Compute $\lim_{x\to 1^+} \lim_{n\to\infty}\frac{\ln(n!)}{n^x} $

Question

May someone give me a hand on this double limit? Does the order of limits impact the result?

$$\lim_{x\to 1^+} \lim_{n\to\infty}\frac{\ln(n!)}{n^x} $$

I showed that the interior of the limits is inferior to the following expression: $$ \frac{\ln(n)}{{n^{x-1}}} $$

Thanks in advance :)

Answer 1

As $n\to \infty,$ $\ln n! \sim n\ln n,$ hence

$$\frac{\ln n!}{n^x}\sim \frac{\ln n}{n^{x-1}}.$$

The limit of the expression on the right is $0$ if $x>1,$ and is $\infty$ if $x\le 1.$ Thus your limit as $x\to1^+$ is $0.$

Answer 2

HINT:

Note that since $\log(n!)=\sum_{k=1}^n \log(k)=\sum_{k=1}^n \log(k/n)+n\log(n)$, we have

$$\frac1{n^x}\log(n!)=\frac{1}{n^{x-1}}\underbrace{\left(\frac1n\sum_{k=1}^n \log(k/n)\right)}_{\text{Riemann Sum of}\,\int_0^1 \log(x)\,dx=-1}+\frac{\log(n)}{n^{x-1}}$$

Answer 3

Note that for convexity

$$0 \le \frac{\ln n!}{n^x}\leq \frac{n}{n^x}\frac{\sum_{k=1}^{n}\ln k}{n}\le\frac{1}{n^{x-1}}\ln\left({\frac{\sum_{k=1}^nk}{n}}\right)=\frac{\ln \left(\frac{n+1}2{}\right)}{n^{x-1}}\to0 \quad \forall x>1$$

thus for squeeze theorem

$$\lim_{x\to 1^+} \lim_{n\to\infty}\frac{\ln n!}{n^x}=0$$

Answer 4

With fixed $x>1$ $$(n+1)^x-n^x=\int_{n}^{n+1}xt^{x-1}dt>n^{x-1}$$ and using Stolz–Cesàro theorem $$\lim_{n\to\infty}\frac{\ln(n!)}{n^x}=\lim_{n\to\infty}\frac{\ln(n+1)}{(n+1)^x-n^x}<\lim_{n\to\infty}\frac{\ln(n+1)}{n^{x-1}}\to0$$ then $$\lim_{x\to 1^+} \lim_{n\to\infty}\frac{\ln(n!)}{n^x}\to0$$