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.
Compact form of the series $\sum\limits_{n=-\infty}^{\infty} {1\over (x-n)^2}$
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0This is not Taylor series, Look up Eisenstein series instead, http://en.wikipedia.org/wiki/Eisenstein_series – 2011-09-19
2 Answers
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.