Question: > 
> 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.

Answer: 
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,

$$\sqrt[n]{n!}=n \cdot \frac{\sqrt[n]{n!}}{n} \xrightarrow{n \to \infty} \infty \cdot \frac{1}{\mathrm{e}}=\infty$$
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\*}