Discuss the convergence of the sequence whose $n$-th term is given by $a_n= \frac{n!}{n^n}$ Discuss the convergence of the sequence whose $n$-th term is given by

$$a_n= \frac{n!}{n^n}$$

I'm just a little confuse on how this can be solved because the factorial is confusing me.
Put

$$a_n:=\frac{n!}{n^n}\implies\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\frac1{\left(1+\frac1n\right)^n}\xrightarrow[n\to\infty]{}\frac1e&lt;1$$

thus the infinite positive series $\;\sum a_n\;$  converges, so $\;a_n\xrightarrow[n\to\infty]{}\;?$
$$a_n  = \frac{{n!}}{{n^n }} \cong \frac{{\sqrt {2\pi n} }}{{e^n }} \Rightarrow \mathop {\lim }\limits_{n \to  + \infty } a_n  \cong \mathop {\lim }\limits_{n \to  + \infty } \frac{{\sqrt {2\pi n} }}{{e^n }} = 0
$$