Evaluate the limit: $\lim\limits_{n\to\infty}\frac{\log_a n!}{n^b},\ n\in\Bbb N$

Evaluate the limit: $$\lim_{n\to\infty}\frac{\log_a n!}{n^b}, \ \ n\in\Bbb N$$

I've tried to consider two cases: $$b < 0$$, $$b \ge 0$$. First $$b < 0$$. This case is simple since the limit becomes: $$\lim_{n\to\infty}\frac{\log_a n!}{n^b} = \lim_{n\to\infty}n^{|b|}\log_a n! = +\infty$$

Now consider the case when $$b \ge 0$$, then we may apply Cesaro-Stolz theorem, then the limit is equal to the following limit: \begin{align} \lim_{n\to\infty} \frac{\log_a n!}{n^b} &= \lim_{n\to\infty} \frac{\log_a (n+1)! - \log_an!}{(n+1)^b - n^b} \\ &= \lim_{n\to\infty} \frac{\log_a(n+1)}{(n+1)^b - n^b} \end{align}

I've shown earlier that: $$\lim_{n\to\infty}\left((n+1)^b - n^b\right) = 0,\ \text{if}\ b\in(0, 1)\\ \lim_{n\to\infty}\left((n+1)^b - n^b\right) = +\infty,\ \text{if}\ b > 1\\$$

For $$b = 1$$ the limit becomes: $$\lim_{n\to\infty}{\log_an!\over n} = \lim_{n\to\infty}\log_a\sqrt[n]{n!} = +\infty$$

So it looks like: $$b \le 1 \implies \lim_{n\to\infty}\frac{\log_a n!}{n^b} = +\infty$$

Here I'm not sure how to handle the case for $$b > 1$$. What are the steps to handle the case for $$b > 1$$? I know the limit is $$0$$, but want to justify that.

• $log n! \sim n \log n$, so you get limit $0$ when $b>1$. So far, you have not completed the case $b=1$. – GEdgar Apr 2 at 16:30
• If you vote down, please provide a reason for that, otherwise, it's not clear what's wrong with the question – roman Apr 8 at 17:46

For $$b>1$$, use MVT for $$f(x)=x^b$$, i.e. $$\exists c\in(n,n+1)$$ and $$n>0$$ s.t. $$(n+1)^b - n^b=bc^{b-1} \Rightarrow \\ b(n+1)^{b-1}>(n+1)^b - n^b > bn^{b-1}$$ and from some $$n$$ onwards, assuming $$\ln{a}>0$$: $$0<\frac{\log_a(n+1)}{b(n+1)^{b-1}}= \frac{\ln(n+1)}{b(n+1)^{b-1}\ln{a}} < \frac{\log_a(n+1)}{(n+1)^b - n^b} < \frac{\log_a(n+1)}{bn^{b-1}}= \frac{\ln(n+1)}{bn^{b-1}\ln{a}}$$ because $$b-1>0$$, RHS goes to $$0$$ and by squeezing, the limit is $$0$$. There are quite a few proofs for RHS going to $$0$$, for example here (proposition 2.2) and here (proposition 2).
For $$\ln{a}<0$$ we have $$0>\frac{\ln(n+1)}{b(n+1)^{b-1}\ln{a}}> \frac{\log_a(n+1)}{(n+1)^b - n^b} > \frac{\ln(n+1)}{bn^{b-1}\ln{a}}$$ we the same result.