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

Question

Prove that

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

I've already considered using l'Hoptials rules but I cannot take the derivative of a factorial (as it is a discrete function).

Thanks

Answer 1

The easiest way is to get an explicit upper bound for the fraction:

$0 < \dfrac{n!}{n^n} = \dfrac{1\cdot 2\cdot ...\cdot n}{n\cdot n\cdot n\cdot...\cdot n} < \dfrac{1}{n}$, and $\dfrac{1}{n} \to 0$. So the answer follows by squeeze theorem.

Answer 2

Write it as $\dfrac{n}{n}\dfrac{n-1}{n}\dfrac{n-2}{n}\cdots\dfrac{2}{n}\dfrac{1}{n}$

Answer 3

Notice that $\frac{3!}{3^3} = \frac{6}{27} = \frac{2}{9} < \frac{3}{9} = \frac{1}{3}$. A naïve conjecture from this observation might be that $\frac{n!}{n^n} < \frac{1}{n}$ for all $n \ge 3$. This is in fact true, as can be proved by a straightforward induction.

Answer 4

Another approach: Stirling's approximation tells you $n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n$. That tells you the thing you're taking the limit of is approximately $\frac{2 \pi n}{ e^n}$. This goes to 0 as $n \to \infty$.

Alternatively, you can use the bound from Stirling's approximation is $n! \leq e \sqrt{n} n^n e^{-n}$. Then, $0 \leq \frac{n!}{n^n} \leq e \sqrt{n} e^{-n} \to 0$.

Answer 5

By using Gosper's approximation, or Stirling's approximation, $n!$ is approximately \begin{align} n! \approx n^{n} \ e^{-n} \ \sqrt{ \left( 2n + \frac{1}{3} \right) \pi } \end{align} which then leads to \begin{align} \frac{n!}{n^{n}} \approx \ e^{-n} \ \sqrt{ \left( 2n + \frac{1}{3} \right) \pi }. \end{align} Now taking the limit as $n \rightarrow \infty$ yields \begin{align} \lim_{n \rightarrow \infty} \ \frac{n!}{n^{n}} = 0. \end{align}