How to compute lim sup and lim inf with ratio tests to $\lim_{x \to \infty} \frac{\sqrt[n]{n!}}{n}$? I am working on real analysis right now. We are using baby rudin as a textbook and I recently was confused by this problem:

Evaluate $\lim_{x \to \infty} \frac{\sqrt[n]{n!}}{n}$

What I was thinking is: take $c_n = \frac{n!}{n^n}$, and then use the ratio test. However, I do not know how to compute the $\limsup_{n \to \infty}  \frac{c_{n+1}}{c_n}$. Could you please help me with this? Thanks!
 A: In THIS ANSWER, I showed that 
$$\bbox[5px,border:2px solid #C0A000]{\liminf_{n\to \infty}\frac{a_{n+1}}{a_n}\le\limsup_{n\to \infty}\sqrt[n]{a_n} \le\limsup_{n\to \infty}\sqrt[n]{a_n}\le \limsup_{n\to \infty}\frac{a_{n+1}}{a_n}}$$
Now, let $a_n=\frac{n!}{n^n}$.  Then, we have 
$$\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{n^n}{(n+1)^{n}}=\frac1e$$
Since the limit, $\lim_{n\to \infty}\frac{a_{n+1}}{a_n}$, exists, then 
$$\liminf_{n\to \infty}\frac{a_{n+1}}{a_n}=\limsup_{n\to \infty}\sqrt[n]{a_n} =\limsup_{n\to \infty}\sqrt[n]{a_n}= \limsup_{n\to \infty}\frac{a_{n+1}}{a_n}=\frac1e$$
Finally, we see that 

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\frac{\sqrt[n]{n!}}{n}=\frac1e}$$

as was to be shown!
A: $$\frac{c_{n+1}}{c_n}=\frac{(n+1)!}{(n+1)^{n+1}}\cdot \frac{n^n}{n!}=\Bigl(\frac{n}{n+1}\Bigr)^n=\frac1{\Bigl(1+\dfrac1n\Bigr)^n}\to\frac1{\mathrm{e}}.$$
A: It is a good idea to switch to logarithms:
$$ \log\frac{\sqrt[n]{n!}}{n} = \frac{1}{n}\sum_{k=1}^{n}\left[\log(k)-\log(n)\right] = \frac{1}{n}\sum_{k=1}^{n}\log\left(\frac{k}{n}\right)\tag{1}$$
and let the wizardry work: $\log(x)$ is a Riemann-integrable function over $(0,1)$, whose integral, by the fundamental theorem of Calculus, equals $-1$. Since the RHS of $(1)$ is a Riemann sum,
$$ \lim_{n\to +\infty}\frac{\sqrt[n]{n!}}{n} = \exp\int_{0}^{1}\log(x)\,dx = \color{red}{\frac{1}{e}}\tag{2} $$
and since we have a finite limit, $\limsup=\liminf=\frac{1}{e}$.
