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.
Best 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 $+\infty$.