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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.

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    $\begingroup$ 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. $\endgroup$ – Daniel Fischer May 12 at 18:16
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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.

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