Calculus- Early Transcendentals, 2021a

358 Sequences and Series
9.6 Absolute Convergence
Roughly speaking there are two ways for a series to converge: As in the case of ∑1/n 2 , the individual terms
get small very quickly, so that the sum of all of them stays finite, or, as in the case of ∑(−1) n−1 /n, the
terms don’t get small fast enough (∑1/n diverges), but a mixture of positive and negative terms provides
enough cancellation to keep the sum finite. You might guess from what we’ve seen that if the terms
get small fast enough that the sum of their absolute values converges, then the series will still converge
regardless of which terms are actually positive or negative.
Theorem 9.39: Absolute Convergence
If
∞
∑
n=0
|a n | converges, then
∞
∑
n=0
a n converges.
Proof. Note that 0 ≤ a n + |a n |≤2|a n | so by the comparison test
∞
∑
n=0
(a n + |a n |) −
∞
∑
n=0
|a n | =
∞
∑
n=0
∞
∞
∑ a n + |a n |−|a n | = ∑ a n
n=0
n=0
(a n + |a n |) converges. Now
converges by Theorem 9.20.
♣
So given a series ∑a n with both positive and negative terms, you should first ask whether ∑|a n | converges.
This may be an easier question to answer, because we have tests that apply specifically to terms
with non-negative terms. If ∑|a n | converges then you know that ∑a n converges as well. If ∑|a n | diverges
then it still may be true that ∑a n converges, but you will need to use other techniques to decide. Intuitively
this results says that it is (potentially) easier for ∑a n to converge than for ∑|a n | to converge, because terms
may partially cancel in the first series.
If ∑|a n | converges we say that ∑a n converges absolutely; to say that ∑a n converges absolutely is to
say that the terms of the series get small (in absolute value) quickly enough to guarantee that the series
converges, regardless of whether any of the terms cancel each other. If ∑a n converges but ∑|a n | does
not, we say that ∑a n converges conditionally. For example
∞
∑
n=1
(−1) n−1 1 converges conditionally.
n
Example 9.40
Does
∞
∑
n=2
sinn
n 2
converge?
Solution. In Example 9.35 we saw that
∞
|sinn|
∑
n=2
n 2
∞
∑
n=1
(−1) n−1 1 converges absolutely, while
n2 converges, so the given series converges absolutely.
♣