Investigating limit of theta like series

Question

Let $v>0$. I want to prove $$\lim_{t \to \infty} \sum_{a \in \mathbb Z} \sum_{b=1}^\infty \exp \left(-v(a/t+bt)^2\right) = 0.$$ It looks quite similar to the theta function $$\theta(z) = \sum_{n \in \mathbb Z }e^{\pi i n^2 z}$$ which is defined on the upper half plane and obeys the transformation law $$\theta\left(- \frac{1}{z}\right) = \sqrt{ \frac{z}{i} } \theta(z).$$ Maybe my original series can somehow be decomposed into a product of two such theta series and then we can use $$\lim_{t \to \infty} \theta(it)=1$$ and $|\theta(it)-1| < C e^{-\pi t}$.

Answer 1

Note that for any $c > 0$ and $a_0 \in \mathbb{R}$, if $\{a_0\} = a_0 - \lfloor a_0 \rfloor$ denotes the fractional part of $a_0$, then

\begin{align*} \sum_{a\in\mathbb{Z}} e^{-c (a - a_0)^2} &= \sum_{a\in\mathbb{Z}} e^{-c (a - \{a_0\})^2} = \left(\sum_{a\in\mathbb{Z}} e^{-c (a - \{a_0\})^2} \right)^{\frac{1}{2}}\left(\sum_{a\in\mathbb{Z}} e^{-c (a + \{a_0\})^2} \right)^{\frac{1}{2}} \\ &\stackrel{\text{(C–S)}}{\geq} \sum_{a\in\mathbb{Z}} e^{-c(a^2 + \{a_0\}^2)} \geq \sum_{a\in\mathbb{Z}} e^{-c(a^2+1)}. \end{align*}

Note that this lower bound does not depend on $a_0$. Therefore,

\begin{align*} \sum_{a\in\mathbb{Z}}\sum_{b=1}^{\infty} e^{-v(a/t + bt)^2} = \sum_{b=1}^{\infty} \sum_{a\in\mathbb{Z}}e^{-(v/t^2)(a + bt^2)^2} \geq \sum_{b=1}^{\infty} e^{-(v/t^2)(a^2+1)} = \infty. \end{align*}

and the limit as $t\to\infty$ also diverges to $\infty$.