osula9a

2022-01-19

Proving limit with $\mathrm{log}\left(n!\right)$
I am trying to calculate the following limits, but I dont

turtletalk75

You can easily show that The first inequality is a very standard induction proof, and the second inequality is straight-forward (you're comparing
From there, since $f\left(n\right)=\mathrm{log}n$ is an increasing function, you have that
$n\mathrm{log}\left(2\right)\le \mathrm{log}\left(n!\right)\le n\mathrm{log}\left(n\right)$
This tells you basically everything you will need. For example, for the first one:
$\underset{n\to \mathrm{\infty }}{lim}\frac{3\sqrt{n}}{n\mathrm{log}n}\le \underset{n\to \mathrm{\infty }}{lim}\frac{3\sqrt{n}}{\mathrm{log}\left(n!\right)}\le \underset{n\to \mathrm{\infty }}{lim}\frac{3\sqrt{n}}{n\mathrm{log}\left(2\right)}$

Andrew Reyes

Stirlings

RizerMix

For the first one, use that $\mathrm{log}\left(n!\right)\sim n\mathrm{log}$ n as $n\to \mathrm{\infty }$ to conclude that $\underset{n\to \mathrm{\infty }}{lim}\frac{3\cdot \sqrt{n}}{\mathrm{log}\left(n!\right)}=\underset{n\to \mathrm{\infty }}{lim}\frac{3\cdot \sqrt{n}}{n\mathrm{log}n}=\underset{n\to \mathrm{\infty }}{lim}\frac{3}{\sqrt{n}\mathrm{log}n}=0$