How to prove $\sum\limits_{n=p+1}^\infty \frac1{n^2-p^2} = \frac1{2p} (1+\frac12 + \cdots + \frac1{2p})$ for all $p \in \mathbb N$? 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.
 A: We have for $N\geq 2p$:
\begin{align*}
\sum_{n=p+1}^{N+p}\frac 1{n^2-p^2}&=\sum_{j=1}^{N}\frac 1{(j+p)^2-p^2}\\
&=\sum_{j=1}^{N}\frac 1{j^2+2jp}\\
&=\frac 1{2p}\sum_{j=1}^{N}\frac {j+2p-j}{j(j+2p)}\\
&=\frac 1{2p}\sum_{j=1}^{N}\frac 1j-\frac 1{j+2p}\\
&=\frac 1{2p}\left(\sum_{j=1}^{2p}\frac 1j+\sum_{j=2p+1}^N\frac 1j-\sum_{j=2p+1}^{N+2p}\frac 1j\right)\\
&=\frac 1{2p}\left(\sum_{j=1}^{2p}\frac 1j-\sum_{j=N+1}^{N+2p}\frac 1j\right)\\
\sum_{n=p+1}^{N+p}\frac 1{n^2-p^2}&=\frac 1{2p}\sum_{j=1}^{2p}\frac 1j-\frac 1{2p}\sum_{k=1}^{2p}\frac 1{k+N}.
\end{align*}
We get the result taking the limit $N\to \infty$.
A: Use partial fractions to write
$$
\begin{align}
\sum_{n=p+1}^N \frac{1}{n^2-p^2}
&=\frac{1}{2p}\sum_{n=p+1}^N \frac{1}{n-p}-\frac{1}{n+p}\\
&=\frac{1}{2p}\sum_{n=1}^{N-p}\frac{1}{n}-\frac{1}{2p}\sum_{n=2p+1}^{N+p}\frac{1}{n}\\
&=\frac{1}{2p}\sum_{n=1}^{2p}\frac{1}{n}-\frac{1}{2p}\sum_{n=N-p+1}^{N+p}\frac{1}{n}\tag{1}
\end{align}
$$
Taking the limit of $(1)$ as $N\to\infty$, we get
$$
\sum_{n=p+1}^\infty \frac{1}{n^2-p^2}=\frac{1}{2p}\sum_{n=1}^{2p}\frac{1}{n}\tag{2}
$$
A: Consider the following form of the sum without loss of generality when the initial sum tends to $\infty$:
$$\frac{1}{2p}\sum_{t=0}^\infty\left(\sum_{n=(2t+1)p+1}^{(2t+3)p}\frac{1}{n-p}-\sum_{n=(2t+1)p+1}^{(2t+3)p}\frac{1}{n+p}\right)$$
We obtain that the last sum goes to $0$, and the other sums excepting the first one cancel. Therefore, we get that:
$$\sum_{n=p+1}^\infty \frac{1}{n^2-p^2}=\frac{1}{2p}\sum_{n=p+1}^{3p}\frac{1}{n-p}=\frac{1}{2p}\sum_{n=1}^{2p}\frac{1}{n}$$
The key observation is to group the sums in 2p terms sums that cancel. So, what i typed above may be simply written as:
$$\frac{1}{2p} [(1+\frac{1}{2}+...+\frac1{2p}) - (\frac{1}{2p+1}+\frac1{2p+2}+...+\frac1{4p}) + (\frac{1}{2p+1}+\frac1{2p+2}+...+\frac1{4p}) - (\frac1{4p+1}+\frac1{4p+2}+...+\frac1{6p}) + (\frac1{4p+1}+\frac1{4p+2}+...+\frac1{6p}) - ... ]$$ 
The proof is complete.
