# Sum the series

## 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 wich 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 wanna know how can be the coth function in the solution!

Thx for the help

## The Attempt at a Solution

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).

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?

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

Thanks again

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.

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