Sum the series

  • Thread starter WarDieS
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  • #1
23
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Homework Statement



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

[tex]\sum\frac{6}{4n^2+1}[/tex] 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
[tex]\frac{3}{2}\left(-2 + \pi coth (\frac{\pi}{2})\right)[/tex]

So i though maybe is a typo and its actually [tex]\sum\frac{6}{4n^2-1}[/tex] 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

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
351
1
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:

[tex] \sum_{k=-\infty}^{\infty} f(n) = -\pi \sum res \left[\cot (\pi z) f(z) \right] [/tex]
at the poles of f(z).
 
  • #3
22,089
3,293
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:

[tex] \sum_{k=-\infty}^{\infty} f(n) = -\pi \sum res \left[\cot (\pi z) f(z) \right] [/tex]
at the poles of f(z).
Wow, that looks interesting! Do you have a reference for that?
 
  • #4
23
0
Thanks, then its a typo for sure because we didnt took that method in calculus I.

Thanks again
 
  • #5
351
1
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 [itex] \pm (N + 1/2) \pm i(N + 1/2) [/itex]. Then you show that the integral of [itex]f(z)cot(\pi z)[/itex] around the square goes to zero as N-> inf. So then you get that [itex]f(z)cot(\pi z)[/itex] has poles at all the integers and at the poles of f(z). At the integers, the residue is [itex]1/ \pi [/itex]. So you get that the integral is equal to [itex]1/ \pi \sum_{k=-N}^N f(n) [/itex] + the residues of [itex]f(z)cot(\pi z)[/itex] 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.
 
  • #6
22,089
3,293
Wow, that's an extremely interesting paper. Thanks a lot!
 

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