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.

We have the series were $n<m$
$\frac 1{2m}(\frac {1}{m + 1} + \frac {1}{m + 2} +\frac {1}{m + 3}+\cdots+ \frac {1}{2m - 1}- (\frac {1}{1-m} + \frac {1}{2-m} +\frac {1}{3-m}+\cdots+ \frac {1}{-1}))$
$\frac 1{2m}(\frac {1}{m + 1} + \frac {1}{m + 2} +\frac {1}{m + 3}+\cdots+ \frac {1}{2m - 1} + \frac {1}{1} + \frac {1}{2} +\frac {1}{3}+\cdots+ \frac {1}{m-1}))$
Then we skip $n = m$ 
And we go from $m<n<2m$
$\frac 1{2m}(\frac {1}{2m + 1} + \frac {1}{2m + 2} +\frac {1}{2m + 3}+\cdots+ \frac {1}{3m - 1}- (\frac {1}{1} + \frac {1}{2} +\frac {1}{3}+\cdots+ \frac {1}{m-1}))$
and we have some stuff that cancels with the first partial sum
$\frac 1{2m}(\frac {1}{m + 1} + \frac {1}{m + 2} +\frac {1}{m + 3}+\cdots+ \frac {1}{2m - 1} + \frac {1}{2m + 1} + \frac {1}{2m + 2} +\frac {1}{2m + 3}+\cdots+ \frac {1}{3m - 1})$
Note that we have skipped $\frac {1}{2m}$
$n\ge 2m$
$\frac 1{2m}(\frac {1}{3m} - \frac {1}{m} + \frac {1}{3m+1} - \frac {1}{m+1}\cdots)$
Now we are subtracting every term from  $\frac {1}{m+1}$ on that we added since the beginning.
$\frac {1}{2m}(-\frac {1}{m}-\frac {1}{2m}) = -\frac {3}{4m^2}$ 
