Show that $\,\,n!<\mathrm{e}\left(\frac{n}{2}\right)^n$

Question

I'd like to prove that $\,\,n!<\mathrm{e}\left(\frac{n}{2}\right)^n$.

What I have so far:

$$\sqrt[n]{n!} = \sqrt[n]{1\cdot 2 \cdot \ldots \cdot n} \leq \frac{1+\ldots +n}{n}=\frac{(n+1)n}{2n}=\frac{(n+1)}{2}.$$

Thus

$$\,\,n!<\mathrm{e}\left(\frac{n}{2}\right)^n.$$

But how do I go from $n+1$ to $n$?

Answer 1

As obtained in the OP: $$ \sqrt[n]{n!}\le \frac{n+1}{2}, $$ and hence $$ n!\le \left(\frac{n+1}{2}\right)^{\!n}=2^{-n}n^n\left(\frac{n+1}{n}\right)^{\!n}= \left(1+\frac{1}{n}\right)^{\!n}\left(\frac{n}{2}\right)^{\!n} <\mathrm{e}\left(\frac{n}{2}\right)^{\!n}, $$ since $$ \left(1+\frac{1}{n}\right)^{\!n}<\mathrm{e}, $$ which is is due to the fact that $$ 1+\frac{1}{n}<\mathrm{e}^{1/n}. $$

Answer 2

In the first line you've shown that

$$ n! \leq \left(\frac{n+1}{2}\right)^n, $$

and the expression on the right is

$$ \left(\frac{n+1}{2}\right)^n = \left(\frac{n}{2}\right)^n \left(1+\frac{1}{n}\right)^n < \left(\frac{n}{2}\right)^n e. $$

Answer 3

$n!<e\bigg(\dfrac{n}{2}\bigg)^n \implies \sqrt{2\pi n}\bigg(\dfrac{n}{e}\bigg)^n<e\bigg(\dfrac{n}{2}\bigg)^n\implies \dfrac{\sqrt{2\pi n}}{e}<\bigg(\dfrac{e}{2}\bigg)^n$

The second step is by Stirling's Approximation. I hope you can prove the last inequality, which is according to me quite trivial to prove. If you are stuck try to use induction or try to use calculus.