Alternating Series Test

Theorem

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

  1. \(a_n \geq a_{n + 1}\) for all \(n \in \mathbb{Z}_{> 0}\)
  2. \(\lim_{n \to \infty} a_n = 0\)

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 \(n \in \mathbb{Z}_{> 0}\)

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

\[ \sum_{n = 1}^m (-1)^{n - 1} a_n = (a_1 - a_2) + \sum_{\substack{n = 3 \\ n \ \text{odd}}}^{m - 1} (a_n - a_{n + 1})\]

if \(m - 1\) is odd, or

\[ \sum_{n = 1}^m (-1)^{n - 1} a_n = (a_1 - a_2) + \sum_{\substack{n = 3 \\ n \ \text{odd}}}^{m - 1} (a_n - a_{n + 1}) + a_m\]

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

\[ \sum_{n = 1}^m (-1)^{n - 1} a_n = (a_1 - a_2 + a_3) + \sum_{\substack{n = 4 \\ n \ \text{even}}}^{m - 1} (- a_n + a_{n + 1})\]

if \(m - 1\) is even, and

\[ \sum_{n = 1}^m (-1)^{n - 1} a_n = (a_1 - a_2 + a_3) + \sum_{\substack{n = 4 \\ n \ \text{even}}}^{m - 1} (- a_n + a_{n + 1}) - a_m\]

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 \(S_{2m}\) and \(S_{2m - 1}\) respectively (for \(m \geq 1\)).

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

\[ \lim_{m \to \infty} S_{2m} \quad \text{and} \quad \lim_{m \to \infty} S_{2m - 1}\]

both exist.

As such we have

\[\begin{align*} \lim_{m \to \infty} S_{2m} - \lim_{m \to \infty} S_{2m - 1} &= \lim_{m \to \infty} (S_{2m} - S_{2m - 1}) \\ &= \lim_{m \to \infty} a_{2m} \\ &= 0 \\ \end{align*}\]

and therefore

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

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 \(\epsilon > 0\) there exists an \(N_0\) and \(N_1\) such that

\[ n > N_0 \implies |S_{2n} - L| < \epsilon \quad \text{and} \quad n > N_1 \implies |S_{2n - 1} - L| < \epsilon\]

so setting \(N = \max\{N_0, N_1\}\), we have that for any \(\epsilon > 0\)

\[ n > N \implies |S_{2n} - L| < \epsilon \quad \text{and} \quad |S_{2n - 1} - L| < \epsilon\]

and given both cases cover any term in the partial sum \(S_n\),

\[ \lim_{n \to \infty} S_n = L.\]