How to evaluate this summation

[atchoh1983]

hello guys,

I have tried to evaluate \Sigma 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 \Sigmaeⁿ 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!

 

[element4]

Im not sure, but I really don't think you can express

\sum_{n=1}^{\infty}e^{-an^2}

in closed form. But if you meant

\sum_{n=1}^{\infty}(e^{-an})^2

you can just do the following steps

\sum_{n=1}^{\infty}(e^{-an})^2 = \sum_{n=1}^{\infty}e^{-2an} = \sum_{n=1}^{\infty}(e^{2a})^{-n} = \frac 1{e^{2n}-1}

 

[peterungar]

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).)