For all $p \in \mathbb N$, I want to prove
$\sum_{n=p+1}^\infty \frac1{n^2-p^2} = \frac1{2p} \left(1+\frac12 + \cdots + \frac1{2p} \right).$
Up to now, I've approached the problem using induction and/or partial fractions. When I use induction, I kind of manage to show the base case for $p = 1$ (I write down "all" terms of the infinite series and we see that all but $1+\frac12$ and the factor $\frac12$ vanish), but I have no idea how to do the induction step. I've tried writing
$\sum_{n=p+2}^\infty \frac1{n^2-p^2} = \frac1{2p} \left(1 + \frac12 + \cdots + \frac1{2p} \right) - \frac1{2p+1}, $
but I fail to see how to simplify this. I also know that I can rewrite e.g. $\frac1{n^2-p^2} = \frac1{2n(p+n)} - \frac1{2n(p-n)},$
but whenever I try to simplify this term, I eventually end up with the initial fraction again...
So my question is: How do I have to approach this problem? Also, can you provide a hint which might bring me to a further point than I am right now?
Thanks in advance.