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Infinite sum help (geometric series)

  1. Sep 30, 2009 #1
    1. The problem statement, all variables and given/known data

    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)


    2. Relevant 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



    3. 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.
     
  2. jcsd
  3. Sep 30, 2009 #2

    Dick

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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).
     
  4. Sep 30, 2009 #3

    tiny-tim

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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:
     
  5. Oct 1, 2009 #4
    Thanks to both of you :0) much appreciated.
     
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