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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?

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2 Answers 2

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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.

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    $\begingroup$ Are you sure about the constant $\pi^{1/4}$? I am simply getting $\pi^{1/2}$ from the Gauss circle problem. $\endgroup$ Dec 23, 2020 at 10:26
  • $\begingroup$ Thanks for pointing this out. I corrected the (rather silly) mistake now. $\endgroup$
    – m.s
    Dec 23, 2020 at 10:46
  • $\begingroup$ Thank you, how to know that $\lim_{t \rightarrow 0}{\theta(\pi/t)} = 1$ ? $\endgroup$
    – hew
    Dec 23, 2020 at 10:50
  • $\begingroup$ 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 $s$. $\endgroup$
    – m.s
    Dec 23, 2020 at 10:54
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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 $x\to 0^+$.

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    $\begingroup$ 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. $\endgroup$ Dec 23, 2020 at 13:36
  • $\begingroup$ The remainder series may still have a singularity at $z=1$, but not worse than $\frac{1}{(1-z)^{3/2}}$. $\endgroup$ Dec 23, 2020 at 13:39

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