Showing that $\frac{\sqrt[n]{n!}}{n}$ $\rightarrow \frac{1}{e}$ Show:$$\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}= \frac{1}{e}$$
So I can expand the numerator by geometric mean. Letting $C_{n}=\left(\ln(a_{1})+...+\ln(a_{n})\right)/n$. Let the numerator be called $a_{n}$ and the denominator be $b_{n}$ Is there a way to use this statement so that I could force the original sequence into the form of $1/\left(1+\frac{1}{n}\right)^n$
 A: It's straightforward if you use Cesaro-Stolz theorem and then the celebre Lalescu's limit. 
$$\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}= \lim_{n\to\infty} \sqrt[n+1]{(n+1)!}-\sqrt[n]{(n)!}=\frac{1}{e}.$$
A: Have not found a way to rewrite your expression to get the desired result. However, here is a suggested approach. 
Maybe rewrite the left-hand side as 
$$\sqrt[n]{\frac{n!}{n^n}}.$$
Take the logarithm. We get
$$\frac{1}{n}\left(\log\left(\frac{1}{n}\right)+ \log\left(\frac{2}{n}\right)+\log\left(\frac{3}{n}\right)+\cdots+\log\left(\frac{n}{n}\right)\right).$$
Now think of the above sum as a Riemann sum for the not quite proper integral
$$\int_0^1 \log x\,dx.$$
A: Here is a rather direct calculation of the limit using squeezing. It needs


*

*$\ln k < \int_k^{k+1}\ln x \;  dx < \ln (k+1)$

*$ \int \ln x \;  dx = x(\ln x - 1) \color{grey}{+C} $

*$\lim_{n \rightarrow \infty}\frac{\ln (n+1)}{n} = 0$


Set
$$
x_n = \ln \frac{\sqrt[n]{n!}}{n} = \frac{1}{n}\sum_{k=1}^n \ln k - \ln n$$ 
So, to show is $\color{blue}{x_n \stackrel{n\rightarrow \infty}{\longrightarrow} -1}$. We have 
$$
\int_1^n \ln x \; dx <  \sum_{k=1}^{n-1} \ln (k+1) =\sum_{k=1}^n \ln k < \int_1^{n+1}\ln x \; dx
$$
We now squeeze:
$$
\color{blue}{L_n} := \frac{1}{n}\int_1^n \ln x \; dx - \ln n  \color{blue}{<x_n < } \frac{1}{n} \int_1^{n+1}\ln x \; dx - \ln n =: \color{blue}{R_n}
$$
\begin{align*}
     \color{blue}{L_n} & = \frac{1}{n}\left( n(\ln n - 1) +1 \right) - \ln n  \\
  & = -1 +\frac{1}{n} \color{blue}{\stackrel{n \rightarrow \infty}{\longrightarrow} -1} \\
  & \\
     \color{blue}{R_n} & = \frac{1}{n}\left( (n+1)(\ln (n+1) - 1) +1 \right) - \ln n  \\
  & = \left( 1 + \frac{1}{n} \right) \ln (n+1) -  \left( 1 + \frac{1}{n} \right) + \frac{1}{n} - \ln n \\
  & = -1 + \ln \left( 1+\frac{1}{n} \right) + \frac{\ln (n+1)}{n} \color{blue}{\stackrel{n \rightarrow \infty}{\longrightarrow} -1}
\end{align*}
A: I would like to use the following lemma: 

If $\lim_{n\to\infty}a_n=a$ and $a_n>0$ for all $n$, then we have 
  $$
\lim_{n\to\infty}\sqrt[n]{a_1a_2\cdots a_n}=a \tag{1}
$$

Let $a_n=(1+\frac{1}{n})^n$, then $a_n>0$ for all $n$ and $\lim_{n\to\infty}a_n=e$. Applying ($*$) we have 
$$
\begin{align}
e&=\lim_{n\to\infty}\sqrt[n]{a_1a_2\cdots a_n}\\
&=\lim_{n\to\infty}\sqrt[n]{\left(\frac{2}{1}\right)^1\left(\frac{3}{2}\right)^2\cdots\left(\frac{n+1}{n}\right)^n}\\
&=\lim_{n\to\infty}\sqrt[n]{\frac{(n+1)^n}{n!}}\\&=
\lim_{n\to\infty}\frac{n+1}{\sqrt[n]{n!}}=\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}+\lim_{n\to\infty}\frac{1}{\sqrt[n]{n!}}=\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}
\end{align}\tag{2}
$$
where we use (1) in the last equality to show that 
$
\lim_{n\to\infty}\frac{1}{\sqrt[n]{n!}}=0.
$
It follows from (2) that 
$$
\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}=\frac{1}{e}.
$$
A: If $a_n \geq 0$, then the following inequality holds:
$$ \liminf_{n\to\infty} \frac{a_{n+1}}{a_n} \leq \liminf_{n\to\infty} \sqrt[n]{a_n} \leq \limsup_{n\to\infty} \sqrt[n]{a_n} \leq \limsup_{n\to\infty} \frac{a_{n+1}}{a_n}. $$
Now let $ a_n = n! / n^n $. Then it follows that
$$ \frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}} = \frac{1}{\left(1+\frac{1}{n}\right)^n},$$
and hence
$$ \liminf_{n\to\infty} \frac{a_{n+1}}{a_n} = \limsup_{n\to\infty} \frac{a_{n+1}}{a_n} = \frac{1}{e}. $$
This proves that $\sqrt[n]{a_n} \to e^{-1}$ .
