I'm trying to prove that the series: $x_n = \sum_{k=1}^{n} (-1)^{k+1}/k$ is Cauchy, i.e. to prove that given $\epsilon > 0$, there is a $N \in \mathbb{N}$ s.t. for all $m,n>N$, $|x_m-x_n| < \epsilon$. WLOG let $m>n$. This means we have to show that
$$|x_m-x_n| = \bigg|\sum_{k=n+1}^{m}\frac{(-1)^{k+1}}{k}\bigg| < \epsilon$$
If the number of terms in the above sum is even, we can rewrite it as
$$\bigg|\frac{(-1)^{n+2}}{n+1}+\frac{(-1)^{n+3}}{n+2}+\ldots+\frac{(-1)^{m+1}}{m}\bigg|$$
$$=\bigg|\frac{(-1)^{n+2}}{(n+1)(n+2)}+\frac{(-1)^{n+4}}{(n+3)(n+4)}+\ldots+ \frac{(-1)^{m}}{(m-1)m}\bigg|$$
$$=\frac{1}{(n+1)(n+2)}+\frac{1}{(n+3)(n+4)}+\ldots+ \frac{1}{(m-1)m}$$
So we have to pick $N$ such that the above sum is less than $\epsilon$. This is where I'm stuck, mainly because the difference between $n$ and $m$ can be as large as possible. Any hints on how to proceed would be appreciated!
HINT: try to show by induction that
$$\left|\sum_{k=n+1}^{m}\frac{(-1)^{k+1}}{k}\right| \le \frac1n$$
for any chosen $n,m\in\Bbb N$.