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?
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, $$ f(q) \!<\! \cdots \!<\! 1 \!+\! 2q \!+\! 2q^4 \!+\! \frac{2q^9}{1\!-\!q^5} \!<\! 1 \!+\! 2q \!+\! \frac{2q^4}{1\!-\!q^3} \!<\! 1 \!+\! \frac{2q}{1\!-\!q}. \tag{4} $$