Cauchy's convergence test can I say that?

Question

Check using Cauchy's convergence test if $a_{n}$ convergent

$$a_n=\sum_{k=1}^n \frac{(-1)^k}{k(k+1)}$$

I want to show that $a_{n}$ is a cauchy sequence meaning there is $m,n\in\mathbb N$ such that $\forall m,n>N$ implies $|a_{m}-a_{n}|<\epsilon$ That's what I tried so far Let $m=n+p$ Then $$ \left| \sum_{k=1}^{n+p}\frac{(-1)^k}{k(k+1)}-\sum_{k=1}^n \frac{(-1)^k}{k(k+1)} \right| = \left|\sum_{k=n+1}^{n+p}\frac{(-1)^k}{k(k+1)}\right|$$ $$=\frac{(-1)^{n+1}}{(n+1)(n+2)}+\frac{(-1)^{n+2}}{(n+2)(n+3)}+ \cdots + \frac{(-1)^{n+p}}{(n+p-1)(n+p)}$$ and I don't know how to go from there I tried to use here a telescoping sum but that got me nowhere..

Answer 1

$\left|\displaystyle\sum_{k=n+1}^{n+p}\dfrac{(-1)^{k}}{k(k+1)}\right|\leq\displaystyle\sum_{k=1}^{k+p}\dfrac{1}{n(n+1)}=\displaystyle \sum_{k=n+1}^{n+p}\left(\dfrac{1}{k}-\dfrac{1}{k+1}\right)=\dfrac{1}{n+1}-\dfrac{1}{n+1+p}<\epsilon$ for $n$ large (uniformly in $p$).

Answer 2

You've lost some absolute values there. Don't be afraid of them, as they can help you get rid of the pesky $(-1)^k$ term. All you need to do is use triangle inequality to get, $$\left|\sum_{k=n+1}^{n+p}\frac{(-1)^k}{k(k+1)} \right| \le \sum_{k=n+1}^{n+p} \left| \frac{(-1)^k}{k(k+1)} \right| = \sum_{k=n+1}^{n+p} \frac{1}{k(k+1)}.$$ The latter sum now telescopes.