Show that $\lim_{n\to\infty}\frac{\ln(n!)}{n} = +\infty$ 
Show that 
  $$
\lim_{n\to\infty}\frac{\ln(n!)}{n} = +\infty
$$

The only way i've been able to show that is using Stirling's approximation:
$$
n! \sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n
$$
Let:
$$
\begin{cases}
x_n = \frac{\ln(n!)}{n}\\
n \in \Bbb N
\end{cases}
$$
So we may rewrite $x_n$ as:
$$
x_n \sim \frac{\ln(2\pi n)}{2n} + \frac{n\ln(\frac{n}{e})}{n}
$$
Now using the fact that $\lim(x_n + y_n) = \lim x_n + \lim y_n$ :
$$
\lim_{n\to\infty}x_n = \lim_{n\to\infty}\frac{\ln(2\pi n)}{2n} + \lim_{n\to\infty}\frac{n\ln(\frac{n}{e})}{n} = 0 + \infty
$$
I'm looking for another way to show this, since Stirling's approximation has not been introduced at the point where i took the exercise from yet.
 A: Another way to show $$\lim_{n\to\infty}\frac{\ln(n!)}{n}=\infty$$ is to consider the following property of logarithms: $$\log(n!)=\log(n)+\log(n-1)+\cdots+\log(2)>\frac{n}{2}\log\left(\frac n2\right).$$ Now $$\frac{\log (n!)}{n}>\frac{\log(n/2)}{2}.$$ As $n\to\infty$, this clearly diverges to $+\infty$.
A: Hint: $$\ln(n!)=\ln 1+\ln2+\cdots+\ln n$$For all but the smallest $n$, most of those terms are larger than $1$, so the sum is larger than $n$.
For somewhat large $n$, most of those terms are larger than $2$, so the sum is larger than $2n$.
For even larger $n$, most of those terms are larger than $3$, so the sum is larger than $3n$.
A: The Cesaro-Stolz criterion is your easiest and cleanest way out of this. It states that given sequences $x, y \in \mathbb{R}^{\mathbb{N}}$ such that $y$ is strictly increasing and unbounded and the sequence of successive increments converges in the extended real line
$$\lim_{n \to \infty} \frac{x_{n+1}-x_{n}}{y_{n+1}-y_{n}}=t \in \overline{\mathbb{R}}$$
then $$\lim_{n \to \infty}\frac{x_{n}}{y_{n}}=t$$
You can apply this to $x=(\mathrm{ln}(n!))_{n \in \mathbb{N}}$ and $y=(n)_{n \in \mathbb{N}}$.
A similar argument would rely on a version of the ratio criterion: if $x \in (0, \infty)^{\mathbb{N}}$ is a sequence of strictly positive reals such that 
$$\lim_{n \to \infty} \frac{x_{n+1}}{x_n}=a \in [0, \infty]$$
then
$$\lim_{n \to \infty} \sqrt [n]{x_{n}}=a$$
This criterion itself can be proved by the Cesaro-Stolz criterion (there are also other methods) and you can apply it to conclude that
$$\sqrt[n]{n!} \xrightarrow{n \to \infty} \infty$$
as $\frac{(n+1)!}{n!}=n+1 \xrightarrow{n \to \infty} \infty$.
In the same vein of employing ratios, one can settle the convergence of the sequence
$$\left(\frac{\sqrt[n]{n!}}{n}\right)_{n \in \mathbb{N}^{*}}=\left(\sqrt[n]{\frac{n!}{n^n}}\right)_{n \in \mathbb{N}^{*}}$$
by studying the sequence of successive ratios:
$$ \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!}=\left(\frac{n}{n+1}\right)^{n} \xrightarrow{n \to \infty} \frac{1}{\mathrm{e}}$$
Hence,
$$\sqrt[n]{n!}=n \cdot \frac{\sqrt[n]{n!}}{n} \xrightarrow{n \to \infty} \infty \cdot \frac{1}{\mathrm{e}}=\infty$$
A: Here is another way considering 


*

*$e^{\frac{\ln n!}{n}}$
\begin{eqnarray*} e^{\frac{\ln n!}{n}}
  & = &  \sqrt[n]{n!}\\
  &  \stackrel{GM-HM}{\geq} & \frac{n}{\frac{1}{1}+\cdots \frac{1}{n}} \\
  &   \stackrel{\sum_{k=1}^n \frac{1}{k} < \ln n + 1}{>}  & \frac{n}{\ln n +1} \\
 & \stackrel{n \to \infty}{\longrightarrow} & +\infty
\end{eqnarray*}
A: We have that 
$$\ln(n!)\ge n \ln n - n$$
then
$$\frac{\ln(n!)}{n}\ge \ln n -1 \to \infty$$
For the proof of the first inequality refer to


*

*Prove that $n \ln(n) - n \le \ln(n!)$ without Stirling
