TITLE: Why this approximation for $\pi$ is so accurate?
QUESTION [21 upvotes]: Berggren and Borwein brothers in "Pi: A Source Book" showed a mysterious approximation for $\pi$ with astonishingly high accuracy:
$$ \left(\frac{1}{10^5}\sum_{n={-\infty}}^\infty e^{-n^2/10^{10}}\right)^2 \approx \pi.$$
In particular, this formula gives $42$ billion (!) correct digits of $\pi$.
I have two questions:
(1) Why this approximation is so much accurate?
(2) Is it possible to prove without computation that this formula is not an identity?

REPLY [17 votes]: It comes from a theta function identity. In particular, if we let
$$\theta(\tau)=\sum_{n=-\infty}^\infty e^{i\pi\tau n^2}$$
for $\tau$ in the upper half plane of complex numbers, then
$$\theta(-1/\tau)=(-i\tau)^{1/2}\theta(\tau)$$
Letting $\tau=i\pi N^2$ with $N=10^5$, we see that
$$\sum_{n=-\infty}^\infty e^{-n^2/N^2}=\theta(-1/(i\pi N^2))=(\pi N^2)^{1/2}\theta(i\pi N^2)=\sqrt\pi N\sum_{n=-\infty}^\infty e^{-\pi^2N^2n^2}\approx\sqrt\pi N$$
since all terms in the final sum other than $n=0$ contribute extremely small amounts. To be somewhat more precise, we have
$$\begin{align}
{1\over N}\sum_{n=-\infty}^\infty e^{-n^2/N^2}
&=\sqrt\pi\left(1+2\sum_{n=1}^\infty e^{-\pi^2N^2n^2}\right)\\
&=\sqrt\pi+2\sqrt\pi\left(e^{-\pi^2N^2}+e^{-4\pi^2N^2}+e^{-9\pi^2N^2}+\cdots \right)\\
&\lt\sqrt\pi+2\sqrt\pi\left(e^{-\pi^2N^2}+e^{-2\pi^2N^2}+e^{-3\pi^2N^2}+\cdots \right)\\
&=\sqrt\pi+{2\sqrt\pi e^{-\pi^2N^2}\over1-e^{-\pi^2N^2}}
\end{align}$$
and for $N=10^5$, we have $e^{-\pi^2N^2}\approx10^{-42{,}863{,}147{,}299.6}$, from which one can see how we get (more than) $42$ billion digits of $\pi$, answering question (1). We also see that the left hand side of the OP's approximation is slightly smaller than $\pi$, so the formula is not an identity, answering question (2).