11.6.1
11.6: Alternating Series, Absolute and Conditional Convergence
So far, the convergence tests we have seen have applied only to series with positive terms. In the
next couple of sections, we will examine tests for series with terms that are not necessarily
positive.
An alternating series is a series whose terms are alternately positive and negative.
∞
Example 1:
∑
n =1
( −1)
n −1
n
2 3 4 5 6
= 1− + − + − +"
2n − 1
3 5 7 9 11
In general, the nth term of an alternating series is of the form
n −1
n
an = ( −1) bn or an = ( −1) bn
where bn = an is positive.
The Alternating Series Test:
If the alternating series
∞
∑ ( −1)
n =1
n −1
bn = b1 − b2 + b3 − b4 + b5 − b6 + "
satisfies
• bn +1 ≤ bn , for all n,
• lim bn = 0
n →∞
then the series is convergent.
∞
Example 2:
Test the alternating harmonic series
∑
n =1
convergence or divergence.
( −1)
n
n −1
1 1 1
= 1 − + − + " for
2 3 4
11.6.2
∞
Example 3:
Test the alternating series
∑
n =1
∞
Example 4:
Test the alternating series
( 5n
2
+ 2n )
n +1
for convergence or divergence.
4n 2
for convergence or divergence.
3n3 + 8
cos nπ
for convergence or divergence.
n!
n =1
∞
Test the series
n
2n 2 + 3n + 4
∑ ( −1)
n =1
Example 5:
( −1)
∑
11.6.3
Absolute and conditional convergence:
Given a series
∑a
n
, we consider the series whose terms are the absolute value of the terms of
the original series.
∞
∑a
n =1
n
= a1 + a2 + a3 + "
A series
∑a
n
is called absolutely convergent if the series of absolute values
∑a
A series
∑a
n
is called conditionally convergent if it is convergent by not absolutely
n
converges.
convergent.
∞
Example 1:
Example 2:
The series
The series
∑
( −1)
3
n =1
n
∞
( −1)
∑
n =1
n
n
1 1
1
1
= −1 + −
+ −
+ " is absolutely convergent.
8 27 64 125
n −1
1 1 1
= 1 − + − + " is conditionally convergent.
2 3 4
Theorem:
If a series
∑a
n
is absolutely convergent, then the series is convergent.
11.6.4
∞
Example 3:
Test the series
sin n sin1 sin 2 sin 3
= 4 + 4 + 4 + " for convergence or divergence.
4
1
2
3
n =1 n
∑
Estimates of sums:
Alternating Series Estimation Theorem:
Suppose the alternating series
∞
∑ ( −1)
n =1
n +1
bn = b1 − b2 + b3 − b4 + b5 − b6 + "
satisfies
• bn ≥ 0 for all n,
• bn +1 ≤ bn , for all n ≥ N for some integer N
• lim bn = 0 .
n →∞
•
Then, the nth partial sum sn = b1 − b2 + ... + (−1) n +1 bn approximates the sum L of the series with
an error whose absolute value is less than bn +1 , the numerical value of the first unused term.
Also, the remainder, L − sn , has the same sign as the first unused term.
11.6.5
(−1) n
by using the 6th partial
n
n =1 3
∞
Example 4:
Estimate the error if we approximate the sum of
∑
sum.
∞
Example 5:
Estimate the error if we approximate the sum of
∑
( −1)
3
n
n
sum. How many terms would be needed to estimate the sum within 10−7 ?
n =1
by using the 6th partial