Let $a_n$ a monotone sequence approaches $0$. Show that for all $m\in\Bbb{N}$:
$$ 0 < (-1)^m\sum\limits_{n=m+1}^{\infty} (-1)^{n+1} {a_n} < a_{m+1} $$
I wanna focus on the RHS inequality:
$$(-1)^m\sum\limits_{n=m+1}^{\infty} (-1)^{n+1} {a_n} = a_{m+1} \color{Blue}{-} \sum\limits_{n=m+2}^{\infty}(-1)^{m+n+1} a_n \\= a_{m+1} - \sum\limits_{k=m+1/2}^{\infty} \underset{>0}{(a_{2k+1} - a_{2k+2})} < a_{m+1}$$
My Question:
Please have a look at the blue minus. Where did it come from? :|
First, the blue minus sign has to be $+$.
Second, your $m+1/2$ should mean $(m+1)/2$. But be careful about the following : $$(m+1)/2\in\mathbb Z\iff \text{$m$ is odd}.$$
$$\begin{align}(-1)^m\sum\limits_{n=m+1}^{\infty} (-1)^{n+1} {a_n} &=\sum_{n=m+1}^{\infty}(-1)^{m+n+1}a_n \\&=(-1)^{2m+2}a_{m+1}+\sum_{n=m+2}^{\infty}(-1)^{m+n+1}a_n\\&= a_{m+1} \color{red}{+} \sum\limits_{n=m+2}^{\infty}(-1)^{m+n+1} a_n \\&= a_{m+1} - \sum\limits_{\color{red}{k=(m+1)/2}}^{\infty} (a_{2k+1}-a_{2k+2})\end{align}$$
I hope this helps!