Section 10.4: Other Convergence Tests
In this section, we learn how to deal with series whose terms are not necessarily positive.
Definition: An alternating series is a series whose terms alternate in sign. That is, a series
of the form
∞
∞
X
X
n
(−1) an
or
(−1)n+1 an ,
n=1
n=1
where an > 0 for all n.
The following is an example of an alternating series:
∞
X
(−1)n+1
n=1
n
=1−
1 1 1
+ − + ··· .
2 3 4
Theorem: (Alternating Series Test)
∞
X
(−1)n an , where an > 0 converges if it satisfies both of the following
The alternating series
n=1
conditions:
1. lim an = 0,
n→∞
2. The sequence {an } is decreasing. That is, an+1 ≤ an .
Example: Determine whether the following series are convergent.
(a)
∞
X
(−1)n
√
n+1
n=1
Check the two conditions of the Alternating Series Test:
1
=0
n→∞
n+1
2. The sequence {an } is decreasing since
1. lim √
an+1 = √
1
1
<√
= an
n+2
n+1
The series converges by the Alternating Series Test.
for n ≥ 1.
(b)
1
1
1
1
1
−
+
−
+
+ ···
ln 2 ln 3 ln 4 ln 5 ln 6
∞
X
(−1)n
.
ln n
Check the two conditions of the Alternating Series Test.
The series can be expressed as
n=2
1
=0
n→∞ ln n
2. The sequence {an } is decreasing since
1. lim
an+1 =
1
1
<
= an
ln(n + 1)
ln n
for n ≥ 2.
The series converges by the Alternating Series Test.
(c)
∞
X
(−1)n 2n
n=1
5n + 1
Check the two conditions of the Alternating Series Test:
2n
2
= 6= 0.
n→∞ 5n + 1
5
lim
The series diverges by the Divergence Test.
(d)
∞
X
(−1)n+1 ln n
n
n=1
Check the two conditions of the Alternating Series Test:
ln n
1
= lim = 0
n→∞ n
n→∞ n
2. It is not clear whether the sequence {an } is decreasing. Consider the corresponding
function f (x) = ln x/x. Then
1. lim
f 0 (x) =
1 − ln x
.
x2
Now f 0 (x) < 0 if and only if 1 − ln x < 0. That is, 1 < ln x or x > e. So the
sequence {an } is eventually decreasing.
The series converges by the Alternating Series Test.
Definition: A series
∞
X
∞
X
an is absolutely convergent if the series
n=1
∞
X
an converges, but
n=1
∞
X
|an | converges. If
n=1
|an | diverges, the series is conditionally convergent.
n=1
Note: If
∞
X
an is a series with positive terms which converges, then it is absolutely convergent.
n=1
Example: Determine whether the following series are absolutely convergent, conditionally
convergent, or divergent.
∞
X
1
(a)
n3
n=1
This is a convergent p-series with p = 3. Since the series has only positive terms, it is
absolutely convergent by the p-Series Test.
(b)
∞
X
(−1)n
√
n
n
n=1
The series is absolutely convergent since
∞
∞
∞ X
X
(−1)n X
1
1
√ =
√ =
n n
3/2
n n n=1 n
n=1
n=1
is a convergent p-series with p = 3/2.
(c)
∞
X
(−1)n+1
n=1
2n + 1
Consider the series
∞ ∞
X
(−1)n+1 X
1
=
.
2n + 1 2n + 1
n=1
Apply the Limit Comparison Test with
n=1
∞
X
n=1
1
.
n
1
n
1
lim 2n + 1 = lim
= ∈ (0, ∞).
1
n→∞
n→∞ 2n + 1
2
n
Since the harmonic series is divergent, the series does not converge absolutely.
Check the two conditions of the Alternating Series Test:
1
=0
n→∞ 2n + 1
2. The sequence {an } is decreasing since
1. lim
an+1 =
1
1
<
= an
2n + 3
2n + 1
for n ≥ 1.
The series is conditionally convergent by the Alternating Series Test.
(d)
∞
X
(−1)n n2
n=1
3n2 + 1
The series is divergent since
1
n2
=
6= 0.
n→∞ 3n2 + 1
3
lim
∞
X
(−1)n an is a convergent alternating series with sum S and the nth partial
Suppose that
n=1
n
X
sum Sn =
(−1)j aj is used to approximate S. How good is this approximation?
j=0
Theorem: (Remainder Estimate for Alternating Series)
∞
X
(−1)n an is a convergent alternating series with sum S, then
If
n=1
|Rn | = |S − Sn | ≤ an+1 .
Example: Consider the alternating series
∞
X
(−1)n
n=1
n2
.
(a) Prove that the series is absolutely convergent.
The series is absolutely convergent since
∞ ∞
X
(−1)n X
1
=
n2 n2
n=1
n=1
is a convergent p-series with p = 2.
(b) Use S6 to approximate the sum of the series and estimate the error.
The 6th partial sum is
S6 = −1 +
1 1
1
1
1
− +
−
+ .
4 9 16 25 36
By the Remainder Estimate for Alternating Series,
1
.
49
|R6 | ≤ a7 =
Example: How many terms of the series
∞
X
(−1)n n
n=1
4n
do we need to add in order to approxi-
mate the sum to within 0.002?
By the Remainder Estimate for Alternating Series,
|Rn | ≤ an+1 =
n+1
.
4n+1
To achieve the indicated accuracy, we must have
n+1
≤ 0.002.
4n+1
If n = 5, then
n+1
6
= 6 ≈ 0.0015 < 0.002.
n+1
4
4
So we need at least n = 5 terms.
Example: Approximate the sum of
X (−1)n
n=0
2n n!
to within four decimal places.
By the Remainder Estimate for Alternating Series,
|Rn | ≤ an+1 =
1
2n+1 (n
+ 1)!
.
To achieve the indicated accuracy, we must have
1
2n+1 (n
≤ 10−4
+ 1)!
104 ≤ 2n+1 (n + 1)!.
If n = 5, then
2n+1 (n + 1)! = 26 · 6! = 46, 080 > 104 .
Then the sum is approximately
1 1
1
1
1
S5 = − + −
+
−
.
2 8 48 384 3840
Theorem: (The Ratio Test)
∞
X
an+1 = L.
Suppose that the series
an satisfies lim n→∞
a
n
n=1
1. If L < 1, then the series is absolutely convergent.
2. If L > 1, then the series is divergent.
3. If L = 1, then the test fails.
Example: Determine whether the following series converge:
(a)
∞
X
(−3)n
n=1
n5n+1
By the Ratio Test,
n+1 (−3)n+1
n5
lim ·
n→∞ (n + 1)5n+2 (−3)n −3n = lim n→∞ 5(n + 1) 3
= − 5
3
=
< 1.
5
The series converges absolutely by the Ratio Test.
an+1 =
lim n→∞ an (b)
∞
X
(n + 2)!
n=1
n!7n
By the Ratio Test,
n (n + 3)!
n!7
lim ·
n→∞ (n + 1)!7n+1 (n + 2)! 7(n + 3) = lim n→∞ (n + 1) = 7 > 1.
an+1 =
lim n→∞ an The series diverges by the Ratio Test.
Example: Consider the series
∞
X
(n + 1)(−5)n
n32n
n=1
.
(a) Prove that this series converges absolutely.
By the Ratio Test,
(n + 2)(−5)n+1
n32n
lim ·
n→∞
(n + 1)32n+2
(n + 1)(−5)n −5n(n + 2) = lim n→∞ (n + 1)(n + 1)32 −5 = lim n→∞ 9
an+1 =
lim
n→∞ an =
5
< 1.
9
The series converges absolutely by the Ratio Test.
(b) Use S3 to approximate the sum of the series and estimate the error.
The 3rd partial sum is
10
75
500
+
−
.
9
162 2187
By the Remainder Estimate for Alternating Series,
S3 = −
|R3 | ≤ a4 =
5 · 54
≈ 0.1191.
4 · 38
(c) Approximate the sum of the series to within two decimal places.
By the Remainder Estimate for Alternating Series,
|Rn | ≤ an+1 =
(n + 2)5n+1
.
(n + 1)32n+2
In order to achieve the indicated accuracy, set
(n + 2)5n+1
≤ 0.01.
(n + 1)32n+2
If n = 8, then
(n + 2)5n+1
≈ 0.0056 < 0.01.
(n + 1)32n+2
Then the sum of the series is approximately S8 .