TITLE: Show that $\lim\limits_{n \to \infty} \frac{(n!)^{1/n}}{n}= \frac{1}{e}$
QUESTION [8 upvotes]: 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?

REPLY [4 votes]: 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.