How can we prove that $\lim_{x\to0^+} \sum_{n=1}^\infty e^{-n^2x}=\infty$?

Question

Consider the function defined on $(0,\infty)$ by $$f(x)=\sum_{n=1}^\infty e^{-n^2x}.$$ How can we prove that $\lim_{x\to0^+} f(x)=\infty$?

Answer 1

For any $N,$ we have

$$f(x)\ge \sum_{n=1}^{N}e^{-n^2x}\implies \liminf_{x\to 0^+}f(x)\ge \liminf_{x\to 0^+}\,(\sum_{n=1}^{N}e^{-n^2x})$$ $$= \lim_{x\to 0^+}(\sum_{n=1}^{N}e^{-n^2x})=N.$$

This is true for all integers $N>0,$ hence the $\liminf$ of $f$ is $\infty.$ Therefore so is $\lim f.$

Answer 2

As suggested in the comments, for all $N\geqslant 1$, we have $$ f\left(\frac{1}{N^2}\right)=\sum_{n=1}^Ne^{-\frac{n^2}{N^2}}+\sum_{n=N+1}^{+\infty}e^{-\frac{n^2}{N^2}}\geqslant Ne^{-\frac{1}{N^2}} $$ Since $\lim\limits_{N\rightarrow +\infty}Ne^{-\frac{1}{N^2}}=+\infty$, then $\lim\limits_{N\rightarrow +\infty}f\left(\frac{1}{N^2}\right)=+\infty$. In the general case, for $x>0$, let $N_x=\left\lfloor\frac{1}{\sqrt{x}}\right\rfloor$, then $N_x\leqslant \frac{1}{\sqrt{x}}$ and thus $x\leqslant\frac{1}{N_x^2}$, since $f$ is decreasing, we have $f(x)\geqslant f\left(\frac{1}{N_x^2}\right)\underset{x\rightarrow 0^+}{\longrightarrow}+\infty$, because $\lim\limits_{x\rightarrow 0^+}N_x=+\infty$. Note that $\displaystyle\vartheta(x):=\sum_{n\in\mathbb{Z}}e^{-\pi n^2 x}$ is Jacobi's $\vartheta$ function which verifies the identity $\vartheta\left(\frac{1}{x}\right)=\sqrt{x}\vartheta(x)$ (this is a consequence of Poisson's summation formula). Since $\lim\limits_{x\rightarrow +\infty}\vartheta(x)=1$, we have $\vartheta(x)\underset{x\rightarrow 0^+}{\sim}\frac{1}{\sqrt{x}}$ and, noticing that $$f(x)=\frac{\vartheta\left(\frac{x}{\pi}\right)-1}{2}$$ We have $$ f(x)\underset{x\rightarrow 0^+}{\sim}\frac{1}{2}\sqrt{\frac{\pi}{x}} $$