Alternating Series Test

Theorem
Let $\sequence {a_n}_{N \mathop \ge 0}$ be a decreasing sequence of positive terms in $\R$ which converges with a limit of zero.

That is, let $\forall n \in \N: a_n \ge 0, a_{n + 1} \le a_n, a_n \to 0$ as $n \to \infty$

Then the series:
 * $\ds \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} a_n = a_1 - a_2 + a_3 - a_4 + \dotsb$

converges.

Proof
First we show that for each $n > m$, we have $0 \le a_{m + 1} - a_{m + 2} + a_{m + 3} - \dotsb \pm a_n \le a_{m + 1}$.

Lemma
Therefore for each $n > m$, we have:
 * $0 \le a_{m + 1} - a_{m + 2} + a_{m + 3} - \dotsb \pm a_n \le a_{m + 1}$

Now, let $\sequence {s_n}$ be the [[Definition:Series|sequence of partial sums of the series:
 * $\ds \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} a_n$]]

Let $\epsilon > 0$.

Since $a_n \to 0$ as $n \to \infty$:
 * $\exists N: \forall n > N: a_n < \epsilon$

But $\forall n > m > N$, we have:

Thus we have shown that $\sequence {s_n}$ is a Cauchy sequence.

The result follows from Convergent Sequence is Cauchy Sequence.

Also known as
The Alternating Series Test is also seen referred to as Leibniz's Alternating Series Test.