how can I find this limit I have the bellow limit, and I know I need to use Cauchy-d’Alembert, and the limit is 1/e but have no idea how to get to it, I get to something like $\frac{\frac{\left(n+1\right)!}{n!}}{n}$, but is not right 
$$\lim _{n\to \infty }\left(\frac{\sqrt[n]{n!}}{n}\right)$$
 A: I would prefer using kotomord's approach.  I disagree that it should be simpler; this is basically equivalent to the coarse version of Stirling's approximation.  
But anyway, rewrite the limit as 
$$
\lim _{n\to \infty}\sqrt[n]{\frac{n!}{n^n}}
$$ the Cauchy-d'Alembert criteria says this should be equal to 
$$
\lim_{n\to \infty}\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}} \\
= \lim_{n\to \infty}\frac{(n+1)n^n}{(n+1)^{n+1}} \\
= \lim_{n\to \infty}\left(\frac{n}{n+1}\right)^n \\
= \lim_{n\to \infty}\left(\frac{1}{1+\frac{1}{n}}\right)^n \\ 
= \frac{1}{e}
$$
A: Try to use Stirling's approximation
https://en.wikipedia.org/wiki/Stirling%27s_approximation
A: Here's a different approach.  
Let $a_n$ be the element of the sequence. We have 
$$ \ln a_n = \frac{1}{n} \ln (n!) - \ln n = \frac{1}{n} \sum_{j=1}^n \ln (j/n).$$   
This is Riemann sum, and converges to $\int_0^1 \ln x dx= -1$ (note it is an improper Riemann integral, so for complete rigor, we need to truncate the summation).  Since $\lim_{n\to\infty} \ln a_n = -1$, it follows that $\lim_{n\to\infty} a_n = e^{-1}$.  
