With $m\in\mathbb{Z^+}$ fixed, is $\sum_{m\ne n\ge1} (n^2-m^2)^{-1}$ evaluable really elementarily? I found this exercise at the begininning of the series section of a calculus workbook, so it shouldn't require machinery like integrals or special functions; merely telescopic summing or some other easy trick, but I can't see what should be used. How to calculate, with $m\in\mathbb{Z^+}$ fixed, the sum $\displaystyle \sum_{m\ne n\ge1}\dfrac1{n^2-m^2}$?

$$\begin{eqnarray*}\sum_{\substack{n\geq 1\\ n\neq m}}\frac{1}{n^2-m^2}&=&\frac{1}{2m}\sum_{\substack{n\geq 1\\ n\neq m}}\left(\frac{1}{n-m}-\frac{1}{n+m}\right)\\&=&\frac{1}{2m}\sum_{\substack{n\in[1,2m-1]\\n\neq m}}\left(\frac{1}{n-m}-\frac{1}{n+m}\right)+\frac{1}{2m}\sum_{n\geq 2m}\left(\frac{1}{n-m}-\frac{1}{n+m}\right)\\&=&\frac{1}{2m}\left(\frac{1}{2m}-H_{3m-1}\right)+\frac{1}{2m}\sum_{h\geq 0}\left(\frac{1}{m+h}-\frac{1}{3m+h}\right)\\&=&\frac{1}{2m}\left(\frac{1}{2m}-H_{3m-1}\right)+\frac{1}{2m}\sum_{t=m}^{3m-1}\frac{1}{t}\\&=&\frac{1}{2m}\left(\frac{1}{2m}-H_{3m-1}+H_{3m-1}-H_{m-1}\right)=\color{red}{\frac{1}{4m^2}-\frac{H_{m-1}}{2m}}.\end{eqnarray*}$$
As usual, $H_n$ is the $n$-th harmonic number, i.e. $\sum_{k=1}^{n}\frac{1}{k}$.
