TITLE: I need help to advance in the resolution of that limit: $ \lim_{n \to \infty}{\sqrt[n]{\frac{n!}{n^n}}} $
QUESTION [7 upvotes]: how I can continue this limit resolution?
The limit is: 
$$ \lim_{n \to \infty}{\sqrt[n]{\frac{n!}{n^n}}} $$
This is that I have done:
I apply this test: $ \lim_{n \to \infty}{\sqrt[n]{a_n}} = \frac{a_{n+1}}{a_n} $
Operating and simplifying I arrive to this point:
$$ \lim_{n \to \infty}{\frac{n^n}{(n+1)^n}} $$
I've done something wrong? Thanks!

REPLY [3 votes]: By Stirling approximation as $n\to \infty$ we have
$$
\left(\frac{n!}{n^n}\right)^{1/n}\sim \left(\sqrt{2\pi n}\frac{(n/e)^n}{n^n}\right)^{1/n} \sim \frac{1}{e}(2\pi n)^{1/2n} \to e^{-1}.
$$
As pointed out by Clement, one should justify why one can take the limit directly inside; the reason is that for any $x< 1<y$ one have definitively
$$
xe^{-1}n^{1/2n}\le \left(\frac{n!}{n^n}\right)^{1/n}\le ye^{-1}n^{1/2n}.
$$
Since $n^{1/2n}\to 1$ and the above inequality holds for all choices of $x$, $y$, we conclude as above :)