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?

Answer 1

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 $$\frac{\sqrt[n]{n!} }{n} =\sqrt[n]{v_n} \to\frac{1}{e} .$$

Answer 2

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.