How do I prove that $\left(\frac{1}{10^5} \sum_{n=-\infty}^{\infty} e^{\frac{-n^2}{10^{10}}}\right)^2 \approx \pi? $

Question

How do I prove that

$$\left(\frac{1}{10^5} \sum_{n=-\infty}^{\infty} e^{\frac{-n^2}{10^{10}}}\right)^2 \approx \pi?$$

I am thinking of start with gaussian function and using the poison summation formula

Does that will work?

Answer 1

We do have

\begin{align*} \int_{-\infty}^{\infty} e^{-y^2} dy &= \sqrt{\pi} \end{align*}

Now, substitute $y = \alpha x$, $dy = \alpha dx$:

\begin{align*} \alpha \int_{-\infty}^{\infty} e^{-\alpha^2 x^2} dx &= \sqrt{\pi} \\ \left( \alpha \int_{-\infty}^{\infty} e^{-\alpha^2 x^2} dx \right)^2 &= \pi \end{align*}

This suggests that we may estimate the given sums by interpreting them as a Riemann sum belonging to the integral we've just computed. However, we have to be careful.

$\sum_{n=1}^{\infty} e^{-\alpha^2 n^2} < \int_{0}^{\infty} e^{-\alpha^2 x^2} dx$

$\sum_{n=-\infty}^{\infty} e^{-\alpha^2 n^2} = 1 + 2 \sum_{n=1}^{\infty} e^{-\alpha^2 n^2} < 1 + \int_{-\infty}^{\infty} e^{-\alpha^2 x^2} \, dx$

And for the other bound use

$\int_0^{\infty} e^{-\alpha^2 x^2} \, dx < \sum_{n=0}^{\infty} e^{-\alpha^2 n^2}$

To find

$\sum_{n=-\infty}^{\infty} e^{-\alpha^2 n^2} = 2 \sum_{n=0}^{\infty} e^{-\alpha^2 n^2} - 1 > \int_{-\infty}^{\infty} e^{-\alpha^2 x^2} dx$

Collect both halves and multiply with ${\alpha}$

$\alpha \left( \int_{-\infty}^{\infty} e^{-\alpha^2 x^2} dx - 1 \right) < \alpha \sum_{n=-\infty}^{\infty} e^{-\alpha^2 n^2}$

$< \alpha \left( \int_{-\infty}^{\infty} e^{-\alpha^2 x^2} dx + 1 \right)$

$\sqrt{\pi} - \alpha < \alpha \sum_{n=-\infty}^{\infty} e^{-\alpha^2 n^2} < \sqrt{\pi} + \alpha$

If $\alpha \to 0$, both sides tend to $\sqrt{\pi}^2$, and the result follows. In fact, we may use this to find

\begin{align*} &\left| \alpha \sum_{n=-\infty}^{\infty} e^{-\alpha^2 n^2} - \sqrt{\pi} \right| < \alpha \\ &\left| \left( \alpha \int_{-\infty}^{\infty} e^{-\alpha^2 n^2} dx \right)^2 - \pi \right| < \alpha(\alpha + 2 \sqrt{\pi}) \approx 2 \alpha \sqrt{\pi}. \end{align*}

Where the approximation is valid for small $α$ similar to the given value $α=1/10^5$

Answer 2

Without integration

Making the problem more general $$A=\sqrt a \sum_{n=-\infty}^{+\infty} e^{-a n^2}=\sqrt a\,\, \vartheta _3\left(0,e^{-a}\right)$$ where appears Jacobi theta function.

Expanding for small values of $a$ $$A=\sqrt \pi \left(1+2 e^{-\frac{\pi ^2}{a}}+\cdots \right)$$ $$A^2=\pi\left(1+4 e^{-\frac{\pi ^2}{a}}+\cdots \right)$$ $$A^2-\pi \sim 4 \pi\,e^{-\frac{\pi ^2}{a}}$$ If you use $a=\frac 1{10} $, the rhs is $1.72\times 10^{-42}$ and for $a=\frac 1{100}$, it is $2.94\times 10^{-428}$.