Proving limit with $\log(n!)$

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!

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

From there, since $f(n) = \log n$ is an increasing function, you have that

$$n\log(2) \leq \log(n!) \leq n\log(n)$$

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

$$ \lim_{n \to \infty} \frac{3 \sqrt{n}}{n\log n} \leq \lim_{n \to \infty}\frac{3 \sqrt{n}}{\log(n!)} \leq \lim_{n \to \infty} \frac{3 \sqrt{n}}{n \log(2)}. $$

If you use the fact that $ (\ln(x!))'= \psi(x+1)$, where $\psi(x)$ is the digamma function, and l'hopital's rule, then the first limit can be evaluated directly

$$ \lim_{n\to\infty}\frac{3\cdot \sqrt{n}}{\log(n!)}=\lim_{n \to \infty} \frac{3}{2}\frac{1}{\sqrt{n}\psi(n+1)}=0 \,,$$

where the fact that $$ \lim_{n\to \infty} \psi(n+1)=\infty \,, $$

has been used.

For the second limit, recalling the asymptotic of $\ln(n!)$ $$\ln(n!)=\sum_{k=1}^{n}\ln(k)\sim \int_{1}^{n} \ln(x)dx \sim n\ln(n)-n+1, $$ we have $$ \frac{\log(n!)}{\log(n)^{\log(n)}}\sim \frac{n\ln(n)-n+1}{\log(n)^{\log(n)}} \,.$$

Making the change of variables $m=\ln(n)$ yields

$$ \frac{\log(n!)}{\log(n)^{\log(n)}}\sim \frac{n\ln(n)-n+1}{\log(n)^{\log(n)}}=\frac{me^m-e^m+1}{m^m} \rightarrow 0 $$

as $m\to \infty,$ since $m^m>me^m \,\,\, \forall m>4. $

Note: we can use $\ln(n!)\sim n\ln(n)-n+1 $ to prove the first limit goes to $0$.

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 $$ \lim_{n\to\infty}\frac{\log(n!)}{\log(n)^{\log(n)}}=\lim_{n\to\infty}\frac{n\log(n)}{\log(n)^{\log(n)}}\stackrel{n\to e^n}{=}\lim_{n\to\infty}\frac{e^nn}{n^n}=\lim_{n\to\infty}\left(\frac{2e}{n}\right)^n\frac{n}{2^n}=0 $$

For the first one you can solve it using stoltz lemma

$$\lim_{n\to\infty}\frac{3\cdot \sqrt{n}}{\log(n!)}= \lim _{n\rightarrow \infty} \frac{3\cdot \sqrt{n+1}-3\cdot \sqrt{n}}{\log((n+1)!)-\log n!}=\lim _ {n\rightarrow \infty }\frac{3 }{(\sqrt{n+1}+\sqrt{n})(\log(n+1))}=0$$

For the second observe you can rewrite the denominator as

$\log(n)^{\log(n)}=e^{\log n\log (\log n)}=n^{\log (\log n)}$

$$\lim_{n\to\infty}\frac{\log(n!)}{\log(n)^{\log(n)}}\leq \lim_{n\to\infty}\frac{n\log(n)}{\log(n)^{\log(n)}} =\lim_{n\to\infty}\frac{\log(n)}{n^{\log (\log n)-1}}$$ Therefore $n>e^{e^3} \Rightarrow \log ( \log n)-1>2\Rightarrow n^{\log ( \log n)-1}>n^{2} \Rightarrow \frac{\log(n)}{n^{\log (\log n)-1}} \leq \frac{\log n}{n^2}$

So finally $$\lim_{n\to\infty}\frac{\log(n!)}{\log(n)^{\log(n)}}\leq \lim _{ n \rightarrow \infty }\frac{\log n}{n^2}=0$$