Computing $\sum_{m \neq n} \frac{1}{n^2-m^2}$ A series arising in perturbation theory in quantum mechanics:
$\sum_{m\neq n} \frac{1}{n^2 - m^2}$, where $n$ is a given positive odd integer and $m$ runs through all odd positive integers different from $n$. I have a hunch that residue methods are applicable here, but I don't know complex analysis.

You can write
$$ \frac{1}{n^2 - m^2} = \frac{1}{2n}
\left\lbrace \frac{1}{m+n} - \frac{1}{m-n} \right\rbrace . \quad (1)$$
Now if we sum up both sides over all odd $m \ne n ,$ taking into account that $n$ is odd, lots of cancelling goes on and we obtain
$$\sum_{m \ne n} \frac{1}{n^2 - m^2} = -\frac{1}{4n^2}.$$
At first sight it appears the cotangent identity could be useful but it's not actually needed.
As a numerical check try summing the following with wolframAlpha
$$1/24 + 1/16 + \sum_{k=3}^\infty 1/(5^2 - (2k+1)^2),$$
you will see that it is $-1/100,$ as expected.
Or try this:
$$1/48 + 1/40 + 1/24 + \sum_{k=4}^\infty 1/(7^2 - (2k+1)^2).$$
You will get $-1/196.$
EDIT: To clarify the cancellation taking place when we sum the RHS of $(1).$
We have
$$\sum_{m \ne n, \,\, m \textrm{ odd} }
\left\lbrace \frac{1}{m+n} - \frac{1}{m-n} \right\rbrace

= \sum_{m \ne n, \,\, m \textrm{ odd} }
\left\lbrace \frac{1}{n+m} + \frac{1}{n-m} \right\rbrace $$
$$= \left\lbrace \left( \frac{1}{n+1} + \frac{1}{n-1} \right) +
\left( \frac{1}{n+3} + \frac{1}{n-3} \right) +
\left( \frac{1}{n+5} + \frac{1}{n-5} \right) + \cdots
+ \left( \frac{1}{2n-2} + \frac{1}{2} \right) \right\rbrace $$
$$+ \left( \frac{1}{2n+2} - \frac{1}{2} \right)
+ \left( \frac{1}{2n+4} - \frac{1}{4} \right)
+ \left( \frac{1}{2n+6} - \frac{1}{6} \right) + \cdots $$
and rearranging all the terms in the braces
$$= \left\lbrace \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n-2} \right\rbrace
+ \left( \frac{1}{2n+2} - \frac{1}{2} \right)
+ \left( \frac{1}{2n+4} - \frac{1}{4} \right) + \cdots $$
$$=\left\lbrace \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n-2} \right\rbrace
- \left\lbrace \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n} \right\rbrace
= - \frac{1}{2n}$$
and hence the result.
