Proof convergence of $\sum\limits_{n = 1}^\infty\frac {n!} {n^n}$ To prove the convergence of $\sum\limits_{n = 1}^\infty\frac {n!} {n^n}$ I used that $\lim\limits_{n\to\infty}|\frac {a_{n+1}} {a_n}|$ has to be $<1$:
$$\lim\limits_{n\to\infty}|\frac {a_{n+1}} {a_n}|$$
$$=\lim\limits_{n\to\infty}|\frac {\frac {(n+1)!} {(n+1)^{n+1}}} {\frac {n!} {n^n}}|$$
$$=\lim\limits_{n\to\infty}|\frac {(n+1)!\cdot n^n} {n!\cdot (n+1)^{n+1}}|$$
$$=\lim\limits_{n\to\infty}|\frac {(n+1)\cdot n^n} {(n+1)^{n+1}}|$$
$$=\lim\limits_{n\to\infty}|\frac {n^n} {(n+1)^n}|$$
$$=\lim\limits_{n\to\infty}|{(\frac {n} {n+1})}^n|=\frac 1 e$$
Okay... problem solved after thankful advice

We can also use the Comparison Test: $$0\le \frac{n!}{n^n}=\frac{1\cdot 2\cdot 3\cdot\ldots\cdot n}{n\cdot n\cdot n\cdot\ldots\cdot n}\le \frac{1\cdot 2\cdot n\cdot\ldots \cdot n}{n\cdot n\cdot n\cdot\ldots\cdot n}=\frac{2}{n^2},$$ and $\displaystyle \sum_{n=1}^{+\infty}\frac{2}{n^2}$is convergent, so the given series is convergent.
