Test $\sum_{n=0}^{\infty}\frac{n!}{2^{n^2}}$ for convergence. Applying the ratio test to $\sum_{n=0}^{\infty}\frac{n!}{2^{n^2}}$ with $x_n=\frac{n!}{2^{n^2}}$ we get
\begin{align*}
\bigg|\frac{x_{n+1}}{x_n}\bigg|&=\frac{\frac{(n+1)!}{2^{(n+1)^2}}}{\frac{n!}{2^{n^2}}}\\&=\frac{(n+1)!}{2^{(n+1)^2}}\frac{2^{n^2}}{n!}\\&=\frac{(n+1)n!\cdot2^{n^2}}{n!\cdot2^{n^2}\cdot2^{2n+1}}\\&=\frac{n+1}{2^{2n+1}}
\end{align*}
Quick question now: Is it sufficient here to state that $2^{2n+1}$ obviously grows faster than $n+1$ (and therefore our term converges towards $0$ and our series passed the test) or do I need to generally prove this first (e.g. by induction)?

$$
0\le\frac{n!}{2^{n^2}} \le \frac{n^n}{2^{n^2}}
$$
by the root test for $n>1$:
$$
\sqrt[n]{\frac{n^n}{2^{n^2}}}=\frac{n}{2^n}\le \frac{1}{2}
$$
hence the series 
$$\sum_{n=0}^{\infty}\frac{n^2}{2^{n^2}}$$
converges and by comparison
$$\sum_{n=0}^{\infty}\frac{n!}{2^{n^2}}$$
converges
