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 $\displaystyle \sum_{m\ne n\ge1}\dfrac1{n^2-m^2}$?

Since $$\frac1{n^2-m^2}=\frac{1}{2m}\left(\frac{1}{n-m}-\frac{1}{n+m}\right)$$the sum does telescope "eventually"; for any specific $m$ you can see that it equals $\frac1{2m}$ times the sum of finitely many terms $$\frac{\mp1}{n\pm m};$$all the other terms cancel. (If that's not clear write out a large number of terms for $m=3$ and see what happens...)
