Radius of Convergence of $\sum_{n=0}^{\infty} e^{-z^{2}\sqrt{n}}$ 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 $|x|&gt;|y|$ which implies that the series does not converge when $|x|\leq |y|$.

Moreover if $|x|&gt;|y|$ 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 $|x|&gt;|y|$.
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 $r&lt;0$, note that for large enough $n$ we have $r\sqrt{n} &lt; -2\log n$, so for large enough $n$ we have $e^{r\sqrt{n}}&lt; n^{-2}$. Thus, by the comparison test the series converges.