Revision as of 10:15, 21 January 2011

How to Evaluate a Complicated Sum

Does somebody know how to calculate the sum

$ \sum_{k=-\infty}^\infty \frac{1}{1+k^2} ? \ $

Wolfram said answer is π * coth(π). is there any easier way to do that? Yimin. Jan 20

Discussion/help

Big hint: See page 189 of the classic book, "Complex Analysis," by Lars Ahlfors and think about letting z=i in formula (11). After that, it will be like falling off a log ... head first.--Steve Bell 12:32, 21 January 2011 (UTC)

Ok, I've got more time now. Here's that famous formula, (11) on p. 189:

$ \pi\cot \pi z =\frac{1}{z}+\sum_{n=1}^\infty\frac{2z}{z^2-n^2} $

where z is a complex number. The cotangent function is the complex cotangent given by

$ \cot z=\frac{\cos z}{\sin z}=i\frac{e^{iz}+e^{-iz}}{e^{iz}-e^{-iz}}. $

You don't have to know much complex analysis to cook up this formula... just Liouville's Theorem and an understanding of isolated singularities and uniform convergence. It follows quite easily from one of the most mind blowing formulas in all of math: The complex sine function satisfies

$ \frac{\pi^2}{\sin^2\pi z}=\sum_{n=-\infty}^\infty\frac{1}{(z-n)^2}. $

--Steve Bell 14:02, 21 January 2011 (UTC)

(I can't resist adding here that writing the formula in Wolfram alpha probably is the easiest way to evaluate the sum!)

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