Explicit form of this series expansion?

Question

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!

Answer 1

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, $$ f(k) = e^{-k} + O(e^{-2k}). $$

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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: $$ \sum_{k = -\infty}^{\infty} e^{-\pi k n^2} = \frac{1}{\sqrt{k}} \sum_{k=-\infty}^{\infty} e^{-\pi n^2/k} $$

Answer 2

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

Answer 3

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