How to find the sum of the series $\sum_{k=2}^\infty \frac{1}{k^2-1}$? I have this problem :
$$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1}$$
My solution
$$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1} = -\frac{1}{2}\sum_{k=1}^\infty \frac{1}{k+1} -\frac{1}{k-1} = -\frac{1}{2}[(\frac{1}{3}-1)+(\frac{1}{4}-\frac{1}{2})+(\frac{1}{5}-\frac{1}{3})+(\frac{1}{6}-\frac{1}{4})+...]$$
I think that the sum should be $\frac{1}{2}$ since the limit of :
$$-\frac{1}{2}[-1+\frac{1}{k}+...+\frac{1}{n}] = \frac{1}{2}$$
But that wrong. Any ideas?

your choice to use the "telescoping sum" technique is fine. what is wrong is just an arithmetic prob.
since the series is absolutely convergent try first rewriting as two sums:
$$
\sum_{k=1}^\infty \frac{1}{(2k)^2-1} + \sum_{k=1}^\infty \frac{1}{(2k+1)^2-1}
$$
and evaluate these separately. 
