[SOLVED] A limit with a factorial

**akolman** · November 29th 2008, 01:42 PM

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

**ThePerfectHacker** · November 29th 2008, 03:33 PM

Originally Posted by akolman

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

Hint: $\frac{n!}{n^n} \leq \frac{1}{n}$ .

**chiph588@** · November 29th 2008, 09:30 PM

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

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

**o_O** · November 29th 2008, 09:43 PM

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

**akolman** · November 29th 2008, 10:05 PM

Thank you all very much for your help!

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