<s>[INST] 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?
 [/INST] 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. 
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