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.

The case of all integers is
$$
\sum_{m=1}^{n-1}\frac{1}{n^2-m^2}+\sum_{m=n+1}^{\infty}\frac{1}{n^2-m^2} = \frac{-3}{4n^2}
$$
and the "odd" case is similar
$$
\sum_{k=1}^{j-1}\frac{1}{(2j-1)^2-(2k-1)^2}
+\sum_{k=j+1}^\infty\frac{1}{(2j-1)^2-(2k-1)^2} =
\frac{-1}{4(2j-1)^2}
$$
These may be computed with information about the "digamma" function $\psi(z) = \Gamma'(z)/\Gamma(z)$, namely:
$$
\sum_{k=u}^v \frac{1}{a+k} = \psi(a+v+1)-\psi(a+u),
\qquad\text{where }u<v\text{ integers and } a\text{ complex}
$$
and
$$
\psi(z) = \log z -\frac{1}{2z} + O\left(\frac{1}{z^2}\right)\text{ as } z \to +\infty
$$
