Upper bound for $\sqrt[n]{n!}$

Question

It is straight forward to show that $\sqrt[n]{n!} < n$ from:

• $2 > \sqrt{2}$

• $\frac{n+1}{n} > \frac{\sqrt[n+1]{n+1}}{\sqrt[n]{n}}$ since $\frac{n+1}{\sqrt[n+1]{n+1}} > \frac{n}{\sqrt[n]{n}}$

Is this the best that can be done? Is it possible to find a better upper bound for $\sqrt[n]{n!}$

Answer 1

by using $AM-GM$ we have $$\frac{1+2+3+...+n}{n}\geq \sqrt[n]{n!}$$ using the well-known formula for the sum of the first $n$ natural numbers we get $$\frac{n(n+1)}{2n}\geq \sqrt[n]{n!}$$ thus we get $$\sqrt[n]{n!}\le \frac{n+1}{2}$$

Answer 2

We find an inequality using Stirling's approximation at mathworld.wolfram.com:

$$n!<\sqrt{2\pi}n^{n+1/2}e^{-n+1/(12n)}$$

Which means

$$\sqrt[n]{n!}<\sqrt[n]{\sqrt{2\pi}n^{n+1/2}e^{-n+1/(12n)}}$$

Which is

$$\sqrt[n]{n!}<\frac{n}{e}\sqrt[n]{\sqrt{2\pi n}\cdot e^{1/(12n)}}$$

This is not as pretty as Dr. Sonnhard Graubner's inequality; it is however a lot better at approximating $\sqrt[n]{n!}$.