Limit As n goes to infinity of $ \sum_{n=1}^\infty e^{- \alpha n^2 }$.

Question

I suspect the following is exactly true ( for positive $\alpha$ )

\begin{equation} \sum_{n=1}^\infty e^{- \alpha n^2 }= \frac{1}{2} \sqrt { \frac{ \pi}{ \alpha} } \end{equation}

If the above is exactly true, then I would like to know a proof of it. I accept showing a particular limit is true, may be far more difficult than just applying a general theorem to show that the limit exists. Also as the result involves $\pi$ this makes me think the proof could well be a long one, BUT … ?

To give some context, the above series crops up in calculating the 'One Particle Translational Partition Function' for the quantum mechanical 'Particle In A Box'.

Answer 1

No doubt that the equality is wrong. For large $\alpha$, the first term dominates and the asymptotic behavior is $e^{-\alpha}$.

No even sure that there exist a value of $\alpha$ such that the expressions are equal.

Answer 2

The infinite sum is $\, (\theta_3(0,e^{-\alpha})-1)/2 \,$ where $\, \theta_3 \,$ is a Jacobi theta function. Only for small values of $\, \alpha \,$ is it approximately $\, \sqrt{\pi /\alpha}/2. \,$ Define $\, f(\alpha) := \theta_3(0,e^{-\alpha}). \,$ Then using modular relations $\, f(\alpha) = \sqrt{\pi/\alpha}f(\pi^2/\alpha). \,$ Since $\, f(\alpha) \to 1 \, $ as $\, \alpha \to +\infty, \,$ this explains the close approximation when $\alpha$ is a small positive number.

Define $\, s(\alpha) := \sum_{n=1}^\infty e^{-\alpha n^2}. \,$ Then the result $\ s(\alpha) = -1/2 + \sqrt{\pi/\alpha}(1/2 + s(\pi^2/\alpha)). \,$ shows the exact relation between $\, s(\alpha) \,$ and $\, s(\pi^2/\alpha). \,$

Answer 3

Consider the Riemannian sum $$\lim_{m\to\infty}\frac 1m\sum_{n=0}^\infty e^{-(n/m)^2}=\int_0^\infty e^{-x^2}dx=\frac{\sqrt\pi}2.$$

Then with the subsitution $m^2\alpha=1$,

$$\lim_{\alpha\to0}\sqrt\alpha\sum_{n=0}^\infty e^{-\alpha n^2}=\frac{\sqrt\pi}2.$$ (Note that the starting index $n=0$ or $n=1$ makes no difference as a finite sum of terms will cancel out when multiplied by $\sqrt\alpha$.)

Answer 4

It is not an exact equality. By the Poisson summation formula, assuming $\alpha>0$,

$$ \sum_{n\in\mathbb{Z}}e^{-\alpha n^2} = \sqrt{\frac{\pi}{\alpha}}\sum_{n\in\mathbb{Z}}e^{-\pi^2 n^2 / \alpha} \tag{1}$$ hence by parity $$ \sum_{n\geq 1}e^{-\alpha n^2} = -\frac{1}{2}+\sqrt{\frac{\pi}{\alpha}}\left(\frac{1}{2}+\sum_{n\geq 1}e^{-\pi^2 n^2 / \alpha} \right).\tag{2} $$ This lemma is usually exploited in the proof of the reflection formula for the Riemann $\zeta$ function.