Alternating Series Test
Let $(a_n)$ be a sequence satisfying,
- $a_1 \geq a_2 \geq a_3 \geq \cdots \geq a_n \geq a_{n+1} \geq \cdots $
- $(a_n) \rightarrow 0$
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.
Best Answer
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.