# Absolute convergence of series

## 1Introduction

We have seen earlier the wonderful news that all reasonable-looking alternating series converge. We have also seen highly worrying examples that show that some alternating series converge to different values depending on how the terms in the series are rearranged. This web page starts a short section on those nice series that converge in such a way that doesn't depend on any such rearrangement.

## 2Absolute convergence

Definition 2.1

A series $∑ n = 1 ∞ a n$ is said to converge absolutely if the series of absolute values $∑ n = 1 ∞ a n$ converges.

This definition, and its rather suggestive use of language begs an important question: does a series that converges absolutely converge at all? Fortunately for whomsoever first thought up this definition, the answer is yes. This is the main point of this web page.

Theorem 2.1 (Absolute Convergence)

Suppose the series $∑ n = 1 ∞ a n$ converges absolutely. Then the series converges.

To prove this we should first recall the Cauchy Property of convergent sequences and restate it for the sequence of partial sums of a series. A sequence $( s n )$ is Cauchy if

and by the completeness of the reals, $( s n )$ is Cauchy if and only if $( s n )$ converges to some $l ∈ ℝ$. In particular, for the sequence $s n = ∑ k = 1 n a k$ of partial sums, we have

$s n - s k = a k + 1 + a k + 2 + … + a n$

Proof

Suppose that $∑ n = 1 ∞ a n$ converges absolutely. Then let $s n = ∑ k = 1 n a k$ and $t n = ∑ k = 1 n a k$. Note that $( t n )$ is Cauchy: we must show that $( s n )$ is Cauchy.

Subproof

Let $ε > 0$ be arbitrary. Then there is $N ∈ ℕ$ such that for all $N ≤ k < n$ we have $t n - t k < ε$.

Subproof

Now let $k , n ∈ ℕ$ be arbitrary such that $N ≤ k < n$. Then we have

$s n - s k = a k + 1 + a k + 2 + … + a n ≤ a k + 1 + a k + 2 + … + a n ≤ a k + 1 + a k + 2 + … + a n$

by the triangle inequality, hence $s n - s k ≤ t n - t k < ε$.

Thus $( s n )$ is Cauchy, as required.

## 3Applications

Clearly, directly from the definition, any convergent series of positive terms converges absolutely.

Recall that our versions of the comparison or ratio tests apply to series of positive terms. We now see how to apply them to more general series $∑ n = 1 ∞ a n$ of terms that may be either positive or negative, by first looking at the series formed from the absolute values of the terms, $∑ n = 1 ∞ a n$, showing this is convergent and hence proving absolute convergence of the original series. The convergence of the original series follows by quoting the above result.

In other words, the comparison and ratio tests prove absolute convergence of series. You should show that you understand that the implication from absolute convergence to convergence is not a complete triviality by quoting it properly when needed.

For example, for any choice of $z n ∈ ℤ$ we have that $∑ n = 1 ∞ ( -1 ) z n n 2$ is absolutely convergent, hence the series is convergent by the theorem on absolute convergence.

On the other hand, not every convergent series converges absolutely: The series $∑ n = 1 ∞ ( -1 ) n n$ gives an example of such a series: it converges by the alternating series test, but not absolutely since the harmonic series diverges.

This web page is available in xhtml, html and pdf. It is copyright and is one of Richard Kaye's Sequences and Series Web Pages. It may be copied under the terms of the Gnu Free Documentation Licence (http://www.gnu.org/copyleft/fdl.html). There is no warranty. Web page design and creation are by GLOSS.