How can we show that $\sqrt{\pi}=\lim_{k\to \infty}{1\over \sqrt{k}}\left(1+2\sum_{n=1}^{\infty}e^{-n^2\over k}\right)?$ 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.
 A: By the Euler Mac-Laurin formula,
$$
S(k)=\lim_{N\rightarrow \infty}\sum_{n=1}^{N} e^{-n^2/k}\sim\lim_{N\rightarrow \infty}\int_1^{N} e^{-n^2/k}dn
$$
it is now not not too difficult to show that the integral asymtotically (large $k$) equals
$$
\frac{\sqrt{\pi k}}{2}-1+\mathcal{O}(k^{-1}) \qquad (1)
$$
Therefore 

$$
\lim_{k\rightarrow\infty}\frac{1}{\sqrt{k}}(1+2S(k))=\lim_{k\rightarrow\infty}\frac{1}{\sqrt{k}}(1-1+\sqrt{\pi k})=\sqrt{\pi}
$$


For a proof of ($1$), write
$$
\lim_{N\rightarrow\infty}\int_1^N e^{-n^2/k}dk=\lim_{N\rightarrow\infty}\int_0^N e^{-n^2/k}dk-\int_0^1 e^{-n^2/k}dk\underbrace{=}_{Gaussian\,integral}\\
\frac{\sqrt{\pi k}}{2}-\sum_{l=0}^{\infty}\int_0^1dn\frac{n^{2l}}{l!k^l}\sim\frac{\sqrt{\pi k}}{2}-1+\mathcal{O}(k^{-1})
$$
A: Note that we have
$$\frac{\sqrt{\pi k}}{2}-e^{-1/k}\le \int_1^\infty e^{-x^2/k}\,dx\le \sum_{n=1}^\infty e^{-n^2/k}\le \int_0^\infty e^{-x^2/k}\,dx=\frac{\sqrt{\pi k}}{2}$$
Hence, we can write
$$\frac{1}{\sqrt k}+\sqrt \pi -\frac{2e^{-1/k}}{\sqrt k}\le \frac1{ \sqrt{k}}\left(1+2\sum_{n=1}^\infty e^{-n^2/k}\right)\le \frac{1}{\sqrt k}+\sqrt \pi$$
whereupon applying the squeeze theorem yields the expected limit
$$\lim_{k\to\infty}\frac1{ \sqrt{k}}\left(1+2\sum_{n=1}^\infty e^{-n^2/k}\right)=\sqrt \pi$$
A: 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
A: For every series of positive decreasing terms $a_n=f(n)$, you can always write
$$
\int_2^\infty f(x)dx\leq \sum_{n=1}^\infty a_n\leq \int_1^\infty f(x)dx
$$
Now, in your case you have $a_n=e^{-\frac{n^2}{k}}$, and you can compute a lower and upper bound to your series. Then, you can compute your limit using the upper bound and lower bound, as $k\to\infty$. If they coincide, thanks to the squeeze theorem, that is also the value of your limit.
A: It's the $\vartheta$-function. Use the functional equation and you get easily the wished result. 
$$\vartheta(iz)=1+2\sum\limits_{n=1}^\infty e^{-\pi z n^2}$$
and
$$\frac{1}{\sqrt{z}}\vartheta(\frac{i}{z})=\vartheta(iz)$$.
Set $z=\pi k$. Then it's done because $e^{-\pi^2 n^2 k}\to 0$ for $k\to \infty$.
