Show that $n! < (n/2)^n$ for all large enough $n$ in as elementary a way as possible

Question

Show that $n! < (n/2)^n$ for large enough $n$ in as elementary a way as possible. Using Stirling's formula is not allowed.

Of, course, what is true, is that $n! < (n/c)^n$ for any $c < e$ for large enough $n$. But this is simpler. For extra credit, determine explicitly $n(c)$ such that $n! < (n/c)^n$ for $n > n(c)$.

My answer:

Suppose that $n! < (n/2)^n$. Then $(n+1)! =n!(n+1) <(n/2)^n(n+1) $. So, we want $(n/2)^n(n+1) <((n+1)/2)^{n+1} $ or $2n^n(n+1) <(n+1)^{n+1} $ or $(1+1/n)^n > 2 $.

By Bernoulli's inequality ($(1+x)^n > 1+nx$ for $n \ge 2$ and $x > 0$), $(1+1/n)^n > 1+n/n =2 $ for $n \ge 2$.

Therefore the induction step will work for $n \ge 2$. We therefore have to find an $n$ for which $n! < (n/2)^n$.

The following table shows that this is true for $n=6$. Therefore $n! < (n/2)^n$ for $n \ge 6$.

$\begin{array}{lll} n & n! & (n/2)^n\\ 2 & 2 & 1\\ 4 & 24 & 16\\ 6 & 720 & 729\\ \end{array} $

Answer 1

My favorite way is to exploit: $$ n = \prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right) $$ in order to have: $$ n! = \prod_{m=2}^{n}\prod_{k=1}^{m-1}\left(1+\frac{1}{k}\right)=\frac{n^n}{\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k}=\frac{n^{n+1}}{\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^{k+1}}\leq\frac{n^{n+1}}{e^{n-1}}$$ so we just need to prove that for any $n$ big enough: $$\frac{en}{e^{n}}\leq\frac{1}{2^n}$$ that is quite trivial since $e>2$.

Answer 2

Consider sequence $a_n=\frac{n!}{(n/2)^n}=\frac{n!2^n}{n^n}$. Now we have $\frac{a_{n+1}}{a_n}=\frac{(n+1)!2^{n+1}}{(n+1)^{n+1}}\frac{n^n}{n!2^n}=\frac{2\cdot n!2^n}{(n+1)^n}\frac{n^n}{n!2^n}=\frac{2}{(\frac{n+1}{n})^n}=\frac{2}{(1+\frac{1}{n})^n}\rightarrow \frac{2}{e}<1$, from which it follows that $a_n\rightarrow 0$ and thus is less than $1$ from some point on.