The fourier series $\sum_{m\neq n} \frac{1}{n^2 - m^2} \cos \frac{m\pi x}{2a}$ A Fourier series arising in perturbation theory in quantum mechanics is
$$\sum_{m\neq n} \frac{1}{n^2 - m^2} \cos \frac{m\pi x}{2a} \, .$$
where $n$ is an odd positive integer and $n$ runs through all odd positive integers other than $n$. (The numbers are odd so that the Fourier terms are zero at $\pm a$.)
I have no idea what kind of function produces this series. Is it familiar to anyone?
 A: Taking $n=1$ and $a= \frac{1}{2}$, wolframalpha gives me $$\displaystyle \sum_{m = \text{odd}}^{\infty} \frac{\cos(m \pi x)}{1-m^2} = \frac{1}{4}(\pi(1-2x)\sin(\pi x) - \cos(\pi x))$$
Taking $n=3$ and $a= \frac{1}{2}$, wolframalpha gives me $$\displaystyle \sum_{m = \text{odd}}^{\infty} \frac{\cos(m \pi x)}{9-m^2} = \frac{1}{36}(3\pi(1-2x)\sin(3\pi x) - \cos(3\pi x))$$
Following this my hunch would be for $a= \frac{1}{2}$
$$\displaystyle \sum_{m = \text{odd}}^{\infty} \frac{\cos(m \pi x)}{n^2-m^2} = \frac{1}{4n^2}(n\pi(1-2x)\sin(n\pi x) - \cos(n\pi x))$$
Playing around a bit more with wolfram alpha, my new guess is 
For $x \in [0,a]$,
$$\displaystyle \sum_{m\neq n} \frac{1}{n^2 - m^2} \cos \frac{m\pi x}{2a} = \frac{1}{4n^2} \left(n\pi \left(1-\frac{x}{a} \right) \sin \left(\frac{n\pi x}{2a} \right) - \cos \left(\frac{n\pi x}{2a} \right) \right)$$
and For $x \in (-a,0]$,
$$\displaystyle \sum_{m\neq n} \frac{1}{n^2 - m^2} \cos \frac{m\pi x}{2a} = \frac{1}{4n^2} \left(n\pi \left(-1-\frac{x}{a} \right) \sin \left(\frac{n\pi x}{2a} \right) - \cos \left(\frac{n\pi x}{2a} \right) \right)$$
A: This, or something similar, is the Fourier series for the fundamental solution for the wave operator (calling time "y") on a product of two circles: $(D_x^2-D_y^2)u=\delta$ (with periodic delta having Fourier expansion with all coefficients 1). The Fourier series for $\delta$ converges in the $-1-\epsilon$ Sobolev space (the point being to legitimize these manipulations!). Since the wave operator is not elliptic, we have no a-priori assurance that the solution $u$ is in any better Sobolev space than is $\delta$, but it may be so by accident.
The manipulations suggested in the other answers are legitimizable as reflecting limits taken in negatively-indexed Sobolev spaces.
A: You can find a differential equation for the expression by differentiating it twice. Just to get rid of the annoying factors, let's define $z={\pi x}/{2a}$. Then
$$\frac{d^2}{dz^2}\sum_{\substack{m \; odd \\ m\neq n}} \frac{1}{n^2 - m^2} \cos (mz) = \sum_{\substack{m \; odd \\ m\neq n}} \frac{-m^2}{n^2 - m^2} \cos (mz) = $$
$$= \sum_{\substack{m\;  odd \\ m\neq n}} \cos (mz) - n^2 \sum_{\substack{m \; odd \\ m\neq n}} \frac{1}{n^2 - m^2} \cos (mz)\; .$$
Or 
$$f''(z) + n^2 f(z) = \sum_{\substack{m \; odd \\ m\neq n}} \cos (mz) = -\cos (nz) \; .$$
The sum of cosines should be easy to perform (replace the cosines by complex exponentials, use the geometric summation formula and take the real part), and solving the differential equation is also not difficult.
