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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.

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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.

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I have to admit this is more straightforward. Cotangent identity is overkill. – Raskolnikov Feb 20 '11 at 10:41
Quite. Thanks to you both! – Jacob Greenfield Feb 20 '11 at 13:42

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.

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Doesn't work - I need $n$ to be an integer, where $\cot(n\pi)$ blows up, not to mention the fact that I'm summing over only odd integers. – Jacob Greenfield Feb 20 '11 at 10:12
Of course it blows up, but that is because you left the term where both integers are equal in the sum. If you pull it out and put it on the other side, you'll get a limit. – Raskolnikov Feb 20 '11 at 10:15
So, to have the sum only over odd integers, some tweaking is needed, but I think that can be done as well. I think it can be done by deriving a formula for even integers and then substract it from the formula for all integers to obtain one only for odd integers. – Raskolnikov Feb 20 '11 at 10:36

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