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: 
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 $+\infty$.
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](https://math.stackexchange.com/q/2681474/505767)