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.


marked as duplicate by Start wearing purple, Norbert, David Mitra, Nick Peterson, Stefan Hamcke Jul 20 '13 at 21:28

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  • 1
    $\begingroup$ Do you know Stirling's approximation? $\endgroup$ – Daniel Fischer Jul 20 '13 at 20:59
  • 1
    $\begingroup$ See this. $\endgroup$ – David Mitra Jul 20 '13 at 21:00
  • $\begingroup$ @DanielFischer : I know it, but I'm not supposed to use it for this. $\endgroup$ – Shookie Jul 20 '13 at 21:03
  • $\begingroup$ You might want to look at this question:math.stackexchange.com/questions/171904/… $\endgroup$ – user84413 Jul 20 '13 at 21:08

Of course, there is a way using Stirling's approximation. If you wish to look at some different method:

$\lim\limits_{n\to\infty} \frac {(n!)^\frac{1}{n}}{n}=\exp \left[\lim\limits_{n\to \infty}\frac{1}{n}\sum_{k=1}^n\log \frac{k}{n}\right]=\exp\left[\int_0^1\log x dx\right]=e^{-1}$

  • $\begingroup$ +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. $\endgroup$ – Pedro Tamaroff Jul 20 '13 at 21:11
  • $\begingroup$ 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? $\endgroup$ – Kunnysan Jul 20 '13 at 21:17
  • $\begingroup$ 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 $\lim_{x \to 0^+} x\log x$.) $\endgroup$ – Tunococ Jul 20 '13 at 21:30
  • $\begingroup$ Yes, @Tunococ, I mean that $\log x$ is improperly integrable, but as I say in my answer, the result holds since $\log$ is monotone. $\endgroup$ – Pedro Tamaroff Jul 20 '13 at 21:31
  • $\begingroup$ 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 =) $\endgroup$ – Pedro Tamaroff Jul 20 '13 at 21:32

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 $$\frac 1n \sum_{k=1}^{n-1} f\left(\frac{k}n\right)\to\int_0^1 f$$


$$\ln \frac {(n!)^\frac{1}{n}}{n} =\frac{\ln(n!)-\ln n^n}{n}$$

By Stolz Cezaro

$$\lim_n \ln \frac {(n!)^\frac{1}{n}}{n} =\lim_n \ln((n+1)!)-\ln (n+1)^{n+1} -\ln(n!)+\ln(n^n)= \lim_n \ln \frac{(n+1)!n^n}{n!(n+1)^{n+1}}$$ $$ =\lim_n \ln \frac{n^n}{(n+1)^{n}}=-\lim_n \ln \frac{(n+1)^{n}}{n^n}=-\lim_n \ln \left( 1+\frac{1}{n}\right)^n=-1$$


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