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 :)

Can you tell us, how you derived that $ln(\frac{n!}{n^x})\leq \frac{ln(n)}{n^{x-1}}$? (for all $n\in\mathbb{N}$ and $x\in (1-\epsilon,1+\epsilon$) I assume) — Mark Neuhaus, Jan 14 '18 at 03:10
$ ln (n !) = the-sum-of-the-ln (k)$. So you can mark up the sum by $n * ln (n)$ — Marine Galantin, Jan 14 '18 at 03:12
I only see $ln(\frac{n!}{n^x})=ln(2)+...+ln(n)-x\cdot ln(n)$. — Mark Neuhaus, Jan 14 '18 at 03:18
Oh yeah you're right ; I ve done a mistake when I first wrote the exercice. I really do apologize. I saw that I miss taped the second line but not the first one. — Marine Galantin, Jan 14 '18 at 03:22
You can use $n!<(\frac n2)^n$, it can be proved easily by induction. (or search the site for numerous proofs). — zwim, Jan 14 '18 at 03:53
@MarineGalantin If you are ok, you can accept the answer and set as solved. Thanks! — user, Jan 16 '18 at 22:43
actually i still don't understand. Everyone says that it is easy it is easily seen but I don't understand so.... I don't know how to manage those two limits at the same time — Marine Galantin, Jan 17 '18 at 17:28
@MarineGalantin Please, if you are ok, you can accept the answer and set it as solved. Thanks! http://cdn.sstatic.net/img/faq/faq-accept-answer.png — user, Feb 06 '18 at 21:59
can you read my message just above yours ? I said it wasnt. PLEASE read also my answers. — Marine Galantin, Feb 09 '18 at 16:06

score 5 · Answer 1 · answered Jan 14 '18 at 04:29

5

HINT:

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

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

answered Jan 14 '18 at 04:29

Mark Viola

179,405

yeah okay thanks but then ? haha – Marine Galantin Jan 14 '18 at 08:54
Use the fact that $x-1>0$ as you take the limit as $x\to 1^+$. – Mark Viola Jan 14 '18 at 13:53

score 3 · Accepted Answer · answered Jan 14 '18 at 04:12

3

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 $0.$

answered Jan 14 '18 at 04:12

zhw.

105,693

The answer is correct but for an assignment the argument that "the right is $0$ if $x>1$..." Will get no points. Use L'hospital's rule to show that identity: $$\lim_{n\to\infty,\x-1=y\to0^+}\frac{\ln n}{n^{y}}=\frac{1/n}{yn^{y-1}}=\lim_{n\to\infty,\y-1=k\to -1^+}\frac{n^{-k}}{yn}=\lim_{n\to\infty,\-(k+1)=u\to0^-}\frac{n^u}y$$for the last part you just use L'hospital's rule again(in respect to $n$) and find that you get $\lim_{n\to\infty,\u\to 0^-,\h\to -1^-}un^h$ this is $0\cdot 0=0$ – ℋolo Jan 14 '18 at 05:15
wut... how can you use the l'hospital's rule when there are more than one variable? could you be more precise in your computation, it is a little bit too fast for me – Marine Galantin Jan 14 '18 at 08:53

score 0 · Answer 3 · answered Jan 14 '18 at 08:44

0

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

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

answered Jan 14 '18 at 08:44

user

154,566

I don't know any theorem for composing limits like this. I see that for zany x > 1 it goes to 0 but what is going to happen when you go near to 1? I don't know – Marine Galantin Jan 14 '18 at 09:01
@MarineGalantin I understand your doubt. You can think to the definition of limit, if you choose any $\epsilon>0$ then you can always find $\delta>0$ and $n_0$ such that $\forall x\in(1,1+\delta)$ and $n\ge n_0$ $$f(x,n)\leq\epsilon$$ – user Jan 14 '18 at 09:14
it is maybe very mathematic but idk what to do with this definition. I don't think it s a good idea to compute limits with epsilon haha – Marine Galantin Jan 14 '18 at 10:54
it's not necessary calculate with $\epsilon$, the important and decisive fact is that for all $x>1$ $$\lim_{n\to\infty}\frac{\ln n!}{n^x}=0$$ – user Jan 14 '18 at 11:09

score 0 · Answer 4 · answered Jan 14 '18 at 09:37

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 $$\lim_{x\to 1^+} \lim_{n\to\infty}\frac{\ln(n!)}{n^x}\to0$$

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

4 Answers4