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How to find the following series' value? $$\sum_{m=1}^{\infty}{\frac{e^{-a m^2}}{m^2}}$$ I know that this series converges. I check it by ratio test or comparison test for $$ a \in \left [ 10^{-15} , 10^{15}\right ]$$ (I need "a" just in this interval.)

thanks.

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Do you know that it has a closed form? –  Antonio Vargas Mar 4 '13 at 6:56
    
Your sum defines a differentiable function in $a$, whose derivative can be expressed in terms of Jacobi Theta functions. Maybe this will lead you to something interesting. –  Ragib Zaman Mar 4 '13 at 7:01
    
I would not expect this series to have a closed form. Why you need it? The convergence is so quick that the value is easy to approximate. –  Emanuele Paolini Mar 4 '13 at 7:06
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When you post a question that's closely related to a question you'd posted previously, please link to the earlier question to avoid duplication of efforts. –  joriki Mar 4 '13 at 7:48
    
I used this series in my calculus for my thesis. Now i have to find this value or find an upper bound and a lower bound for it. This value or upper bound and lower bound should be very simple, because my calculus is very complicated. If this series doesn't have any closed form, are there any upper bound and lower bound for it?? –  hasti Mar 4 '13 at 8:15

1 Answer 1

There is no 'closed-form expression' for the series $$f(a):=\sum_{m=1}^{\infty}{\frac{e^{-a\,m^2}}{m^2}}$$ since this would imply that the derivative, essentially a Jacobi theta function, admits a closed-form too!

Concerning lower and upper bounds you may consider for example : $$e^{-a}\ <\ f(a)\ \le\ \frac{\pi^2}6+\frac a2-\sqrt{\pi\,a}\quad\quad \text{for}\ a\ge 0$$ The lower bound is very precise for $a\gg 1$ (for more precision use $\,e^{-a}+\frac 14 e^{-4a}\left[+\frac 19e^{-9a}+\cdots\right]$)

The upper bound is very precise for $a\ll 1$ (the error is of order $\,\left(\frac a{\pi}\right)^{\frac 32}e^{-\frac{\pi^2}a}$ as we will see) and may be obtained from the previous lower bound using the functional equation from the derivative $f'(a)=-\psi\bigl(\frac a{\pi}\bigr)$ where $\psi$ is the Jacobi theta function defined by (with Riemann's notation) : $$\psi(x)=\sum_{n=1}^\infty e^{-\pi\, x\, n^2}$$ and the functional equation (previous reference or tag $(43)$ from the link) is : $$1+2\,\psi(x)=\frac 1{\sqrt{x}}\left(1+2\,\psi\left(\frac 1x\right)\right)$$ Since the lower bound is straightforward let's obtain the upper bound : \begin{align} f(a)&=C-\int \psi\left(\frac a{\pi}\right) da=C+\frac 12\int 1-\sqrt{\frac{\pi}a}\left(1+2\,\psi\left(\frac {\pi}a\right)\right) da\\ &=C+\frac 12\int 1-\sqrt{\frac{\pi}a}\left(1+2 \sum_{m=1}^\infty\frac{e^{-\frac {\pi^2\,m^2}a}}{m^2}\right) da\\ &=\frac{\pi^2}6+\frac a2-\sqrt{\pi\,a}+2\sum_{m=1}^\infty \left(\frac{\pi^2}m\operatorname{erfc}\left(\frac{\pi\,m}{\sqrt{a}}\right)-\sqrt{\pi\,a}\frac{e^{-\frac {\pi^2\,m^2}a}}{m^2}\right)\\ &\sim \frac{\pi^2}6+\frac a2-\sqrt{\pi\,a}+2\sqrt{\pi\,a}\sum_{m=1}^\infty \frac{e^{-\frac {\pi^2\,m^2}a}}{m^2}\sum_{n>0}(-1)^n\,(2n-1)!!\left(\frac a{2\pi^2}\right)^n\\ \end{align}

using the asymptotic expansion of the complementary error function '$\operatorname{erfc}$' (where the $n=0$ term disappears and only a few $n>0$ terms may be used).

The double sum at the right is negative so that the upper bound is indeed : $$\frac{\pi^2}6+\frac a2-\sqrt{\pi\,a}$$

The most important contribution from the double sum (for $m=n=1$) would subtract $\,\displaystyle\left(\frac a{\pi}\right)^{3/2}e^{-\frac {\pi^2}a}$ to this expression.

Hoping all this this helped more,

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