Does $x_n \colon= \frac{ \sqrt[n]{n!} }{ n }$ converge? If so, then how to find the limit? For every natural number $n$, let 
$$ x_n \colon= \frac{ \sqrt[n]{n!} }{ n }. $$
Then does the sequence $\left( x_n \right)_{n \in \mathbb{N} }$ converge or diverge? And, how to find the limit?
My Attempt: 

We find that, for any $n \in \mathbb{N}$, 
  $$ \frac{ x_{n+1} }{ x_n } = \frac{ \frac{ \sqrt[n+1]{ (n+1)!} }{ n + 1 }  }{ \frac{ \sqrt[n]{n!} }{ n } } = \frac{ \left( n! \right)^{ \frac{1}{n+1} - \frac{1}{n} } }{ 1 + \frac{1}{n} } \sqrt[n+1]{n+1} = . . .  $$

What next? I was hoping to be able to apply the so-called "ratio" test for sequences, but there seems to be no such possibility available. 
 A: Consider $a_n = x_n^n =  \dfrac{n!}{n^n}$. Then
$$
\frac{a_{n+1}}{a_n}
= \left(\frac{n}{n+1}\right)^n
= \left(\frac{1}{1+\frac1n}\right)^n
\to \frac1e
$$
Now, it is true that if $\lim \frac{a_{n+1}}{a_n}$ exists, then so does $\lim \sqrt[n]{a_n}$ and they are equal (see a proof here).
Therefore,
$$
\lim x_n = \lim \sqrt[n]{a_n} = \lim \frac{a_{n+1}}{a_n} = \frac1e
$$
A: Using Stirling's approximation:
$$n!\sim\sqrt{2\pi n} \,\left(\frac{n}{e}\right)^n$$
You are looking for 
\begin{align}
\lim_{n\to\infty} x_n&=\lim_{n\to\infty}\frac{\sqrt[n]{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}}{n}\\&=\lim_{n\to\infty} e^{-1} \left(2\pi n\right)^{1/(2n)} \\
&=\frac{1}{e}\lim_{n\to\infty} (2\pi)^{1/2n}\sqrt{n^{1/n}}\\
&=\frac{1}{e}\lim_{n\to\infty} (2\pi)^{1/2n} \cdot \sqrt{\lim_{n\to\infty} \sqrt[n]{n}}\\
&=\frac{1}{e}
\end{align}
So your sequences converges.
A: Another way is to take logarithms.  $$
\ln \frac{\sqrt[n]{n!}}{n}= \frac{1}{n}\sum_{k=2}^{n}{\ln k}-\ln n
$$
Then using $$
\int_2^{n+1}{\ln x}\,\mathrm dx\ge \sum_{k=2}^{n}{\ln k} \ge \int_1^{n}{\ln x}\,\mathrm dx \\
\text{ and }\int{\ln x}\,\mathrm dx=x\ln x - x,
$$
it's easy to see the logarithm goes to $-1$.
A: Here is an easy trick, without sterling. 
If $a_n\to L$ then $$\sqrt[n]{a_1a_2 \cdots a_n}\to L$$ apply this to $$a_n=\left(1+\frac{1}{n}\right)^n$$
A: By Stirling approximation 
$$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$$
$$ x_n \colon= \frac{ \sqrt[n]{n!} }{ n }\sim \frac{ \sqrt[n]{\sqrt{2 \pi n}} }{ e }\to \frac1e$$
or as an alternative proceed by ratio-root criteria.
A: Using Stirling's approximation:
$$x_n\implies\frac{\sqrt[n]{n!}}{n}\sim\frac{(\tau n)^\frac1{2n}\left(\frac{n}{e}\right)^\frac{n}{n}}{n}\implies\frac{1}{e}(\tau n)^\frac1{2n}$$
$$\rightarrow\lim_{n\to\infty}\left(\frac{1}{e}(\tau n)^\frac1{2n}\right)=\boxed{\frac{1}{e}}$$
