# Explicit form of this series expansion?

I am considering the following series expansion: $$f(k):=\sum_{n\geq 1} e^{-k n^2}$$ with $k>0$ a fixed parameter. Is there a possibility to either find a closed form expression for $f(k)$? Or at least an upperbound of the type $|f(k)|\leq \frac{C}{k^p}$ for some constant $C$ and power $p$?

Thanks in advance!

• The upper bound is pretty easy: $f(k) \leq \int_0^\infty e^{-kx^2} \, dx = \frac{1}{2} \sqrt{\pi/k}$ since $f(k)$ is concave up. I doubt there's a simple closed form expression in elementary functions, but there could be using special functions. – Jair Taylor May 21 '15 at 19:45
• Aha of course, I overlooked that possibility. I am actually interested in a better bound than $1/\sqrt{k}$. Something like $1/k^{1/2+\varepsilon}$ with $\varepsilon \in (0,1/2)$. But I don't know if it trivially seen. – Martingalo May 21 '15 at 19:55
• Do you want the estimate for all $k>0$ or for, say, $k\ge 1$? – zhw. May 21 '15 at 19:58
• if it works for $k\geq 1$ fine :) Actually yes, because I am interested in the decay, for large $k$, so actually I need $|f(k)|\leq C/k^{1/2+\varepsilon}$ for large enough $k$. – Martingalo May 21 '15 at 19:59
• Well, the first term will tend to dominate: $f(k) \approx e^{-k}$ as $k$ gets large. For $k = 6$, $f(k) = e^{-k}$ up to several digits. – Jair Taylor May 21 '15 at 20:11

## 3 Answers

A slight refinement: $$0 < f(k) - e^{-k} = \sum_{k=2}^{\infty} e^{-kn^2} < \sum_{k=2}^{\infty} e^{-kn} = \frac{e^{-2k}}{1-e^{-k}},$$ which is bounded by, say, $2e^{-2k}$ for $1-e^{-k}>1/2$, i.e. $k>\log{2}$. Hence, in the Poincaré-type asymptotics, $$f(k) = e^{-k} + O(e^{-2k}).$$

Aside: The $(\pi/k)^{1/2}$ bound is useful when $k$ is positive and close to zero: it can be understood in terms of the Poisson summation formula: $$\sum_{k = -\infty}^{\infty} e^{-\pi k n^2} = \frac{1}{\sqrt{k}} \sum_{k=-\infty}^{\infty} e^{-\pi n^2/k}$$

There is no way to express in elementary functions. In fact, go to http://mathworld.wolfram.com/JacobiThetaFunctions.html and you will find $$f(k)=\frac12(-1+\vartheta_3(0,e^{-k})).$$

• Aha! Interesting. Is there literature about the decay of such functions when $k$ is large? – Martingalo May 21 '15 at 20:04

For $k\ge 1,$ $$\sum_{n=1}^{\infty}e^{-kn^2} \le \sum_{n=1}^{\infty}e^{-kn} = \frac{e^{-k}}{1-e^{-k}}\le \frac{1}{1-e^{-1}}\cdot e^{-k}.$$