Compute the limit $\lim_{n \to \infty} \frac{n!}{n^n}$

Question

I am trying to calculate the following limit without Stirling's relation. \begin{equation} \lim_{n \to \infty} \dfrac{n!}{n^n} \end{equation} I tried every trick I know but nothing works. Thank you very much.

Answer 1

By estimating all the factors in $n!$ except the first one, we get: $$0 \leq \lim_{n \rightarrow \infty} \frac{n!}{n^n} \leq \lim_{n \rightarrow \infty} \frac{n^{n-1}}{n^n} = \lim_{n \rightarrow \infty} \frac{1}{n} = 0$$

Answer 2

Hint:

We have that $n!=1\cdot 2\cdot ...\cdot n<n\cdot n \cdot ...\cdot n$ ,$n-1$ times.

Answer 3

Consider the series $$\sum_{n=1}^\infty \frac{n!}{n^n} $$ of positive terms. The ratio of two consecutive terms is $$\frac{a_{n+1}}{a_n}=\frac{(n+1)!/(n+1)^{n+1}}{n!/n^n}= \left( \frac{n}{n+1} \right)^n=\left[ \left(1+\frac{1}{n} \right)^n \right]^{-1} $$ which tends to $e^{-1}<1$. It follows from the ratio test that the series converges, and by the necessary condition for convergence of series the limit is obtained. We have $$\lim_{n \to \infty} \frac{n!}{n^n}=0. $$