# evaluation of $\sum_{n=-\infty}^{\infty}e^{-2\pi^{2}z^{2}n^{2}}$ for large z

I want to evaluate the following

$$$$\sum_{n=-\infty}^{\infty}e^{-2\pi^{2}z^{2}n^{2}}$$$$ I know for $$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?

• If $z$ is large, the terms with $n\neq 0$ are very small. – metamorphy Apr 24 '20 at 11:00
• @metamorphy yes but how does this help? – Jason Apr 24 '20 at 11:01
• 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.) – metamorphy Apr 24 '20 at 11:05
• As @metamorphy points out, the natural approximation is just the first terms of the series: $S \approx 1 +2[\exp{(-2 \pi^2 z^2) } + \exp{(-2 \pi^2 z^2 4)} + \cdots]$ – leonbloy Apr 24 '20 at 11:14
• @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 – Jason Apr 24 '20 at 11:15

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 $$\,0 This implies that truncating the infinite sum will result in increasing lower bounds such as $$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}$$