Sum of a series $\frac {1}{n^2 - m^2}$ m and n odd, $m \ne n$ I was working on a physics problem, where I encountered the following summation problem:
$$ \sum_{m = 1}^\infty \frac{1}{n^2 - m^2}$$ where m doesn't equal n, and both are odd. n is a fixed constant
We could alternatively write:
$$ \sum_{j = 1}^\infty \frac{1}{(2i - 1)^2 - (2j - 1)^2}$$ where i doesn't equal j. I used a numerical solver to solve for a few values of n, and the solution seems to be $-1/(4n^2)$ which fits with the problem, but I'd like to find a rigorous solution.
My attempts so far have involved simplifying the term in the sum in various was to see if I could find a way solve the problem, but I've had no luck.
Ex. The second equation can be simplified to
$$\sum_{j=1}^\infty \frac {1} {4(i + j - 1)(i - j)} $$
but I'm not sure where to go from here.
I've also tried splitting the problem into two sections, one where m < n, and one where m > n, but again, I'm not sure how to continue.

Assume $N>i\geq1$. Starting by a partial fraction decomposition, one may write
$$
\begin{align}
&\sum_{j=1,\,j\neq i}^N\frac1{(2i-1)^2-(2j-1)^2}
\\\\&=\frac1{4(2i-1)}\sum_{j=1,\,j\neq i}^N\left(\frac1{j+i-1}-\frac1{j-i}\right)
\\\\&=\frac1{4(2i-1)}\left(\sum_{j=1}^N\frac1{j+i-1}-\frac1{2i-1}-\sum_{j=1,\,j\neq i}^N\frac1{j-i}\right)
\\\\&=\frac1{4(2i-1)}\left(\sum_{\ell=i}^{N+i-1}\frac1{\ell}-\frac1{2i-1}-\sum_{j=1}^{i-1}\frac1{j-i}-\sum_{j=i+1}^N\frac1{j-i}\right)
\\\\&=\frac1{4(2i-1)}\left(H_{N+i-1}-H_{i-1}-\frac1{2i-1}+H_{i-1}-H_{N-i}\right)
\\\\&=-\frac1{4(2i-1)^2}+\frac1{4(2i-1)}\left(H_{N+i-1}-H_{N-i}\right)
\end{align}
$$ then, as $N \to \infty$,

$$
\sum_{j=1,\,j\neq i}^\infty\frac1{(2i-1)^2-(2j-1)^2}=-\frac1{4(2i-1)^2}
$$ 

where $H_N$ is the harmonic number and where we have used
$$
H_{N+i-1}-H_{N-i}=O\left(\frac1{N}\right)
$$ as $N \to \infty$.
