How to show that $(n!/n^n)_{n\geq 0}$ is a null sequence?

Question

How to show that $\left(\frac{n!}{n^n}\right)_{n\geq 0}$ is a null sequence, meaning it tends to zero.

i tried proving that $n^n\geq n!$ for $n\in \mathbb{N}:n>0$. Is that a correct method?

Base case: $1^1\geq 1! \quad \checkmark$

hypothesis: $\exists n \in \mathbb{N}: n^n\geq n!$

inductive step: $$(n+1)^{n+1}\geq (n+1)! \\ (n+1)^1\cdot (n+1)^n\geq (n+1)! \\ (n+1)^n\geq n!$$ How can I continue...?

Answer 1

Note that $n^n\ge n!$ is not enoug in order to show $\frac{n!}{n^n}\to 0$. For example, $n+1\ge n$, but $\frac{n}{n+1}\not\to 0$.

Instead, you might note that for $n\ge 1$ $$ \frac{n!}{n^n}=\frac{\prod_{k=1}^nk}{\prod_{k=1}^nn}=\prod_{k=1}^n\frac kn=\frac1n\cdot \prod_{k=2}^n\frac kn\le \frac1n$$ (In fact, $\frac{n!}{n^n}$ goes to $0$ so fast that there is a lot of leeway for proving it)

Answer 2

Using the Stirling approximation, $n!\sim C\left(\frac{n}{e}\right)^n$ where $e$ is Euler's constant, and $C$ is a "small" error term (asymptotically $\sqrt{2\pi n}$). Thus $\frac{n!}{n^n}$ is just $\frac{C}{e^n}$ which can easily be proven to go to 0.

Answer 3

$$\frac{\dfrac{(n+1)!}{(n+1)^{n+1}}}{\dfrac{n!}{n^n}}=\frac{n^n(n+1)}{(n+1)^{n+1}}=\left(1-\frac1n\right)^n\to e^{-1}<1$$

By Ratio Test for series, $$\sum_{n=1}^\infty \dfrac{n!}{n^n}<\infty$$ Therefore, the limit of the sequence is $0$.