osula9a

2022-01-19

Proving limit with $\mathrm{log}(n!)$

I am trying to calculate the following limits, but I dont

I am trying to calculate the following limits, but I dont

turtletalk75

Beginner2022-01-19Added 29 answers

You can easily show that ${2}^{n}\le n!\le {n}^{n}\text{}\text{for}\text{}n\ge 4$ The first inequality is a very standard induction proof, and the second inequality is straight-forward (you're comparing $1\times 2\times \cdots \times n\text{}\text{with}\text{}n\times n\times \cdots \times n).$

From there, since$f\left(n\right)=\mathrm{log}n$ is an increasing function, you have that

$n\mathrm{log}\left(2\right)\le \mathrm{log}(n!)\le n\mathrm{log}\left(n\right)$

This tells you basically everything you will need. For example, for the first one:

$\underset{n\to \mathrm{\infty}}{lim}\frac{3\sqrt{n}}{n\mathrm{log}n}\le \underset{n\to \mathrm{\infty}}{lim}\frac{3\sqrt{n}}{\mathrm{log}(n!)}\le \underset{n\to \mathrm{\infty}}{lim}\frac{3\sqrt{n}}{n\mathrm{log}\left(2\right)}$

From there, since

This tells you basically everything you will need. For example, for the first one:

Andrew Reyes

Beginner2022-01-20Added 24 answers

Stirlings

RizerMix

Skilled2022-01-27Added 437 answers

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