Why is $\sum\limits_{n=1}^{\infty}e^{-(n/10)^2}$ almost equal to $5\sqrt\pi-\frac12$ (agreeing up to $427$ digits)?

Question

The following I saw as an exercise in a text on modular forms. I am lacking the understanding for why the following is true, but it is nevertheless astonishing:

Why is the numerical value of

$$x:=\sum_{n=1}^{\infty} \exp(-(n/10)^2)$$

so ridiculously close to the value of

$$y:=5\sqrt\pi-\frac12$$

while (and this is the truly surprising aspect) not being equal to it?

And by "ridiculously close" I mean that $x$ and $y$ agree up to the 427-th digit after the decimal point:

\begin{align} x=8. &3622692545275801364908374167057259139877472806\\ &1193564106903894926455642295516090687475328369\\ &2723327081134118121412853331180764328622113012\\ &6254685480139353423101884932655256142496258651\\ &4475413114466047689633981400087319507675739860\\ &2583500950926170092927234872474563201569608877\\ &6295310820270966625045319920380686673873757671\\ &6833994894682925918204397725582580869380029533\\ &6967158956664049274231240924510273274260978066\\ &257808237337\color{red}{62} \end{align}

They first disagree in the red digits.

Question: Can someone explain this in "simple" words? It does not have to be elementary, but I want to understand what machinery has to interact to arrive at this. And how would one even come up with such an example, or similar ones?

Answer 1

Define $$\psi(q)\equiv-\frac12+\frac12\vartheta_3(0,e^{-\pi q})=\sum_{n\ge1}e^{-\pi n^2q}$$ where $\vartheta_3(z,q)$ is Jacobi's third theta function. An application of the Poisson summation formula leads to the identity $$\vartheta_3(0,e^{-\pi/q})=\sqrt q\,\vartheta_3\left(0,e^{-\pi q}\right)$$ whose proof is outlined in Jacobi (1828). Substituting $q:=(\pi k^2)^{-1}$ yields $$\sum_{n\ge1}e^{-n^2/k^2}=-\frac12+\frac12\vartheta_3(0,e^{-1/k^2})=-\frac12+\frac12\cdot\sqrt{\pi k^2}\vartheta_3(0,e^{-\pi^2k^2})$$ so that $$\sum_{n\ge1}e^{-n^2/k^2}=-\frac12+\frac k2\sqrt\pi(1+2\psi(\pi k^2))=-\frac12+\frac k2\sqrt\pi+k\sqrt\pi\sum_{n\ge1}e^{-(\pi kn)^2}.$$ When $k=10$ the error has an order of magnitude of $$\log_{10}\left(10\sqrt\pi e^{-100\pi^2}\right)=1+\log_{10}\sqrt\pi-100\pi^2\log_{10}e=\bf-427$$ which corresponds to your observation.