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!
For reference, another approach starting from scratch:
Using the fact that $\ln n! = n\ln n - n +o(n)$ (see e.g. this proof of it), we can rewrite the quantities considered in the exponential form (almost always helpful) to obtain that $$ \sqrt[n]{\frac{n!}{n^n}} = e^{\frac{1}{n}\ln \frac{n!}{n^n}} = e^{\frac{1}{n}\left(\sum_{k=1}^n \ln k - n\ln n\right)} = e^{\frac{1}{n}\left(n\ln n - n +o(n) - n\ln n\right)} = e^{\frac{1}{n}\left(- n +o(n)\right)} = e^{-1 +o(1)} $$ so the limit is $e^{-1}$.
$$\frac{n^n}{(n+1)^n} = \left(\frac{n}{n+1}\right)^n = \left(\frac{1}{\frac{n+1}{n}}\right)^n = \frac{1}{(1+1/n)^n} \to1/e$$ as $n\to\infty$.
Let $A=\lim_{n\to\infty}\sqrt[n]{\dfrac{n!}{n^n}}$
$\ln A=\lim_{n\to\infty}\dfrac1n\sum_{r=1}^n\ln\dfrac rn$
Like The limit of a sum $\sum_{k=1}^n \frac{n}{n^2+k^2}$, this equals to $$\int_0^1\ln x\ dx$$
Hint Observe that $$\frac{n}{n+1} = \frac{n + 1 - 1}{n + 1} = 1 - \frac1{n + 1}$$
Now rewrite the limit argument as $\exp(\ln \cdot)$...
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 :)
HINT $$\lim_{n\to\infty }\left(\frac{n}{n+1}\right)^n=\lim_{n\to\infty }\frac{1}{\left(1+\frac{1}{n}\right)^n}$$
