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I want to evaluate the following

\begin{equation} \sum_{n=-\infty}^{\infty}e^{-2\pi^{2}z^{2}n^{2}} \end{equation} I know for $z\ll 1$ we can use Euler-Maclaurin formula but in my case z is quite large ($z\gg 1$). can anyone give me a hint of how can I evaluate this or at least be able to approximate it? or even an almost tight upper bound?

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  • $\begingroup$ If $z$ is large, the terms with $n\neq 0$ are very small. $\endgroup$ – metamorphy Apr 24 '20 at 11:00
  • $\begingroup$ @metamorphy yes but how does this help? $\endgroup$ – Jason Apr 24 '20 at 11:01
  • $\begingroup$ I just don't understand what kind of approximation do you need. The series itself is a very good one ;) (BTW, how do you use E.M. for small $z$? It is done much better by modular transformations of theta functions.) $\endgroup$ – metamorphy Apr 24 '20 at 11:05
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    $\begingroup$ As @metamorphy points out, the natural approximation is just the first terms of the series: $S \approx 1 +2[\exp{(-2 \pi^2 z^2) } + \exp{(-2 \pi^2 z^2 4)} + \cdots]$ $\endgroup$ – leonbloy Apr 24 '20 at 11:14
  • $\begingroup$ @metamorphy in fact I used Mathematica to evaluate this sum for (z=2 as an example) and I found the answer to be 1. I just need some sort of analytical approximation to confirm that value of Mathematica $\endgroup$ – Jason Apr 24 '20 at 11:15
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Define the null theta funtion $$ f(q) := \sum_{n=-\infty}^\infty q^{n^2} = 1 + 2\sum_{n=1}^{\infty} q^{n^2} \tag{1} $$ where $\,|q|<1\,$ is needed for convergence. Assume that further $\,0<q<1.\,$ This implies that truncating the infinite sum will result in increasing lower bounds such as $$ 1 < 1 + 2q < 1 + 2q + 2q^4 < \cdots < f(q). \tag{2} $$ However, comparing it to a geometric series gives an upper bound $$ f(q) < 1 + 2\sum_{n=1}^\infty q^n = 1 + \frac{2q}{1-q} = \frac{1+q}{1-q} \tag{3} $$ although better upper bounds exist. For example, $$ f(q) < 1 + 2\sum_{n=0}^\infty q^{3n+1} = 1+2q+\frac{2q^4}{1-q^3} \tag{3} $$ and so on. Thus, $$ f(q) \!<\! \cdots \!<\! 1 \!+\! 2q \!+\! 2q^4 \!+\! \frac{2q^9}{1\!-\!q^5} \!<\! 1 \!+\! 2q \!+\! \frac{2q^4}{1\!-\!q^3} \!<\! 1 \!+\! \frac{2q}{1\!-\!q}. \tag{4} $$

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