Sum of series $\sum_{\underset{m \neq n}{n = 1}}^{\infty} \frac{1}{m^2 - n^2}$ I was trying to solve this series and I have an exam in a week. I can't understand how to find its sum although I managed to rework it by transforming $\frac{1}{m^2 - n^2}$ into $\frac{1}{2m}(\frac{1}{m+n} - \frac{1}{n-m})$.
I'm sure there is a way, as my book explicitly confirms this as being convergent and having a sum.
Any help is appreciated.

I get
$-\frac{3}{4m^2}
$.
$\begin{array}\\
s(m)
&=\sum_{\underset{m \neq n}{n = 1}}^{\infty} \frac{1}{m^2 - n^2}\\
&=\sum_{\underset{m \neq n}{n = 1}}^{\infty} \frac{1}{2m}(\frac{1}{m+n} + \frac{1}{m-n})\\
&= \frac{1}{2m}\sum_{\underset{m \neq n}{n = 1}}^{\infty}(\frac{1}{m+n} + \frac{1}{m-n})\\
&= \frac{1}{2m}(\sum_{n = 1}^{m-1}(\frac{1}{m+n} + \frac{1}{m-n})
+\sum_{n = m+1}^{\infty}(\frac{1}{m+n} + \frac{1}{m-n}))\\
&= \frac{1}{2m}(t(m)+u(m))\\
t(m)
&=\sum_{n = 1}^{m-1}(\frac{1}{m+n} +\frac{1}{m-n})\\
&= \sum_{n = 1}^{m-1}\frac{1}{m+n} +\sum_{n = 1}^{m-1}\frac{1}{m-n}\\
&= \sum_{n = m+1}^{2m-1}\frac{1}{n} + \sum_{n = 1}^{m-1}\frac{1}{n}\\
&=H_{2m-1}-H_m+H_{m-1}\\
&=H_{2m-1}-\dfrac1{m}\\
u(m)
&=\sum_{n = m+1}^{\infty}(\frac{1}{m+n} + \frac{1}{m-n})\\
&=\lim_{k \to \infty}\sum_{n = m+1}^{k}(\frac{1}{m+n} + \frac{1}{m-n})\\
&=\lim_{k \to \infty}\left(\sum_{n = m+1}^{k}\frac{1}{m+n} +\sum_{n = m+1}^{k}\frac{1}{m-n}\right)\\
&=\lim_{k \to \infty}\left(\sum_{n = m+1}^{k}\frac{1}{m+n} -\sum_{n = m+1}^{k}\frac{1}{n-m}\right)\\
&=\lim_{k \to \infty}\left(\sum_{n = 2m+1}^{k+m}\frac{1}{n} - \sum_{n = 1}^{k-m}\frac{1}{n}\right)\\
&=\lim_{k \to \infty}\left(\sum_{n = 2m+1}^{k-m}\frac{1}{n}+\sum_{n = k-m+1}^{k+m}\frac{1}{n} - (\sum_{n = 1}^{2m}\frac{1}{n}+\sum_{n = 2m+1}^{k-m}\frac{1}{n})\right)\\
&=\lim_{k \to \infty}\left(\sum_{n = k-m+1}^{k+m}\frac{1}{n} - \sum_{n = 1}^{2m}\frac{1}{n}\right)\\
&=-H_{2m}
\qquad\text{since }\sum_{n = k-m+1}^{k+m}\frac{1}{n}<\frac{2m}{k-m+1} \to 0\\
\text{so}\\
s(m)
&=\frac1{2m}(H_{2m-1}-\dfrac1{m}-H_{2m})\\
&=\frac1{2m}(-\dfrac1{m}-\dfrac1{2m})\\
&=-\frac{3}{4m^2}\\
\end{array}
$
