Why is $\lim\limits_{n \to \infty} \ln \left(\frac{n}{(n!)^{\frac{1}{n}}}\right)=1$?

Question

Why is $$\lim_{n \to \infty} \ln \left(\frac{n}{(n!)^{\frac{1}{n}}}\right)=1?$$

I see from looking at the graph that it goes to $1$ but I am not too sure how to prove this algebraically.

The only way I can see this function going to $1$ is if $(n!)^{\frac{1}{n}}>n$ but I am not too sure if that is true.

score 4 · Answer 1 · edited Mar 08 '18 at 22:21

4

It is $$\lim_{n\to\infty}\Bigl(\ln (n)-\frac {1}{n}\sum_{k=1}^{n}\ln (k) \Bigr)$$

But $$\ln (n)-\frac {1}{n}\sum_{k=1}^n\ln (k)=$$

$$-\frac {1}{n}\sum_{k=1}^n\ln \left(\frac {k}{n}\right) $$

It is a Riemann sum, its limit is $$-\int_0^1\ln (x)dx=-\Big[x\ln (x)-x\Big]_0^1=1$$

edited Mar 08 '18 at 22:21

Mark Viola

answered Mar 08 '18 at 22:17

hamam_Abdallah

Why is its limit what you wrote? – zhw. Mar 08 '18 at 22:52
@zhw. The improper integral is convergent. – hamam_Abdallah Mar 08 '18 at 23:07
But such sums need not converge to the value of the integral if the improper integral exists. – zhw. Mar 08 '18 at 23:32
@zhw. Do you have a counterexample – hamam_Abdallah Mar 08 '18 at 23:52
Yes: Tall thin triangles marching to $0.$ Your proof idea still goes through, because if $f$ is monotone, everything works. But that needs to be understood, and is worth mentioning in your proof I think. – zhw. Mar 09 '18 at 17:53
@zhw. Thanks for your remark. – hamam_Abdallah Mar 09 '18 at 20:27

score 3 · Answer 2 · answered Mar 08 '18 at 22:15

3

It is equivalent to proving that $\left(\frac{n^n}{n!}\right)^{1/n}\to e$.

The limit is easier to compute using that, after simplification, $\frac{a_{n+1}}{a_n}=(1+1/n)^n\to e$.

answered Mar 08 '18 at 22:15

crivair

I also use often the ratio-root criteria (+1), so I had to use Casaro-Stolz in this case! (+1) – user Mar 08 '18 at 22:25
@gimusi They are the same theorem written in two forms: $a_n^{1/n}$ is the geometric mean of $a_n/a_{n-1}, a_{n-1}/a_{n-2},...,a_{2}/a_1, a_1/1$, and a geometric mean is the same as an arithmetic after taking logarithm. – crivair Mar 08 '18 at 22:32
@crivair Yes I know they are strictly related, indeed they lead to the same limit, but the set up and resolution come out different. – user Mar 08 '18 at 22:38

user · Accepted Answer · 2018-03-08T22:27:21.597

2

Note that

$$\ln\left(\frac{n}{(n!)^{\frac{1}{n}}}\right)=\frac1n\ln\left(\frac{n^n}{n!}\right)=\frac{a_n}n$$

$$\frac{a_{n+1}-a_n}{n+1-n}=\ln\left(\frac{(n+1)^{n+1}}{(n+1)!}\right)-\ln\left(\frac{n^n}{n!}\right)=\ln\left(1+\frac1n\right)^n\to 1$$

edited Mar 08 '18 at 22:27

answered Mar 08 '18 at 22:24

user

Thanks Mark, ratio-root criteria has been already used thus I had to find something else! I should also become confident with Riemann sum, I'm not skilled about that method. – user Mar 08 '18 at 22:29
I like this one! Thanks – bigfocalchord Mar 09 '18 at 09:58
1

@dydxx You are welcome, it's plenty of very good answer here! It should be possible to accept multiple answers in these cases. Bye – user Mar 09 '18 at 10:03

M. Hennes · Answer 4 · 2018-03-08T22:46:48.287

0

Use the properties of the logarithm:

$\ln \frac{a}{b} = \ln a - \ln b$,

and

$\ln a^b = b \ln a$.

So we get:

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

where we used Stirlings approximation: $n! \approx n \ln n - n$ for large $n$, see http://mathworld.wolfram.com/StirlingsApproximation.html.

edited Mar 08 '18 at 22:46

answered Mar 08 '18 at 22:38

M. Hennes

Maybe you should fix a littel bit the notaton and approximation used in order to make it more clear and rigorous. – user Mar 08 '18 at 22:40
fixed! Notation should be clear now. – M. Hennes Mar 08 '18 at 22:56
Using limit notation you should write
$$\lim_{n\to\infty} \ln\left(\frac{n}{(n!)^{1/n}}\right)= \lim_{n\to\infty} \left(\ln n - \ln (n!^{1/n})\right) = \lim_{n\to\infty}\left(\ln n - \frac{1}{n} \ln n!\right) = \lim_{n\to\infty} \left(\ln n - \frac{1}{n} n \ln n + \frac{1}{n} n\right) = 1$$

or in short notation

$$\ln\left(\frac{n}{(n!)^{1/n}}\right)= \ln n - \ln (n!^{1/n}) = \ln n - \frac{1}{n} \ln n! \sim \ln n - \frac{1}{n} n \ln n + \frac{1}{n} n \to 1$$
– user Mar 08 '18 at 23:09

4 Answers4