Limit with factorial (Stolz & Stirling) We have this limit:
$\lim\limits_{n \to \infty} (\ln(n!))/n \to \infty$
I know that it should be answered with Stolz or Stirling, but I don't know how.
 A: HINT
By Stolz-Cesaro with $\frac{a_n}{b_n}=\frac{\ln n!}{n}$ we obtain
$$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\frac{\ln (n+1)!-\ln n!}{n+1-n}$$
then recall that $\log a-\log b =\log \frac a b$.
As a simpler alternative by ratio test
$$\frac{\frac{\ln (n+1)!}{n+1}}{\frac{\ln n!}{n}}=\frac n {n+1}\frac{\ln (n+1)!}{\ln n!}=\frac n {n+1}\frac{\ln (n+1)+\ln n!}{\ln n!}$$
A: By Stirling,
$$ \frac{\ln n!}{n} = \frac{n\ln n -n + O(\ln n)}{n} = \ln n - 1 + \frac{O(\ln n)}{n} \to \infty, \quad \text{as } n\to\infty. $$
EDIT: By request, using the Stirling formula and the continuity of $\ln$,
\begin{align*}
\lim_{n\to\infty} \frac{\ln n!}{n} &= \lim_{n\to\infty} \frac{\ln (\sqrt{2\pi n} \;e^{-n}n^n)}{n} = \lim_{n\to\infty} \ln \left( (2\pi n)^{1/2n} e^{-1} n \right) 
= \ln \left( e^{-1}\lim_{n\to\infty} n(2\pi n)^{1/2n}  \right) = \infty,
\end{align*}
since $(2\pi n)^{1/2n} \to 1$ as $n\to \infty$.
A: Stirling's approximation tells us that the following is true:
$$ln(n!) = n \cdot ln(n) - n + O\left(ln(n)\right)$$
If we divide this by $n$, we have:
$$\frac{ln(n!)}{n} = ln(n) - 1 + O\left(\frac{ln(n)}{n}\right)$$
As $n \rightarrow \infty, \frac{ln(n!)}{n} \rightarrow \infty$.
A: With Stolz:
$$ \lim_{n\to +\infty}\frac{\log(n)}{1}=+\infty\quad\Longrightarrow\quad \lim_{n\to +\infty}\frac{\sum_{k=1}^{n}\log(k)}{\sum_{k=1}^{n}1}=+\infty.$$
With Stirling:
$$ n!\geq \left(\frac{n}{e}\right)^n \sqrt{2\pi n}\quad\Longrightarrow\quad \log(n!)\geq n\log(n)-n.$$
With Hermite-Hadamard ($\log x$ is concave on $\mathbb{R}^+$):
$$ \log(n!)=\sum_{k=1}^{n}\log(k)\geq \frac{1}{n}\int_{1/2}^{n+1/2}\log(x)\,dx=\left(n+\tfrac{1}{2}\right)\log(2n+1)-(1+\log 2)n. $$
