This series came up when working out a physics problem, and I have been unable to derive the sum with any rigor. Here is the series and it's probable evaluation...
$$\sum_{odd m\neq n}^{\infty}\frac{1}{n^2-m^2}=\frac{-1}{4n^2}$$
To clarify, the sum is being taken over all odd $m$ except the specific term where $m$ is equal to $n$. $n$ is being held constant at an arbitrary odd integer. I arrived at the result somewhat empirically after looking at the first couple $n$ and seeing that it is a collapsing sum. I hope that someone can give me a more justifiable derivation, and/or check/correct my result.
Hint: $\displaystyle\sum_{m=\text{odd}}\frac1{m^2-x^2}=\sum_{k=0}^\infty\frac1{(2k+1)^2-x^2}=\frac\pi4\cdot\frac{\tan\bigg(\dfrac\pi2x\bigg)}x.~$ Then your sum becomes
$\displaystyle\lim_{x\to n}\bigg[\frac\pi4\cdot\frac{\tan\bigg(\dfrac\pi2x\bigg)}x-\frac1{n^2-x^2}\bigg]$, where odd $n=2p+1$.