factorial inequality with $(\frac{n}{2})^n$

Question

I came across a problem with proving an equality:

$n! < e(\frac{n}{2})^n$

I tried to simplify the RHS using the limit of $(1+\frac{1}{n})^n$ as $n$ goes to infinity (to use with $sandwich$), but to no avail. I also thought about using induction, but couldn't prove the inductive step.

Answer 1

Since $\left(1+\frac1n\right)^{n+1}$ is decreasing, we have $$ \begin{align} \frac{2^nn!/n^n}{2^{n+1}(n+1)!/(n+1)^{n+1}} &=\frac12\left(1+\frac1n\right)^n\\ &\ge\frac{en/2}{n+1} \end{align} $$ Therefore, $$ \begin{align} \frac{n!}{n^n} &\le\frac{n}{e^{n-1}}\\ &\le\frac{e}{2^n} \end{align} $$ where the last inequality is valid for $n\ge6$. We can verify $0\le n\lt6$ manually.

Answer 2

Induction works easily. It is true for $n=1$. The induction step goes as follows:

$(n+1)! = (n+1)n! < (n+1)e(\frac{n}{2})^n$ which is less than $e(\frac{n+1}{2})^{n+1}$ if and only if $2n^n < (n+1)^n \iff 2 < (1+\frac{1}{n})^n$ which is true by Binomial Theorem.