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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.

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

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Hint:

Use Stirling approximation.

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