Is $\ln(x!)/x$ convergent?

Question

Hello I have a question:

Is $limx→∞log(x!) / x$ convergent?

It seems like it gets stuck around $3$. But I am not so sure about this function. Does anyone have an idea?

Thanks

Not sure what you mean by "it gets stuck around $3$". Anyway, look at Stirling's approximation. — lulu, Nov 06 '18 at 14:48
Hello and welcome. You are asking whether the average of the numbers $\ln 1, \ln 2, \dots, \ln x$ remains bounded as $x$ becomes large. But $\ln x \to \infty$ as $x \to \infty$. So ...? — Hans Engler, Nov 06 '18 at 14:49
But $\log(x!)$ is growing faster than $x$, even though it is slow. — edm, Nov 06 '18 at 14:51
@HansEngler Yes but I thought n was getting bigger faster than the sum of lnx (for x values starting from 1.) You think I was wrong? — bebe, Nov 06 '18 at 14:51
@bebe - Numerical approximations can be deceiving. Pick any $k$ as large as you want and consider $x = e^{k+5}$, for example. Then more than 99% of the numbers between 1 and $x$ are larger than $e^k$. Each of these contributes at least $k$ to $\ln x!$. Now estimate the fraction. — Hans Engler, Nov 06 '18 at 14:56
@bebe I suggest that you clarify the OP. Certainly $\frac{\log(x!)}{x}$ is continuous for all real values of $x$ with $x>0$. As such, $\lim_{x\to a}\frac{\log(x!)}{x}=\frac{\log(a!)}{a}$ converges. Are you trying to evaluate the limit as $a\to 0$? Are you trying to evaluate the limit as $a\to \infty$? — Mark Viola, Nov 06 '18 at 16:46
@bebe Please modify your OP to state that you're seeking $\lim_{x\to \infty}\frac{\log(x!)}{x}$. — Mark Viola, Nov 06 '18 at 18:34

score 5 · Accepted Answer · answered Nov 06 '18 at 14:50

Stirling's approximation states $\ln x!\approx \ln \sqrt{2\pi}+(x+\frac{1}{2})\ln x-x$, so $\frac{\ln x!}{x}\sim\ln x$. This diverges as $x\to\infty$, albeit slowly enough you might not have noticed it with numerical experiments. Judging by your conjecture that the function converges to a value approximating $3$, I'm guessing you tried values of $x$ up to approximately $\exp 3\approx 20$.

score 2 · Answer 2 · answered Nov 06 '18 at 14:54

2

As shown here

Prove that $n \ln(n) - n \le \ln(n!)$ without Stirling

we have that

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

then

$$\frac{\ln n!}{n}\ge \ln n -1$$

answered Nov 06 '18 at 14:54

user

162,563

score 2 · Answer 3 · answered Nov 06 '18 at 15:13

2

No, the sequence does not converge.

We have $$ \frac {\ln (n!)}{n} = \frac {1}{n} \sum_1^n \ln k$$

That is a Riemann's sum for the integral $$\int _0^1 \ln(nx)dx = \ln(n) -1$$Which diverges to $\infty$

answered Nov 06 '18 at 15:13

Mohammad Riazi-Kermani

69,891

Is $\ln(x!)/x$ convergent?

3 Answers3