How to prove $\lim_{n\to \infty} \frac{n!}{n^n}$ is zero?

Question

I am struggling in this limit: $$\lim_{n\to \infty} \frac{n!}{n^n}$$ I really worked on this for few days: clearly, it goes to zero but I couln't do it rigorously. I'd really appreciate your help. Thanks from now.

Answer 1

You may use the quotient criterion for series:

$$a_n = \frac{n!}{n^n}\Rightarrow \frac{a_{n+1}}{a_n}=\frac 1{\left(1+\frac 1n\right)^n}\stackrel{n\to\infty}{\longrightarrow}\frac 1e<1$$

Hence, the series $\sum a_n$ is convergent and hence $\lim_{n\to\infty} a_n = 0$.

Answer 2

$\begin{aligned} y=& \lim _{n \rightarrow \infty} \frac{n !}{n^{n}} \\ \ln y &=\lim _{n \rightarrow \infty} \ln \frac{n !}{n^{n}} \\ &=\lim _{n \rightarrow \infty}(n \ln n-n-n \ln n) \\ &=\lim _{n \rightarrow \infty}-n \\ &\\ \ln y \rightarrow-\infty \\ \Rightarrow & y \rightarrow 0 \end{aligned}$

Answer 3

A quick and dirty way to show it goes to 0 really fast: $$ n! \leq n^{\lfloor n/2\rfloor}\cdot (n/2)^{\lceil n/2\rceil} $$ So $$ \frac{n!}{n^n} \leq \frac{1}{2^{\lceil n/2\rceil}} $$