Evaluate $\lim_{t\to\infty}\sum_{n=1}^{\infty}e^{-\pi tn^2}$ My idea is to prove the series converges uniformly, then change the order of limit and summation so that 
$$\lim_{t\to\infty}\sum_{n=1}^{\infty}e^{-\pi tn^2} = \sum_{n=1}^{\infty}\lim_{t\to\infty}e^{-\pi tn^2} = 0$$
But I don't know how to prove the series converges uniformly. Is there any way to solve or is there a better approach?
 A: Proving uniformly convergence is a good approach and is not hard to show: Note that for any $t \geq 1$ we always have $e^{-\pi tn^2} \leq e^{-\pi n^2}$ and thus
$$ \sum_{k=m}^n e^{-\pi t n^2} \leq  \sum_{k=m}^n e^{-\pi n^2}.$$
Choosing $n \geq N$ such that the last sum is $< \varepsilon$ for all $n >m \geq N$ gives uniformly convergence. (Note that the choice of $N$ is independent of $t \geq 1$.)
A: Hint:
$$0\leq\int_0^{\infty}e^{-y^2}dy<\infty$$
Apply substitution $y=x\sqrt{\pi t}$ and compare integral and sum.
A: Alternative approach: it is pretty clear that
$$ f(t) = \sum_{n\geq 1} e^{-\pi n^2 t} $$
is a continuous, positive and decreasing function on $\mathbb{R}^+$. Since $\int_{0}^{+\infty}f(t)\,dt =\frac{\pi}{6} $, 
$$\lim_{t\to +\infty}f(t)=0.$$

Addendum: in terms of Jacobi theta functions we have $f(t)=\frac{1}{2}\left(-1+\vartheta_3(e^{-\pi t})\right)$ and
$$g(z)=\sum_{n\geq 1} z^{n^2} $$
is clearly holomorphic in $\|z\|<1$, with $\lim_{t\to +\infty}f(t) = \lim_{z\to 0^+}g(z) = g(0)=0.$
We also have $f(t)\in L^p(\mathbb{R}^+)$ for any $1\leq p < 2$, and due to the Poisson summation formula
$$ 2 f(t)+1 = \frac{1}{\sqrt{t}}\left( 2 f\left(\tfrac{1}{t}\right) + 1 \right) $$
holds for any $t>0$, so $f(t)$ behaves like $\frac{1}{2\sqrt{t}}-\frac{1}{2}$ in a right neighbourhood of $t=0$.
A: If $n\geq1$ then $n^2-1=(n+1)(n-1)\geq 2(n-1)$. It follows that
$$0<\sum_{n=1}^\infty e^{-\pi t n^2}=e^{-\pi t}\sum_{n=1}^\infty e^{-\pi t (n^2-1)}=e^{-\pi t}\sum_{n'=0}^\infty e^{-2n'\pi t}={e^{-\pi t}\over 1-e^{-2\pi t}}\to0\qquad(t\to\infty)\ .$$
