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How to evaluate this summation

  1. Jan 17, 2009 #1
    hello guys,

    I have tried to evaluate [tex]\Sigma[/tex] e-an2 so many times, but I didn't get it.

    where a is just a constant and summation begin from n=1 to infinity.

    I know that [tex]\Sigma[/tex]en is just geometric series which is equal 1/(1-e)

    But when n changes to be n2, I have no ideas how to do that.

    If anyone know, please give me a suggestion.

    with much thanks and appreciate!!
     
  2. jcsd
  3. Jan 17, 2009 #2
    Im not sure, but I really dont think you can express

    [tex]\sum_{n=1}^{\infty}e^{-an^2}[/tex]

    in closed form. But if you meant

    [tex]\sum_{n=1}^{\infty}(e^{-an})^2[/tex]

    you can just do the following steps

    [tex]\sum_{n=1}^{\infty}(e^{-an})^2[/tex] = [tex]\sum_{n=1}^{\infty}e^{-2an}[/tex] = [tex]\sum_{n=1}^{\infty}(e^{2a})^{-n}[/tex] = [tex]\frac 1{e^{2n}-1}[/tex]
     
  4. Jan 17, 2009 #3
    This is an elliptic theta function, see Handboook of Mathematical Functions by Milton Abramowitz and Irene Stegun, p. 576, 16.27.3 (Put z=1 and q=exp(-a).)
     
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