Math 152 – Spring 2016
Section 10.4
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Section 10.4 – Other Convergence Tests
Alternating Series. An alternating series is a series whose terms are alternately
∞
P
postive and negative. The terms an of an alternating series
an can be written as
n=1
an = (−1)n bn or an = (−1)n−1 bn with each bn > 0.
1. 1 −
1
2
2. −1 +
1
3
−
−
1
9
+
1
4
1
4
+
+ ··· =
∞
P
(−1)n−1 n1
n=1
1
16
−
1
25
∞
P
+
n=1
(−1)n n12
Note. It must change sign with every other term. The following is NOT an alternating
series.
√
√
√
√
∞
X
sin nπ
2 1
2
2
2
1
4
=
+
+
+
0
−
−
−
+ 0 + ···
2
n
2
4
18
50
36
98
n=1
The Alternating Series Test. If the alternating series
∞
X
(−1)n−1 bn = b1 − b2 + b3 − b4 + b5 − b6 + · · ·
n=1
satisfies
1. lim bn = 0
n→∞
2. bn+1 ≤ bn
for all n
then the series is convergent.
bn > 0
Math 152 – Spring 2016
Section 10.4
Example 1. Show the following series are convergent or divergent.
(a) 1 −
(b)
∞
P
n=5
(c)
∞
P
n=1
1
2
+
1
3
−
1
4
(−1)n 3n2
2n(n−4)
(−1)n−1 e2n
e3n −en
+ ··· =
∞
P
(−1)n−1 n1
n=1
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Math 152 – Spring 2016
Section 10.4
Note. What if an alternating series
∞
P
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(−1)n−1 bn fails the conditions of the alternating
n=1
series test?
• When an alternating series has lim bn 6= 0 or the limit does not exist, use the
n→∞
Test for Divergence to show the series diverges (if lim bn 6= 0 or the limit does
n→∞
not exist, then lim (−1)n−1 bn does not exist).
n→∞
• There are alternating series such that bn+1 ≤ bn is not true for all n, but the
series still converges. The following series clearly fails the condition bn+1 ≤ bn :
1−
where bn =
∞
X
1
1
1
1
1
1 1
+ −
+
−
+
−
+ ··· =
(−1)n−1 bn
8 9 64 25 216 49 512
n=1
1
n2
1
n3
when n is odd and bn =
when n is even. This series is absolutely
∞
P
1
convergent (use the comparison test with
n2 ), and thus, it is convergent.
n=1
Alternating Series Estimation Theorem. If s =
∞
P
(−1)n−1 bn with each bn > 0
n=1
is the sum of an alternating series that satisfies
• lim bn = 0
n→∞
• bn+1 ≤ bn
for all n
then
|Rn | = |s − sn | ≤ bn+1
Note. This equation Rn ≤ bn ONLY applies to Alternating Series.
Example 2. Find the sum of the series
∞
P
n=0
Example 3. How many terms of the series
of the series with error less than 0.0001?
(−1)n
n!
accurate to three decimal places.
∞
X
(−1)n 5
should we use to find the sum
7n
n=1
Math 152 – Spring 2016
Section 10.4
Absolute Convergence. A series
series of absolute values
∞
P
∞
P
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an is called absolutely convergent if the
n=1
|an | is convergent.
n=1
Note. If the series has only positive terms, then absolute convergence is the same as
convergence.
Example 4. Determine if the following series are absolutely convergent.
(a) − 10
3 +
(b)
∞
P
n=1
20
9
−
40
27
+
80
81
−
160
243
+ ···
(−1)n−1
n
Theorem.
1. If a series
∞
P
an is absolutely convergent, then it is convergent.
n=1
2. If a series is convergent, then it may or may not be absolutely convergent.
Math 152 – Spring 2016
Section 10.4
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Example 5. Determine if the following series are absolutely convergent, conditionally
convergent, or divergent. Conditionally convergent means convergent, but not absolutely convergent.
(a)
∞
P
n=1
(b)
∞
P
n=1
(c)
∞
P
n=1
(d)
∞
P
n=1
(−3)n−1
.
23n
(−1)n arctan n
n
sin( nπ
4 )
n2
(−1)n+1 2n
n4
Math 152 – Spring 2016
Section 10.4
6 of 6
The Ratio Test.
• If lim | aan+1
| = L < 1, then the series
n
n→∞
∞
P
an is absolutely convergent (and
n=1
therefore convergent).
| = L > 1 or lim | aan+1
| = ∞, then the series
• If lim | aan+1
n
n
n→∞
n→∞
∞
P
an is divergent.
n=1
• If lim | aan+1
| = 1, then the series may be convergent or divergent. The ratio test
n
n→∞
fails and you must use a different test. (Ex: all p-series fall in this case, but a
p-series may converge or diverge).
Example 6. Determine if the following series is absolutely convergent.
(a)
∞
P
n=1
(b)
∞
P
n=1
(−3)n
n!
(−1)n+1 5n−1
(n+1)2 4n+2