prove that $\lim_{n\to∞} \frac{n!}{n^n}=0$

Question

any sugestions?, by Stolz–Cesàro?. The trouble says that first try with $k<n$ and then for a particular $k<\frac{n}{2}$ I try by definition of $\epsilon$ but it doesn't help

Answer 1

Every number in the product $n!$ is less than or equal to $n$. In particular, one of them is $1$. Can you use this to show $n! / n^n \leq 1/n$?

Answer 2

You can also use Stirling's approximation: $n!\sim \left(\dfrac{n}{e}\right)^n\sqrt{2\pi n}$ as $n\to \infty$

Answer 3

There is a very simple approach: by the AM-GM inequality, $$ n!^2 \leq \left(\frac{n+1}{2}\right)^{2n},$$ hence the wanted limit is clearly $\color{red}{0}$.

Answer 4

$0\leq{n!}\leq{n^{n-1}}\implies$

$\frac{0}{n^n}\leq\frac{n!}{n^n}\leq\frac{n^{n-1}}{n^n}\implies$

$0\leq\frac{n!}{n^n}\leq\frac{1}{n}\implies$

$\lim\limits_{n\to\infty}0\leq\lim\limits_{n\to\infty}\frac{n!}{n^n}\leq\lim\limits_{n\to\infty}\frac{1}{n}\implies$

$0\leq\lim\limits_{n\to\infty}\frac{n!}{n^n}\leq0\implies$

$\lim\limits_{n\to\infty}\frac{n!}{n^n}=0$