<s>[INST] 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 $1/\left(1+\frac{1}{n}\right)^n$
 [/INST] 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*}
</s>