Convergence of $a_n= \frac{n!}{n^n}$? [duplicate]

$\displaystyle \frac{n!}{n^n} = \frac{n\cdot(n-1)\cdots1}{n\cdot n\cdots n}=\frac{n}{n}\frac{n-1}{n}\frac{n-2}{n}\cdots\frac{1}{n}\leq 1\cdot1\cdot1\cdots\frac{1}{n}=\frac{1}{n}$

To answer the first part of your question: Yes: $\frac{n!}{n^n}\leq\frac{1}{n}$ for $n>2$ and since you are looking at the limit $n\rightarrow\infty$, $n$ really is larger than $2$! Just try it out with some values.

For the rest:

$\displaystyle \frac{n!}{n^n} = \frac{n\cdot(n-1)\cdot\dots \cdot 1}{n\cdot n\cdot \dots \cdot n}=\frac{n}{n}\cdot\underbrace{\frac{n-1}{n}}_{<1}\cdot\underbrace{\frac{n-2}{n}}_{<1}\cdot \dots \cdot \frac{1}{n}\leq 1\cdot1\cdot1\cdot \dots \cdot\frac{1}{n}=\frac{1}{n}$