I am trying to understand this summation: $$\sum_{m \gt 0, m\neq n} \frac{1}{m^2-n^2}=\frac{1}{2n} \sum_{m \gt 0, m\neq n } \left(\frac{1}{m-n}- \frac{1}{m+n}\right)=\frac{3}{4n^2}$$
But I can't see why the following step is right: $$\sum_{m \gt 0, m\neq n } \left(\frac{1}{m-n}- \frac{1}{m+n}\right)=\frac{3}{2n}$$
For $N>n$:
$$\sum_{m \gt 0, m\neq n, m\le N }\left( \frac{1}{m-n}- \frac{1}{m+n}\right)=[H_{N-n}-H_{n-1}]-[H_{N+n}-H_n-\frac{1}{2n}]$$
because:
$$\sum_{m \gt 0, m\neq n, m\le N }\left( \frac{1}{m-n}\right)=\frac{1}{1-n}+\frac{1}{2-n}+\cdots+\frac{1}{-1}+\frac{1}{1}+\frac{1}{2}+\cdots\frac{1}{N-n}=-\left(\frac{1}{n-1}+\frac{1}{n-2}+\cdots+\frac{1}{1}\right)+\left(\frac{1}{1}+\frac{1}{2}+\cdots\frac{1}{N-n}\right)$$
$$\sum_{m \gt 0, m\neq n, m\le N } \left(\frac{1}{m+n}\right)=\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+N}\right)-\frac{1}{n+n}$$
Now, noting that $H_k=\log k +\gamma + o(1)$, we see that the sum up to $N$ will be
$$\log\left(\frac{N-n}{N+n}\right)+o(1)+H_n-H_{n-1}+\frac{1}{2n}$$
$$=\frac{3}{2n}+o(1)\to \frac{3}{2n}$$
(noting that $\log\left(\frac{N-n}{N+n}\right)\to0$ as $n\to\infty$).