I need to find the limit for:
$ \lim_{n\to\infty} \frac {(n!)^\frac{1}{n}}{n} $
I know the answer is $\frac {1}{e}$ but I have no idea how to get that answer. I'd appreciate some help.
Asked
Active
Viewed 283 times
1
-
1Do you know Stirling's approximation? – Daniel Fischer Jul 20 '13 at 20:59
-
1See this. – David Mitra Jul 20 '13 at 21:00
-
@DanielFischer : I know it, but I'm not supposed to use it for this. – matanc1 Jul 20 '13 at 21:03
-
You might want to look at this question:http://math.stackexchange.com/questions/171904/limit-of-a-sequence-involving-root-of-a-factorial-lim-n-to-infty-fracn – user84413 Jul 20 '13 at 21:08
3 Answers
6
Of course, there is a way using Stirling's approximation. If you wish to look at some different method:
$\lim\limits_{n\to\infty} \frac {(n!)^\frac{1}{n}}{n}=\exp \left[\lim\limits_{n\to \infty}\frac{1}{n}\sum_{k=1}^n\log \frac{k}{n}\right]=\exp\left[\int_0^1\log x dx\right]=e^{-1}$
-
+1 The solution is interesting, but the justification as to why the sum approaches the integral is different from that in usual proper Riemann integrals. – Pedro Jul 20 '13 at 21:11
-
I used $\lim\limits_{n\to \infty}\frac{1}{n}\sum\limits_{k=1}^nf(\frac{k}{n})=\int\limits_0^1f(x)dx$. And this will be true for any Riemann integrable function. Isn't it? – Kunnysan Jul 20 '13 at 21:17
-
Probably Peter meant the function $x \mapsto \log x$ is not defined at $x = 0$, so it is not Riemann integrable on $[0, 1]$, but it is Riemann integrable on $[\epsilon, 1]$ for any $\epsilon \in (0, 1)$. (That may be why the evaluation of the integral involves a not-so-trivial limit $\lim_{x \to 0^+} x\log x$.) – Tunococ Jul 20 '13 at 21:30
-
Yes, @Tunococ, I mean that $\log x$ is improperly integrable, but as I say in my answer, the result holds since $\log$ is monotone. – Pedro Jul 20 '13 at 21:31
-
Kunnysan: note your sum avoids $x=0$, but we need that any partition of $[0,1]$ includes that number. You're effectively taking a limit $\epsilon \to 0^{+}$ of $\int_\epsilon^1$ in some sense =) – Pedro Jul 20 '13 at 21:32
-
Yes, of course, there is a 'hidden' limit $\epsilon \to 0$. I thought that it would be understood by the OP and that's why I didn't mention it. – Kunnysan Jul 20 '13 at 21:35
-
The monotone (or dominated) convergence theorem should do the trick. Give up Riemann, go Lebesgue. – Julien Jul 20 '13 at 21:37
-
-
@PeterTamaroff Consider $f_n=\sum_{k=1}^nf(k/n)1_{((k-1)/n,k/n)}$. With $f(x)=-\log x$ in this case. – Julien Jul 20 '13 at 21:50
-
5
Let $a_n=\dfrac{n!}{n^n}$. This is a sequence of positive numbers. A well known theorem says that in this case, if $\ell =\lim \dfrac{a_{n+1}}{a_n}$ exists then so does $\ell'=\lim a_n^{1/n}$ and both are equal. You should be able to obtain what $\ell$ is.
One can shed some light on Kunnysan's solution.
If $f:[a,b]\to\Bbb R$ is properly integrable on $[a,b]$; define $$f_{kn}=f(a+k\delta_n)\; ;\;\delta_n=\frac{b-a}n$$
Then $$\lim\limits_{n\to\infty}\left(f_{1n}f_{2n}\cdots f_{nn}\right)^{1/n}\to \exp\left(\frac{1}{b-a}\int_a^b \log f(x) dx\right)$$
provided $\sup\limits_{[a,b]} f>0$.
In the case of $f(x)=x$, the result holds, but rather because
If $f$ is monotone on $(0,1)$ and the (improper) integral $\int_0^1 f$ exists, then $$\frac 1n \sum_{k=1}^{n-1} f\left(\frac{k}n\right)\to\int_0^1 f$$
Pedro
- 122,002
2
$$\ln \frac {(n!)^\frac{1}{n}}{n} =\frac{\ln(n!)-\ln n^n}{n}$$
By Stolz Cezaro
$$\lim_n \ln \frac {(n!)^\frac{1}{n}}{n} =\lim_n \ln((n+1)!)-\ln (n+1)^{n+1} -\ln(n!)+\ln(n^n)= \lim_n \ln \frac{(n+1)!n^n}{n!(n+1)^{n+1}}$$ $$ =\lim_n \ln \frac{n^n}{(n+1)^{n}}=-\lim_n \ln \frac{(n+1)^{n}}{n^n}=-\lim_n \ln \left( 1+\frac{1}{n}\right)^n=-1$$
N. S.
- 132,525