money?

Determine which of the series in Exercises 1–8 are 4n 2 1
10.
geometric series. Find the sums of the geometric se- n 1
3n
ries.
2n 5n
2n 11.
1. 4n 9
n 1
3n n 1
n 2
n3
1 12. 1
2. 5n 1
n 1
n2 n 0
1
3.
n2 2n
n 1 Find the sums of the series in Exercises 13–16.
3n 3n 5n
4. 13.
42n 1 n 1
7n
n 1
1n 14. en π n
5.
n 0
2n n 1
2
9 2n
5n 15.
6. 4
32n
n 0
5n n 1
2 2
3n 4n
5n 16.
7. 4 n 1
6n
n 1
25n
3n
8.
n 0
n! Determine which values of x the series in Exer-
cises 17–20 converge for. When these series con-
verge, they define functions of x. What are these
functions?
Determine which of the geometric series (or sums of
xn
geometric series) in Exercises 9–12 converge. Find 17.
the sums of the convergent series. n 1
3n
32n 2x 1 n
9. 18.
n 0
2n n 1
3n
19. 5n 2 n
x shown below.
n 1
x 3n
20.
n 1
2n 1
For Exercises 21–24, use geometric series to approx-

imate these reciprocals, as in Example 6.
21. 1 99
22. 1 102
27. Write a series representing the area of the Koch

23. 2 99
snowflake, and find its value.
28. Write a series representing the perimeter of the
24. 1 24
Koch snowflake. Does this series converge?
29. Two 50% marksmen decide to fight in a duel
in which they exchange shots until one is hit. What
are the odds in favor of the man who shoots first?
For Exercises 25 and 26, suppose that an is a ge- 30. In the decimal system, some numbers have
n 1
more than one expansion. Verify this by showing
ometric series such that the sum of the first 3 terms
that 2.35999 . . . 2.36.
is 3875 and the sum of the first 6 terms is 3906.
31. The negadecimal number system is like the
25. What is a6 ? decimal system except that the base is 10. So,
for example, 12.43 10 1 10 2 2 10
4 10 1 3 10 2 97.63 in base 10. Prove
26. What is an ? that (like the decimal system) there are non-unique
n 1 expansions in the negadecimal system by showing
that 1.909090 . . . 10 0.090909 . . . 10 .
Exercises 27 and 28 consider the Koch snowflake. In-

troduced in 1904 by the Swedish mathematician Exercises 32 and 33 ask you to prove that there are
Helge von Koch (1870–1924), the Koch snowflake infinitely many primes, following a proof of Euler.
is one of the earliest fractals to have been described. Exercise 34 provides the classic proof, due to Euclid.
We start with an equilateral triangle. Then we di-
vide each of the three sides into three equal line 32. Prove, using the fact that every positive inte-
segments, and replace the middle portion with a ger n has a unique prime factorization , that
smaller equilateral triangle. We then iterate this
1 1 1 1
construction, dividing each of the line segments of
n 2n 3n 5n
the new figure into thirds and replacing the mid-
dle with an equilateral triangle, and then iterate this where the right-hand side is the product of all series
again and again, forever. The first four iterations are of the form p1n for primes p.
a
This refers to the fact that every positive integer n can be written as a product n pa1 1 pa2 2 pkk for
primes p1 , p2 , . . . , pk and nonnegative integers a1 , a2 , . . . , ak .
33. Use Exercise 32 to prove that there are in- 37. Suppose the gambler alters his game once
finitely many primes. Hint: the left hand-side di- more. If the coin lands tails an even number of times
verges to , while every sum on the right hand-side before landing heads, you must pay him $1. How-
converges. ever, if the coin lands tails an odd number of times
34. (Euclid’s proof) Suppose to the contrary before landing heads, he pays you $2. Would you
that there are only finitely many prime numbers, be willing to play this game? Why or why not?
p1 , p2 , . . . , pm . Draw a contradiction by consid-
ering the number n p1 p2 pm 1. 38. Suppose that after a few flips, you grow
suspicious of the gambler’s coin because it seems
n2 to land heads more than 50% of the time (but less
35. Compute .
n 1
2n than 100% of the time). Design a procedure which
Hint: consider the derivative of x 1 x 2 . For now, will nevertheless produce 50-50 odds using his un-
assume that you can differentiate this series as done fair coin. (The simplest solution to the problem is
in Example 8. attributed to John von Neumann (1903–1957).)
36. Suppose the gambler from Example 8 alters
his game as follows. If the coin lands tails an even 39. Consider the series 1 n where the sum
number of times before landing heads, you must is taken over all positive integers n which do not
pay him $1. However, if the coin lands tails an odd contain a 9 in their decimal expansion. Due to
number of times before landing heads, he pays you A.J. Kempner in 1914, series like this are referred
$1. The gambler argues that there are just as many to as depleted harmonic series. Show that this series
even numbers as odd numbers, so the game is fair. converges. Hint: how many terms have denomina-
Would you be willing to play this game? Why or tors between 1 and 9? Between 10 and 99? More
why not? (Note that 0 is an even number.) generally, between 10n 1 and 10n 1?

23
1. Geometric series, converges to 2.
1 23
3. Not a geometric series.

1
5. Geometric series, converges to 2 3.
1 12
5 255
7. Geometric series, converges to 4 9765625 .
1 5 25
9. Diverges; the ratio is 9 2 1.

11. Diverges.
37 57
13. The sum is 13 4.
1 37 1 57
92 23
15. The sum is 92 22 .
1 12
x 3
17. Converges for 3 x 3, to the function .
1 x3
53 x
19. Converges for 15 x 1 5, to the function .
1 5x
2.4. I MPROPER I NTEGRALS AND T HE I NTEGRAL T EST

In this section we discuss a very simple, but powerful, idea: in order to prove that certain
series converge or diverge, we may compare them to integrals. There are a few important
caveats with this comparison, which we will make note of when we present the Integral
Test formally. To motivate this test, we return to the harmonic series 1 n. In Section 2.2,
we saw Nicolas Oresme’s classic proof that the harmonic series diverges. Here, we present
another proof, which will generalize to handle many more series.
To begin, let us represent the series 1 n as the total area contained in an infinite se-
quence of 1 1 n rectangles. Beginning with the first rectangle stretching from x 1 to
x 2 and placing the rectangles next to each other, we get the following.
0 ...
0 1 2 3 4 5 6 7 8 9 10
We now approximate the area under these rectangles. In this case, we only have to
observe that the function 1 x lies below the tops of these rectangles for x 1, as shown
below.
0 ...
0 1 2 3 4 5 6 7 8 9 10
Therefore, there is more area under the rectangles than under the function 1 x. As we
know that area under a curve is given by an integral, to find the area under 1 x for x 1,
we need to evaluate
1
dx.
1 x
This type of an integral may be unfamiliar because it involves infinity, and for this
reason, integrals of this type are called improper integrals . Since we can’t simply take the
anti-derivative of 1 x and plug in , we do the next best thing. We define the improper
integral as a limit:
b
1 1
dx lim dx.
1 x b 1 x
These are not the only type of improper integrals. Others involve integrating near a vertical asymptote of
a function. See Exercises 36–43 for more examples of improper integrals.
S ECTION 2.4 I MPROPER I NTEGRALS AND T HE I NTEGRAL T EST 59
Using this definition, we have another argument for why 1n diverges:
1 1
dx
n 1
n 1 x
b
1
lim dx
b 1 x
lim ln x b1
b
lim ln b ln 1
b
.
In general, if we have a function f x defined from x a to x , we define
b
f x dx lim f x dx,
a b a
and we say that this improper integral converges if the limit converges, and that it diverges
if the limit diverges. The reader should note the similarity between this definition and the
definition we made for the convergence of a series.
Comparing series to integrals can also be used to show that they converge, as we illus-
trate in the next example.
Example 1. Show that 1 n2 converges by comparing it to an integral.
Solution. We do roughly the same thing as we did with the harmonic series, but here,
since we are to show that the series converges, we want the area in our rectangles to be
less than the area under the curve. For this reason, we begin by placing 1 1 n2 rectangles
starting at x 0:
0 ...
0 1 2 3 4 5 6 7 8 9 10
Now we have exactly what we want, because the area under these rectangles is strictly
less than the area under 1 x2 for x 0:
0 ...
0 1 2 3 4 5 6 7 8 9 10
This suggests that we should integrate 1 x2 from x 0 to x to get an upper bound

on 1 1
n . However, since x has a vertical asymptote at x
2 2 0, such an integral would be
doubly improper and actually diverges, as shown by Exercises 37 and 38, so can’t be used
to bound 1 n2 from above. To get around this problem, we can simply pull off the first
term of the series, and compare the rest to the integral of 1 x2 from x 1 to x :
1 1
1
n 1
n2 n 2
n2
1
1 2
dx
1 x
b
1
1 lim 2
dx
b 1 x
b
1
1 lim
b x 1
1
2 lim
b b
2.
It is tempting to conclude right now that 1 n2 converges, because we know that it is

at most 2, but this would be reckless. Remember in Example 6 of Section 2.2 we presented
a series — the series 1 n 1 — which has bounded partial sums but does not converge.
However, in this case, 1 n2 is positive for all n, so the partial sums sn of 1 n2 are
monotonically increasing. By the above argument, these partial sums are bounded: sn lies
between 0 and 2 for all n, and therefore we know by the Monotone Convergence Theorem
that the sequence sn converges to a limit, and thus 1 n2 converges.
While we have shown that 1 n2 converges, we have not computed its value. For
series that aren’t geometric, such questions are generally extremely difficult, and 1 n2 is
no exception. Finding 1 n2 became known as the Basel problem after it was posed by
Pietro Mengoli (1626–1686) in 1644. In 1735, at the age of twenty-eight, Leohnard Euler
showed that
1 π2
2
,
n 1
n 6
one of the first major results in what would be a marvelous career. We discuss his first
proof (which has been incredibly influential despite the fact that it contains a flaw) in Exer-
cises 54–58 of Section 3.3. Euler began his exploration of the Basel problem by computing
the sum to 17 decimal places (which in itself was quite a feat, accomplished by viewing
the series as an integral), a bit like we did in Section 2.2 to guess the sum of the alternating
harmonic series. Amazingly, without any aid like the Inverse Symbolic Calculator, Euler rec-
ognized that this approximation looked like π2 6! This gave Euler a significant advantage
in finding the solution, since he knew what the answer should be.
Euler went on to find formulas for 1 np for all even integers p. But what about the
odd integers? For a very long time, mathematicians could not even prove that 1 n3 was
irrational, let alone express it in terms of well-known constants. In 1978, Roger Apéry
(1916–1994) announced that he had a proof of this result. However, Apéry was not well-
known and there were significant doubts that his proof could be correct. Apéry fed this
suspicion by giving a strange talk announcing his proof, one of the key ingredients of
which was the equation
1 5 1 n 1 n! 2
3
.
n 1
n 2n 1
n3 2n !
When asked how he derived this equation, Apéry is alleged to have replied “they grow
in my garden.” Nevertheless, he completed his proof, stunning the mathematical estab-
lishment. The analogous question about the irrationality of 1 np for odd integers p 5
remains unsolved.
It is now time to generalize our two examples and make a test out of them. First we
must decide what was special about 1 n and 1 n2 that allowed us to make the com-
parisons we made. In the case of 1 n, we needed that the function 1 x lies below the
rectangles we formed. This relies on the fact that 1 x is decreasing. Similarly, in the case
of 1 n2 we needed that the function 1 x2 lies above the rectangles we formed. Because
these rectangles were slid over by one unit, this too relies on the fact that 1 x2 is decreasing.
We also used, in the 1 n2 case, the fact that 1 x2 is positive. Finally, we need to be able to
evaluate the integrals. We could add this as a hypothesis, but in the interest of simplicity,
we simply require that our functions be continuous, which guarantees that they can be
integrated. Under these conditions, we have the following test.
The Integral Test. Suppose that f is a positive, decreasing, and
continuous function, and that an f n . Then an converges

n 1
if and only if the improper integral f x dx converges.

1
Proof. As in our two examples, we can sandwich the partial sums sn between two improper
integrals:
n n 1
f x dx sn a1 f x dx.
1 1
Now since we are proving an “if and only if” statement, we have two things to prove. First, suppose
that
f x dx
1
converges. Then, by using the upper-bound, we have

n 1
an lim sn lim a1 f x dx .
n n 1
n 1
Because an f n is positive, we know that the partial sums sn are monotonically increasing, so
since the above inequality shows that the sequence sn is bounded, the Monotone Convergence
Theorem implies that sn has a limit. This proves that an converges if the improper integral
f x dx converges.
1
Now suppose that
f x dx
1
diverges. Using the lower-bound, we have

n
an lim sn lim f x dx .
n n 1
n 1
Therefore, the sequence sn of partial sums diverges to , so the series an diverges.
The Integral Test is a very powerful tool, but it has a serious drawback: we must be
able to evaluate the improper integrals it requires. For example, how could we use the
Integral Test to determine whether 4n n! converges? Nevertheless, there are numerous
examples of series to which it applies.
1
Example 2. Does the series converge or diverge?
n 1
n2 1
Solution. We first evaluate the improper integral in the Integral Test:

b
dx dx
lim lim arctan b arctan 1.
1 x2 1 b 1 x2 1 b
To evaluate this limit, it may be helpful to recall the plot of arctan:

π2
π2
Therefore, we have
lim arctan b arctan 1 π2 π 4,
b
so the series 1 n2 1 converges by the Integral Test.

We began the section by considering 1 n and 1 n2 . What about 1 np for other val-
ues of p? We can evaluate the integral of 1 xp , so the Integral Test can be used to determine
which of these series converge. Because series of this form occur so often, we record this
fact as its own test.
The p-Series Test. The series 1 np converges if and only if p 1.
Proof. When p 1, we already know that the series diverges ( 1 n is the example we began the
section with). For other values of p, we simply integrate the improper integral from the Integral
Test:
b
1 p
dx lim x dx
1 xp b 1
b
1
lim x p 1
b p 1 1
1 1
lim b p 1
b p 1 p 1
If p 1 then the function x p 1 decreases to 0 (the exponent is negative), so in this case the limit
above converges to 1 p 1 1 1 p. Therefore, by the Integral Test, 1 np converges if p 1. On
the other hand, if p 1 then the function x p 1 increases without bound, so in this case the limit
above diverges to , and so 1 np diverges if p 1.
We conclude this section with error estimates. Since improper integrals can be used to
bound series, they can also be used to bound the tails of series, i.e., the error in a partial
sum:
The Integral Test Remainder Estimates. Suppose that f is a posi-

tive, decreasing, and continuous function, and that an f n . Then
the error in the nth partial sum of an is bounded by an improper
integral:
sn an f x dx.
n 1 n
The proof of the Integral Test Remainder Estimate is almost identical to the proof of the
Integral Test itself, so we content ourselves with an example.
Example 3. Bound the error in using the fourth partial sum s4 to approximate 1 n2 .
n 1
Solution. The error in this case is the difference between sn and the true value of the
series:
1
Error sn
n 1
n2
By the remainder estimates, we have:
1
Error 2
dx
4 x
b
1
lim 2
dx
b 4 x
b
1
lim
b x 4
1
.
4
This is not a very good bound. As we mentioned earlier, Euler approximated the value of
this series to within 17 decimal places. How many terms would we need to take to get the
upper bound on the error from the Integral Test Remainder Estimates under 10 17 ?

Arrange the quantities in Exercises 1–4 in order 1
6. 3
from least to greatest. n 1
n 3
1 dx dx cos n
1. 1 7.
n 1
n2 1 x2 2 x2 n2
n 1
1 dx dx 1n 1
2. 8.
n 9
n2 1 10 x2 1 9 x2 1 n
n 1
10 11 10
1 dx dx 4n
3. 9.
n 1 n7 2 1 x7 2 0 x7 2 n 1
2n2 3 2
10 10 11 3
4. n x dx x dx 10.
n 1
2 5n
n 1 0 1
n n
11.
n 1
n
For Exercises 5–20, use the Integral Test to deter-
1
mine if the series converge or diverge, or indicate 12.
why the Integral Test cannot be used. n 1
n2 sin n
1 n
5. 13.
n 1
n3 n 0
n2 1
ln n 26. Is the total circumference inside all of the cir-

14.
n 2
n cles finite? (Note that you are not asked to find this
total.)
2n 3
15.
n 1
n
5 Exercises 27–32 require integration by parts. Use

16.
n n the Integral Test to determine if the series converge
n 1
or diverge.
1 ln n
17. 27.
n 2
n ln n n2
n 1
1 ln n
18. 2 28.
n 2
n ln n n7 6
n 1
1 n2
19. 29.
n ln n ln ln n n 1
en
n 2
1 e1 n
20. 30.
n ln n ln ln n 2
n 1
n2
n 2
1
31.
n 1
ne1 n
By the Integral Test Remainder Estimates, how n

many terms would you need to use to approximate 32.
n 1
2n
the sums in Exercises 21–24 to within 1 100?
33. Use the Integral Test to verify that the geomet-
1
21.
n 1
n4 ric series ar n converges for 0 r 1.
n 1
1 34. In the Integral Test, we began both the se-
22.
n 1
n5 ries and the integral at 1 (technically, n 1 and
x 1, respectively). Show that this is not necessary
1 by proving the following. Suppose that f is a posi-
23. 2
n 1
n ln n tive, decreasing, and continuous function, and that
an f n . Then
5
24. an
n 1
n n
n 1
converges if and only if there is some N so that the
improper integral
Exercises 25 and 26 concern an infinite sequence of f x dx
circles which do not overlap and have radii 1, 1 2, N
1 3, . . . , as shown below. converges.
35. Use the fact that
1
... 1n 1 1 1
1
n 1
n 2 3 4
1 1
n 1
2n 1 2n
25. Is the total area inside all of the circles finite? to prove that the alternating harmonic series con-
(Note that you are not asked to find this total.) verges using the Integral Test.
We have focused on only one type of improper in- We then evaluate each of these improper integrals
tegrals, which are called improper because their do- using the rules above. Using this definition, evalu-
mains are infinite. However, there is another type, ate the integrals in Exercises 42–43.
which are called improper because their integrands 2
1
have vertical asymptotes. To begin with, suppose 42. 2
dx
0 1 x
that f x is continuous on the interval a, b but dis-
1
continuous at x a. Then we define the integral of 43. x 3
dx
f x from a to b as a limit: 1
b b
f x dx lim f x dx.
a c a c
Euler’s constant γ is defined as
Use this definition to evaluate the integrals in Exer- 1 1 1
cises 36–39. γ lim 1 ln n.
n 2 3 n
1
36. x dx It is not clear a priori that this limit exists, and so Ex-
0 ercises 44 and 45 show how to prove that it does ex-
1 ist. Its value is approximately 0.577215, and a very
37. x2 dx readable account of research related to γ is given by
0
Julian Havil in his book Gamma.
38. x2 dx 44. Show that
0 1 1 1
ln n 1 1 ln n,
1 2 3 n
39. x1.001 dx and conclude from this that the sequence bn
0
1 12 13 1n ln n is bounded both above
and below.
45. Show that
If instead f x is continuous on the interval a, b
1
but discontinuous at x b, then we consider the ln n 1 ln n ,
n 1
limit as the upper-bound of the integral approaches
b from below: and use this to conclude that the sequence bn from
Exercise 44 is decreasing. This will imply that bn
b c
f x dx lim f x dx. is a decreasing bounded sequence, so its limit, γ, ex-
a c b a ists by the Monotone Sequence Theorem.
Use this definition to evaluate the integrals in Exer-

cises 40–41.
1
1 Exercises 46 and 47 show one way to sum the al-
40. dx
0 1 x ternating harmonic series, using Euler’s constant γ
discussed in Exercises 44 and 45.
1
1 46. Let sn denote the nth partial sum of the alter-
41. dx
0 1 x2 1n 1
nating harmonic series . Verify that
n 1
n
1 1 1
Finally, it could be the case that f x has a verti- s2n 1
2 3 2n
cal asymptote between the bounds a and b, say at
x c for a c b. In this case, assuming that f x 1 1 1
1 .
is continuous on both a, c and c, b , we break the 2 3 n
integral in two:
b c b
47. Use Exercise 46 to show that
f x dx f x dx f x dx. s2n ln 2 s2n ln 2n γ ln n γ 0,
a a c
1
1n there are 24 time zones in the world, show that the
proving that ln 2.
n 1
n radius of the Earth is about 4000 miles.
52. Two masses attract each other with a force
whose magnitude is proportional to the product of
Exercises 48–50 ask you to develop estimates for n!. the masses divided by the square of the distance be-
A more precise estimate is named for James Stirling tween them. So for a mass m, the magnitude of the
(1692–1770). force of gravity is GmM r 2 , where G is a constant,
M is the mass of the Earth, and r is the distance
n! to the center of the Earth. Since work is equal to
Stirling’s Formula. lim n 1. force times distance, show that the amount of work
n 2πn n e
needed to lift a mass m from the surface of the Earth
48. Using the fact that ln n! ln 2 ln 3 ln n, to altitude R is
prove that R
GmM
dr,
n
4000 r2
ln n 1! ln x dx ln n!
1 and calculate this integral. Then let R to show
that GmM 4000 is the most work you can do to lift
Note that ln x dx x ln x x C. the mass m to anywhere in the universe (disregard-
ing all objects besides Earth, of course).
49. Prove that n! e n e n.
53. Using the fact that kinetic energy is mv 2 2, com-
50. Prove that n! en n e n.
pute how much kinetic energy we would need to
supply to lift the mass m to anywhere in the uni-
verse.
Exercises 51–55 consider a back-of-the-envelope cal- 54. Use the fact that acceleration due to gravity on
culation of the escape velocity from Earth using im- the surface of the Earth is about 32 ft/sec2 , which is
proper integration. These exercises are due to Pro- equal to GmM 40002 to solve for GM .
fessor Stephen Greenfield. 55. Use the answers to Exercises 53 and 54 to show
51. The continental US is about 3400 miles wide (at that the escape velocity from the Earth is about 7
its widest point) and contains 4 time zones. Since miles per second.

dx 1 dx
1. 1
2 x2 n 1
n2 1 x2
11 10 10
dx 1 dx
3.
1 x7 2 n 1 n7 2 0 x7 2
5. Converges by the Integral Test.
7. The Integral Test does not apply (some terms are negative).
9. Converges by the Integral Test (make a u-substitution).
11. The Integral Test does not apply (the series is not decreasing). However, this series diverges by the Test
for Divergence.
13. Diverges by the Integral Test (make a u-substitution).
15. The Integral Test does not apply (the series is not decreasing). However, this series diverges by the Test
for Divergence.
17. Diverges by the Integral Test (set u ln x to evaluate the integral).
19. Diverges by the Integral Test (set u ln ln x to evaluate the integral).
21. 4 terms suffice.
23. e100 terms are enough (note that this is 2.688 1043 )
25. The total area inside the circles is finite.
S ECTION 2.5 T HE C OMPARISON T EST 69
2.5. T HE C OMPARISON T EST

We began our systematic study of series with geometric series, proving the
• Geometric Series Test: ar n converges if and only if r 1.
Then in the last section we compared series to integrals in order to determine if they con-
verge or diverge, and established the
• p-Series Test: 1 np converges if and only if p 1.
In this section we study another type of comparison where we compare series to other
series to determine convergence. The general principle is this:
• if a positive series is bigger than a positive divergent series, then it diverges, and
• if a positive series is smaller than a positive convergent series, then it converges.
For example, in the last section (Example 1) we showed that 1 n2 converges using the
Integral Test. Then we used the Integral Test again (Example 2) to show that 1 n2 1
converges. But, 1 n2 1 is smaller than 1 n2 for all n 1, so the convergence of 1 n2 1
is guaranteed by the convergence of 1 n2 . While this approach should seem intuitively
clear and simple, we caution the reader that it takes a lot of practice to become comfortable
with comparisons. We state the formal test below.
The Comparison Test. Suppose that 0 an bn for sufficiently

large n.
• If an diverges, then bn also diverges.
• If bn converges, then an also converges.
Before presenting the proof of the Comparison Test, note the phrase “sufficiently large
n”. By “0 an bn for sufficiently large n”, we mean that there is some number N such
that 0 an bn for all n N . This is just a formal way to say that we only care about
tails, and should remind the reader of the Tail Observation from Section 2.2.
Proof. Suppose that for all n, 0 an bn . This seems slightly weaker than the result we have
claimed, but the full result will then follow either by the Tail Observation of Section 2.2 or by an
easy adaptation of this proof.
Let sn denote the nth partial number of an and tn denote the nth partial sum of bn , so
sn a1 a2 an ,
tn b1 b2 bn .
From our hypotheses (that 0 an bn for all n), we know that sn tn for all n.
First suppose that an diverges. Because the terms an are nonnegative, the only way that an
can diverge is if sn as n (why?). Therefore the larger partial sums tn must also tend to
as n , so the series bn diverges as well.
Now suppose that bn converges, which implies by our definitions that tn bn as n .
The sequence sn is nonnegative and monotonically increasing because an 0 for all n, and
0 sn tn bn
so the sequence sn has a limit by the Monotone Convergence Theorem. This shows (again, by the
definition of series summation) that the series an converges.
The Comparison Test leaves open the question of what to compare series with. In
practice, however, this choice is usually obvious, and we will almost always compare with
a geometric series or a p-series. Our next four examples demonstrate the general technique.
1
Example 1. Does the series converge or di- sn
n 2
ln n 6
verge? 4
2
an
Solution. First note that we probably shouldn’t try to
apply the Integral Test in this example — the function 1 2 3 4 5 6 7 8 9 10
1 ln x has an antiderivative, but it has been proved
that its antiderivative cannot be expressed in terms of elementary functions.
However, the Comparison Test is easy to apply in this case. Note that
ln n n for n 2, so
1 ln n 1n for n 2.
1 1
Since is a divergent p-series, diverges by comparison.
n 1
n n 2
ln n
ln n
n 1
n
Solution. This example can be done with the Integral Test, but it’s easier to use the
Comparison Test. We know that ln n 1 for n 3, so
ln n n 1n for n 3.
1 ln n
Since diverges, must also diverge.
n 1
n n 1
n
2
cos n
n 1
n
Solution. We can write this series as cos2 n n2 . The numerator of this fraction, cos2 n , is
nonnegative for all n (this is important since we can’t apply the Comparison Test to series
with negative terms) and bounded by 1, so
cos2 n 1
for n 1.
n2 n2
2
cos n
Therefore since 1 n2 converges (it is a convergent p-series), the smaller series
n 1
n
must converge as well.
1
n 1
nln n
Solution. For n e2 7.39, ln n 2, so for these values of n,
1 nln n 1 n2 .
1 n2
1
Since is a convergent p-series, ln n
converges by comparison.
n 1
n
Sometimes the inequalities we need to apply the Comparison Test seem to go the wrong
way. Consider for example the series
1
n 1
2n 1
We would like to compare this series with the divergent series
1
,
n 1
2n
but the terms in our series seem to be smaller than the terms of 1 2n. Therefore we cannot
naively apply the Comparison Test in this case.
1
Example 5. Show that the series diverges.
n 1
2n 1
Solution. We have that

1 1 1 1
n 1
2n 1 3 5 7
1 1 1 1 1 1
,
4 6 8 n 2
2n 2n 2
n
so the series diverges by comparison to 1 n.
Our next example displays a similar phenomenon. Note that 1 n2 1 1 n2 , but we are
still able to compare the series.
1
Example 6. Show that the series converges.
n 2
n2 1
Solution. Because n2 1 n 1 2 , we have that
1 1 1 1
n 2
n2 1 3 8 15
1 1 1 1
,
1 4 9 n 1
n2
so the series converges by comparison to 1 n2 .
n
Example 7. Show that the series diverges.
n 1 n4 7
Solution. We should expect this series to diverge, because the numerator is n and the
denominator behaves like n2 , but the inequality goes the wrong way. By giving up a bit in
the denominator, however, we get the desired conclusion:
n n 1
,
n4 7 n4 7n4 8n
so the series we are interested in diverges by comparison to the harmonic series.
In Examples 5–7, we are really reindexing the series. This procedure is demonstrated
more formally in the example below and in Exercises 25–28. Another method for dealing
with such problems, known as the Limit Comparison Test, is discussed in Exercises 42–50.
n2 3
Example 8. Show that the series converges by reindexing the series with the
n 2
n4 2
substitution m n 1.
Solution. We want to compare this series to the series given by its leading terms, n2 n4
(or some multiple of this), but the comparison seems to go the wrong way. By setting
m n 1, which is equivalent to n m 1, we have
n2 3 m 1 2 3 m2 2m 4
.
n 2
n4 2 m 1
m 1 4 2 m 1
m4 4m2 6m2 4m 1
(Note here the change in the lower bound, as in the previous example.) The inequality in
the numerators (we want to compare m2 2m 4 with m2 ) still goes the wrong way, but
we can take care of this by using a slightly different inequality:
m2 2m 4 m2 2m2 4m2 7m2
for m 1. The inequality in the denominators does go the right way:
m4 4m2 6m2 4m 1 m4 .
Since we have made the numerators larger and the denominators smaller, we have made
the fractions larger, and thus
m2 2m 4 7m2 7
,
m 1
m4 4m2 6m2 4m 1 m 1
m4 m 1
m2
which implies by the Comparison Test that the series in question converges, because 7 m2
7 1 m2 is a convergent p-series.
Our next example doesn’t require reindexing, but does require a clever bound for ln n.
So far we have used the facts that ln n n for n 2 (in Example 1) and ln n 1 for n 3
(in Example 2). In fact, a much stronger upper bound holds. Let p be any positive real
number. Then by l’Hôpital’s Rule, we have
ln x 1x 1
lim lim lim 0.
x xp x pxp 1 x pxp
Recalling a fact from Section 2.1, this means that
ln n
lim 0
n np
for every p 0. This in turn means that for every p 0, ln n np for sufficiently large n, a
handy fact to have around for comparisons, as we demonstrate next.
n ln n
n 3 5
n 1
Solution. As we showed above, ln n n1 4 for sufficiently large n (we could give a

smaller bound, but 1 4 is good enough here) and n 3 5 n5 , we can use the comparison
n ln n n1 1 4 1
.
n 3 5 n5 2 n5 4
n ln n
Because 1 n5 4 is a convergent p-series, converges by the Comparison
n 3 5
n 1
Test.
Our last example is considerably trickier than the previous examples. The reader
should pay attention to the two themes it demonstrates: first, when dealing with a vari-
able in an exponent, it is a good idea to use e and natural log, and second, no matter how
slowly a function (such as ln ln n) goes to infinity, it must eventually grow larger than 2!
1
Example 10. Does the series ln n
converge or diverge?
n 2
ln n
Solution. As the terms have a variable in the exponent, we first manipulate the using e
and ln:
ln n
ln n ln n eln ln n eln n ln ln n nln ln n .
1
We now need to test for convergence. The approach from here on is similar to
n 2
nln ln n
2
Example 4: for n ee 1618.18 (i.e., for large n), we have ln ln n 2, so
1 1
,
nln ln n n2
1 1 n2 .
and thus ln n
converges by comparison to the convergent p-series
n 2
ln n
If a series converges by the Comparison Test, then we have the following remainder
estimate, which we conclude the section with.
The Comparison Test Remainder Estimate. Let an and bn be

series with positive terms such that an bn for n N . Then for
n N , the error in the nth partial sum of an , sn , is bounded by
bn 1 bn 2 :
sn an bn 1 bn 2 .
n 1
Proof. By definition,
sn an an 1 an 2 .
n 1
Now because the terms of an are positive, this is an 1 an 2 , and since we have assumed
that n N ,
an 1 an 2 bn 1 bn 2 ,
proving the estimate.
1
Example 11. How many terms are needed to approximate to within 1 10?
n 1
2n n
Solution. We use the comparison

n
1 1
2n n 2
1
for all n 1 to bound the error in approximating . The first partial sum may not
2n n
be a good enough approximation:
2 3 1 2
1 1 1 2 1
s1 1 .
n 1
2n n 2 2 1 2
2
The second and third partial sums are also not guaranteed to be as close to the true sum as
required:
1 3
1 1 3 1 4 2 1
s2 n 1 ,
n 1
2 n 2 2 1 2 4
4 5 1 4
1 1 1 2 1
s3 1 ,
n 1
2n n 2 2 1 2
8
but the fourth partial sum is within 1 10:
5 6 1 5
1 1 1 2 1
s4 1 .
n 1
2n n 2 2 1 2
16
Therefore the answer is that 4 terms will certainly approximate the series within 1 10.

In Exercises 1–4, assume that an and bn are 3n
16.
both series with positive terms. n 1
2n 5n
1. If an bn for sufficiently large n and bn is 2
n 1
convergent, what can you say about an ? 17.
n 1
3n2
2. If an bn for sufficiently large n and bn is e1 n
divergent, what can you say about an ? 18.
n 1
n
3. If an bn for sufficiently large n and bn is
convergent, what can you say about an ?
4. If an bn for sufficiently large n and bn is Suppose that an is a convergent series with pos-
divergent, what can you say about an ? itive terms. Determine whether the series listed
in Exercises 19–22 necessarily converge. If a se-
ries doesn’t necessarily converge, give an example
of a convergent series an with positive terms for
Determine if the series in Exercises 5–18 converge which it diverges. It may be helpful to remember
or diverge. that there are only finitely many values of an at least
1, so these have no affect on the convergence of the
n 3
5. series.
n 1
n3
an
19.
1 n 1
n
6.
2n 7
n 1 n 1
20. an
n
1 n 1
7.
n 1 n3 4
21. nan
4 n 1
8. 3
n 1 n5 8
22. an sin n
n 1
n 1
9.
n2 n
n 1 23. a2n
n 1
9
10.
n 1
3n 1 24. an
n 1
arctan n
11.
n 1 n3 1
n3 1 In Exercises 25–28, use reindexing like we did in Ex-

12.
n5 1 amples 5–8 to determine if the given series converge
n 1
or diverge.
1 1
13. 25.
n 1
n 1! n2 3
n 2
2 sin n 1
14. 26.
n 1
n5 4 2n 1
n 1
1 n 2
15. 27.
n 2
n 1 n 1
n2
1 Another way to deal with problems like Exer-

28.
n 1
n2 2n n cises 25–28 is to apply the following test.
The Limit Comparison Test. Let an and bn

an
be series with positive terms. If lim is a finite
Using the Comparison Test, determine if the series n bn
in Exercises 29–38 converge or diverge. number, then an and bn both converge or both
diverge.
1
29.
n 1
n n 1
1 Exercise 42 leads you through the proof of the

30. Limit Comparison Test. After that, Exercises 43–
n 1 n n 1
48 present applications, while Exercises 49 and 50
n 2 extend the Limit Comparison Theorem to the case
n
31. where the limit is 0 or .
n 1
3n 1 an
42. Suppose that lim c where c is a finite
n bn
2n ln n 4 number. Therefore there are positive numbers m
32. an
n 1 n4 4 and M with m c M such that m M for
bn
n
all large n. Use this inequality and the Comparison
n Test to derivate the Limit Comparison Test.
33.
n 1
n
43. Show that the series
n ln n
34. n2 3
nn 1 n
n 1
n 2
n4 2
1 from Example 8 converges using the Limit Compar-
35. ln n
n 1
ln ln n ison Test.
1 2n2 3n
36. 44. Does converge or diverge?
ln n ln ln n
n 1 5 n5
n 1
4n n2 2n 1
37. 45. Does 5
converge or diverge?
n! n 1 n1 1 11n
n 1
n2 2n 1
n! 46. Does converge or diverge?
38. 5
n9 11n
n 1
nn n 1
47. Suppose that an 0 and an 0. Show that

39. Construct an example showing that the Com-
sin an converges if and only if an converges.
parison Test need not hold if an and bn are not
required to have positive terms. 48. Suppose that 0 an 1 for all n. Prove that
arcsin an converges if and only if an converges.
40. If an , bn 0 and and a2n
both converge, b2n
show that the series an bn converges.
49. Let an and bn be series with positive
41. Prove that the series terms. If an bn 0 and bn converges, prove that
an converges.
1
np sin n 50. Let an and bn be series with positive
n 1
terms. If an bn and bn diverges, prove that
converges for p 2. What about when p 2? an diverges.

1. an converges, by the Comparison Test
3. You cannot conclude anything
5. Converges by comparison to n n3 1 n2
7. Converges by comparison to 1 n3 1 n3 2
9. Converges by comparison to n n5 2 1 n3 2
π 1
11. Converges by comparison to
2 n3 2
13. Converges by comparison to 1 n2
15. Diverges by comparison to 1 n
17. Converges by comparison to 1 9n2
an
19. Converges by the Comparison Test: an
n
21. Need not converge, consider taking an 1 n2
23. Since an converges, an 1 for sufficiently large n. For these values of n, a2n an , so a2n converges
by the Comparison Test
S ECTION 2.6 T HE R ATIO T EST 79
2.6. T HE R ATIO T EST

We now know how to handle series which we can integrate (the Integral Test), and series
which are similar to geometric or p-series (the Comparison Test), but of course there are a
great many series for which these two tests are not ideally suited, for example, the series
4n
.
n 1
n!
Integrating the terms of this series would be difficult, especially since the first step would
be to find a continuous function which agrees with n! (this can be done, but the solution is
not easy). We could try a comparison, but again, the solution is not particular obvious (in-
deed, those readers who solved Exercise 37 of the last section should feel proud). Instead,
the simplest approach to such a series is the following test due to Jean le Rond d’Alembert
(1717–1783).
The Ratio Test. Suppose that an is a series with positive terms

and let L lim an 1 an .
n
• If L 1 then an converges.
• If L 1 then an diverges.
• If L 1 or the limit does not exist then the Ratio Test is incon-
clusive.
You shold think of the Ratio Test as a generalization of the Geometric Series Test. For
example, if an ar n is a geometric sequence then
an 1
lim r,
n an
and we know these series converge if and only if r 1. (Note that we will only consider
positive series here; we deal with mixed series in the next section.) In fact, the proof of the
Ratio Test is little more than an application on the Comparison Test.
Proof. If L 1 then the sequence an is increasing (for sufficiently large n), and therefore the
series diverges by the Test for Divergence.
Now suppose that L 1. Choose a number r sandwiched between L and 1: L r 1. Because
an 1 an L, there is some integer N such that
0 an 1 an r
for all n N . Set a aN . Then we have
aN 1 raN ar,
and
aN 2 raN 1 ar2 ,
and in general, aN k ark . Therefore for sufficiently large n (namely, n N ), the terms of the
series an are bounded by the terms of a convergent geometric series (since 0 r 1), and so
an converges by the Comparison Test.
Since the Ratio Test involves a ratio, it is particularly effective when series contain
factorials, as our first example does.
4n
Example 1. Does the series converge or di- 50 sn
n 1
n!
verge? 25
Solution. First we compute L: an

1 2 3 4 5 6 7 8 9 10
4n 1
an 1 n 1! 4n 1 n! 4
L lim lim lim lim 0.
n an n 4n n 4n n 1! n n 1
n!
Since L 0, this series converges by the Ratio Test.
It is important to note that the Ratio Test is always inconclusive for series of the form
polynomial 1 n2 .
polynomial . As an example, we consider the harmonic series and
Example 2. Show that the Ratio Test is inconclusive for 1n and 1 n2 .
Solution. For the harmonic series, we have

1
n 1 n
L lim lim .
n 1 n n 1
n
In order to evaluate this limit, remember that we factor out the highest order term:
n 1
L lim 1 1,
n n 1 n
so the test is inconclusive.

The series 1 n2 fails similarly:
1
n 1 2 n2
L lim lim ,
n 1 n n 12
n2
and again we factor out the highest order term, leaving

n2 1
L lim 1 2 1,
n n2 1 n
so neither series can be handled by the Ratio Test.
As Example 2 demonstrates, knowing that an 1 an 1 is not enough to conclude that

the sequence converges; we must know that the limit of this ratio is less than 1.
10n
n 1
n42n 1
Solution. The ratio between consecutive terms is

10n 1
an 1 n 1 42n 3 10n 10
an 10n 42 n 1 16
n42n 1
as n . Since this limit is less than 1, we can conclude that the series converges by the
Ratio Test.
The last example could also be handled by the Comparison Test, since
n
10n 10n 1 10
,
n42n 1 42n 1 4 16
so the series converges by comparison with a convergent geometric series. However, what
if we moved the n from the denominator to the numerator:
n10n
?
n 1
42n 1
Now the inequality in the comparison goes the wrong way, making the Comparison Test
much harder to use. On the other hand, the limit in the Ratio Test is unchanged (you
should check this for yourself). In general, it is usually a good idea to try the Ratio Test on
all series with exponentials (like 10n ) or factorials.
2n !
n 1
2n n!
Solution. Here the ratio between consecutive terms is

2n 2!
an 1 2n 1 n 1! 2n 2 2n 1
2n 1
an 2n ! 2n 1
2n n!
as n . Since this limit is greater than 1 (or any other number, for that matter), the series
diverges by the Ratio Test.
Our last example could be done using the Comparison Test (how?), but it is (probably)
easier to use the Ratio Test.
n2 2n 1
Example 5. Does the series converge 3
n 1
3n 2
2
or diverge?
1
Solution. In this case the ratio between consecutive
terms is 1 2 3 4 5 6 7 8 9 10
n 1 2 2n 1 1
an 1 3n 1 2
an n2 2n 1
3n 2
n 12 2n 1 1 3n 2
,
n2 2n 1 3n 1 2
so pulling out the highest order terms, we have
1 2 1 1 1 2
an 1 n2 1 n 2 n n2 n2 3n 1 3n 1
an n2 1 2 n1 1
n2
3n 1 1 3n
2
1 3
as n . Because this limit is less than 1, the series converges by the Ratio Test.

Exercises 1–4 give various values of In Exercises 5–16, first compute
an 1 an 1
L lim . L lim ,
n an n an
In each case, state what you conclude from the Ratio and then use the Ratio Test to determine if the given
Test about the series an . series converge or diverge.
2n 5
1. L 2 5.
n 1
3n
2. L 1 7n 2
6.
n 1
2n6n
3. L 12
n3n
7.
4. L n 1
n 2
n3n Use the Root Test to determine if the series in Exer-

8.
n 1
n 4n cises 20–26 converge or diverge.
4n
1 3n
9. 20.
n! n 1
5n
n 1
2n n n2 1
n
10. 21.
n 1
n! 2n2 n
n 1
n! 2
11. n
2n ! 22. n
n 1
n 2
ln n
2n !
12. 1
n
n 1
n! 2 23. 1
n 1
n
n!
13. n
n! 1
n 1 24. 1
n 1
n
n!
14.
99n n! n2
n 1 1
25. 1
nn n 1
n
15.
n!
n 1 n2
1
n! 26. 1
16. n 1
n
n 1
nn
17. Find a sequence an of positive (in particular,
nonzero) numbers such that both an and 1 an
diverge. Exercises 27 and 28 show that the Root Test is a
stronger test than the Ratio Test.
18. Is there a sequence an satisfying the condi-
tions of the previous problem such that 27. Show that the Root Test can handle any se-
ries that the Ratio Test can handle by proving that if
an 1
lim L lim an 1 an exists then lim n an L.
n an n n
exists and is not equal to 1? 28. Show that there are series that the Root Test
can handle but that the Ratio Test cannot handle by
considering the series an where
A stronger test than the Ratio Test, proved by Au- n 2n if n is odd,

an
gustin Louis Cauchy (1789–1857), is the following. 1 2n if n is even.
The Root Test. Suppose that an 0 for all n and let
L lim n
an . The series an converges if L 1
n
n 1 In some cases where the ratio and root tests are in-
and diverges if L 1. (If L 1 then the Root Test conclusive, the following test due to Joseph Raabe
is inconclusive.) (1801–1859) can prove useful.
Raabe’s Test. Suppose that an is a positive series.

Our first task is to prove this result. If there is some choice of p 1 such that
19. Copying the beginning of the proof of the Ra- an 1 p
tio Test, give a proof of the Root Test. 1
an n
for all large n, then an converges. where bn 1 n 1 p.

n 1
31. Rewrite the inequality derived in Exercise 30
as
an 1 bn 1
Exercises 29–31 ask you to prove Raabe’s Test, while ,
Exercises 32 and 33 consider an application of the bn 1 bn
test. use this to show that an M bn for some positive
29. Show that if p 1 and 0 x 1 then number M and all large n, and use this to prove
1 px 1 x p. Raabe’s Test.
This is called Bernoulli’s inequality, after Johann
Bernoulli (1667–1748). Hint: Set f x px 1 x p. 32. Show that the Ratio Test is inconclusive for the
Show that f 0 1 and f x 0 for 0 x 1. series
Conclude from this that f x 1 for all 0 x 1. 1 3 5 2k 1
.
n 1
4 6 8 2k 2
30. Assuming that the hypotheses of Raabe’s Test
hold and using Exercise 29, show that
p 33. Use Raabe’s Test to prove that the series in
an 1 1 bn 1
1 Exercise 32 converges.
an n bn

1. The series diverges
3. The series converges
5. L 2 3, so the series converges by the Ratio Test.
7. L 3, so the series diverges by the Ratio Test.
an 1 1
9. 0 as n , so the series converges by the Ratio Test.
an n 1
an 1 n 1 n 1 1
11. as n , so the series converges by the Ratio Test.
an 2n 2 2n 1 4
an 1
13. n 1 as n , so the series diverges by the Ratio Test.
an
15. The ratio here is
n 1 n
an 1 n 1 n! n 1n 1 nn 1
1 .
an nn n 1! n 1 nn n 1 n n
Recall from Example 11 of Section 2.1 that the limit of this ratio is L e, so the series diverges by the
Ratio Test because e 1.
2.7. T HE A LTERNATING S ERIES T EST

We have focused almost exclusively on series with positive terms up to this point. In
this short section we begin to delve into series with both positive and negative terms,
presenting a test which works for many series whose terms alternate in sign.
The Alternating Series Test. Suppose that the sequence bn sat-

isfies the three conditions:
• bn 0 for sufficiently large n,
• bn 1 bn for sufficiently large n (i.e., bn is monotonically

decreasing), and
• bn 0 as n .
Then the alternating series
n 1
1 bn b1 b2 b3 b4
n 1
converges.
While we have stated the test with 1 n 1 , it of course applies if the terms involve
1 n instead (or cos nπ, since this is just a convoluted way to write 1 n ). Also, notice
that the Alternating Series Test can not be used to show that a series diverges (see Exam-
ple 2).
Proof of the Alternating Series Test. Assume that the sequence bn is positive and decreasing
for all n, and that it has limit 0. By the Tail Observation of Section 2.2, if we can prove that these
series converge, the full Alternating Series Test will follow.
Let sn denote the nth partial sum of this series. We have
s2n b1 b2 b3 b4 b2n 1 b2n .
Because bn is monotonically decreasing, b2n 1 b2n 0 for all n, so this shows that s2n is mono-
tonically increasing. We can also write
s2n b1 b2 b3 b4 b5 b2n 2 b2n 1 b2n ,
so since b2n 2 b2n 1 0, s2n b1 . Thus the sequence s2n has a limit by the Monotone Con-
vergence Theorem. Let L lim s2n . Now we consider the odd partial sums: s2n 1 s2n b2n 1 ,
n
so
lim s2n 1 lim s2n lim bn L,
n n n
S ECTION 2.7 T HE A LTERNATING S ERIES T EST 87
because bn 0 by our hypotheses. Since both the even and odd partial sums converge to the same
value, the sum of the series exists.
Example 1 (The Alternating Harmonic Series, again). Show that the alternating harmonic
series
1n 1
n 1
n
converges using the Alternating Series Test.
1
Solution. The sequence 1 n is positive, monoton- sn
ically decreasing, and has limit 0, so the alternating
harmonic series converges by the Alternating Series an
Test.
n 1 2n 3
Example 2. Does the series 1 con-
n 1
3n 4
verge or diverge?
Solution. This series does alternate in sign, and 2n 3 3n 4 is decreasing, but
2n 3 3n 4 23 0,
so the series diverges by the Test for Divergence.
Note that in the solution of Example 2, we did not appeal to the Alternating Series
Test, but instead used the Test for Divergence. The Alternating Series Test never shows that
series diverge.
n 4n2 4n2
Example 3. Show that the series 1 con- 1
n 1
n3 9 n3 9
verges.
bn
Solution. This series alternates in sign, and 4n2 n3
9 0, but it is not immediately obvious that the se- 1 2 3 4 5 6 7 8 9 10
quence 4n2 n3 9 is decreasing. Indeed, its first
three terms are increasing, as indicated in the plot. Of course, we only need the sequence
to be monotonically decreasing for large n. To check this condition, we take a derivative:
d 4x2 x3 9 8x 4x2 3x2 4x4 72x

.
dx x3 9 x3 9 2 x3 9 2
This fraction is negative for large x, so the sequence 4n2 n3 9 is decreasing for large
n. Therefore the series converges by the Alternating Series Test.
The proof of the Alternating Series Test implies the following very simple bound on
remainders of these series.
The Alternating Series Remainder Estimates. Suppose that the

sequence bn satisfies the three conditions of the Alternating Series
Test:
• bn 0,
• bn 1 bn , and
• bn 0 as n
for all n N . Then if n N , the error in the nth partial sum of

1 n 1 bn is bounded by bn 1 :
n 1
sn 1 bn bn 1.
n 1
Example 4. How many terms of the alternating series must we add to approximate the
true sum with error less than 1 10000?
Solution. Since the alternating harmonic series 1 n 1 n satisfies the conditions of the
Alternating Series Test for all n 0, the Remainder Estimates show that
1n 1 1
sn .
n 1
n n 1
Therefore, if we want the error to be less than 1 10000, we need

1 1
Error ,
n 1 10000
so we need n 9999, or in other words, n 10000.

In Exercises 1–12, determine if the given series con- 1n
15.
verge or diverge. n 1
n! 2
n
1 1 2
1. 1n
n4 7 16.
n 1
n 1
n
1 nn
2.
n 1
n!
2n Exercises 17–19 verify that the hypotheses of the Al-

3. ternating Series Test are all necessary, in the sense
n 1
3n
that if any of them is removed, then the statement
3n becomes false.
4.
n 1
2n 17. Construct a sequence bn which is monotoni-
cally decreasing with limit 0 such that 1 n 1 bn
n 1n 1 diverges. (I.e., bn needn’t be positive.)
5. 1
n 1
n
18. Construct a sequence bn which is positive and
1 n 1 monotonically decreasing such that 1 n 1 bn
6. diverges. (I.e., bn needn’t have limit 0.)
n 2
ln n
n 19. Construct a sequence bn which is positive
1 ln n
7. with limit 0 such that 1 n 1 bn diverges. (I.e.,
n
n 2 bn needn’t be monotonically decreasing.)
n
n
8.
n 1
n2
Dirichlet’s Test, due to Johann Peter Gustav Leje-
nn
9. une Dirichlet (1805–1859), is a strengthening of the
n3n
n 1 Alternating Series Test (as shown in Exercise 23).
1n
10. Dirichlet’s Test. If bn is a positive, eventually
n 1
arctan n
monotonically decreasing sequence with limit 0 and
n 1 n2 the partial sums of the series an are bounded,
11. 1 then an bn converges.
n 1
n3 1
3n
2
12.
n
n 1 Exercises 20–22 ask you to develop the proof of this
theorem, while Exercises 23–27 ask you to apply the
test.
For Exercises 13–16, first determine if the given se- 20. Let sn a1 a2 an . Use the fact that
ries satisfies the conditions of the Alternating Series sn sn 1 an to prove
Test. Then, if the series does satisfy the conditions,
decide how many terms need to be added in order a n bn s m bm 1 s n bn bn 1 .
to approximate the sum to within 1 1000. n m 1 n m 1
1 n sin n (This formula is often referred to as summation by

13.
n 1
n6 1 parts.)
n
1 21. Let an and bn be sequences satisfying the
14.
n 1
n 1 hypotheses of Dirichlet’s Test. Use Exercise 20 to
show that if the partial sums of the sequence an

are at most M then
26. Use the identity derived in Exercise 25 to show
a n bn 2M bm 1 . that
n m 1
m m
2 sin π 4 sin n cos n π 4 cos n π 4 ,
22. Use Exercise 21 to prove Dirichlet’s Test. n 1 n 1
23. Show that Dirichlet’s Test implies the Alter-

then show that this is equal to cos π 4 cos m π 4 .
nating Sign Test.
24. Suppose that an 2, 4, 1, 3, 2, 4, 1, 3, . . .
and that bn 1 n. Does an bn diverge, converge 27. Use Exercise 26 to show that the partial sums
absolutely, or converge conditionally? of sin n are bounded, and then conclude from
25. Use the angle addition identity Dirichlet’s Test that the series
cos α β cos α cos β sin α sin β. sin n

n
to derive the identity n 1
2 sin α sin β cos α β cos α β . converges.


1. Converges by the Alternating Series Test
5. Diverges by the Test for Divergence:
n 1
1 0
n
as n .
11. Converges by the Alternating Series Test. To see that bn is decreasing for sufficiently large n, take a
derivative.
13. Alternating Series Test not applicable.
15. The Alternating Series Test is applicable. Using n 4 will work to approximate the sum to within
1 1000, because
1 1 1
.
5! 2 14400 1000
2.8. A BSOLUTE VS . C ONDITIONAL C ONVERGENCE

We are now ready to examine the strange behavior of the alternating harmonic series we
first observed in Section 2.2. Remember that we showed that the alternating harmonic
series converged and then we went on to bound its sum. For a lower bound, we grouped
the terms in pairs, observing that
1n 1 1 1 1 1 1 1 1
1
n 1
n 2 3 4 5 6 7 8
1 1 1 1 1 1 1
1
2 3 4 5 6 7 8
1
1
2
1
.
2
While for an upper bound, we group the terms in different pairs, showing that
1n 1 1 1 1 1 1 1 1
1
n 1
n 2 3 4 5 6 7 8
1 1 1 1 1 1
1
2 3 4 5 6 7
1 1
1
2 3
5
0.8333 . . . .
6
(In fact, that true sum is ln 2 0.69315, see Exercises 46 and 47 of Section 2.4 or Exercise 24
of Section 3.2.)
Then we showed in Example 5 of Section 2.2 that by rearranging the terms of this series,
we could get it to converge to a different sum:
1 1 1 1 1 1 1 1 389 9
1 .
n 1
4n 3 4n 1 2n 3 2 5 7 4 420 10
Our first order of business in this section is to explore this phenomenon:

When are we allowed to rearrange the terms of a series without changing the sum?
We begin by looking at series with positive terms. If an is a convergent series with
positive terms, are we allowed to rearrange the terms without changing the sum? Suppose
bn is such a rearrangement, and consider the partial sums of each series,
sn a1 a2 a3 ,
tn b1 b2 b3 .
S ECTION 2.8 A BSOLUTE VS . C ONDITIONAL C ONVERGENCE 93
We would like to figure out if an bn which, by the very definition of series summa-
tion, is equivalent to lim sn lim tn . Because an and bn are positive sequences, the
n n
sequences of partial sums sn and tn are both increasing. Now consider any value of
n. Since the sequence bn is a rearrangement of the sequence an , there must be some
number N so that each of the terms
a1 , a2 , . . . , an
occurs in the list

b1 , b2 , . . . , bN .
Since all the terms are positive, for this value of N , we have
sn a1 a2 an b1 b2 bN tN .
This shows that every partial sum of an is less than or equal to some partial sum of bn .
Of course, the same argument works with the roles of an and bn interchanged, so every
partial sum of bn is less than or equal to some partial some of an . This implies that the
two series converge to the same value.
So we have made some progress: convergent series with positive terms can be rear-
ranged without affecting their sums, but rearranging the alternating harmonic series can
affect its sum. What is the difference between these two examples?
Intuitively, there are two different ways for a series to converge. First, the terms could
just be really small. Indeed, this is the only way that a series with positive terms can
converge. But then there is a second way, illustrated by the alternating harmonic series:
the terms could cancel each other out. Our next definition attempts to make precise the
notion of series that converge “because their terms are really small.”
Absolute Convergence. The series an is said to converge abso-

lutely if an converges.
The first thing we should verify is that absolutely convergent series actually, well, con-
verge. Our next theorem says even more: rearrangements don’t affect the sum of an abso-
lutely convergent series.
The Absolute Convergence Theorem. If an converges abso-

lutely, then an converges. Moreover, every rearrangement of an
converges to the same sum.
This first part of this theorem — that absolutely convergent series converge — follows
from the Comparison Test and some basic facts about series, see Exercises 25–26. The
second part is more complicated, and we omit its proof.
While we have defined absolute convergence in order to investigate rearranging series,
this notion is very useful on its own. Consider the series
1n 1
.
n 1
n2
Even though this series is very much like 1 n2 , it is not a p-series, so we can’t apply the p-
series Test to it. Similarly, we can’t apply the Integral Test or the Comparison Test, because
those tests require series to have positive terms. However, it is easy to see that this series
is absolutely convergent, from which it follows that the series converges by the Absolute
Convergence Theorem:
1n 1
Example 1. Show that the series converges absolutely.
n 1
n2
Solution. The series

1n 1
n 1
n2 n 1
n2
1 n
is a convergent p-series, so converges absolutely by the Absolute Convergence
n 1
n2
Theorem.
Our next example is another stereotypical use of the Absolute Convergence Theorem.
In general when trigonometric functions appear in a series, we need to test for absolute
convergence and then make a comparison.
sin n
Example 2. Show that the series converges absolutely.
n 1 n3 1
Solution. First we take the absolute values of the terms,
sin n sin n
.
n 1 n3 1 n 1 n3 1
We may use any test we like on this series (although some, like the Ratio Test in this exam-
ple, might not tell us anything). Because sin n 1, n3 1 n3 n3 2 , and the terms
of this series are positive, we can compare it:
sin n 1
.
n 1 n3 1 n 1
n3 2
sin n sin n
Therefore, is convergent by comparison to a convergent p-series, so
n 1 n3 1 n 1 n3 1
is absolutely convergent by the Absolute Convergence Theorem.
We’ve identified a special type of convergent series, the absolutely convergent series.
But what about the others? Intuitively, these are the series which converge only because
their terms happen to cancel each other out. These series are called conditionally conver-
gent.
Conditional Convergence. The series an is said to converge con-

ditionally if an converges but an diverges.
If you want to show that the series an is conditionally convergent, it is important

to note that this requires two steps. First you must show that an converges, and sec-
ond, you must show that an is not absolutely convergent (in other words, that an
diverges). Our first example of a conditionally convergent series should not come as a
surprise.
Example 3 (The Alternating Harmonic Series, last time). Show that the alternating har-
monic series
1n 1
n 1
n
is conditionally convergent.
1
Solution. The alternating harmonic series converges sn
by the Alternating Series Test because the sequence
1 n is monotonically decreasing, positive, and has
an
limit 0.
The alternating harmonic series does not converge
absolutely because
1n 1 1
n 1
n n 1
n
(the harmonic series) diverges. Therefore the alternating harmonic series is conditionally
convergent.
n 4n2
Example 4. Show that the series 1 converges conditionally.
n 1
n3 9
Solution. We saw in Example 3 of the previous section that this series converges, so we
only need to show that it does not converge absolutely. To test for absolute convergence,
we take the absolute value:
n 4n2 4n2
1 .
n 1
n3 9 n 1
n3 9
There are at least two different ways to show that this series diverges.
With the Integral Test: We must evaluate the integral
4x2 b
4x2
dx lim dx.
1 x3 9 b 1 x3 9
Setting u x3 9 gives du 3x2 dx, so dx du 3x2 . Making these substitutions leaves

us with
x b x b
4 4 4 4
lim du lim ln u lim ln b3 9 ln 10 ,
b x 1 3u b 3 x 1 b 3 3
so the series diverges by the Integral Test.

With the Comparison Test: Here we can use the bound
4x2 4x2 4
x3 9 x3 9x3 10x
to see that the series diverges by comparison to 4 10 1 n.
We’ve seen one example of how by rearranging the terms of the alternating harmonic
series we can change its sum. What if we wanted to rearrange the series to make it sum
to a specific number? Would that be possible? Yes! We begin with a specific example, and
then discuss how to generalize this example.
Example 5. Rearrange the terms of the alternating harmonic series to get a series which
converges to 1.
Solution. The positive terms of this series are
1 13 15 17 19
while the negative terms are
12 14 16 18 1 10 .
Note that both of these series diverge. By our Tail Observation of Section 2.2, this means
that all tails of these series diverge as well.
Now, how are we going to rearrange the series to make it sum to 1? First, we make the
series sum to more than 1:
1 13 1.3333 . . . 1.
Next we use negative terms to make the series sum to less than 1:
1 13 12 0.8333 . . . 1.
Then we use as many of the positive terms that we haven’t used yet to make the series sum
to more than 1 again:
1 13 12 15 1.0333 . . . 1,
and then use negative terms to make it sum to less than 1:
1 13 12 15 14 0.7833 . . . 1.
In doing so we obtain with the rearrangement

1
1 13 12 15 14 17 19 16 ,
1
1
1
but does this rearrangement really sum to 1? Mightn’t we get stuck at some point and not
be able to continue the construction?
We certainly won’t get stuck under 1. No matter how many of the positive terms we
have used up to that point, the positive terms that we have remaining will sum to (they
are a tail of the divergent series 1 1 3 1 5 ). Similarly, we can’t get stuck over 1.
Therefore, we will be able to create partial sums which are alternatively greater than 1 and
less than 1, but will they converge to 1? This follows because the terms we are using are
getting smaller. If we add the term 1 93 to get a partial sum over 1, that means that our
previous partial sum was under 1, which means that the new partial sum is within 1 93 of
1. As we use up the larger terms of the series, we will have no choice but to get closer and
closer to 1. Therefore this construction (if we carried it out forever) would indeed yield a
sum of 1.
Now we know we can rearrange the alternating harmonic series to sum to 1, but what
was so special about 1? Absolutely nothing, in fact. If you replace the number 1 in the
previous argument with any other number S, everything works just fine. Now, what was
so special about the alternating harmonic series? First, we needed that the positive terms
formed a divergent series and that the negative terms formed a divergent series (so that
our partial sums wouldn’t get stuck under or above 1). This fact is actually true for all
conditionally convergent series though (why?). Then we needed that the terms get in-
creasingly small, to prove that the limit of the partial sums was really 1. But if the terms
didn’t get close to 0, then the series would diverge by the Test for Divergence, so this is
true for all conditionally convergent series as well.
We have just sketched the proof of a famous theorem of Bernhard Riemann (1826–1866).
Reimann’s Rearrangement Theorem. If an is a conditionally

convergent series and S is any real number, then there is a rear-
rangment of sn which converges to S.
We conclude with a more formal proof.
Proof of Reimann’s Rearrangement Theorem We begin by dividing the terms of the sequence
an into two groups. Let bn denote the sequence which contains the positive terms of an and
cn denote the sequence which contains the negative terms of an .
Clearly an bn cn , so since an is not absolutely convergent, at least one of bn or
cn must diverge. But an is conditionally convergent, so if bn diverges (to ), cn must also
diverge (to ), and vice versa. Therefore both bn and cn diverge, to and , respectively.
Suppose that a target sum S is given. Choose N1 to be the minimal integer such that
b1 bN1 S
(note that if S is negative, then N1 will be 0). We can be certain that N1 exists because bn diverges
to . Note that, because b1 bN1 1 S, b1 bN1 is within bN1 of S. Next choose M1
minimal so that
b1 bN1 c1 cM1 S.
Again, M1 must exist because cn diverges to . Note that any partial sum of the form b1 bN1
c1 cn where n M1 must be within bN1 of S. Next choose N2 so that
b1 bN1 c1 cM1 bN1 1 bN2 S.
Next we choose M2 so that
b1 bN1 c1 cM1 bN1 1 bN2 cM1 1 cM2 S.
Continuing in this manner, define N3 , M3 , . . . . At each stage, our partial sums will be within bNi or
cMi of S for some i, and so since bn and cn (why?) we obtain a rearrangement that sums
to S, as desired.

n n
For Exercises 1–12, determine if the given series 1 x 2
16.
converge absolutely, converge conditionally, or di- n 1
3n n
verge. Note that these exercises may require the use
of all the tests we have learned thus far.
cos n
1. Determine whether the series in Exercises 17–20
n 1
n2
converge at x 2 and at x 4.
n n n2 x 1
n
2. 1 17.
n 1 3
n 1
n 0 n9 5
n
2
3. 2 n3
2n 1! 18. x 1
n 1
n 0
3n n7 2n
1n
4. n2 1 n
n 1
n ln n 19. x 1
n 1
2n 3n
1
5. 2 sin n x 1
n
n 2
n ln n 20.
n 0
n3 3
1
6.
n 1
n ln n ln ln n
n
1 Determine whether the series in Exercises 21–24
7.
n 1
n 4 converge at x 0 and at x 4.
n x 2n
8. 1 cos 1n 21.
n 1
2n n3 2
n 1
n
n 1 n 1 x 2
9. 1 22.
n 1
n2 1 n 1
n ln n 2
cos n4 sin n5 2 x n
10. 23.
n 1
n2 n2n 2n
n 0
n 4
1 n x 2 n sin n
11. 24.
n 1
en 2n
n 0
1n
12.
n 1
ln ln ln ln ln n
Exercises 25 and 26 prove the first part of the Abso-
lute Convergence Theorem: absolutely convergent
series converge.
Determine if the series in Exercises 13–16 converge
at x 1 and at x 5. 25. Verify the inequality
x 2n 0 an an 2 an
13.
n 1
3n
and use this to prove that if an is absolutely con-
x 2n vergent, then the series an an converges.
14.
n 1
3n n
26. Use the conclusion of Exercise 25 and Exer-
x 2n cise 27 from Section 2.2 to prove that all absolutely
15.
n 1
3n n2 convergent series converge.

1. Absolutely convergent (use a comparison on the absolute values)
3. Absolutely convergent (use the Ratio Test on the absolute values)
5. Absolutely convergent (use the Integral Test)
7. Conditionally convergent (use the Alternating Series Test, and then use a comparison on the absolute
values)
9. Conditionally convergent (use the Alternating Series Test, and then use a comparison on the absolute
values)
11. Absolutely convergent (use the Ratio Test on the absolute values)
13. Diverges at both x 1 and x 5
15. Converges at x 1, diverges at x 5
3. P OWER S ERIES
3.1. S ERIES AS F UNCTIONS

At the end of Chapter 1 we saw that Taylor polynomials of “infinite degree” might be
valuable for approximating functions. First we needed to understand what it meant to
add infinitely many numbers together, a notion that we formalized and studied in Chap-
ter 2. With these prerequisites covered, we return to the analysis of functions, beginning
by studying power series, which are series involving powers of x, such as
cn xn c0 c1 x c2 x2 c3 x3 .
n 0
This power series is centered at x 0. Our definition below is slightly more general.
Power Series. A power series centered at x a is a series of the form
n 2 3
cn x a c0 c1 x a c2 x a c3 x a .
n 0
The first question we should ask is:

Given a power series, for what values of x does it converge?
As the next three examples show, the techniques we have developed to analyze series are
capable of answering this question as well.
n
Example 1. Find the values of x for which the power series n! x 2 converges and
n 0
plot them on a number line.
The reader may have noticed that we have switched (mostly) from sums starting at n 1 to sums starting
at n 0. This is because when dealing with regular series, it is natural to index the first term as a1 , while with
power series it is more convenient to index the terms based on the power of x.
101
102 C HAPTER 3 P OWER S ERIES
Solution. We test the series for absolute convergence using the Ratio Test:
an 1 n 1!x 2n 1
n 1 x 2 unless x 2.
an n! x 2 n
Therefore the series converges only when x 2, so our plot is a single point,
4 3 2 1 0
x
showing that the power series converges only at x 2.
x 1 n
Example 2. For what values of x does the power series converge?
n 0
n!
Solution. We again test the series for absolute convergence using the Ratio Test:
x 1 n 1
an 1 n 1 ! x 1
x 1 n
0 for all x.
an n 1
n!
Therefore this series converges (absolutely) for every x, so our number line contains all
real numbers,
2 1 0 1 2
x
We can also write this as the interval , , or we may simply express it as the set of all
real numbers, R.
Our third and final example is a bit more interesting.
x 3n
Example 3. For what values of x does the power series converge?
n 0
n 1 4n
Solution. Again we begin by testing the series for absolute convergence with the Ratio
Test, although in this case we will need to work more afterward:
x 3 n 1
an 1 n 2 4n 1 x 3 n 2 x 3
x 3 n
.
an 4 n 1 4
n 1 4n
x 3
For what values of x is 1? This inequality can be rewritten as
4
x 3
1 1,
4
S ECTION 3.1 S ERIES AS F UNCTIONS 103
or, simplifying,
4 x 3 4,
so the given power series converges by the Ratio Test if 7 x 1, or in other words,
if x lies in the interval 7, 1 . The power series diverges if x 7 or x 1. But
when x 7 or x 1, the Ratio Test is inconclusive, so we have to test these endpoints
individually. This is typical for power series, not specific to this example.
Plugging in x 7, our series simplifies to
4n 1 n
.
n 0
n 1 4n n 0
n 1
Since this is the alternating harmonic series, we know that it converges (conditionally). So,
our power series converges at x 7.
Plugging in x 1, our series simplifies to
4n 1
.
n 0
n 1 4n n 0
n 1
This is the harmonic series which we know diverges, so our power series diverges at x 1.
Putting this all together, the given power series converges if and only if 7 x 1,
which we can also write as the interval 7, 1 . The number line plot of this interval is:
7 6 5 4 3 2 1 0 1
x
(Here the closed circle means that x 7 is included, while the open circle means that
x 1 is excluded.)
These examples have demonstrated three different types of convergence. As our next
theorem shows, every power series exhibits one of these three behaviors.
Radius Theorem. Every power series cn x a n satisfies one of

the following:
(1) The series converges only when x a, and this convergence

is absolute.
(2) The series converges for all x, and this convergence is abso-
lute.
(3) There is a number R 0 such that the series converges abso-

lutely when x a R and diverges when x a R. Note
that the series may converge absolutely, converge condition-
ally, or diverge when x a R.
The proof of the Radius Theorem is outlined in Exercises 40–45.

Case (1) of this theorem holds when the coefficients cn are “large”, while case (2) holds
when these coefficients are “small”. Case (3) holds for coefficients which lie somewhere
in between these two extremes. Note that for series which satisfy case (3), the interval of
convergence is centered at x a. We call the number R in this case the radius of convergence.
(In case (1) we might say that the radius of convergence is 0, while in case (2) we might say
that it is .)
When case (3) holds, there are four possibilities for the interval of convergence:
a R, a R, a R, a R, a R, a R, a R, a R.
In order to decide which of these is the interval of convergence, we must test the endpoints
one-by-one, as we did in Example 3. Therefore the general procedure for determining the
interval of convergence of a given power series is:
1. Identity the center of the power series.
2. Use the Ratio Test to determine the radius of convergence. (There are rare instances
in which the Ratio Test is not sufficient, in which case the Root Test should be used
instead, see Exercises 20–26 of Section 2.6.)
3. If the series has a positive, finite radius of convergence (case (3)), then we need to
test the endpoints a R and a R. These two series may be tested with any method
from the last chapter.
When a power series converges, it defines a function of x, so our next question is:
What can we say about functions defined as power series?
The short answer is that inside its radius of convergence, a power series can be treated like
a long polynomial. In particular, we can differentiate power series like polynomials:
Term-by-Term Differentiation. Suppose that
n 2 3
f x cn x a c0 c1 x a c2 x a c3 x a
n 0
converges for all x in the interval a R, a R . Then f is differen-

tiable for all values of x in the interval a R, a R , and
n 1 2 3
f x ncn x a c1 2c2 x a 3c3 x a 4c4 x a .
n 1
n 1
In particular, the radius of convergence of f x ncn x a
n 1
is at least R.
This theorem says quite a lot about the behavior of a power series inside its radius of
convergence. Not only can we differentiate such a power series, but the derivative has
at least as large a radius of convergence. Well then, there’s nothing stopping us from taking
the derivative of this derivative, and so on. Therefore, inside its radius of convergence,
a power series defines an infinitely differentiable, or smooth, function of x. Such functions
are extremely well-behaved. For one, remember that in order to be differentiable, a func-
tion must first be continuous, so inside its radius of convergence, a power series defines a
continuous function.
Integration can be handled the same way, by treating a power series as a long polyno-
mial inside its radius of convergence.
Term-by-Term Integration. Suppose that
n 2
f x cn x a c0 c1 x a c2 x a
n 0
converges for all x in the interval a R, a R . Then
x a n 1
f x dx cn C
n 1
n 1
x a 2 x a 3
c0 x a c1 c2 C,
2 3
and this series converges for all x in the interval a R, a R.
Term-by-term differentiation and integration should not seem obvious, and their justi-
fication takes quite a bit of work, even in more advanced courses. For now, we take them
for granted.

Find the intervals of convergence of the series in Ex- 18. 3, 7
ercises 1–10.
19. ,2
xn
1.
n 1
7
n 20. 7
3n xn
2.
n 0
n 23
Rewrite the expressions in Exercises 21–24 as series
3n x 2 n in which the generic term involves xn .
3.
n 0
n 23
2
21. nn 1 cn xn
n
4x n 2
4.
n4 3
n 0
22. cn xn
n n 0
4x
5.
n4 1
n 0 23. ncn xn 2x an xn
n n 1 n 0
4x 2
6.
n4 2
n 0 24. x nn 1 cn xn an xn
n 2 n 0
7. xn n!
n 0
x 2 n n! Explain why none of the functions plotted in Exer-

8.
n 0
2n cises 25–28 are equal to power series on the interval
3, 3 .
n
x 2 25.
9.
n 0
3n
1 3 5 2n 1
10. xn
n 0
n! 1
In Exercises 11–20, construct a power series with 2 1 1 2

the given intervals of converge, or explain why one
does not exist.
11. 2, 2
26.
12. 4, 0
13. 0, 2
1
14. ,
15. 0,
2 1 1 2
16. 1,
17. 3, 7
27.
31. Show that this series converges for all x.
32. Does this series define a periodic function of x,
1 as the plot above seems to demonstrate?
33. Verify that the actual series is within 1 100 of the
partial sum plotted above for all values of x.
2 1 1 2 34. Suppose you know, from using the Ratio Test,
that the radius of convergence of cn xn is R 6.
What is the radius of convergence of cn n3 xn ?
35. Suppose you know, from using the Ratio Test,
28. that the radius of convergence of cn xn is R 6.
What is the radius of convergence of cn xn 3n ?
36. Suppose that the radius of convergence of
cn xn is R 1. Then what is the radius of con-
1
vergence of sn xn where sn c0 c1 cn ?
37. Suppose that the radius of convergence of

2 1 1 2 cn xn is R 1. Then what is the radius of con-
vergence of sn xn where sn c0 c1 cn ?
29. Explain why the radius of convergence of the

power series for f x tan x centered at a 0 is at
most π 2. Use the bounds given by Exercises 50 and 50 of Sec-
tion 2.4 to find the radii of convergence of the series
30. Explain why a power series can converge con- in Exercises 38 and 39.
ditionally for at most two points. n! n
38. x
n 0
nn
2n!
Exercises 31–33 concern the series 39. xn
n 0 3n2 2n 1
2n cos nx
.
n 0
n!
Note that this is not a power series. Below, the 100th Exercises 40–45 detail the proof of the Radius The-
partial sum, orem. For simplicity, we assume that the series is
100 n centered at 0, that is, that a 0, but the proof eas-
2 cos nx
ily extends to other centers by making a change of
n!
n 0 variables, setting y x a.
of this series is plotted. 40. Suppose that the power series cn xn converges
at x s. Prove that there is an integer N so that
6 cn 1 s n for all n N .
41. With N as in the previous exercise, prove that if
x s and n N , cn xn x s n.
3
42. Using the previous two exercises and the Com-
parison Test, prove that if the power series cn xn
converges at x s then it converges absolutely
6 12 18 whenever x s.
43. Prove that if the power series cn xn diverges 45. Use the fact that every bounded set of real num-
at x t then it diverges whenever x t. bers has a least upper bound (this is called the Com-
44. Define C to be the set of values of x for which pleteness Property) to prove the Radius Theorem.
cn xn converges. Prove that either C contains all (The least upper bound b of the set C is the least
real numbers or C is bounded. number such that c b for all c C.)

1. 1, 1
3. 7 3, 53
5. 1 4, 1 4
7. Converges only for x 0

9. Converges for all real numbers
xn
11. One example is
n 0
2n
n
x 1
13. One example is
n 0
n2
15. No such power series exists, by the Radius Theorem
x 5n
17. One example is
n 0
n2n
19. No such power series exists, by the Radius Theorem
n
21. n 2 n 1 cn 2x
n 0
n n
23. n 1 cn 1x 2an 1x
n 0 n 1
25. Because the function has a discontinuity at x 2

27. Because the function has a sharp corner at x 1 (so its first derivative is not defined there)
29. Because tan x has a vertical asymptote at x π 2
3.2. M ANIPULATION AND D ERIVATION

We concluded the previous section by noting that power series can be differentiated and
integrated term-by-term (within their radii of convergence). This is quite a strong prop-
erty of power series (which does not hold for series in general). We begin this section by
showing that term-by-term differentiation and integration can be used to find power series.
Our starting point will always (in this section, at least) be a geometric series of some
kind, the simplest example of such being
1
xn for x 1,
n 0
1 x
which we know from our study of geometric series in Section 2.3. By differentiating (term-
by-term) both sides of this equation, we obtain our first new power series:
1 d 1 1
nxn 2
for x 1.
n 0
dx 1 x 1 x
This is one of the three basic forms of power series we derive in this section. The other two
are given in Examples 1 and 2.
Example 1. Find the power series centered at x 0 for ln 1 x and its radius of conver-
gence.
Solution. Recall that ln 1 x is the antiderivative of 1 1 x:
1
dx ln 1 x.
1 x
Furthermore, we can express 1 1 x as a geometric power series (for x 1):
1 1 n
x 1 n xn .
1 x 1 x n 0 n 0
Therefore, all we have to do to get the power series for ln 1 x is integrate this series
term-by-term,
1
ln 1 x dx
1 x
1 n xn
n 0
n xn 1
1 C.
n 0
n 1
S ECTION 3.2 M ANIPULATION AND D ERIVATION 111
But what is C, the constant of integration? To find C we substitute x 0 into both sides.
We know that ln 1 0 ln 1 0, so C 0. This gives
n xn 1
1 .
n 0
n 1
The geometric series we integrated had radius of convergence R 1, so the radius of

convergence of this series for ln 1 x is also R 1.
The power series for ln 1 x that we found in Example 1 is known as the Mercator
series, after Nicholas Mercator (1620–1687). Note that by substituting x 1 into this series,
we obtain
1n 1n 1
ln 2 ?
n 0
n 1 n 1
n
This seems to indicate that the sum of the alternating harmonic series is ln 2. However,
there is a problem with this line of reasoning: term-by-term integration is only guaranteed
to work inside the interval of convergence, and x 1 is an endpoint of the interval of
convergence for the Mercator series. Nevertheless, this computation can be made rigorous,
as shown by Abel, see Exercise 33. (Another proof of this result, using Euler’s constant γ
is given in Exercises 46 and 47 of Section 2.4.)
We move on to another example of using integration to derive a power series.
Example 2. Find the power series centered at x 0 for arctan x and its radius of conver-
gence.
Solution. For this we need to recall that

1
arctan x dx.
1 x2
Again, we can write 1 1 x2 as a geometric power series (for x 1):
1 1
x2 n
1 n x2n .
1 x2 1 x2 n 0 n 0
Now we integrate this series term-by-term:
1
arctan x dx
1 x2
1 n x2n
n 0
n x2n 1
1 C.
n 0
2n 1
Finally, we substitute x 0 into both sides of this equation to see that C 0, giving
n x2n 1
arctan x 1 for x 1.
n 0
2n 1
The geometric series we integrated had radius of convergence R 1, so the radius of

convergence of this series for arctan x is also R 1.
As in Example 1, this series suggests an intriguing equality:
π 1 n
arctan 1 ?
4 n 0
2n 1
But once again, x 1 lies on the endpoint of the interval of convergence for this series, so
this equation does not necessarily follow from what we have done. Nevertheless, as with
the previous example, this can be made precise using Abel’s Theorem, see Exercise 34.
(This is called the GregoryLeibniz formula for π, after Gottfried Leibniz (1646–1716) and
James Gregory (1638–1675)).
But what if we wanted power series for ln 1 3x2 or arctan 2x3 ? We could write them
as integrals and then integrate some form of geometric power series as in Examples 1 and
2, but this is tedious and error-prone. More worryingly, what about more complicated
functions like ln 1 x 1 2x ?
Just as with differentiation and integration, it turns out that within their radii of conver-
gence we may treat power series just like polynomials when
• substituting,
• multiplying, and
• dividing.
Here even stating the theorems is technical; we instead illustrate the point with examples.
1
Example 3. Find the power series centered at x 0 for .
1 8x3 2
Solution. We know from the beginning of the section that
1 1
2
nxn for x 1,
1 x n 0
so to get the power series for 1 1 8x3 2 , we simply replace x with 8x3 :
1 n 1
n 8x3 n8n 1 3n 3
x .
1 8x3 2
n 0 n 0
As the power series for 1 1 x 2 held when x 1, this power series holds when 8x3 1,
which simplifies to x 1 2.
Example 4. Find the power series centered at x 0 for ln 4 3x2 .
Solution. First we need to get the function in the form ln 1 something :
3x2 3x2
ln 4 3x2 ln 4 1 ln 4 ln 1 .
4 4
Now we substitute 3x2 4 into the power series we found for ln 1 x in Example 1:
n 1
3x2
3x2 n 4 n 3n 1 x2n 2
ln 4 ln 1 ln 4 1 ln 4 1 .
4 n 0
n 1 n 0
4n 1 n 1
Note that our series for ln 1 x was valid for x 1, so this new series is valid for
3x2 4 1, or x 4 3.
arctan 2x3
Example 5. Find the power series centered at x 0 for dx.
x3
Solution. First we substitute 2x3 into our power series for arctan x from Example 2 to
find a power series for arctan 2x3 :
2n 1
3 n 2x3 n2
2n 1 x6n 3
arctan 2x 1 1 .
n 0
2n 1 n 0
2n 1
Now we divide each of the terms of this series by x3 and integrate term-by-term:
2n 1 x6n 3
n2
1
arctan 2x3 n 0
2n 1
dx dx
x3 x3
2n 1 x6n
n2
1 dx
n 0
2n 1
n 22n 1 x6n 1
1 C.
n 0
2n 1 6n 1
Since our series for arctan x was valid for x 1, this series is valid for 2x3 1, or
3 1
x 2.
In practice, it can be quite tedious to find many coefficients by multiplication. However,

the first few coefficients are the most important.
ln 1 x
Example 6. Compute the first four nonzero terms of the power series for .
1 2x
Solution. We simply need to multiply the geometric power series for 1 1 2x with the
power series for ln 1 x that we found in Example 1:
ln 1 x x2 x3 x4
1 2x 4x2 8x3 x
1 2x 2 3 4
5x2 16x3 131x4
x
2 3 12
Since the series for 1 1 2x has radius of convergence 1 2 and the series for ln 1 x
has radius of convergence 1, the radius of convergence of their product is the minimum of
these two values, 1 2.
In the next section we use division to derive the power series for tan x. In order to show
how division works in this section, we repeat the previous example, dividing instead of
multiplying.
Example 7. Use division to compute the first four nonzero terms of the power series for
ln 1 x
the function f x .
1 2x
Solution. We use long division to divide the power series for ln 1 x by 1 2x:
x 5x2 2 16x3 3 131x4 12
1 2x x x2 2 x3 3 x4 4
2
x 2x
5x2 2 x3 3 x4 4
5x2 2 3
5x
16x3 3 x4 4
16x3 3 32x4 3
131x4 12
Note that this agrees with our computation in the previous example.

Find power series centered at x 0 for the func- Exercises 19 and 20 explore the Fibonacci numbers.
tions in Exercises 1–16 and give their radii of con- 19. Define
vergence.
f x fn xn ,
1
1. f x n 0
1 x where fn denotes the nth Fibonacci number. Use
2 the recurrence relation and initial conditions for
2. f x fn to show that
3 x
3 f x 1 xf x x2 f x ,
3. f x
1 x3 and derive from this that
4 1
4. f x f x .
2x3 3 1 x x2
2
5. f x
1 x 2 20. Use Exercise 19, partial fractions, and geometric
series to derive Binet’s formula.
x
6. f x 2
1 x
x2
7. f x Use partial fractions and geometric series to find
1 x5 2
formulas for the coefficients of the functions in Ex-
2 ercises 21–24.
8. f x x
4 2x2 2
21. f x
1 5x 6x2
9. f x ln 1 x 2 5x
22. f x
1 5x 6x2
10. f x ln 1 2x3 x
23. f x
1 3x 2x2
11. f x ln e e2 x 2 244 246x
24. f x
1 4x 3x2
12. f x x2 arctan 3x3
25. Using the fact that
13. f x arctan x 2 xn
ln 1 x for x 1,
n
arctan x n 1
14. f x
1 x show that
xn
15. f x 1 x ln 1 x lim ,
x 1
n 1
n
1 x 1 x
16. f x ln 1 x ln 1 x thereby proving (again) that the harmonic series di-
verges.
1
17. Show that n
ln 2. Hint: try substituting 26. Let an be a series with partial sums sn and
n 1
n2
an appropriate value of x into the series for ln 1 x . suppose that the function
f x an xn
18. Use division of power series to give another n 0
proof that converges for x 1. Prove that
1
xn f x
1 x n 0 sn x n .
n 0
1 x
for x 1.
27. The harmonic numbers Hn are defined by (this is also proved in Exercises 46 and 47 of Sec-
Hn 1 12 13 1 n. Show that tion 2.4), while Exercise 17 gives a different series
that converges to ln 2. Which of these two series
1 1
Hn xn ln . converges “faster”?
n 1
1 x 1 x
In 1995, after a several month long search by com-

The sequence an is said to be Abel summable to L puter, Bailey, Borwein, and Plouffe discovered a re-
if an xn converges on (at least) the interval 0, 1 markable formula for π which allows one to com-
and pute any binary or hexadecimal digit of π without
lim an xn L. computing any of the digits that come before it. For
x 1 details about how the formula was discovered, we
(Note that the first term of these sequences is a0 .) refer the reader to the book Mathematics by Experi-
In Exercises 28–32 consider Abel summability. This ment by Borwein and Bailey. Exercises 36–38 estab-
concept is due to Niels Henrik Abel (1802–1829). lish the formula.
28. Show that the sequence 2 n is Abel
36. Show that
summable to 2.
8x3 4 2x4 8x5
1 2
29. Show that 1 n is Abel summable to 1 2. 4 2
dx π.
(C.f. Exercise 50 from Section 2.2.) 0 1 x8
n
1
30. Show that is Abel summable to ln 2. Hint: Once we substitute u 2x, the integral be-
n 1
comes
1
31. Show that 1 n 1 n is Abel summable to 1 4. 16u 16
4
du,
(C.f. Exercise 51 in Section 2.2.) 0 u 2u3 4u 4
32. Show that 1 nn n 1 is Abel summable which, using partial fractions, is equal to
to 1 4.
1 1
4u 4u 8
du du.
0 u2 2 0 u2 2u 2
Abel’s Theorem guarantees that convergent series The first integral requires another substitution,
are Abel summable to their true values: while the second must be split into two integrals.
One can be evaluated by substitution, the other can
Abel’s Theorem. If the series an converges to a be done by writing u2 2u 2 as u 1 2 1.
finite value L, then an is Abel summable to L.
37. Let k be a fixed integer less than 8. Show that
xk 1
1 2
1 1
Assume the truth of Abel’s Theorem in Exercises 33 dx .
0 1 x8 2k 2 16n 8n k
and 34. n 0
1n 1
33. Show that ln 2.
n 1
n
38. Use Exercises 36 and 37 to show that π is equal
n
1 to
34. Show that 4 π.
n 0
2n 1
1 4 2 1 1
35. Exercise 33 shows that .
n 0
16n 8n 1 8n 4 8n 5 8n 6
n 1
1
ln 2
n 1
n This is the Bailey-Borwein-Plouffe formula for π.

1. 1 n xn
n 0
3. 3x3n
n 0
1
5. 2n 1 n xn
n 0
7. 1 n nx5n 3
n 0
xn 1
9.
n 0
n 1
1 2n 2
en x
11. 1
n 0
n 1
n x2n 1
13. 1
n 0
22n 1 2n 1
3.3. TAYLOR S ERIES

Having explored sequences, series, and power series, we are now ready to return to our
original motivation: Taylor polynomials. Recall that the Taylor polynomial of degree n for
f x centered at x a, Tn x , is the unique polynomial of degree n which matches f x
and its first n derivatives at x a.
Back in Section 1.1, we observed that higher degree Taylor polynomials tended to give
better approximations to f x . Then in Section 1.2 we made this observation precise with
the Remainder Theorem.
The Remainder Theorem. Suppose that f is n 1 times differen-

tiable and let Rn denote the difference between f x and the Taylor
polynomial of degree n for f x centered at a. Then
f n 1 c n 1
Rn x f x Tn x x a
n 1!
for some c between a and x.
Now that we have established the notion of series, we are ready to consider Taylor
polynomials of “infinite degree”, which we will call Taylor series . These are a special type
of power series.
Taylor Series. Suppose that the function f x is infinitely differen-

tiable (smooth) at x a. The Taylor series for f x centered at x a
is then
f n a
x a n.
n 0
n!
Assuming that we can compute all the derivatives of f x at x a, this definition is

very concrete, but it leaves some unanswered questions:
Where does the Taylor series for f x converge?

When the Taylor series for f x does converge, does it converge to f x ?
In general, these two questions must be answered on a function-by-function basis, and

such answers revolve around the Remainder Theorem.
Taylor series centered at x 0 are sometimes referred to as Maclaurin series after the Scottish mathemati-
cian Colin Maclaurin (1698–1746).
S ECTION 3.3 TAYLOR S ERIES 119
For our first case study, we consider the function f x sin x, centered at x 0.
Since the derivatives of sin x repeat in the pattern cos x, sin x, cos x, sin x, the sequence
of derivatives evaluated at x 0 is 1, 0, 1, 0, . . . . Therefore the Taylor series for sin x
centered at x 0 is:
n x2n 1 x3 x5 x7
1 x .
n 0
2n 1 ! 3! 5! 7!
We could apply the Ratio Test to this series (as in Section 3.1) to show that it converges for
all x, but there is a more important question: Does this series converge to sin x for all x?
By the Remainder Theorem, the difference between sin x and its Taylor polynomial of
degree n centered at x 0, Tn x , is
sin n 1 c n 1
sin x Tn x Rn x x
n 1!
for some c between 0 and x. Since the derivatives of sin x only take on values between 1
and 1, we have that
xn 1
Rn x .
n 1!
We would like to show that the Taylor series for sin x converges to sin x for all x, which is
equivalent to showing that the limit of the remainders Rn x is 0 (as n ). This is really
just an exercise from Section 2.1, although we have delayed it now.
xn
Fact 1. For all values of x, lim 0.
n n!
Proof. First let us consider a special case, where x 4. In this case, we are interested in the
sequence 4n n! . For the first 4 terms, this sequence is actually increasing:
n 4n n!
0 1
1 4
2 42 2! 8
3 43 3! 10.6666 . . .
4 44 4! 10.6666 . . .
But then the sequence starts decreasing. To get the n 5 term, we multiply the n 4 term by 4 5;
to get the n 6 term we multiply this by 4 6; to get the n 7 term we multiply this by 4 7, and so
on. For n 5, we therefore have the bound
n 4
4n 4 44
0 as n .
n! 5 4!
Our proof for generic x follows the same general approach. Choose an integer m x . We
have, for n m,
xn xn xn
.
n! n n 1 m 1 m m 1! mn m m 1!
n m terms, all m x
Ideally we would like to have mn in the denominator, so we multiply top and bottom by mm :
xn x n mm xn mm
.
n! mn m 1 ! mn m 1!
Our goal is to prove that the sequence x n n! converges to 0 by sandwiching it between the
constant sequence 0 and the sequence above.
Remember that m is not changing, so mm m 1 ! is just a constant. Since m x 0, x m
is between 0 and 1, so Example 5 of Section 2.1 shows that the geometric sequence x m n 0.
n
Therefore, by the Sandwich Theorem, x n! 0 as n for all values of x.
Our next example collects these observations to show that the Taylor series for sin x
centered at x 0 converges to sin x.
Example 2. Show that the Taylor series centered at x 0 for sin x converges to sin x for all
values of x.
Solution. Proving this statement is equivalent to showing that for any fixed value of x, the
remainder term Rn x tends to 0 as n . Fix a value of x. By the Remainder Theorem,
our previous bounds, and Fact 1,
xn 1
Rn x sin x Tn x 0 as n ,
n 1!
which is precisely what we wanted to show.
Example 2 is in some sense completely bizarre. Remember that we defined Taylor

polynomials (and therefore series as well) by mathcing derivatives at x a (the center).
Therefore, this example shows that if a power series matches all of the derivatives of sin x
at x 0, then the series is equal to sin x for all values of x. Why should sin x be completely
determined by its derivatives at 0? With our limited tools, the best answer we will be able
to provide here is that sin x is a very well-behaved function; the technical term for such
functions is analytic.
The same approach would work for the function f x cos x, but there is an easier
way. Now that we know that the Taylor series for sin x centered at x 0 converges to sin x
This type of convergence is known as pointwise convergence, because the Taylor series converges to sin x at
each individual point. There is a stronger, more global type of convergence known as uniform convergence.
for all values of x, we may differentiate it term-by-term to obtain:
d
cos x sin x
dx
d n x2n 1
1
dx n 0 2n 1 !
n x2n
1 2n 1
n 0
2n 1 !
n x2n
1
n 0
2n !
x2 x4 x6
1 .
2 4! 6!
For our next case study, we consider the function f x ex , again centered at x 0.
x x
Since every derivative of e is e itself, f n 0 1 for every n, and thus the Taylor series
centered at x 0 for ex is
xn x2 x3 x4
1 x .
n 0
n! 2 3! 4!
Example 3. Show that the Taylor series centered at x 0 for ex converges to ex for all
values of x.
Solution. We want to show that for fixed x, Rn x 0 as n . The Remainder

Theorem shows that
ec 1
Rn x ex Tn x xn
n 1!
for some c between 0 and x (here we used the fact that the n 1st derivative of ex is ex
itself). We have to be a little careful here about giving a bound for ec , since the bound
changes depending on whether x is positive or not:
ex if x 0,
ec
1 if x 0.
Still, since we are considering a fixed value of x, our bound on ec is just a constant, and
Fact 1 shows that
xn 1
0 as n .
n 1!
Therefore Rn x 0 as n , proving that the Taylor series for ex centered at x 0
converges to ex for all values of x.
Substituting x 1 into the Taylor series from Example 3 gives an appealing identity:
1 1 1 1
e 1 .
n 0
n! 1! 2! 3!
Because n! grows so quickly, this series converges to e exceptionally fast. For example,
using just the first 10 terms, we get e correct to six decimal places:
e 2.7182818 . . .
9
1
2.7182815 . . .
n 0
n!
The speed of this convergence (and it really is fast — at n 59, we are adding the reciprocal
of the estimated atoms in the universe, about 10 80 ) allows us to prove that e is irrational
in Exercises 50–52.
As with general power series, we are allowed to substitute into Taylor series and to
multiply and divide them. This can be quite useful for expressing integrals as power series
(our next example), evaluating limits (Example 5), computing derivatives (Example 6), and
finding Taylor series for quotients (Example 7).
Example 4. Express sin x2 dx as a power series.
Solution. We know from Example 2 that
n x2n 1
sin x 1 ,
n 0
2n 1 !
so
x2 2n 1 x4n 2
sin x2 1 n
1 n
.
n 0
2n 1! n 0
2n 1 !
All that remains is to integrate this series term-by-term:
x4n 2 x4n 3
sin x2 1 n
1 n
C.
n 0
2n 1 ! n 0
4n 3 2n 1!
It has been proved that the antiderivative of sin x2 dx is not an “elementary function”, so
this one of the nicest possible ways to express this integral.
cos x 1 x2 2
Example 5. Evaluate lim .
x 0 x4
Solution. While this example could be solved with l’Hôpital’s Rule, it is easier to use
power series. We know that
n x2n x2 x4 x6
cos x 1 1 ,
n 0
2n ! 2 4! 6!
so
x2 x2 x4 x6 x2
cos x 1 1 1
2 2 4! 6! 2
x4 x4
x4 x6
4! 6!
x4
1 x2
.
4! 6!
As x 0, this quantity approaches 1 4! 1 24, so this is the limit.
3
Example 6. Compute the 102nd derivative of f x e2x at x 0.
Solution. Example 3 shows that

xn
ex ,
n 0
n!
so by substitution, we have that
n
3 2x3 2n x3n
e2x .
n 0
n! n 0
n!
3
Since this is the Taylor series for f x e2x centered at x 0, by the definition of Taylor
series, we know that it is equal to
f n 0
xn .
n 0
n!
Therefore if we are interested in f 102 0 , we need only look at the coefficient of x102 . We
get the coefficient of x102 in
2n x3n
n 0
n!
by substituting n 102 3 34, so
234 f 102 0
.
34! 102!
Solving for f 102 0 shows that
102 234 102!

f 0 .
34!
We could have solved this problem simply by taking 102 derivatives, but finding the Taylor
series is much easier.
Example 7. Find the first four nonzero terms of the Taylor series centered at x 0 for the
function f x tan x.
Solution. It would be possible to solve this problem by taking derivatives of tan x, but
this approach gets quite messy. An easier method is to use the series we have for sin x and
cos x together with the Division Theorem from Section 3.2. Since
sin x x x3 3! x5 5! x7 7!
tan x x2 2! x4 4! x6 6!
,
cos x 1
we can use long division
x x3 3 2x5 15 17x7 315
1 x2 2 x4 24 x6 720 x x3 6 x5 120 x7 5040
x x3 2 x5 24 x7 720
x3 3 x5 30 x7 840
x3 3 x5 6 x7 72
2x5 15 4x7 315

2x5 15 x7 15
17x7 315
to find that the first four nonzero terms of the Taylor series centered at x 0 for tan x are
x3 2x5 17x7
x .
3 15 315
This is also therefore the Taylor polynomial of tan x of degree 7, centered at x 0.
We can also use the Taylor series for ex to define the function ex for more quantities
than just real numbers. For example, we can define ex for complex numbers x. Recall that
a complex number is a number of the form a bi where a and b are real numbers, and
i is the (imaginary) square-root of 1. This leads to the following formula, which the
Nobel Prize winning physicist Richard Feynman (1918–1988) referred to as “one of the
most remarkable, almost astounding, formulas in all of mathematics”:
Example 8 (Euler’s Formula). Prove Euler’s Formula eiθ cos θ i sin θ.

Solution. We begin by substituting iθ into the Taylor series for ex :
in θ n i2n θ 2n i2n 1 θ 2n 1
eiθ .
n 0
n! n 0
2n ! n 0
2n 1 !
Now we need to compute the powers of i:
i2 1,
3
i i i2 i,
4 2 2
i i i 1 1 1,
5 4
i i i i.
This shows that the powers of i form a periodic sequence i, 1, i, 1, . . . , which allows
us to simplify our series for eiθ above:
1 n θ 2n 1 n θ 2n 1
eiθ i cos θ i sin θ.
n 0
2n ! n 0
2n 1 !
An alternative proof, using derivatives, is outlined in Exercise 45.
A particularly wondrous special case of Euler’s Formula, known as Euler’s Identity, is,
eπi 1 0,
an identity relating five of the most important numbers in all of mathematics. In Exer-
cises 46–49 we use a similar approach to define eM for a matrix M .
Euler’s formula is also a very convenient way to prove various trigonometric identities.
For example, the angle addition formulas, which show how to evaluate sin α β and
cos α β in terms of sin α, sin β, cos α, and cos β, have a particularly straightforward
derivation using Euler’s formula. As the French mathematician Jacques Hadamard (1865–
1963) wrote, “the shortest route between two truths in the real domain sometimes passes
through the complex domain.”
Example 9. Derive the angle addition formulas.
Solution. We begin with the trigonometric functions we are interested in, convert to
exponentials, simply the expression, and then convert back to trigonometric functions:
cos α β i sin α β eα β i
eαi eβi
cos α i sin α cos β i sin β
cos α cos β sin α sin β i sin α cos β cos α sin β .
Now equate real and imaginary parts of both sides to see that
cos α β cos α cos β sin α sin β,

sin α β sin α cos β cos α sin β,
as desired.
We know from Section 3.1 that power series define infinitely differentiable (smooth)
functions when they converge, so not every function is equal to its Taylor series (in fact,
a function much be infinitely differentiable for us to even define its Taylor series). Our
success in this section and the last with finding Taylor series for infinitely differentiable
functions suggests a final question:
Is every infinitely differentiable function equal to its Taylor seres?
The answer is no. A counterexample is provided by the function
e 1 x2 if x 0,
f x
0 if x 0,
whose plot is shown below.
2 1 1 2
It is possible (using l’Hôpital’s Rule) to establish that every derivative of this function
at x 0 is 0, so therefore its Taylor series is simply 0, which does not converge to this
function. Therefore, infinite differentiability is a necessary condition for a function to equal
its Taylor series, but it is not a sufficient condition.

n
Derive Taylor series for the functions in Exercises 1– nπ
19. 1
8 at the specified centers. n 0
n!
1. f x sin 2x centered at x 0 π 2n
n
20. 7
2. f x cos x centered at x π 2 n 0
2n !
3. f x x2 cos 2x3 centered at x 0

2
4. f x ex centered at x 0 Derive Taylor series for the integrals in Exer-
x cises 21–24 (centered at x 0). Note that none of
5. f x 2xe centered at x 0
these integrals has an answer in terms of elemen-
6. f x 2xex centered at x 1 tary functions.
cos x3 1
ex e x
21. dx
7. f x sinh x centered at x 0 x2
2
sin x
ex e x
22. dx
8. f x cosh x centered at x 0 x
2
x2 2
23. e dx
Compute the limits in Exercises 9–16 using Taylor ln x 1

24. dx
series. x
sin x x
9. lim
x 0 x3
x arctan x
10. lim Use the Remainder Theorem in Exercises 25–30.
x 0 x3
25. Give an upper bound on the error when us-
ex e x
11. lim ing T3 x 1 x x2 2 x3 6 to approximate
x 0 x x
f x e for 0 x 1 2.
sin x arctan x
12. lim 26. Give an upper bound on the error when us-
x 0 x3
3 ing T1 x x to approximate f x tan x for
2 2 cos x
13. lim π 4 x π 4.
x 0 x6
ln 1 x3 27. Give an upper bound on the error when using
14. lim T3 x 1 2x 1 3 x 1 2 4 x 1 3 to
x 0 x3
approximate f x 1 x2 for 3 4 x 5 4.
sin x x 2
15. lim 28. Give an upper bound on the error when using
x 0 cos 5x 1 3
T2 x 2 x 4 4 x 4 2 64 to approximate
sin 3x 3x 2 f x x for 4 x 4.1.
16. lim
x 0 e2x 1 2x 3
29. Give an upper bound on the error when using
T3 x x 1 x 12 2 x 1 3 6 to approx-
imate f x x ln x for 2 x 3 2.
1
Use Taylor series of known functions to evaluate the

30. Give an upper bound on the error when using
sums in Exercises 17–20.
T4 x x x2 2 x3 3 x4 4 to approximate
π 2n
17. 1n f x ln 1 x for 0 x 1.
n 0
2n !
31. Prove that xn n! 0 as n for every fixed
2n 1
n π value of x. (Note that this is a bit stronger than
18. 1
n 0
2n 2n 1! Fact 1.)
For Exercises 32–35, give the Taylor polynomials of 39. Give a bound on the error involved in the esti-
degree 8 (centered at x 0) for the specified func- mate in Exercise 38 by observing that
tions.
4 12 4 12 5 12 6
32. f x ecos x R3 12
4 5 6
33. f x 1 x x2 sin x3 12 4
1 12 12 2 .
sin x 4
34. f x
1 2x
35. f x ln cos x
Exercises 40 and 41 concern the series

Exercises 36–39 investigate another way to bound
the error in using Taylor polynomials to approxi- πn
.
mate functions. Consider first the example of using n 2
n!
T3 x 1 x x2 2 x3 6 to approximate e1 2 . We
could of course use the Remainder Theorem for this
40. Compute the 10th partial sum of this series.
problem (as is requested in Exercise 25), but since
we know that 41. Prove that the sum of this series is not 19.
12 2 12 3 12 n
e1 2
1 12 , 42. Use Euler’s formula to prove the two identities
2 6 n 4
n!
eiθ e iθ
another way to estimate this error is simply to cos θ ,
2
bound the tail by comparing it to a geometric series:
eiθ e iθ
12 n 12 4 12 5 12 6
sin θ .
2i
n 4
n! 4! 5! 6!
12 4 12 5 12 6
43. Use Euler’s formula to prove De Moivre’s For-
4! 4 4! 42 4! mula,
12 4 12 12
n
1 cos θ i sin θ cos nθ i sin nθ.
4! 4 42
12 4 1 for all integers n.
.
4! 1
12
44. Give an example showing that De Moivre’s
4
(This gives an error estimate of about 0.003, Formula does not necessarily hold when n is not an
whereas the Remainder Theorem shows that the er- integer.
ror is at most 0.004.) 45. Define
36. Bound the error involved in using the Taylor iθ
polynomial of degree 5 to approximate sin x near f θ cos θ i sin θ e .
x 1 2 without using the Remainder Theorem.
Compute the derivative of f θ and use this to give
37. Bound the error involved in using the Taylor another proof of Euler’s Formula.
polynomial of degree 5 to approximate
1
x2 2
e dx
0
As with complex numbers, we also use the Taylor
without using the Remainder Theorem. series for ex to define eM for matrices square M as
38. Show that the Remainder Theorem cannot give
M2 M3
any practical bounds on the error involved in ap- I M ,
2! 3!
proximating f x ln 1 x at x 1 2 by its Tay-
lor polynomial of degree 3, T4 x x x2 2 x3 3. where I denotes the identity matrix (which has 1s
along the diagonal and 0s everywhere else). Use
this to compute eM for the matrices in Exercises 46– polynomials to power series (see Exercise 58). The
49. fact he assumed, however, does not hold for ar-
1 0 bitrary series, as Exercise 59 shows. It was not
46. M
0 1 until about 150 years later that Karl Weierstrass
(1815–1897) “corrected” this proof by proving the
1 0
47. M Weierstrass Factorization (or, Product) Theorem. It
0 1
should also be noted that Euler gave two other cor-
1 2 rect proofs for this calculation in the same paper,
48. M
0 1 and a fourth proof in 1741, but his “incorrect” proof
is the one that is best remembered. As the American
1 1 author Henry Mencken (1880–1956) wrote, “For ev-
49. M
0 1 ery problem, there is one solution which is simple,
neat, and wrong.”
55. Find a series centered at 0 which is equal to
Suppose that e were a rational number, so e ab
for positive integers a and b. In Exercises 50–52 we sin x

f x
will draw a contradiction, thereby proving that e is x
irrational. This proof is often attributed to Charles
for x 0.
Hermite (1822–1901).
50. Define 56. Find all the roots of the power series from Ex-
n ercise 55. This should involve two steps; first show
1 that this series has no negative roots, and then find
an n! e ,
k 0
k! all the positive roots, which are (by the previous
and prove that an is an integer for all n b. problem) also roots of
1 1 sin x
51. Show that an . f x .
n 1 n 1 n 2 x
52. Show that 0 an 1, and therefore an can not
be an integer, contradicting our assumption that e is
rational. 57. If a polynomial of degree 3 has roots r1 , r2 , and
r3 , then it is given by p x c x r1 x r2 x r3
for some constant c. By expanding this product, ver-
ify that
Is there an irrational number which raised to an ir-
rational power is rational? Exercise 54 proves that 1 1 1 coefficient of x
.
there is, although it doesn’t tell us precisely what it r1 r2 r3 constant term
is. I thank Professor Steven Landsburg for making (This generalizes to polynomials of any degree.)
me aware of these problems.
2
2 58. Assume, as Euler did, that Exercise 57 holds for
53. Verify that 2 2. (One method is to take series to show that
the logarithm of both sides.)
1 1 1 1
54. Using the result of the previous exercise and the r1 r2 r3 6
fact that 2 is irrational, prove that there is an irra-
tional number which raised to an irrational number where r1 , r2 , r3 , . . . are the roots from Exercise 56.
is rational. 1 π2
Conclude that 2
.
n 1
n 6
59. The function f x 2 1 1 x has a sin-

Exercises 55–59 present Euler’s original computa- gle root, x 1 2. Derive its power series (centered
tion of 1 n2 , from 1735. While this proof was at x 0) and conclude that, contrary to Euler’s as-
considered a breakthrough at the time, Euler mis- sumption, Exercise 57 cannot be applied to arbitrary
takenly assumed that he could apply a fact about series.

2n 1 2n 1 2n 1
n 2x n2 x
1. 1 1
n 0
2n 1! n 0
2n 1!
3 2n n 6n n 6n 2
2x n2 x n2 x
3. x2 1 n
x2 1 1
n 0
2n ! n 0
2n ! n 0
2n !
xn xn 1
5. 2x 2
n 0
n! n 0
n!
1 xn 1 x n
x2n 1
7. 2
n 0
n! 2 n 0
n! n 0
2n 1 !
2 4
x x 1
9. The series is , so the limit is 1 3! 1 6.
5! 3!
7!
2x2 2x4
11. The series is 2 , so the limit is 2.
3! 5!
x2
13. The series is 1 , so the limit is 1.
4
2 41x2
15. The series is , so the limit is 2 140625.
140625 468750
17. cos π 0
π
19. e
6n 2
nx n x6n 1
21. 1 dx 1 C
n 1
2n ! n 1
6n 1 2n !
n x2n n x2n 1
23. 1 dx 1 C
n 0
2n n! n 0
2n 2n 1 n!

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2.

2.1. A P RE -R EQUISITE : S EQUENCES

an 2, 4, 6, 8, 10, 12, 14, . . . ,

an 0, 4, 3, 2, 11, 29, 54, 59, 35, 41, 46, . . . .

Example 1. Guess a formula for the general term of the sequence

A sequence grows geometrically (or exponentially) if each term is obtained by multiplying

Wolfram Alpha (http://www.wolframalpha.com/) is an incredible resource for performing calculations

Does the sequence converge or diverge?

0.3, 0.33, 0.333, 0.3333, . . .

converges to 1 3, while the sequence

converges to 0. Slightly more formally, the sequence an converges to the number L if

Converges to L. The sequence an is said to converge to L if for

To indicate that the sequence an converges to L, then we may write

Example 2. Show, using the definition, that the sequence 1n converges to 0.

Solution. Let ! 0 be any positive number. We want to show that

Example 4. Determine the limit as n of the sequence n 2 n .

Diverges to Infinity. The sequence an is said to diverge to infinity

Example 5. Showing, using only the definition of convergence, that:

(b) lim r n 0 if 1 r 1 (i.e., if r 1),

(c) lim r n does not exist if r 1.

for all n N . Again taking logarithms, this holds if n ln r ln !. Since r 1, its

and the initial condition a1 1.

A more complicated recurrence relation is provided by the Fibonacci numbers fn n 0,

fn n 0 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, . . . .

Monotone Sequences. The sequence an is (weakly) increasing if

It is good to practice identifying monotone sequences. Here are a few examples to

Bounded Sequences. The sequence an is bounded if there are

For monotone sequences, boundedness implies convergence:

The Monotone Convergence Theorem. Every bounded mono-

Example 6. Prove that the sequence defined recursively by a1 2 and

Solution. First we ask: for which values of n is an an 1? Substituting the definition of

The Sandwich Theorem. Suppose the sequences an , "n , and

Theorem. If an is a function satisfying an f n and lim f x

Why might we want to switch from sequences to functions? Generally we do so in

l’Hôpital’s Rule. Suppose that c is a real number or c , and

The Continuous Function Limits Theorem. Suppose the se-

We conclude the section with two examples using this theorem.

Our final example is less straight-forward, but more important.

Applying l’Hôpital’s Rule, we see that

so the solution to the example is e.

E XERCISES FOR S ECTION 2.1

4. 5 2, 7 5, 9 10, 11 17, 13 26, . . . 7n3 5

7. 1 2, 2 4, 6 8, 24 16, 120 32, . . . 19. an 2n 2 2n

Prove the limit laws states in Exercises 31–34.

46. Define the sequence an recursively by

34. If an is a convergent sequence, then

lim can c lim an

with this property. Determine the periods of the se-

Compute the sequence f 0 , f 0 , f 0 , f 3 0 , . . .

68. Define the sequence an recursively by

Compute this sequence when a1 13 and a1 24.

69. Compute the first 10 terms of the sequence

77. A partition of the integer n is a way of writing 83. 3, 3, 5, 4, 4, 3, 5, 5, . . .

The distinguished Israeli mathematician Noga Alon recounts:

8 Secondly, we might chop through a 2 1 domino,

The number of n 1 tilings is fn , so there are fn2

f2n fn2 fn2 1. 94. Prove the Sandwich Theorem.

A NSWERS TO S ELECTED E XERCISES , S ECTION 2.1