Find all $z \in \mathbb{C}$ such that the series
$$ F(z)=\sum_{n=0}^{\infty} e^{-z^{2}\sqrt{n}}$$
converges.
I tried the ratio test and root test, but both were inconclusive. I considered the Cauchy-Hadamard formula, but could not see how to apply it to this problem.
Any guidance would be much appreciated.
Let $z=x+iy$ then $z^2=(x^2-y^2)+i(2xy)$ and $$e^{-z^{2}\sqrt{n}}=e^{(y^2-x^2)\sqrt{n}}\cdot e^{-2xyi\sqrt{n}}.$$ Note that $|e^{-z^{2}\sqrt{n}}|=e^{(y^2-x^2)\sqrt{n}}\to 0$ iff $|x|>|y|$ which implies that the series does not converge when $|x|\leq |y|$.
Moreover if $|x|>|y|$ then $$n^2|e^{-z^{2}\sqrt{n}}|=n^2e^{(y^2-x^2)\sqrt{n}}=e^{(y^2-x^2)\sqrt{n}+2\ln n}\to 0$$ because $(y^2-x^2)\sqrt{n}+2\ln n\to -\infty$ as $n\to+\infty$. Hence eventually $|e^{-z^{2}\sqrt{n}}|\leq \frac{1}{n^2}$. Since the series $\sum\frac{1}{n^2}$ is convergent, we conclude that that series converges iff $|x|>|y|$.
I will look at the convergence of $\sum\limits_{n=0}^{\infty}{e^{r\sqrt{n}}}$ for $r$ real. Can you see how it's related to your problem?
For $r\ge 0$, we have $e^{r\sqrt{n}}\ge 1$ for all $n$, and hence the series diverges.
For $r<0$, note that for large enough $n$ we have $r\sqrt{n} < -2\log n$, so for large enough $n$ we have $e^{r\sqrt{n}}< n^{-2}$. Thus, by the comparison test the series converges.
