Prove $\lim_{n\to \infty}n/\log(n!)=0$ without using Stirling's approximation

Question

It is convincing that the limit is 0 since $n$ has one fixed increasing step 1 but $\log(n!)$ has one strictly increasing step $\log n$. This is by seeing the limit as $\lim_{n\to \infty}\frac{\sum_{k=1}^{n}1}{\sum_{k=1}^{n}\log k}$.

Wolfram Alpha tells me to use Stirling's approximation. Then is there one way to prove $\lim_{n\to \infty}\frac{n}{\log(n!)}=0$ without using Stirling's approximation? By the above step inspect, it seems that we can use Squeeze theorem where one direction $\frac{n}{\log(n!)}\ge 0$ is trivial.

Answer 1

We have that $$\dfrac{n}{\ln(n!)}=\dfrac{n}{\ln(2)+\ln(3)+\cdots+\ln(n)}$$

where both $a_n=n$ and $b_n=\ln(2)+\ln(3)+\cdots+\ln(n)$ are strictly increasing sequences that diverge to $+\infty$.

We can see that $$\dfrac{a_{n+1}-a_n}{b_{n+1}-b_n}=\dfrac{1}{\ln(n+1)}\to 0$$

By the Stolz-Cesàro theorem, $\dfrac{a_n}{b_n}\to 0$.

Answer 2

Here is an elementary argument. Fix $k$ with $\ln k >\frac 1 {\epsilon}$. Then $\frac{n}{\ln(2)+\ln(3)+\cdots+\ln(n)}\le \dfrac{n}{\ln(k)+\ln(k+1)+\cdots+\ln(n)}<\frac n {(n-k+1)\ln k}\to \frac 1 {\ln k} <\epsilon. $

Answer 3

Without Stirling's approximation one can prove that

$$\ln(n!)\ge n \ln(n) - n $$

then

$$\frac n{\ln(n!)} \le \frac{n}{n \ln(n) - n} =\frac{1}{ \ln(n) - 1}\to 0$$

---

As an alternative by this other bound $n! \ge {\left(\frac n2\right)}^{\frac n2}$ we have

$$\frac n{\ln(n!)} \le \frac{n}{\ln \left( \left( \frac n2 \right) ^{\frac n2} \right)} =\frac{2}{ \ln \left( \frac n2 \right) }\to 0$$