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 $\displaystyle \sum_{m\ne n\ge1}\dfrac1{n^2-m^2}$?
 A: 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...)
A: $$\begin{eqnarray*}\sum_{\substack{n\geq 1\\ n\neq m}}\frac{1}{n^2-m^2}&=&\frac{1}{2m}\sum_{\substack{n\geq 1\\ n\neq m}}\left(\frac{1}{n-m}-\frac{1}{n+m}\right)\\&=&\frac{1}{2m}\sum_{\substack{n\in[1,2m-1]\\n\neq m}}\left(\frac{1}{n-m}-\frac{1}{n+m}\right)+\frac{1}{2m}\sum_{n\geq 2m}\left(\frac{1}{n-m}-\frac{1}{n+m}\right)\\&=&\frac{1}{2m}\left(\frac{1}{2m}-H_{3m-1}\right)+\frac{1}{2m}\sum_{h\geq 0}\left(\frac{1}{m+h}-\frac{1}{3m+h}\right)\\&=&\frac{1}{2m}\left(\frac{1}{2m}-H_{3m-1}\right)+\frac{1}{2m}\sum_{t=m}^{3m-1}\frac{1}{t}\\&=&\frac{1}{2m}\left(\frac{1}{2m}-H_{3m-1}+H_{3m-1}-H_{m-1}\right)=\color{red}{\frac{1}{4m^2}-\frac{H_{m-1}}{2m}}.\end{eqnarray*}$$
As usual, $H_n$ is the $n$-th harmonic number, i.e. $\sum_{k=1}^{n}\frac{1}{k}$.
A: We have that
$$
\begin{align*}
\sum_{\substack{j\geqslant 1\\j\neq k}}\frac1{j^2-k^2}&=\frac1{2k}\sum_{\substack{j\geqslant 1\\j\neq k}}\left(\frac1{j-k}-\frac1{j+k}\right)
=\frac1{2k}\lim_{\substack{N\to \infty \\N>k}}\sum_{\substack{j=1\\j\neq k}}^N\left(\frac1{j-k}-\frac1{j+k}\right)\\
&=\frac1{2k}\lim_{\substack{N\to \infty \\N>k}}\left(\sum_{\substack{j=1\\j\neq k}}^N\frac1{j-k}-\sum_{j=1}^N \frac1{j+k}+\frac1{2k}\right)\\
&=\frac1{2k}\lim_{\substack{N\to \infty \\N>k}}\left(\sum_{j=1}^{k-1}\frac1{j-k}+\sum_{j=k+1}^N \frac1{j-k}-\sum_{j=1}^N \frac1{j+k}+\frac1{2k}\right)\\
&=\frac1{2k}\lim_{\substack{N\to \infty \\N>k}}\left(-H_{k-1}+\sum_{h=1}^{N-k} \frac1h-\sum_{h=k+1}^{N+k} \frac1h+\frac1{2k}\right)\\
&=\frac1{2k}\lim_{\substack{N\to \infty \\N>k}}\left(-H_{k-1}+H_k+\sum_{h=k+1}^{N-k} \frac1h-\sum_{h=k+1}^{N+k} \frac1h+\frac1{2k}\right)\\
&=\frac1{2k}\lim_{\substack{N\to \infty \\N>k}}\left(\frac1{k}-\sum_{h=N-k+1}^{N+k} \frac1h+\frac1{2k}\right)\\
&=\frac1{2k}\lim_{\substack{N\to \infty \\N>k}}\left(\frac3{2k}-\sum_{s=1}^{2k} \frac1{s+N-k}\right)\\
&=\frac{3}{4k^2}
\end{align*}
$$
