# Alternating series

## 1. Introduction

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.

## 2. Alternating series

Definition.

An alternating series is one of the form $∑ n = 1 ∞ ( -1 ) n + 1 a n$ where $( a n )$ is a sequence of positive terms.

For example, the series $∑ n = 1 ∞ ( -1 ) n + 1 1 n$ is an alternating series. The next theorem shows that this series converges.

Alternating Series Test.

Let $∑ n = 1 ∞ ( -1 ) n + 1 a n$ be an alternating series, where the sequence $( a n )$ is a decreasing sequence of positive terms converging to 0. Then $∑ n = 1 ∞ ( -1 ) n + 1 a n$ converges.

Proof.

Let $n ∈ ℕ$ be arbitrary. We consider the partial sums $s 2 n = a 1 - a 2 + a 3 - … + a 2 n - 1 - a 2 n$ and $s 2 n + 1 = a 1 - a 2 + a 3 - … + a 2 n - 1 - a 2 n + a 2 n + 1$.

Subproof.

By bracketing one way,

$s 2 n = ( a 1 - a 2 ) + ( a 3 - a 4 ) + … + ( a 2 n - 1 - a 2 n ) > 0$

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

$s 2 n = a 1 - ( a 2 - a 3 ) - ( a 4 - a 5 ) - … - ( a 2 n - 2 - a 2 n - 1 ) - a 2 n$

so $s 2 n < a 1$ for all $n$. Thus $( s 2 n )$ is bounded and by the Monotone Convergence Theorem converges to some limit $l 1 ∈ ℝ$.

Subproof.

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

$s 2 n + 1 = a 1 - ( a 2 - a 3 ) - ( a 4 - a 5 ) - … - ( a 2 n - 2 - a 2 n - 1 ) - ( a 2 n - a 2 n + 1 ) < s 2 n - 1$

so $( s 2 n + 1 )$ is monotonic decreasing, and on the other we have

$s 2 n + 1 = ( a 1 - a 2 ) + ( a 3 - a 4 ) + … + + ( a 2 n - 1 - a 2 n ) + a 2 n + 1$

which is a sum of positive terms hence positive. Thus $( s 2 n + 1 )$ is decreasing and bounded below by 0, and hence by the Monotone Convergence Theorem converges to some $l 2 ∈ ℝ$.

Subproof.

Finally we show $l 1 = l 2$. Note that $s 2 n + 1 - s 2 n = a 2 n + 1 → 0$ as $n → ∞$, as $( a 2 n + 1 )$ is a subsequence of the null sequence $( a n )$. But by continuity of $-$ we also have $s 2 n + 1 - s 2 n → l 2 - l 1$ 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 $s n → l 1 = l 2$ as $n → ∞$, as required.

Example.

The series $∑ n = 1 ∞ ( -1 ) n + 1 1 n$ and $∑ n = 1 ∞ ( -1 ) n + 1 1 log n$ converge by the alternating series test, even though the corresponding terms of positive terms, $∑ n = 1 ∞ 1 n$ and $∑ n = 1 ∞ 1 log n$, do not converge. (One is the harmonic series; the other can be proved divergent by comparison with the harmonic series.)

We saw the series $∑ n = 1 ∞ ( -1 ) n + 1 1 n$ 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 $∑ n = 1 ∞ a n$ which is convergent but for which the series $∑ n = 1 ∞ | a n |$ diverges is said to be conditionally convergent.

So for example, $∑ n = 1 ∞ ( -1 ) n + 1 1 n$ 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 $∑ n = 1 ∞ ( -1 ) n + 1 1 n$ 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.

## 3. Conclusion

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.