# Thread: [SOLVED] A limit with a factorial

1. ## [SOLVED] A limit with a factorial

I know that $\lim \limits_{n \to \infty } \frac{n!}{{n^n }} = 0$
since $\lim \limits_{n \to \infty } \frac{n!}{{n^n }} =
\lim_{n \to \infty} \left( \frac{1}{n} \cdot \frac{2}{n} \cdot \ \cdots \ \cdot \frac{n-1}{n} \cdot \frac{n}{n}\right)$
and each limit of this product goes to 0. However, how would you write an $\varepsilon$-proof for this limit?

2. Originally Posted by akolman
I know that $\lim \limits_{n \to \infty } \frac{n!}{{n^n }} = 0$
since $\lim \limits_{n \to \infty } \frac{n!}{{n^n }} =
\lim_{n \to \infty} \left( \frac{1}{n} \cdot \frac{2}{n} \cdot \ \cdots \ \cdot \frac{n-1}{n} \cdot \frac{n}{n}\right)$
and each limit of this product goes to 0. However, how would you write an $\varepsilon$-proof for this limit?
Hint: $\frac{n!}{n^n} \leq \frac{1}{n}$.

3. this means $n^{n-1} \geq n!$

How do we know this is true? (for large enough $n$ of course)

4. $\frac{n!}{n^n} \ = \ \frac{n{\color{red}(n-1)}{\color{blue}(n-2)}{\color{magenta}(n-3)} \cdots (2)}{n^{n-1}} \cdot \frac{1}{n} \ \leq \ \frac{\overbrace{n \cdot {\color{red}n} \cdot {\color{blue}n} \cdot {\color{magenta}n} \cdots n}^{n-1 \ \text{times}}}{n^{n-1}} \cdot \frac{1}{n} \ = \ \frac{n^{n-1}}{n^{n-1}} \cdot \frac{1}{n} \ = \ \frac{1}{n}$

5. Thank you all very much for your help!