# 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?

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.

• Are you sure about the constant $\pi^{1/4}$? I am simply getting $\pi^{1/2}$ from the Gauss circle problem. Dec 23, 2020 at 10:26
• Thanks for pointing this out. I corrected the (rather silly) mistake now.
– m.s
Dec 23, 2020 at 10:46
• Thank you, how to know that $\lim_{t \rightarrow 0}{\theta(\pi/t)} = 1$ ?
– hew
Dec 23, 2020 at 10:50
• This is equivalent to $\lim_{s \rightarrow \infty}{\sum_{n=1}^{\infty}{e^{- n^2 s}}} = 0$, which can be proved in many ways, for example by factoring out $e^{-s/2}$ from the sum and bounding the rest uniformly in $s$.
– m.s
Dec 23, 2020 at 10:54

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 $$x\to 0^+$$.

• Jack D'Aurizio, why we may be sure that $\sum_n O(\sqrt{n})z^n$ is "good"? I think it's better to mention that in our case we may write $\theta(n) \le 100 \sqrt{n}$ instead of $O(\sqrt{n})$ - I mean that all $O(\sqrt{n})$ have the same constant. Dec 23, 2020 at 13:36
• The remainder series may still have a singularity at $z=1$, but not worse than $\frac{1}{(1-z)^{3/2}}$. Dec 23, 2020 at 13:39