11.7: Absolute Convergence

Roughly speaking, one could say that there are two ways for a series to converge. On the one hand, as in the case of \(\sum 1/n^2\), the individual terms may become small enough sufficiently quickly so that the sum of all of them stays finite. Alternatively, as in the case of a series of mixed terms such as \(\sum(-1)^{n-1}/n\), if the terms were all made to be positive it may happen in that case that they would not reduce in size fast enough for their sum to converge (\(\sum 1/n\) diverges), but leaving them as a mixture of positive and negative terms does provide enough cancellation to keep their sum finite. So from what we've seen so far, you might correctly guess that if making the terms of a mixed series all positive would still leave the resulting series convergent, then whether or not some of its terms are negative and some are positive, the original series will also converge.

Theorem \(\PageIndex{1}\)

If \(\displaystyle\sum_{n=0}^\infty |a_n|\) converges, then \(\displaystyle\sum_{n=0}^\infty a_n\) converges.

Proof

Note that \(0\le a_n+|a_n|\le 2|a_n|\), so by the comparison test \(\displaystyle\sum_{n=0}^\infty (a_n+|a_n|)\) converges. Now \[\sum_{n=0}^\infty (a_n+|a_n|)- \sum_{n=0}^\infty |a_n| = \sum_{n=0}^\infty a_n+|a_n|-|a_n| = \sum_{n=0}^\infty a_n\nonumber\] converges by theorem 11.3.2.

So, given a series \(\sum a_n\) with both positive and negative terms, you should first ask whether \(\sum |a_n|\) converges. This may be an easier question to answer, because we have tests that apply specifically to terms with non-negative terms. If \(\sum |a_n|\) converges then you know that \(\sum a_n\) converges as well. If \(\sum |a_n|\) diverges then it may still be true that \(\sum a_n\) converges---you will have to do more work to decide the question. Another way to think of this result is: it is (potentially) easier for \(\sum a_n\) to be convergent than for \(\sum |a_n|\), because the latter series cannot take advantage of cancellation.

If \(\sum |a_n|\) converges we say that \(\sum a_n\) converges absolutely; to say that \(\sum a_n\) converges absolutely is to say that any cancellation between terms that happens to be present is not really needed (as far as convergence is concerned), as the terms already get small enough quickly enough that convergence is guaranteed by that alone. If \(\sum a_n\) converges but \(\sum |a_n|\) does not, we say that \(\sum a_n\) converges conditionally. For example, \(\displaystyle\sum_{n=1}^\infty (-1)^{n-1} {1\over n^2}\) converges absolutely, while \(\displaystyle\sum_{n=1}^\infty (-1)^{n-1} {1\over n}\) converges conditionally.

Example \(\PageIndex{2}\)

Does \(\displaystyle\sum_{n=2}^\infty \dfrac{\sin n}{n^2}\) converge?

Solution

In example 11.6.2 we saw that \(\displaystyle\sum_{n=2}^\infty \dfrac{|\sin n|}{n^2}\) converges, so the given series converges absolutely.

Example \(\PageIndex{3}\)

Does \(\displaystyle\sum_{n=0}^\infty (-1)^{n}\dfrac{3n+4}{2n^2+3n+5}\) converge?

Solution

Taking the absolute values, \(\displaystyle \sum_{n=0}^\infty \dfrac{3n+4}{2n^2+3n+5}\) diverges by comparison to \(\displaystyle\sum_{n=1}^\infty \dfrac{3}{10n}\) which is divergent, so if the given series converges it does so conditionally. It is true that \(\displaystyle\lim_{n\to\infty}\dfrac{3n+4}{2n^2+3n+5}=0\), so to apply the alternating series test we need to know whether the terms are decreasing. If we let \(f(x)=\dfrac{3x+4}{2x^2+3x+5}\) then \(f'(x)=-\dfrac{6x^2+16x-3}{(2x^2+3x+5)^2}\), and it is not hard to see that this is negative for \(x\ge 1\), so the terms of the series are decreasing in absolute magnitude and by the alternating series test the series converges.

Exercises \(\PageIndex{}\)

Determine whether each series converges absolutely, converges conditionally, or diverges.

Exercise \(\PageIndex{1}\)

\(\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{1}{2n^2+3n+5}\)

Answer

Converges absolutely

Exercise \(\PageIndex{2}\)

\(\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{3n^2+4}{2n^2+3n+5}\)

Answer

Diverges

Exercise \(\PageIndex{3}\)

\(\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{\ln n}{n}\)

Answer

Converges conditionally

Exercise \(\PageIndex{4}\)

\(\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{\ln n}{n^3}\)

Answer

Converges absolutely

Exercise \(\PageIndex{5}\)

\(\displaystyle\sum_{n=2}^{\infty}(-1)^n\dfrac{1}{\ln n}\)

Answer

Converges conditionally

Exercise \(\PageIndex{6}\)

\(\displaystyle\sum_{n=0}^{\infty}(-1)^n\dfrac{3^n}{2^n+5^n}\)

Answer

Converges absolutely

Exercise \(\PageIndex{7}\)

\(\displaystyle\sum_{n=0}^{\infty}(-1)^n\dfrac{3^n}{2^n+3^n}\)

Answer

Diverges

Exercise \(\PageIndex{8}\)

\(\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{\arctan n}{n}\)

Answer

Converges conditionally

Contributors

• David Guichard (Whitman College)

• Integrated by Justin Marshall.