Answer
Converges to $0$
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{n!}{n^n}$
Since, $\lim\limits_{n \to \infty} \sqrt[n] {n}=1$ and $\lim\limits_{n \to \infty} x^{1/n}=1$ when $x \gt 0$
If the sequence $a_{n}$ is both bounded and monotonic then the sequence converges.
So, $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{n!}{n^n} \leq \lim\limits_{n \to \infty} (\dfrac{1}{n}) =0$
Hence, $\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} is Convergent and converges to $0$
