How to evaluate $\sum_{n=2}^\infty\frac{(-1)^n}{n^2-n}$ 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.
 A: Hint
We have
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=\ln2$$
and we can prove it by several ways.
A: \begin{align}
& \sum_{n=2}^\infty (-1)^n\left(\frac{1}{n-1}-\frac{1}{n}\right) \\[10pt]
= {} & \underbrace{\left(1-\frac 1 2 \right) - \left( \frac 1 2\right.}_{=\,0} - \left.\frac 1 3 \right) + \left( \frac 1 3 - \frac 1 4 \right) - \left( \frac 1 4 - \frac 1 5\right) + \left( \frac 1 5 - \frac 1 6 \right) - \cdots \\[10pt]
= {} & \frac 2 3 - \frac 2 4 + \frac 2 5 - \frac 2 6 + \cdots \\[10pt]
= {} & 2\int_0^1 \left( u^2 - u^3 + u^4 - u^5 + \cdots \right) \,du = 2\int_0^1 \frac{u^2}{1+u} \, du \\[10pt]
= {} & 2 \int_0^1 \left( u - 1 + \frac 1 {u+1} \right)\,du = \text{etc.}
\end{align}
A: $$\sum_{n=2}^{+\infty}\frac{(-1)^n}{n^2-n}=\sum_{n=0}^{+\infty}\frac{(-1)^n}{(n+1)(n+2)}$$
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:
$$\sum_{n=2}^{+\infty}\frac{(-1)^n}{n^2-n}=\color{red}{-1+\log 4}.$$
