# Alternating Series Test proof (induction)

Alternating Series Test

Let $$(a_n)$$ be a sequence satisfying,

1. $$a_1 \geq a_2 \geq a_3 \geq \cdots \geq a_n \geq a_{n+1} \geq \cdots$$
2. $$(a_n) \rightarrow 0$$

Then, the alternating series $$\sum_{n=1}^{\infty} (-1)^{n+1} a_n$$ converges.

In the book I'm reading the proof tries to show that $$s_n = \sum_{k=1}^{n} (-1)^{k+1} a_k$$ is a Cauchy sequence. So it's trying to prove that given $$\epsilon > 0$$ there is a $$N$$ that when $$n, m > N$$ then:

$$|a_{m+1} - a_{m+2} + \cdots \pm a_{n}| < \epsilon$$

This would imply it converges by the Cauchy Criterion.

The part I don't understand is when it says that by an induction argument it is possible to show that:

$$|a_{m+1} - a_{m+2} + \cdots \pm a_{n}| < |a_{m+1}|$$

Base case is clear.

Now assume that:

$$|a_{m+1} - a_{m+2} + \cdots \pm a_{n}| < |a_{m+1}|$$

We want to prove that:

$$|a_{m+1} - a_{m+2} + \cdots \pm a_{n+1}| < |a_{m+1}|$$

I could say $$|(a_{m+1} - a_{m+2} + \cdots \pm a_{n}) \pm a_{n+1}| \leq |a_{m+1} - a_{m+2} + \cdots \pm a_{n}| + |a_{n+1}| < |a_{m+1}| + |a_{n +1}|$$

But this is not the result I'm looking for.

I could try with the terms $$a_{n+2},\ldots,a_n$$ but can't justify my argument with the absolute values.

• You can get rid of the absolute values if you set out to prove $$0 \leqslant a_{m+1} - a_{m+2} + \ldots \pm a_n \leqslant a_{m+1}\,.$$ Give that a try. – Daniel Fischer May 12 at 18:16

If I were following the same general route, I would prove by induction on $$n$$ that

$$0\le\sum_{k=1}^n(-1)^{k+1}a_{m+k}\le a_{m+1}$$

for each $$m\in\Bbb Z^+$$; since $$a_1\ge a_2\ge\ldots\ge 0$$, there is no need for absolute value signs.

This is clearly true for $$n=1$$. Suppose that it is true for some $$n\ge 1$$; we want to show that

$$0\le\sum_{k=1}^{n+1}(-1)^{k+1}a_{m+k}\le a_{m+1}\tag{0}$$

for any $$m\in\Bbb Z^+$$. By the induction hypothesis (applied to $$m+1$$ instead of $$m$$) we know that

$$0\le\sum_{k=1}^n(-1)^{k+1}a_{(m+1)+k}\le a_{m+2}\;,$$

which can be rewritten as

$$0\le\sum_{k=2}^{n+1}(-1)^ka_{m+k}\le a_{m+2}\;.\tag{1}$$

Now

\begin{align*} \sum_{k=1}^{n+1}(-1)^{k+1}a_{m+k}&=a_{m+1}+\sum_{k=2}^{n+1}(-1)^{k+1}a_{m+k}\\ &=a_{m+1}-\sum_{k=2}^{n+1}(-1)^ka_{m+k}\;, \end{align*}

and it follows from $$(1)$$ that

$$a_{m+1}-a_{m+2}\le a_{m+1}-\sum_{k=2}^{n+1}(-1)^ka_{m+k}\le a_{m+1}\;.$$

And since $$a_{m+1}\ge a_{m+2}$$, $$(0)$$ follows immediately.