Compute $\lim\limits_{x \to \infty} \frac{\ln (x!)}{x \ln (x)}$

Question

I was just playing with logarithms and factorials, and then realized that $f(x) = \frac{\ln (x!)}{\ln(x)}$ is slower than $f(x) = x$. From there I got $\frac{\ln x!}{(x \ln (x))}$. In most calculators, I can't find the limit, and when I put a ridiculously large number it doesn't go below 0.6. So is the limit 0?

Answer 1

If the limit is considered along $\mathbb{N}$ it is enough to invoke Cesàro-Stolz: $$\lim_{n\to +\infty}\frac{\log(n!)}{n\log n}=\lim_{n\to +\infty}\frac{\log(n+1)}{(n+1)\log(n+1)-n\log n}=\lim_{n\to +\infty}\frac{\log(n+1)}{\log(n+1)+\underbrace{n\log\left(1+\frac{1}{n}\right)}_{\to 1}}=1. $$ If the limit is considered along $\mathbb{R}$ it is enough to invoke De l'Hospital: $$ \lim_{x\to +\infty}\frac{\log\Gamma(x+1)}{x\log x} = \lim_{x \to +\infty}\frac{\psi(x+1)}{1+\log x}=\lim_{x\to +\infty} x\cdot\psi'(x+1)=\lim_{x\to +\infty}x\sum_{n\geq 1}\frac{1}{(n+x)^2} $$ where $$\sum_{n\geq 1}\frac{1}{(n+x)^2}>\sum_{n\geq 1}\frac{1}{(n+x)(n+x+1)}=\frac{1}{x+1}, $$ $$\sum_{n\geq 1}\frac{1}{(n+x)^2}<\frac{1}{(x+1)^2}+\sum_{n\geq 1}\frac{1}{(n+x)(n+x+1)}=\frac{x+2}{(x+1)^2}.$$

Answer 2

As per Sterling approximation for large $n$, $n! \sim n^{n} e^{-n}$. So $$L=\lim_{n \rightarrow \infty} \frac{n!}{n\ln n}= \lim_{n \rightarrow \infty} \frac{n^n e^{-n}}{n \ln n}=\lim_{n \rightarrow}\frac{e^{n\ln n-n}}{n\ln n}= \lim_{n \rightarrow} \frac{e^{n\ln n-n}}{n\ln n-n} \left( 1-\frac{1}{\ln n}\right)=\lim_{u \rightarrow \infty} \frac{e^u}{u}. 1$$ $$= \lim_{u \rightarrow \infty} \frac{1}{ue^{-u}}.1= \frac{1}{0}= \infty.$$

Answer 3

I assume you see $x$ here as a natural number (otherwise we would have to deal with the Gamma function):

Note that $\ln n! = \sum_{k=1}^n\ln k = \sum_{k=\color{blue}{2}}^n\ln k = \sum_{k=\color{blue}{1}}^{n-1}\ln (k+1)$

You can squeeze the fraction $\frac{\ln n!}{n\ln n}$ as follows using

• $\ln k \leq \int_k^{k+1}\ln x\; dx \leq \ln (k+1)$

$$\int_1^n\ln x\;dx \leq \sum_{k=\color{blue}{1}}^{n-1}\ln (k+1)=\sum_{k=\color{blue}{2}}^n\ln k \leq \int_2^{n+1}\ln x\;dx$$

• We have $\int \ln x\; dx = x(\ln x - 1) (+C)$.

So, you get $$\underbrace{\frac{1}{n\ln n}\left(n \ln n - n +1\right)}_{\stackrel{n \to \infty}{\longrightarrow}1}\leq\frac{\ln n!}{n\ln n} \leq \underbrace{\frac{1}{n\ln n}\left((n+1) \ln (n+1) - (n+1) +2(1-\ln 2)\right)}_{\stackrel{n \to \infty}{\longrightarrow}1}$$ $$\Rightarrow \boxed{\lim_{n\to \infty}\frac{\ln n!}{n\ln n} = 1}$$