Alternating Series Test

Theorem

Suppose ${a_{n}}_{n = 1}^{\infty}$ is a sequence of non-negative real numbers satisfying

$a_{n} \geq a_{n + 1}$ for all $n \in Z_{> 0}$
$lim_{n \to \infty} a_{n} = 0$

Then the alternating series

\sum_{n = 1}^{\infty} (- 1)^{n - 1} a_{n} = a_{1} - a_{2} + a_{3} - a_{4} + \dots

converges.

Note that the same result holds if the sequence is all non-positive, or if the power of $- 1$ is $n$ instead of $n - 1$ . This just follows from a substitution of the terms in the series with their negatives, and a reindexing of the sum with leading terms excluded.

Lemma

For $m \geq 3$ , and a series ${a_{n}}_{n = 1}^{\infty}$ which is non-negative and satisfies

$a_{n} \geq a_{n + 1}$ for all $n \in Z_{> 0}$

we have that

a_{1} - a_{2} \leq \sum_{n = 1}^{m} (- 1)^{n - 1} a_{n} \leq a_{1} - a_{2} + a_{3} .

Proof

This result follows by grouping the terms in pairs. Namely we first note

\sum_{n = 1}^{m} (- 1)^{n - 1} a_{n} = (a_{1} - a_{2}) + \sum_{n = 3}^{m} (- 1)^{n - 1} a_{n} .

Then because $a_{n} \geq a_{n + 1}$ , we have $a_{n} - a_{n + 1} \geq 0$ . Hence we can group the tail of the series into pairs which are strictly non-negative, and then we have potentially one remaining non-negative term. That is we either have

\sum_{n = 1}^{m} (- 1)^{n - 1} a_{n} = (a_{1} - a_{2}) + \sum_{\begin{matrix} n = 3 \\ n odd \end{matrix}}^{m - 1} (a_{n} - a_{n + 1})

if $m - 1$ is odd, or

\sum_{n = 1}^{m} (- 1)^{n - 1} a_{n} = (a_{1} - a_{2}) + \sum_{\begin{matrix} n = 3 \\ n odd \end{matrix}}^{m - 1} (a_{n} - a_{n + 1}) + a_{m}

if $m - 1$ is even.

Either way, this expression is $a_{1} - a_{2}$ plus non-negative terms. Therefore we can conclude that

a_{1} - a_{2} \leq \sum_{n = 1}^{m} (- 1)^{n - 1} a_{n} .

A similar argument follows for the other inequality, where we write

\sum_{n = 1}^{m} (- 1)^{n - 1} a_{n} = (a_{1} - a_{2} + a_{3}) + \sum_{n = 4}^{m} (- 1)^{n - 1} a_{n}

and then group in pairs with negative terms. In particular we have

\sum_{n = 1}^{m} (- 1)^{n - 1} a_{n} = (a_{1} - a_{2} + a_{3}) + \sum_{\begin{matrix} n = 4 \\ n even \end{matrix}}^{m - 1} (- a_{n} + a_{n + 1})

if $m - 1$ is even, and

\sum_{n = 1}^{m} (- 1)^{n - 1} a_{n} = (a_{1} - a_{2} + a_{3}) + \sum_{\begin{matrix} n = 4 \\ n even \end{matrix}}^{m - 1} (- a_{n} + a_{n + 1}) - a_{m}

if $m - 1$ is odd.

In this case, we have $a_{1} - a_{2} + a_{3}$ plus non-positive terms, and conclude, with the above inequality, that

a_{1} - a_{2} \leq \sum_{n = 1}^{m} (- 1)^{n - 1} a_{n} \leq a_{1} - a_{2} + a_{3} .

This lemma will now allow us to prove the main result.

Proof

Define the sequence

S_{m} = \sum_{n = 1}^{m} (- 1)^{n - 1} a_{n}

of partial sums such that we can define the partial sums of an even number of terms and the partial sum of an odd number of terms as $S_{2 m}$ and $S_{2 m - 1}$ respectively (for $m \geq 1$ ).

With the same grouping in pairs argument from our lemma, it is clear that $S_{2 m}$ is monotonic increasing and $S_{2 m - 1}$ is monotonic decreasing. Thus with the bounds from this lemma and the monotone convergence theorem, we know that

lim_{m \to \infty} S_{2 m} and lim_{m \to \infty} S_{2 m - 1}

both exist.

As such we have

\begin{aligned} lim_{m \to \infty} S_{2 m} - lim_{m \to \infty} S_{2 m - 1} & = lim_{m \to \infty} (S_{2 m} - S_{2 m - 1}) \\ = lim_{m \to \infty} a_{2 m} \\ = 0 \end{aligned}

and therefore

L = lim_{m \to \infty} S_{2 m} = lim_{m \to \infty} S_{2 m - 1} .

Then, the limit of partial sums of the desired series must be equal to the shared limit between the even and odd terms.

In essence, for any $ϵ > 0$ there exists an $N_{0}$ and $N_{1}$ such that

n > N_{0} ⟹ | S_{2 n} - L | < ϵ and n > N_{1} ⟹ | S_{2 n - 1} - L | < ϵ

so setting $N = max {N_{0}, N_{1}}$ , we have that for any $ϵ > 0$

n > N ⟹ | S_{2 n} - L | < ϵ and | S_{2 n - 1} - L | < ϵ

and given both cases cover any term in the partial sum $S_{n}$ ,

lim_{n \to \infty} S_{n} = L .