Show that $\lim\limits_{n \to \infty} \frac{(n!)^{1/n}}{n}= \frac{1}{e}$

Question

Show that $$\lim_{n \to \infty} \left\{\frac{(n!)^{1/n}}{n}\right\} = \frac{1}{e}$$

What I did is to let $U_n = \dfrac{(n!)^{\frac{1}{n}}}{n}$ and $U_{n+1} = \dfrac{(n+1)!^{\frac{1}{n+1}}}{n+1}$. Then

$$\frac{ U_{n+1} }{U_n } = \frac{\frac{(n+1)!^{\frac{1}{n+1}}}{n+1}}{\frac{(n!)^{\frac{1}{n}}}{n}}$$

Next I just got stuck. Am I on the right track, or am I wrong doing this type of sequence?

One option is to use Stirling's Approximation and its (rather tight) error estimates. http://en.wikipedia.org/wiki/Stirling%27s_approximation — Travis Willse, Jun 04 '15 at 06:58
@Travis The basest version of Stirling (with no error control) yields the result. — Did, Jun 04 '15 at 19:21
@Did Surely if one uses an approximation to evaluate a limit one must use error bounds for that approximation to justify that the computation, no? — Travis Willse, Jun 05 '15 at 02:35
@Travis Actually no, due to the power $1/n$. Note that $n!\sim\sqrt{2\pi n}(n/e)^n$ implies $(n!)^{1/n}\sim n/e$, that is, $(n!)^{1/n}/n\to e$. — Did, Jun 05 '15 at 05:06
@Travis Again, no, since $x_n\sim y_n$ (when $y_n\ne0$) means that $x_n/y_n\to1$, with no error bound whatsoever. — Did, Jun 05 '15 at 06:35

score 17 · Accepted Answer · answered Jun 04 '15 at 07:04

17

Let $v_n = \frac{n!}{n^n } $ then $$ \frac{v_{n+1}}{v_n } =\frac{(n+1)! }{(n+1)^{n+1}} \cdot \frac{n^n }{n!} =\frac{n^n}{(n+1)^n }=\frac{1}{\left(1+\frac{1}{n}\right)^n}\to \frac{1}{e}$$ hence $$\frac{\sqrt[n]{n!} }{n} =\sqrt[n]{v_n} \to\frac{1}{e} .$$

answered Jun 04 '15 at 07:04

I approve of this. +1 – Chinny84 Jun 04 '15 at 07:05
how does (n+1)! and n! vanish.? – Sam Christopher Jun 04 '15 at 07:09
@sam $$(n+1)n! = (n+1)!$$ – Chinny84 Jun 04 '15 at 07:12
Yes..there should be (n+1) in the numerator I suppose....!!!when applying infinity it should go to infinit..Am i right? – Sam Christopher Jun 04 '15 at 07:18
@sam but the $(n+1)$ in the numerator reduces the power by one in the denominator(which yeilds his last equality before taking the lim). It is quite a nice trick which my old eyes didn't see when I first saw this question. – Chinny84 Jun 04 '15 at 07:44
Yep..You are right.. – Sam Christopher Jun 04 '15 at 08:05

Claude Leibovici · Answer 2 · 2015-06-04T10:49:40.527

4

As Travis suggested, consider $$A=\frac{(n!)^{1/n}}{n}$$ Now, use Stirling approximation $$n!= \sqrt{2\pi n}\,\Big(\frac n e\Big)^n \Big(1+\frac 1 {12n}+\cdots\Big)$$ So $$A=(2\pi n)^{\frac 1 {2n}}\, \Big(1+\frac 1 {12n}+\cdots\Big)^{\frac 1 n} \frac 1e$$ The first two terms tends to $1$ when $n$ becomes large and then the limit.

Pushing the developments, you could arrive, for large values of $n$, to $$A\approx \frac{1}{e}+\frac{\log (2 \pi n)}{2\, e \, n}+\cdots$$

If you do not use to practice Stirling approximation, try to remember it; it is extremely useful every time you have to find the limit or the asymptotics of any function containing factorials.

edited Jun 04 '15 at 10:49

answered Jun 04 '15 at 08:27

Claude Leibovici

260,315

Several times now, you posted an answer based on some very precise expansions. When the last terms of the expansion are unneeded to reach the result asked for by the OP, the fact should be mentioned. Here the $1/(12n)$ is not necessary, only the equivalent is (and actually even the $\sqrt{2\pi n}$ prefactor in Stirling is unnecessary, but this becomes more subtle). – Did Jun 05 '15 at 07:20

Show that $\lim\limits_{n \to \infty} \frac{(n!)^{1/n}}{n}= \frac{1}{e}$

2 Answers2