Infinite Series of Positive and Negative Terms
Infinite Series of Positive and Negative Terms
Infinite Series of Positive and Negative Terms
Illustration 1
An example of and alternating series of the form 1, where the first term is positive is
+∞
1 1 1 1 1
∑(−1)n+1 = 1 − + − + ⋯ + (−1)n+1 + ⋯
n 2 3 4 n
n=1
An example of and alternating series of the form 2, where the first term is negative is
+∞
1 1 1 1 1
∑(−1)n = −1 + − + + ⋯ + (−1)n + ⋯
n! 2! 3! 4! n!
n=1
Example 1
Prove that the following alternating series is convergent:
+∞
1 1 1 1 1
∑(−1)n+1 = 1 − + − + ⋯ + (−1)n+1 + ⋯
n 2 3 4 n
n=1
Example 2
n+2 n2
Determine whether the series is convergent or divergent: ∑+∞
n=1(−1)
n and ∑+∞
n=1(−1)
n .
n(n+1) n2 +5
Illustration 4
In Example 1 we proved that the series
+∞
1
∑(−1)n+1
n
n=1
is convergent. This series is not absolutely convergent because the series of absolute values is the harmonic series, which
is divergent.
The series in Illustration 4, sometimes called alternating harmonic series, is an example of a conditionally convergent
series.
Theorem
If the series ∑+∞ +∞
n=1 |un | is convergent, then the series ∑n=1 un is convergent.
Example 3
nπ
cos 3
Determine whether the series is convergent or divergent: ∑+∞
n=1(−1)
n+1 .
n2
+∞ 1 1 1 1 1 nπ
1 1 cos
2 2 2 2 2
∑ un = 2 − 2 − 2 − 2 − 2 − 2 − 2 − ⋯ + 3 −⋯
1 2 3 4 5 6 7 n 2
n=1
+∞
1 1 1 1 1 1 1
∑ un = − − − + + + −⋯
2 8 9 32 50 36 98
n=1
This is a series of positive and negative terms. We can prove this series is convergent if we can show that it is absolutely
convergent.
+∞ +∞ nπ
cos
∑ |un | = ∑ 3
n2
n=1 n=1
nπ
Because |cos | ≤ 1 for all n
3
nπ
|cos |
3 ≤ 1 for all positive integers n
n 2 n2
1
The series ∑+∞
n=1 is the p series, with p = 2, and is therefore convergent. So by the comparison test ∑+∞
n=1 |un | is
n2
convergent. The given series is therefore absolutely convergent; hence, it is convergent.
The ratio test, given in the next theorem, is used frequently to determine whether a given series is absolutely
convergent.
un+1
i) if lim | | = L < 1, the series is absolutely convergent;
n→+∞ un
un+1 un+1
ii) if lim | | = L > 1 or if lim | | = +∞, the series is absolutely divergent;
n→+∞ un n→+∞ un
u
iii) if lim | n+1| = 1 no conclusion may be made from this test.
n→+∞ un
Example 4
n
Determine whether the series is convergent or divergent: ∑+∞
n=1(−1)
n+1 .
2n
n+1
n n+1 u 2n+1 n+1
Solution: un = (−1)n+1 and un+1 = (−1)n+2 . Therefore | n+1| = n =
2n 2n+1 un 2n
2n
un+1 n+1 1
So lim | | = lim = < 1.
n→+∞ un n→+∞ 2n 2
Therefore, by the ration test, the given series is absolutely convergent and hence, it is convergent.
Example 5
n+2
In example 2 we showed that the series ∑+∞
n=1(−1)
n is convergent. Is this series absolutely convergent or
n(n+1)
conditionally convergent?
n2 + 3n
= lim
n→+∞ n2 + 4n + 4
=1
So the ratio test fails. Because
n+2 n+2 1 1
|un | = = ⋅ >
n(n + 1) n + 1 n n
1
the comparison test can be applied. And because the series ∑+∞
n=1 is the harmonic series, which diverges, we conclude
n
that the series ∑+∞ +∞
n=1|un | is divergent and hence ∑n=1 un is not absolutely convergent. Therefore the series is not
conditionally convergent.
un+1
Note that the ratio test does not include all possibilities for lim | | because it is possible that the limit does not exist
n→+∞ un
and is not +∞.
Example 6
32n+1
Apply the root test to determine whether the series is convergent or divergent: ∑+∞
n=1(−1)
n
.
n2n
Example 7
1
Determine whether the series is convergent or divergent: ∑+∞
n=1 .
[ln(n+1)]n
Exercises
Determine whether the alternating test is convergent or divergent.
1
1. ∑+∞
n=1(−1)
n+1
2n
1
2. ∑+∞
n=1(−1)
n
n2
∑+∞ n 3
3. n=1(−1) 𝑛2 +1
∑+∞ n 1
4. n=1(−1) ln 𝑛
𝑛
∑+∞ n+1 3
5. n=1(−1) 𝑛2