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.
 A: Use the cotangent indentity at the very end of the page.
There are several ways to derive this identity, some use residues, other the Euler product formula for the sine.
EDIT: So we have the identity:
$$\frac{\pi z \cot (\pi z) - 1}{2z^2} = \sum_{m=1}^{\infty} \frac{1}{z^2-m^2}$$
Now, we want to compute the sum $\sum_{n\neq m} \frac{1}{n^2-m^2}$, so we rewrite the identity as follows:
$$\sum_{\substack{m=1, \\ n\neq m}}^{\infty} \frac{1}{z^2-m^2} = \frac{\pi z \cot (\pi z) - 1}{2z^2} - \frac{1}{z^2-n^2}$$
And we take the limit for $z \to n$ for both sides to get the desired sum.
EDIT 2: After some work rearranging, I arrived at the following identity
$$-\frac{\pi \tan \left(\pi \frac{z}{2}\right)}{4z} = \sum_{n \; odd} \frac{1}{z^2-n^2}$$
which can be used along the lines explained in the previous edit.
A: 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.
