See if a sequence $\frac{n!}{n^n}$ has a limit and find it

Question

I need to find a candidate for the limit and then prove it is indeed one using the definition of a sequence's limit. That is, to solve inequality with a parameter $\epsilon > 0$. I suppose the limit is $0$(for intuitive reasons that $n^n$ ''grows faster'' than $n!$).

Now I need to solve the following inequality:

$\frac{n!}{n^n} < \epsilon$

Well, it is easy for $\epsilon \geq 1$ $n! < n^n \leq \epsilon n^n \ \ \forall n>1$, obviously.

If $\epsilon \in (0,1)$ we have $\epsilon n < n^n$.

By induction if $n! < \epsilon n^n$ for some $n$, then $(n+1)! < \epsilon (n+1)^(n+1) \Leftarrow !n < \epsilon n < \epsilon (n+1)^n$.

So now we need to prove that $\forall \epsilon \in (0,1)\ \ \exists n \in \mathbb{N}: n! < \epsilon n^n$

How should I approach that?

Answer 1

First rewrite the sequence

$$\frac{1\cdot2\cdot3\cdots n}{n\cdot n\cdot n \cdots n}$$

We can clearly see

$$0\leq \frac{1\cdot2\cdot3\cdots n}{n\cdot n\cdot n \cdots n}\leq \frac{n^{n-1}}{n^n}=\frac{1}{n}$$

Convergence can be conluded using the squeeze theorem.

Answer 2

$$ \frac{n!}{n^n} = \frac{1}{n} \frac{2}{n} \cdots \frac{n}{n} < \frac{1}{n}. $$ That is if you take $n > \frac{1}{\epsilon}$ $$ \frac{n!}{n^n} < \frac{1}{n} < \epsilon $$