# Compute the limit $\lim_{n \to \infty} \frac{n!}{n^n}$ [duplicate]

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

• Note that $n! \leq n^{n-1}$. Nov 24, 2013 at 22:36
• It would converge to $0$ as for a very large $n$, $n!$ is puny compared to $n^n$ Nov 24, 2013 at 22:39
• Hint:$$\frac{n!}{n^n}=\frac{n}{n}\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdot\frac{n-3}{n}\cdot\ldots\cdot\frac{3}{n}\cdot\frac{2}{n}\cdot\frac{1}{n}$$From here you can "see" that one is certainly _losing_ as $n\to\infty$... Nov 24, 2013 at 22:40
• May 24, 2015 at 7:30
• Dec 14, 2016 at 12:40

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

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 $$\lim_{n \to \infty} \frac{n!}{n^n}=0.$$

Hint:

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

• You probably want to have $n-1$ times in there not $n$ because that doesn't tell you anything. Nov 24, 2013 at 22:43
• Haha,correct,wrote it wrong.thnx
– Haha
Nov 24, 2013 at 23:02