TITLE: How can we show that $\sqrt{\pi}=\lim_{k\to \infty}{1\over \sqrt{k}}\left(1+2\sum_{n=1}^{\infty}e^{-n^2\over k}\right)?$
QUESTION [5 upvotes]: Limit

$$\sqrt{\pi}=\lim_{k\to \infty}{1\over \sqrt{k}}\left(1+2\sum_{n=1}^{\infty}e^{-n^2\over k}\right)?\tag1$$
$$\sqrt{\pi}=\lim_{k\to \infty}{1\over \sqrt{k}}+\lim_{k\to\infty}{2\over \sqrt{k}}\sum_{n=1}^{\infty}e^{-n^2\over k}?\tag2$$
$$\sqrt{\pi}=\lim_{k\to\infty}{2\over \sqrt{k}}\sum_{n=1}^{\infty}e^{-n^2\over k}?\tag3$$

I checked via a sum calculator it seem to converges, but don't know how to verify it.

REPLY [4 votes]: Here is a solution using Riemann sum 
your limit $$\lim \limits_{k\to \infty }\frac{2}{\sqrt{k} } \sum \limits^{k}_{n=1}e^{\frac{-n^{2}}{k} }$$
Now let $k=u^2$ we get the limit$$\lim \limits_{u\to \infty }\frac{2}{u} \sum \limits^{u^{2}}_{n=1}e^{-\left( \frac{n}{u} \right) ^{2}}=2\int \limits^{\infty }_{0}e^{-y^{2}}dy=\sqrt{\pi } $$
The last integral is known as Gaussian intintegral