Alternating Series Test

Theorem
Let $\left \langle {a_n} \right \rangle$ 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:
 * $\displaystyle \sum_{n \mathop = 1}^\infty \left({-1}\right)^{n-1} a_n = a_1 - a_2 + a_3 - a_4 + \ldots$

converges.

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

This will be achieved by means of the Second Principle of Mathematical Induction.

For all $n \in \N_{>0}$, let $P_n$ be the proposition:
 * $0 \le a_{m+1} - a_{m+2} + a_{m+3} - \cdots \pm a_n \le a_{m+1}$

Basis for the Induction
$P_1$ holds, as $a_{m+1} \ge 0$ by definition and $a_{m+1} \le a_{m+1}$.

$P_2$ also holds, as $a_{m+2} \le a_{m+1}$ and so:
 * $0 \le a_{m+1} - a_{m+2} \le a_{m+1}$

This is the basis for the induction.

Induction Hypothesis
In order to simplify the algebra, let:
 * $b_k = a_{m+1} - a_{m+2} + a_{m+3} - \cdots \pm a_k$

The $\pm$ signifies the fact that $a_k$ is positive for $k$ odd and negative for $k$ even.

Suppose that $\forall j \le k: P \left({j}\right)$ holds:
 * $0 \le b_j \le a_{m+1}$

This is the induction hypothesis.

We now show that $\forall k: P \left({k}\right) \implies P \left({k+1}\right)$.

Induction Step
This is the induction step:

Suppose $k$ is odd.

By the induction hypothesis:
 * $0 \le b_k \le a_{m+1}$

Because $P \left({k}\right)$ holds:
 * $0 \le b_{k-1} + a_k \le a_{m+1}$

But as $a_k \ge a_{k+1}$:
 * $a_k - a_{k+1} \ge 0$

and so:
 * $0 \le b_{k-1} + \left({a_k - a_{k+1}}\right) = b_{k+1}$

But as $b_k \le a_{m+1}$:
 * $b_k - a_{k+1} = b_{k+1} \le a_{m+1}$

So:
 * $0 \le b_{k+1} \le a_{m+1}$

or:
 * $0 \le a_{m+1} - a_{m+2} + a_{m+3} - \cdots - a_{k+1} \le a_{m+1}$

Thus for odd $k$ it follows that $P \left({k}\right) \implies P \left({k+1}\right)$.

Now suppose $k$ is even.

By the induction hypothesis:
 * $0 \le b_k \le a_{m+1}$

Then:
 * $0 \le b_k + a_{k+1} = b_{k+1}$

Because $P \left({k}\right)$ holds:
 * $0 \le b_{k-1} - a_k \le a_{m+1}$

But as $a_k \ge a_{k+1}$:
 * $a_k - a_{k+1} \ge 0$

and so:
 * $b_{k+1} = b_{k-1} - a_k + a_{k+1} = b_{k-1} - \left({a_k - a_{k+1}}\right) = b_{k+1} \le a_{m+1}$

So:
 * $0 \le b_{k+1} \le a_{m+1}$

or:
 * $0 \le a_{m+1} - a_{m+2} + a_{m+3} - \cdots + a_{k+1} \le a_{m+1}$

Thus for even $k$ it follows that $P \left({k}\right) \implies P \left({k+1}\right)$.

So for both even and odd $k$ it follows that $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Second Principle of Mathematical Induction.

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

Now, let $\left \langle {s_n} \right \rangle$ be the [[Definition:Series|sequence of partial sums of the series:
 * $\displaystyle \sum_{n \mathop = 1}^\infty \left({-1}\right)^{n-1} a_n$]]

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 $\left \langle {s_n} \right \rangle$ is a Cauchy sequence.

The result follows from the fact that a Cauchy sequence is convergent.