CHAPTER 4

Taylor series. Taylor expansions

4.1. Taylor series

Definition 4.1. Let an , ∀n ∈ N and x0 be real numbers. The Taylor series

with the coefficients an , n ∈ N and centered at x0 is the function series
∑
an (x − x0 )n = a0 + a1 (x − x0 ) + a2 (x − x0 )2 + .... (4.1)
n≥0

Remark that the set of converge of a Taylor series is non-empty, since it

always contains its center, x0 .

Definition 4.2. A Maclaurin series (or, simply, a power series) is a

Taylor series having the center x0 = 0, that is,
∑
an xn = a0 + a1 x + a2 x2 + ...,
n≥0
the real numbers an , n ∈ N being called the coefficients of the series, too.

∑
Definition 4.3. Let n≥0 an (x − x0 )n be a Taylor series with the coefficients
an , n ∈ N and centered at x0 . With the convention 1/∞ = 0 and 1/0 = ∞, the
number r ≥ 0 defined by
1
r := √ (4.2)
lim sup n |an |
n→∞
is called the radius of convergence of the Taylor series. Relation (4.2) is
called the Cauchy-Hadamard formula.

∑
Theorem 4.1. (Abel-Cauchy-Hadamard Theorem). Let n≥0 an (x − x0 )n be
a Taylor series with the coefficients an , n ∈ N, and center x0 .
a) If r = 0, then the only point of convergence is x0 ;
b) If r > 0, then the series is absolutely convergent on the interval (x0 − r,
x0 + r) and it is divergent on (−∞, x0 − r) ∪ (x0 + r, ∞) ;
51
52 4. TAYLOR SERIES. TAYLOR EXPANSIONS

c) If r > 0, then the series is uniformly convergent on every interval [a, b] ⊂

(x0 − r, x0 + r) ;
d) If the power series converges for x = x0 + r (or x = x0 − r), then its
sum is continuous in x0 + r (respectively x0 − r);
e) If r > 0, then the sum of the series admits derivatives of any order in
the interval (x0 − r, x0 + r), and these derivatives can be determined by differ-
entiating term by term the series;
f ) If r > 0, then the series can be integrated term by term on every interval
[a, b] ⊂ (x0 − r, x0 + r).

 
 
Remark 4.1. Due to a property of the sequences, if there exists limn→∞  an+1an ,
√
then it also exists
 lim
 n→∞
n
|an | and the two limits are equal. Hence, if there
 an+1 
exists limn→∞  an , the radius of convergence can be determined as
 
 an 

r = lim  ,
n→∞ an+1 

with the same convention, 1/∞ = 0 and 1/0 = ∞.

Example 4.1. Find the set of convergence and the sum of the Taylor series
∑ n+1 xn
n≥1 (−1) n , x ∈ R.
By applying Cauchy-Hadamard formula (4.2), we get that the radius of
convergence is
1
r= √  = 1.
 xn 
lim sup n (−1)n+1 n 
n→∞

Therefore, by applying Abel-Cauchy-Hadamard Theorem, the series is abso-

∑ on (−∞, −1) ∪ (1, ∞) .
lutely convergent on (−1, 1) and divergent
For x = −1, the series becomes − n≥1 n1 , and it is divergent. For x = 1,
∑
the series becomes n≥1 (−1)n+1 n1 , and it is convergent, since Leibniz’s test.
So, the set of convergence is (−1, 1]. Let S : (−1, 1] → R,
∞
∑ xn
S (x) = (−1)n+1 .
n
n=1

By applying again Theorem 4.1, we deduce that

∞
∑ 1
S ′ (x) = (−x)n−1 = , ∀x ∈ (−1, 1) .
1+x
n=1
 4.1. TAYLOR SERIES 53

Since S (0) = 0, we obtain that S (x) = ln (1 + x) , ∀x ∈ (−1, 1). But the sum
of the series is also continuous in x = 1, due to Theorem 4.1. Hence,
S (1) = lim S (x) = lim ln (1 + x) = ln 2.
x→1 x→1
Let us notice that in Theorem 4.1 we can deduce properties of the sum of
the series, if we know the coefficients an , n ∈ N, and the center x0 .
The main problem that rises is the converse. How can we find a Taylor se-
ries whose sum is a given function ? Such a function will be called expandable
in Taylor series.

Definition 4.4. Let f : I → R be a function and x0 ∈ I, where I is an open

interval. f is called exapandable in Taylor series about the point x0 iff
there exists (an )n∈N and ε > 0, with the following properties:
1) (x0 − ε,∑x0 + ε) ⊆ I, ε ≤ r, where r is the convergence radius of the
Taylor series n≥0 an (x − x0 )n ;
∑
2) ∀x ∈ (x0 − ε, x0 + ε), we have f (x) = ∞ n
n=0 an (x − x0 ) .
Therefore the following two issues appear.
A) In which conditions a given function is expandable in Taylor series ?
B) How can we effectively determine the coefficients an , n ∈ N, by knowing
function f ?
Regarding the second issue, we have the following result.

Theorem 4.2. Let I be an open interval, f : I → R be a function, and x0 ∈ I

be fixed. If f is expandable in Taylor series about the point x0 , then f does
admit derivatives of any order and
f (n) (x0 )
an = , ∀n ∈ N. (4.3)
n!
Due to Theorem 4.2, the existence of the all derivatives represents a nec-
essary condition for a function to be exapandable in Taylor series. But this
condition is not sufficient. Indeed, if we consider the function f : R → R,
defined by
{
e− x2 , if x ̸= 0,
1

f (x) =
0, if x = 0,
then f is differentiable of any order, with
f (0) = f ′ (0) = ...f (n) (0) = ... = 0, ∀n ∈ N.
Hence, if f would be expandable in Taylor series about 0, then we would have
∞
∑ ∞
∑ f (n) (x0 ) n
f (x) = an xn = x = 0, ∀x ∈ (−ε, ε) .
n!
n=0 n=0
54 4. TAYLOR SERIES. TAYLOR EXPANSIONS

Remark 4.2. By Theorem 4.2 we deduce that if f is expandable in Taylor series

about the point x0 , then there exists a single Taylor series whose sum equals f
(n)
on (x0 − ε, x0 + ε) , i.e. the Taylor series having the coefficients an = f n!(x0 ) ,
∀n ∈ N. This series is also called the Taylor series associated to f on the
interval (x0 − ε, x0 + ε) about the point x0 .
Therefore, regarding the first issue, it is enough to establish the condi-
tions for an infinitely differentiable function to be the sum of the Taylor series
associated on (x0 − ε, x0 + ε) .

4.2. Taylor’s formula

Let f : I → R be a n + 1 times differentiable function on the open interval
I. If x0 ∈ I, then the Taylor polynomial of order n, centered at x0 is
∑
n
f (k) (x0 )
Tn (x) := (x − x0 )k
k!
k=0
and the difference
Rn (x) = f (x) − Tn (x)
is called the Lagrange remainder of the Taylor’s formula. Taylor’s formula
is
f (x) = Tn (x) + Rn (x)
that represents the link between the function and its Taylor polynomial, and
gives us an estimate of the remainder.
Recalling the Mean Value Theorem from the high school, for every x0 ,
x ∈ I, there exists a c between x0 and x, such that
f (x) − f (x0 ) = (x − x0 ) f ′ (c) .
In this situation, n = 0, f is continuous on I and differentiable on the interior
of I (which is I, too), Tn (x) = f (x0 ), and Rn (x) = (x − x0 ) f ′ (c) .
The Taylor’s formula has many applications in practice. An example of
Taylor’s formula of order 2 is the formula in kinematics, that describes the
linear motion of an object having uniform acceleration,
at2
x (t) = x (0) + v0 t + .
2
In the case of differentiable functions of higher order, we have the following
general result.

Theorem 4.3. (Taylor’s formula). Let f : I → R be a function which is n + 1

 4.2. TAYLOR’S FORMULA 55

times differentiable on the open interval I. Then, for every x0 , x ∈ I, there is

a c between x0 and x, such that
f ′ (x0 ) f (n) (x0 ) f (n+1) (c)
f (x) = f (x0 ) + (x − x0 ) + ... + (x − x0 )n + (x − x0 )n+1 .
1! n! (n + 1)!

Remark 4.3. For a n + 1 times differentiable function on the open interval I,

the Taylor’s formula has the form
(x − x0 )n
f (x) = Tn (x) + ω (x) ,
n!
where ω : I → R is a function such that limx→x0 ω (x) = ω (x0 ) = 0.

Remark 4.4. Let us notice that the coefficients an , n ∈ N in (4.3) are exactly
the ones from the Taylor’s formula. Therefore, the Taylor polynomial Tn (x)
equals the partial sum of rank n of the Taylor series associated to f about the
point x0 .
And our first issue is geting now an answer.

Theorem 4.4. Let f : I → R be an infinitely differentiable function on the

open interval I and x0 ∈ I. Then f is expandable in Taylor series about the
point x0 if and only if for every x ∈ I, limx→x0 Rn (x) = 0.
In application, the following test will be of high importance.

Theorem 4.5. (Cauchy’s test). Let I be an open interval and f : I → R be an

infinitely differentiable function. Suppose that there exist M > 0 and δ > 0,
such that
 
 
sup f (n) (x) ≤ M δ n n!, ∀n ∈ N.
x∈I

Then
∞
∑ f (n) (x0 )
f (x) = (x − x0 )n ,
n!
n=0

for all x0 , x ∈ I with |x − x0 | < 1δ .

Another test, that is equivalent to the Cauchy’s test is the following.

Theorem 4.6. Let I be an interval and f : I → R be a function, infinitely

56 4. TAYLOR SERIES. TAYLOR EXPANSIONS

differentiable on a neighborhood V of x0 ∈ I. Suppose that there exists M > 0

such that
 
 
sup f (n) (x) ≤ M, ∀n ∈ N.
x∈I
Then
∞
∑ f (n) (x0 )
f (x) = (x − x0 )n , ∀x ∈ V.
n!
n=0
∑
Theorem 4.7. If n≥0 an (x − x0 )n is convergent for x = x1 , then the series
converges uniformly for every z in the segment from x0 to x1 , i.e. for z ∈
{(1 − t) x0 + tx1 , t ∈ [0, 1]}.

Example 4.2. We prove that the Taylor expansion about 0 (or the Maclaurin
series) of the the function f (x) = ln (1 + x) is
x2 x3
ln (1 + x) = x − + + ..., ∀x ∈ (−1, 1]. (4.4)
2 3
Estimating its derivatives, we get, by mathematical induction,
(n − 1)! (−1)n+1
f (n) (x) = , ∀n ≥ 1.
(1 + x)n
Hence, ∀δ ∈ (0, 1) ,
 
 (n)  n!
sup f (x) ≤ , ∀n ∈ N,
x∈(−δ,δ) (1 − δ)n
and so, from Theorem 4.5 we deduce that (4.4) is fulfilled ∀x ∈ (−1, 1) . For
∑ n
x = 1, Leibnitz’s test ensures us that the Maclaurin series, n≥1 (−1)n+1 xn is
also convergent. Since f and the sum S of the series are continuous on (−1, 1],
and f (x) = S (x) , ∀x ∈ (−1, 1), by passing to limit as x → 1, we get that (4.4)
is also fulfilled for x = 1.

4.3. Exercises
(1) Find the sets of convergence of the following Taylor series:
∑ ∑ x2n−1
a) n≥1 n!xn , x ∈ R; b) n≥1 (−1)n+1 (2n−1)(2n−1)! , x ∈ R;
∑ xn ∑ ( n+1 )n2 n
c) n≥1 n(n+1) , x ∈ R; d) n≥1 n x , x ∈ R;
∑ n ( )
e) n≥0 (−1) n √ 2 tgn x, x ∈ − 2 , 2 .
1 π π
32 n +1

A: a) r = 0, the set of convergence is {0} ;

b) r = ∞, the set of convergence is R;
c) r = 1, the set of convergence is [−1, 1] ;
 4.3. EXERCISES 57

d) r = 1/e, the set of convergence is [−1/e, 1/e] ;

e) r = π3 , the set of convergence is [− π3 , π3 ].
(2) Find the sets of convergence and the sums of the following Taylor
series: ∑ ∑
4n+1 2n+1
a) n≥0 x4n+1 , x ∈ R; b) n≥0 (−1)n x2n+1 , x ∈ R;
∑ xn+1
c) n≥1 (−1)n+1 n(n+1) , x ∈ R.
{
0, if n = 4k, 4k + 2, or 4k + 3,
A: a) The coefficients are an = 1
n , if n = 4k + 1.
By applying Cauchy-Hadamard formula (4.2), we get that the
radius of convergence is

1
r= √ = 1.
lim sup n |an |
n→∞

Therefore, by applying Abel-Cauchy-Hadamard Theorem, the se-

ries is absolutely convergent on (−1, 1) and divergent on (−∞, −1) ∪
(1, ∞) . ∑ −1
For x = −1, the series becomes n≥1 4n+1 , and it is divergent.
∑ 1
For x = 1, the series becomes n≥1 4n+1 , and it is divergent. So, the
set of converge is (−1, 1).
∑
Let S : (−1, 1) → R, S (x) = ∞ x4n+1
n=0 4n+1 . By applying Theorem
4.1, we deduce successively

∞
∑ ∞
∑
′
( )n 1
S (x) = x4n
= x4 = , ∀x ∈ (−1, 1) .
1 − x4
n=0 n=0

Since S (0) = 0, we obtain that

1 1 1+x
S (x) = arctg x + ln , ∀x ∈ (−1, 1);
2 4 1−x

b) The set of convergence is [−1, 1] and the sum of the series is

S (x) = arctg x, ∀x ∈ [−1, 1] ;
c) The set of convergence is (−1, 1) and the sum of the series is
S (x) = (x + 1) ln (x + 1) − x, ∀x ∈ (−1, 1) .
(3) Determine the following sums, by using Taylor series:
∑ ∑ ∑
a) ∞ n=1 (−1)
n+1 1
; b) ∞ n=1 (−1)
n+1 1
; c) ∞ n 1
n=0 (−1) 6n+1 ;
∑∞ 4n−3 ∑ 3n−2
d) n=1 (−1)n+1 n(2n−1)1
; e) ∞ 1
n=0 (4n+1)(4n+3) .
58 4. TAYLOR SERIES. TAYLOR EXPANSIONS

A: a) By applying Theorem 4.1, we deduce successively

∑∞ ∑∞ ∫ 1
n+1 1 n+1
(−1) = (−1) x4n−4 dx
4n − 3 0
n=1 n=1
∫ 1 (∑∞
)
n+1 4n−4
= (−1) x dx
0 n=1
∫ (∞
) ∫
1 ∑ ( ) 1
1
4 n
= −x dx = dx
0 0 1 + x4
n=0
√ ( √ )
2 2+ 2
= π + ln √ ;
8 2− 2
∑ ( )
b) Analogously, we obtain ∞ n=1 (−1) n+1 1
3n−2 = 1 √
3
π
3
+ ln 2 ;
c) By applying Theorem 4.1, we deduce successively
(∞ )′ ∞ ( ) ∞
∑ 6n+1 ∑ 6n+1 ′ ∑
n x n x
(−1) = (−1) = (−1)n x6n
6n + 1 6n + 1
n=0 n=0 n=0
∑∞
( )n 1
= −x6 = , ∀x ∈ (−1, 1) .
1 + x6
n=0
So
∞
∑ ∫
x6n+1
n 1
S (x) : = (−1) = dx
6n + 1 1 + x6
n=0
= F (x) + c, ∀x ∈ (−1, 1) ,
where
1 [ ( √ ) ( √ )]
F (x) = 2 arctg x + arctg 2x − 3 + arctg 2x + 3
6√ √
3 x2 + 3x + 1
+ ln √ .
12 x2 − 3x + 1
Since S (0) = 0, we deduce that c = 0, and hence S (x) = F (x) ,
∀x ∈ (−1, 1) ;
∑ ∑∞ ∑∞
d) ∞ n=1 (−1)
n+1 1
n(2n−1) = 2 n=1 (−1)
n+1 1
2n−1 − n=1 (−1)
n+1 1
n;
∑∞ 1 ∑∞ 1 ∑∞
e) n=0 (4n+1)(4n+3) = 2 n=0 4n+1 − 2 n=0 4n+3 .
1 1 1

(4) Expand in Taylor series about 0 the following functions, by specifying

their sets of convergence:
a) f (x) = cos x, x ∈ R; b) f (x) = sin x, x ∈ R; c) f (x) = ex ,
x ∈ R;
d) f (x) = (1 + x)α , x > −1, where α ∈ R;
 4.3. EXERCISES 59

( √ )
e) f (x) = ln x + 1 + x2 , x ∈ R;
f) f (x) = 2x
(x−1)2
, x ∈ R \ {1} ; g) f (x) = 2x+1
x2 −3x+2
, x ∈ R \ {1, 2};
h) f (x) = x2 e−x , x ∈ R; i) f (x) = sin 2x + x cos 2x, x ∈ R;
j) f (x) = ln 1+x
1−x , x ∈ (−1, 1); k) f (x) = sin x cos x, x ∈ R;
2
l) f (x) = (1 + e ) , x ∈ R.
x

A: a) By mathematical induction, we prove that

( nπ )
f (n) (x) = cos x + , ∀x ∈ R, ∀n ∈ N.
2
Since
   ( nπ )
 (n)  
 f (x)  
= cos x +  ≤ 1, ∀x ∈ R, ∀n ∈ N,
2
by Theorem 4.6 we get that f is expandable in Taylor series about 0
on R and the following Taylor expansion about 0:
∞
∑
x2 x4 x2n x2n
cos x = 1 − + − ... + (−1)n + ... = (−1)n , ∀x ∈ R.
2! 4! (2n)! (2n)!
n=0
b) We deduce similarly that f is expandable in Taylor series about
0 on R and has the following Taylor expansion about 0:
∞
∑
x3 x5 x2n+1 x2n+1
sin x = x − + − ... + (−1)n + ... = (−1)n , ∀x ∈ R.
3! 5! (2n + 1)! (2n + 1)!
n=0

c) Since f (n) (x) = ex , ∀x ∈ R, ∀n ∈ N, we deduce that

 
 (n) 
f (x) ≤ eα , ∀x ∈ [−α, α] , ∀α > 0.
So f is expandable in Talor series about 0 on each interval [−α, α] ,
∀α > 0, hence f is expandable in Talor series about 0 on R. We easily
deduce the following Taylor expansion about 0:
∞
∑
x x2 xn xn
ex = 1 + + + ... + + ... = , ∀x ∈ R.
1! 2! n! n!
n=0
d) We get the following Taylor expansion about 0:
∑∞ ( )
α α
(1 + x) = xn , ∀x with |x| < 1, (4.5)
n
n=0
( )
α
where denotes the generalized combinations
n
( )
α α (α − 1) ... (α − n + 1)
:= .
n n!
60 4. TAYLOR SERIES. TAYLOR EXPANSIONS

Relation (4.5) is also called binomial expansion of power α and

generalizes the well-known
√ Newton’s binomial formula.
′
e) Since f (x) = 1/ 1 + x2 , x ∈ R, for |x| < 1 we get, by using
the binomial expansion of power −1/2,
( √ ) ∑∞
(2n − 1)!!
2
ln x + 1 + x = x + (−1)n n x2n+1 , ∀x ∈ (−1, 1) ;
2 (2n + 1) · n!
n=1

f) Since f (x) = 2x−2+2 2 , ∀x ∈ R\ {1}, we easily

2 2
(x−1)2
= x−1 + (x−1)
deduce the Taylor expansions about 0:
∑∞ ∑∞ ( )
−2 α
f (x) = + 2 (1 − x)2 = −2 xn + 2 xn
1−x n
n=0 n=0
∞
∑ [( ) ]
α
= 2 − 1 xn , ∀x ∈ (−1, 1) ;
n
n=0
g) By decomposing f into simple ratios, we get
2x + 1 −3 5
f (x) = 2 = + , ∀x ∈ R\ {1, 2} .
x − 3x + 2 x−1 x−2
Hence, we deduce
3 −5/2
f (x) = +
1 − x 1 − x/2
∞ ( )
5 ∑ ( x )n ∑
∑∞ ∞
5
= 3 x −
n
= 3 − n+1 xn ,
2 2 2
n=0 n=0 n=0
for all x ∈ (−1, 1) ;
∑ (−x)n ∑ (−1)n n+2
h) x2 e−x = x2 ∞ n=0 n! = ∞ n=0 n! x , ∀x ∈ R;
i)
∑∞ ∑∞
(2x)2n+1 (2x)2n
sin 2x + x cos 2x = (−1)n +x (−1)n
(2n + 1)! (2n)!
n=0 n=0
∑∞
2n + 3 2n+1
= (−1)n 22n x , ∀x ∈ R;
(2n + 1)!
n=0
∑∞
j) f (x) = ln 1+x
1−x = 2
n=0 2n+1 x
2n+1 , x ∈ (−1, 1) ;
k)
∞
1 1∑ (2x)2n+1
f (x) = sin 2x = (−1)n
2 2 (2n + 1)!
n=0
∞
∑ (−1)n 22n
= x2n+1 , ∀x ∈ R;
(2n + 1)!
n=0
 4.3. EXERCISES 61

l) We calculate the first derivative

∞
∑ xn
f ′ (x) = 2 (1 + ex ) = 2 + 2 , ∀x ∈ R
n!
n=0
so, by Theorem 4.1, we deduce
∞
∑ xn+1
f (x) = 2x + 2 + C, C ∈ R, ∀x ∈ R.
(n + 1)!
n=0
Since f (0) = 4, it follows that C = 4. Therefore,
∑∞
xn+1
(1 + ex )2 = 4 + 2x + 2 , ∀x ∈ R.
(n + 1)!
n=0