TITLE: How to evaluate $\sum_{n=2}^\infty\frac{(-1)^n}{n^2-n}$
QUESTION [5 upvotes]: How would you go about evaluating:$$\sum_{n=2}^\infty\frac{(-1)^n}{n^2-n}$$
I split it up to $$\sum_{n=2}^\infty\left[(-1)^n\left(\frac{1}{n-1}-\frac{1}{n}\right)\right]$$
but I'm not sure what to do from here.  If the $(-1)^n$ term wasn't there then it would be a simple telescoping series but the alternating bit causes trouble.

REPLY [2 votes]: $$\sum_{n=2}^{+\infty}\frac{(-1)^n}{n^2-n}=\sum_{n=0}^{+\infty}\frac{(-1)^n}{(n+1)(n+2)}$$
and if we define $A_N$ as:
$$ A_N = \sum_{n=0}^{N}\frac{(-1)^n}{n+1}$$
we have:
$$\sum_{n=0}^{N}\frac{(-1)^n}{(n+1)(n+2)}=\sum_{n=0}^{N}(-1)^n\left(\frac{1}{n+1}-\frac{1}{n+2}\right)=A_N+A_{N+1}-1,$$
but since $A_N\to \log 2$ as $N\to +\infty$, it follows that:
$$\sum_{n=2}^{+\infty}\frac{(-1)^n}{n^2-n}=\color{red}{-1+\log 4}.$$