Finding the convergence interval of $\sum_{n=0}^\infty\frac{n!x^n}{n^n}$.

Since $\displaystyle e^x > \frac{x^n}{n!}$ for all $x>0, n \in \mathbb{N}$, letting $x=n$ gives $\displaystyle \frac{n!e^n}{n^n}>1$ so the terms don't even approach zero.

This may be overkill but we have $$n!\approx n^{n+1/2}e^{-n}$$ by Stiriling's formula when $n$ goes very large. Thus the original sum becomes approximately $$n^{1/2}\frac{x^{n}}{e^{n}}$$ So if $x\ge e$ this series obviously will not converge. When $0<x<e$ name $\frac{x}{e}=k<0$, then we have $$\sum n^{1/2}k^{n}$$ the ratio test give $$(\frac{n+1}{n})^{1/2}k$$ which becomes less than 1 when $n$ goes very large.

The case that $x$ is negative is similar by the alternating series test.