If $f(t)=\sum_{n=1}^{\infty} e^{-n^2t}t^a$ , then $ \lim_{t \to 0}f(t) = 0$ Let $$ f(t)=\sum_{n=1}^{\infty} e^{-n^2t}t^a , \hspace{5mm} t >0$$ for some $a>0$
Then show that $$ \lim_{t \to 0}f(t) = 0$$

[My attempt]
I think  $$ \lim_{t \to 0}f(t)=\lim_{t \to 0}\sum_{n=1}^{\infty} e^{-n^2t}t^a=\sum_{n=1}^{\infty}\lim_{t \to 0} e^{-n^2t}t^a=\sum_{n=1}^{\infty}0=0$$
However, for changing limit and series , we need to uniform convergence of $f$
I tried to use weierstrass M - test but it didn't work.
I'm stuck here, how that solve this?
 A: The limit is not always zero. It depends on $a$.
Let $\theta(t) := \sum_{n \in \mathbb{Z}}e^{-\pi  n^2 t}$ for $t >0$. It is well-known (and not hard to prove via Poisson summation) that
$$
\theta(t) = t^{-1/2}\theta(1/t)
$$
for all $t >0$. We can write
$$
f(t) = t^{a}\frac{\theta(t/\pi)-1}{2}  = t^{a}\frac{\pi^{1/2}t^{-1/2}\theta(\pi/t)-1}{2}.
$$
We have $\lim_{t \rightarrow 0}{\theta(\pi/t)} = 1$ and so
$$
f(t) \sim t^{a}\frac{\pi^{1/2}t^{-1/2}-1}{2} \sim  \frac{\pi^{1/2}}{2 }t^{a-1/2},
$$
where $g(t)\sim h(t)$ means that $g(t)/h(t) \rightarrow 1$ as $t \rightarrow 0$. In summary, for $a \in \mathbb{R}$,
$$
\lim_{t \rightarrow 0}{f(t)} = \begin{cases}
\infty & a < 1/2\\
\frac{\pi^{1/2}}{2 } &a =1/2\\
0 & a>1/2
\end{cases}
$$
The last case can also easily be shown via a geometric series estimate.
A: The question relates to the Gauss circle problem. Indeed by considering
$$ \Theta(x)=\sum_{n\in\mathbb{Z}}e^{-n^2 x}$$
we have
$$ \Theta(-\log z)^2 = \sum_{N\geq 0} r_2(N) z^N $$
where $ r_2(N)=\left|\left\{(a,b)\in\mathbb{Z}^2:a^2+b^2=N\right\}\right| $. By Gauss geometric estimations we have
$$ \sum_{N=0}^{M}r_2(N) = \pi M + O(\sqrt{M}) $$
hence
$$ \frac{\Theta^2(-\log z)}{1-z}=\frac{\pi z}{(1-z)^2}+O\left(\sum_{n\geq 0}\sqrt{n}z^n\right) $$
and
$$ \Theta(-\log z)\sim\sqrt{\frac{\pi z}{1-z}} $$
as $z\to 1^-$. This implies
$$ \sum_{n\geq 0}e^{-n^2 x}\sim -\frac{1}{2}+\frac{1}{2}\sqrt{\frac{\pi e^{-x}}{1-e^{-x}}}\sim -\frac{1}{2}+\frac{\sqrt{\pi}}{2\sqrt{x}} $$
as $x\to 0^+$.
