TITLE: Proving limit with $\log(n!)$
QUESTION [12 upvotes]: I am trying to calculate the following limits, but I don't know how:
$$\lim_{n\to\infty}\frac{3\cdot \sqrt{n}}{\log(n!)}$$
And the second one is
$$\lim_{n\to\infty}\frac{\log(n!)}{\log(n)^{\log(n)}}$$
I don't need to show a formal proof, and any tool can be used.
Thanks!

REPLY [2 votes]: Stirling's approximation yields
$$
\log(n!)=\left(n+\frac12\right)\left(\log(n)-1\vphantom{\frac12}\right)+\frac12\log(2\pi e)+O\left(\frac1n\right)
$$
which implies
$$
\lim_{n\to\infty}\frac{\log(n!)}{n\log(n)}=1
$$
Then for the first limit
$$
\lim_{n\to\infty}\frac{3\sqrt n}{\log(n!)}=\lim_{n\to\infty}\frac3{\sqrt{n}\log(n)}=0
$$
For the second limit
$$
\lim_{n\to\infty}\frac{\log(n!)}{\log(n)^{\log(n)}}=\lim_{n\to\infty}\frac{n\log(n)}{\log(n)^{\log(n)}}\stackrel{n\to e^n}{=}\lim_{n\to\infty}\frac{e^nn}{n^n}=\lim_{n\to\infty}\left(\frac{2e}{n}\right)^n\frac{n}{2^n}=0
$$