So far we have looked mainly at series consisting of positive terms,
and we have derived and used the comparison tests
and ratio test for these.
But many series have positive and negative terms, and we also need to
look at these. This page discusses a particular case of these,
*alternating series*. Some aspects of alternating series are easy
and you will see why. Other aspects are rather frightening!
To put this into context you will need to also read the pages on
absolute convergence that come later.

Definition.

An alternating series is one of the form $\sum _{n=1}^{\infty}{\left(-1\right)}^{n+1}{a}_{n}$ where $\left({a}_{n}\right)$ is a sequence of positive terms.

For example, the series
$\sum _{n=1}^{\infty}{\left(-1\right)}^{n+1}\frac{1}{n}$
is an alternating series. The next theorem shows that this series
*converges*.

Alternating Series Test.

Let $\sum _{n=1}^{\infty}{\left(-1\right)}^{n+1}{a}_{n}$ be an alternating series, where the sequence $\left({a}_{n}\right)$ is a decreasing sequence of positive terms converging to 0. Then $\sum _{n=1}^{\infty}{\left(-1\right)}^{n+1}{a}_{n}$ converges.

**Proof.**

Let $n\in \mathbb{N}$ be arbitrary. We consider the partial sums ${s}_{2n}={a}_{1}-{a}_{2}+{a}_{3}-\dots +{a}_{2n-1}-{a}_{2n}$ and ${s}_{2n+1}={a}_{1}-{a}_{2}+{a}_{3}-\dots +{a}_{2n-1}-{a}_{2n}+{a}_{2n+1}$.

**Subproof.**

By bracketing one way,

(since $\left({a}_{n}\right)$ is a decreasing sequence) so that ${s}_{2n}$ is a sum of positive termsand hence the sequence $\left({s}_{2n}\right)$ is monotonic increasing. Also,

so ${s}_{2n}<{a}_{1}$ for all $n$. Thus $\left({s}_{2n}\right)$ is bounded and by the Monotone Convergence Theorem converges to some limit ${l}_{1}\in \mathbb{R}$.

**Subproof.**

Now consider ${s}_{2n+1}$. On the one hand we have

so $\left({s}_{2n+1}\right)$ is monotonic decreasing, and on the other we have

which is a sum of positive terms hence positive. Thus $\left({s}_{2n+1}\right)$ is decreasing and bounded below by 0, and hence by the Monotone Convergence Theorem converges to some ${l}_{2}\in \mathbb{R}$.

**Subproof.**

Finally we show ${l}_{1}={l}_{2}$. Note that ${s}_{2n+1}-{s}_{2n}={a}_{2n+1}\to 0$ as $n\to \infty $, as $\left({a}_{2n+1}\right)$ is a subsequence of the null sequence $\left({a}_{n}\right)$. But by continuity of $-$ we also have ${s}_{2n+1}-{s}_{2n}\to {l}_{2}-{l}_{1}$ so ${l}_{2}-{l}_{1}=0$ by uniqueness of limits, so ${l}_{1}={l}_{2}$. Even numbered terms and odd numbered terms of ${s}_{n}$ converge to the same limit ${l}_{1}={l}_{2}$. This means by the Theorem on Covering by Subsequences that ${s}_{n}\to {l}_{1}={l}_{2}$ as $n\to \infty $, as required.

Example.

The series
$\sum _{n=1}^{\infty}{\left(-1\right)}^{n+1}\frac{1}{n}$
and
$\sum _{n=1}^{\infty}{\left(-1\right)}^{n+1}\frac{1}{logn}$
converge by the alternating series test, even though the
corresponding terms of positive terms,
$\sum _{n=1}^{\infty}\frac{1}{n}$
and
$\sum _{n=1}^{\infty}\frac{1}{logn}$,
do *not* converge. (One is the harmonic series;
the other can be proved divergent by comparison with the
harmonic series.)

We saw the series $\sum _{n=1}^{\infty}{\left(-1\right)}^{n+1}\frac{1}{n}$ a long time ago at the very beginning of the course, where it caused some consternation because its value seemed to be greater than a half, but on rearranging its terms it got a value less than a quarter.

Definition.

A series $\sum _{n=1}^{\infty}{a}_{n}$ which is convergent but for which the series $\sum _{n=1}^{\infty}\left|{a}_{n}\right|$ diverges is said to be conditionally convergent.

So for example, $\sum _{n=1}^{\infty}{\left(-1\right)}^{n+1}\frac{1}{n}$
is conditionally convergent. The thing we observed
about rearrangments of terms for this series is, unfortunately, in fact true
more generally for *all* conditionally convergent series.
Conditionally convergent series are particularly badly behaved,
should be avoided if possible, and *always* need great care. The results
we have proved about this one *are* correct though:
according to the definitions we have given,
the sum of the series $\sum _{n=1}^{\infty}{\left(-1\right)}^{n+1}\frac{1}{n}$
*does* exist and *is* greater than a half. But a rearrangement of the
series can make the sum less than a half, or in fact, any number you
like. To avoid such bad

cases you need to look at
absolutely convergent series, which are
the subject of the next web page.

You have seen the definition of alternating series, the alternating series test, and some examples. You have also seen how many such series behave very badly with respect to rearrangement of their terms.