What is the limit $ \underset{n\to\infty}{\lim} \frac {{n!}^{1/n}}{n} $

Question

What is the following limit equal to and how do I prove it?

$$ \underset{n\to\infty}{\lim} \frac {{n!}^{1/n}}{n}. $$

I've been trying for a while and I can't seem to get it.

Answer 1

Using Stirling's approximantion: $n! \sim (n/e)^n \sqrt{2 \pi n}$ you get

$\lim_{n\to \infty} \frac{n!^{1/n}}{n} = \lim_{n\to \infty} \frac{n (2\pi n)^{\frac{1}{2n}}}{e n} = \lim_{n \to \infty} \frac{(2 \pi n)^{\frac{1}{2n}}}{e}$

Can you go on from here?

Answer 2

I would just use Stirling:

$$n! \sim n^n e^{-n} \sqrt{2 \pi n} \quad (n \to \infty)$$

From this, you should see that the limit is merely $e^{-1}$.

Answer 3

Credit to David Mitra:

We know that if $\lim\limits_{n\rightarrow\infty}{a_{n+1}\over a_n}$ exists than it's equal to $\lim\limits_{n\rightarrow\infty}{\root n\of {a_n}}$

We can apply this to $ a_n= {n!\over n^n} $ and see that $ \lim\limits_{n\rightarrow\infty}{a_{n+1}\over a_n} = \frac{1}{e}$ and so $\lim\limits_{n\rightarrow\infty}{\root n\of {a_n}} = \underset{n\to\infty}{lim} \frac {{n!}^{1/n}}{n} = \frac{1}{e}$