# Compact form of the series $\sum\limits_{n=-\infty}^{\infty} {1\over (x-n)^2}$

What is the compact form of $$\sum\limits_{n=-\infty}^{\infty} {1\over (x-n)^2}?$$ If it isn't a standard result, could someone please explain how they spot the compact form? Thanks.

-
This is not Taylor series, Look up Eisenstein series instead, en.wikipedia.org/wiki/Eisenstein_series – Arjang Sep 19 '11 at 7:05

If you google for "cotangent Herglotz" you will find explanations for why $$\pi\cot(\pi x) = \frac{1}{x} + \sum_{n=1}^{\infty} \left( \frac{1}{x+n} + \frac{1}{x-n} \right).$$ Taking derivatives on both sides of this identity shows that your sum equals $$\frac{\pi^2}{\sin^2(\pi x)}.$$
This looks like the partial fraction decomposition of a function with a double pole at each integer. The first such function that comes to mind is $$f(x):={\pi^2\over \sin^2(\pi x)}\ .$$ In order to prove that this function is indeed the solution of your problem you may take off from the formula $$\cot(\pi x)={1\over\pi}\left({1\over x}+\sum_{k=1}^\infty\Bigl({1\over x-k}+{1\over x+k}\Bigr)\right)\qquad(*)$$ which can be found in many textbooks. You arrive at $(*)$, e.g., by developing the function $$g(t):=\cos(x\ t)\qquad(-\pi\leq t\leq \pi)$$ for fixed $x$ into a Fourier series with respect to $t$ and put $t:=\pi$ at the end.