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 $$\sum_{n=1}^\infty \left({-1}\right)^{n-1} a_n = a_1 - a_2 + a_3 - a_4 + \ldots$$ 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 $$n \in \N^*$$, let $$P \left({n}\right)$$ be the proposition $$0 \le a_{m+1} - a_{m+2} + a_{m+3} - \cdots \pm a_n \le a_{m+1}$$.

It certainly holds for $$n = 1$$ as $$a_{m+1} \ge 0$$ by definition and $$a_{m+1} \le a_{m+1}$$.

It also holds for $$n = 2$$ as $$a_{m+2} \le a_{m+1}$$ and so $$0 \le a_{m+1} - a_{m+2} \le a_{m+1}$$.

These are the base cases.

Let $$b_k = a_{m+1} - a_{m+2} + a_{m+3} - \cdots \pm a_{k}$$ (this simplifies the algebra). 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, that is, that $$0 \le b_j \le a_{m+1}$$.

This is our induction hypothesis.

We now show that $$\forall k: P \left({k}\right) \Longrightarrow P \left({k+1}\right)$$. This is the induction step.


 * Suppose $$k$$ is odd.

We have by induction hypothesis that $$0 \le b_k \le a_{m+1}$$.

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 $$b_k \le a_{m+1}$$ it follows that $$b_k - a_{k+1} = b_{k+1} \le a_{m+1}$$.

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 $$P \left({k}\right) \Longrightarrow P \left({k+1}\right)$$.


 * Now suppose $$k$$ is even.

We have by induction hypothesis that $$0 \le b_k \le a_{m+1}$$.

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 $$P \left({k}\right) \Longrightarrow P \left({k+1}\right)$$.

So for both even and odd $$k$$ it follows that $$P \left({k}\right) \Longrightarrow P \left({k+1}\right)$$ 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 sequence of partial sums of the series $\sum_{n=1}^\infty \left({-1}\right)^{n-1} a_n$.

Let $$\epsilon > 0$$.

Since $$a_n \to 0$$ as $$n \to \infty$$, it follows that $$\exists N: \forall n > N: a_n < \epsilon$$.

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.