Solving the Series: \sum\frac{6}{4n^2+1}

[WarDieS]

Homework Statement

Hi, i had to solve a some series and i had no problem except for this one

\sum\frac{6}{4n^2+1} from 1 to infinity

The Attempt at a Solution

i've tried to do it by simple fractions didn't work, it's no geometric,hipergeometric, telescopic.

All exercices had "nice" solutions meaning they were round numbers or fractions, so i solved this one with mathematica and gave me an "ugly" solution which is
\frac{3}{2}\left(-2 + \pi coth (\frac{\pi}{2})\right)

So i though maybe is a typo and its actually \sum\frac{6}{4n^2-1} from 1 to infinity, so i did this one by simple fractions and gave me 3, but i still want to learn how to solve the original, i want to know how can be the coth function in the solution!

Thx for the help

 

[hgfalling]

This series can be summed using the method of complex residues.

Basically if you have a function f(z) that satisfies some weak criteria and an associated series as a function of n in the integers, you get:

\sum_{k=-\infty}^{\infty} f(n) = -\pi \sum res \left[\cot (\pi z) f(z) \right]
at the poles of f(z).

 

[micromass]

This series can be summed using the method of complex residues.

Basically if you have a function f(z) that satisfies some weak criteria and an associated series as a function of n in the integers, you get:

\sum_{k=-\infty}^{\infty} f(n) = -\pi \sum res \left[\cot (\pi z) f(z) \right]
at the poles of f(z).

Wow, that looks interesting! Do you have a reference for that?

 

[WarDieS]

Thanks, then its a typo for sure because we didnt took that method in calculus I.

Thanks again

 

[hgfalling]

Wow, that looks interesting! Do you have a reference for that?

http://scipp.ucsc.edu/~haber/archives/physics116A06/Sixways.pdf p.12-15 outlines the method.

The quick version:
Basically you draw a square with vertices \pm (N + 1/2) \pm i(N + 1/2). Then you show that the integral of f(z)cot(\pi z) around the square goes to zero as N-> inf. So then you get that f(z)cot(\pi z) has poles at all the integers and at the poles of f(z). At the integers, the residue is 1/ \pi. So you get that the integral is equal to 1/ \pi \sum_{k=-N}^N f(n) + the residues of f(z)cot(\pi z) at the poles of f(z). Since the integral is zero, those two things must be equal, and the first one is the sum of the series as N->inf.

 

[micromass]

Wow, that's an extremely interesting paper. Thanks a lot!