Bounds on Gaussian infinite sum What are some good upper and lower bounds on the following sum?
$$S=\sum_{n=-\infty}^{+\infty}\dfrac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{1}{2}\left(\frac{n}{\sigma}\right)^2}$$
I am looking for something better than $1<S<2$.
 A: If fact, there is an explicit solution to the expression
$$S=\sum_{n=-\infty}^{+\infty}\dfrac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{1}{2}\left(\frac{n}{\sigma}\right)^2}=\frac{1}{\sigma\sqrt{2 \pi }  }\,\vartheta _3\left(0,e^{-\frac{1}{2 \sigma ^2}}\right)$$ where appears the elliptic theta function.
The function decreases asymptotically to $1$ but it goes to infinity for small values of $\sigma$.
I give you below some values (if you need more, just ask me)
$$\left(
\begin{array}{cc}
\sigma & S \\
 0.10 & 3.98942 \\
 0.15 & 2.65962 \\
 0.20 & 1.99473 \\
 0.25 & 1.59684 \\
 0.30 & 1.34009 \\
 0.35 & 1.17832 \\
 0.40 & 1.08500 \\
 0.45 & 1.03673 \\
 0.50 & 1.01438
\end{array}
\right)$$
Edit
Using $x=e^{-\frac{1}{2 \sigma ^2}}$, we can expand (around $x=0$) $$\vartheta _3\left(0,x\right)=1+2\sum_{i=1}^\infty x^{n^2}$$ which will converge very fast since $x<1$. For any value of $\sigma$, the function approximation will be extremely accurate with very few terms.
A: Let $f(x)=e^{-x^2/2\sigma^2}$. This is an even function, so that all its derivatives of odd order are odd; they vanish at zero and infinity.
Using the Euler-McLaurin summation formula,
$$\sum_{k=1}^\infty f(k)=\int_0^\infty f(x)\,dx+B_1(f(\infty)-f(0))+\sum_{k=1}^\infty\frac{B_{2k}}{(2k)!}\left(f^{(2k-1)}(\infty)-f^{(2k-1)}(0)\right)\\
=\sqrt{\frac\pi2}\sigma-\frac12.$$
I can't believe my eyes, what's wrong ?
A: A lower bound is $S(\sigma) \gt \max\left\{1, \dfrac{1}{\sigma\sqrt{2\pi}}  \right\}$, based on the integral and the $n=0$ term. For large and small $\sigma$, the result is close to this lower bound.  
An upper bound is the lower bound multiplied by $S\left(\frac{1}{\sqrt{2 \pi}}\right) \approx S(0.3989423) \approx 1.086435$ with equality when $\sigma=\frac{1}{\sqrt{2 \pi}}$.
