Limit $\lim_{n \to \infty} \frac{n!}{n^n}$

Question

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

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Attempt:

$$n!=n\left( n-1 \right)\left( n-2 \right)...\left( n-\left( n-1 \right) \right)=n^{n}+...$$

$$\lim_{n \to \infty} \frac{n!}{n^{n}}=\lim_{n \to \infty} \frac{n^{n}+...}{n^{n}}=\lim_{n \to \infty} \frac{1+...}{1}=1+\lim_{n \to \infty} \ ...=1$$

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I based this on the fact that $n$ is the highest power. The answer's meant to be $0$ though.

Answer 1

Here is an elementary way using Sandwich Theorem:

$$0 \leq \lim_{n \rightarrow \infty} \frac{n!}{n^n} = \lim_{n \rightarrow \infty} \frac {n(n-1)...3.2.1}{n^n} \leq \lim_{n \rightarrow \infty} \frac{1}{n} = 0$$

Edit: $$ \frac{n!}{n^n} = \frac{n \times (n - 1) \times \cdots \times 1}{n \times n \times \cdots \times n} = 1 \times \frac{n - 1}{n} \times \cdots \times \frac{1}{n} \leq 1 \times 1 \times \cdots \times \frac{1}{n} = \frac{1}{n}.$$

Answer 2

You can use series, for example:

$$a_n:=\frac{n!}{n^n}\implies\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\frac1{\left(1+\frac1n\right)^n}\xrightarrow[n\to\infty]{}\frac1e<1\implies$$

$$\implies\text{the series}\;\;\sum_{n=1}^\infty a_n\;\;\;\text{converges}\;\implies \lim_{n\to\infty}a_n=0$$