Infinite sum help (geometric series)

[TheMightyJ]

Homework Statement

Well, the original question is to solve this ...

\sum 1/(a² + x²)

the sum goes from x=-infinity to infinity (i wasnt sure how to show this with the latex??)

and the answer i am supposed to show is \pi/a + (2*\pi/a) * (1/(e^2*\pi*a - 1)

Homework Equations

So, using Poisson resummation and some results from previous exercises, i get the new problem to being

\sum (\pi/a) * (e^-2*\pi*a*|v|)

the sum now from v=-infinity to infinity

The Attempt at a Solution

so i can split this into three.

the easy bit is to take the v=o part, this gives me the \pi/a required in the answer.

then i have two parts, the sum with v=1 to infinity and the sum with v=-infinity to -1

these two parts are identical, due to the |v|.

so i have 2*\sum (\pi/a) * (e^-2*\pi*a*|v|)

i can take the 2*\pi/a outside of the sum, as it has no v element.

so now i am left having to show that

\sum (e^-2*\pi*a*|v|) = (1/(e^2*\pi*a - 1)

which, i realize is just a (i think) simple geometric series sum ... and yet I am not sure how to show these two equate?

any help would be brilliant, thank you.

 

[Dick]

The sum from 0 to infinity is 1/(1-exp(-2*pi*a)), geometric series, right? So the sum from 1 to infinity is [1/(1-exp(-2*pi*a))]-1. Do some algebra including multiply numerator and denominator by exp(2*pi*a).

 

[tiny-tim]

Hi TheMightyJ!

(have a pi: π and an infinity: ∞ and a sigma: ∑ )

With LaTeX, it's \sum_{-\infty}^{\infty}, and without, it's just ∑_-∞^∞

 

[TheMightyJ]

Thanks to both of you :0) much appreciated.