How to find the value $\sum_{n\geq2}^{\infty}(-1)^{n+1}\frac{n}{n^2-1}$ converges to? How to find the value this sum converges to?$$\sum_{n\geq2}^{\infty}(-1)^{n+1}\frac{n}{n^2-1} $$
I've tried writing it like this
$$\sum_{n\geq2}^{\infty}(-1)^{n+1}·n·\Bigg(\frac{1/2}{n-1}-\frac{1/2}{n+1}\Bigg) $$
and writing a few terms, but they won't cancel and I ended up with no ideas, any hint?
I haven't learnt integration nor differentiation FYI.

Try writing $$\begin{align}n\left({1\over n-1}-{1\over n+1}\right)&=
n\left({1\over n-1}-{1\over n} + {1\over n} -{1\over n+1}\right)\\
&=n\left({1\over n(n-1)}+{1\over n(n+1)}\right)\\&={1\over n-1}+{1\over n+1}\end{align}$$
