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Infinite sum help (geometric series)

  • Thread starter TheMightyJ
  • Start date
  • #1

Homework Statement



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

[tex]\sum[/tex] 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 [tex]\pi[/tex]/a + (2*[tex]\pi[/tex]/a) * (1/(e2*[tex]\pi[/tex]*a - 1)


Homework Equations



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

[tex]\sum[/tex] ([tex]\pi[/tex]/a) * (e-2*[tex]\pi[/tex]*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 [tex]\pi[/tex]/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*[tex]\sum[/tex] ([tex]\pi[/tex]/a) * (e-2*[tex]\pi[/tex]*a*|v|)

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

so now i am left having to show that

[tex]\sum[/tex] (e-2*[tex]\pi[/tex]*a*|v|) = (1/(e2*[tex]\pi[/tex]*a - 1)

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

any help would be brilliant, thank you.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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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).
 
  • #3
tiny-tim
Science Advisor
Homework Helper
25,832
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Hi TheMightyJ! :smile:

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

With LaTeX, it's \sum_{-\infty}^{\infty}, and without, it's just ∑-∞ :smile:
 
  • #4
Thanks to both of you :0) much appreciated.
 

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