Prove central binomial coefficient upper bound I am trying to prove that $\binom{2n}{n} < \frac{4^n}{\sqrt{2n}}$. I tried induction, but with no effect (all I can get to is $(2n+1)(2n+2) < 4\sqrt{n(n+1)}$ which is false)

$ n^n e^{-n}\sqrt{2\pi n} < n! < n^n e^{-n} \sqrt{2\pi n} (1+\frac{1}{8n}) $ this is Stirling approximation
So $\binom{2n}{n}  \leq \frac{(2n)^{2n} e^{-2n} \sqrt{4\pi n} (1+\frac{1}{16n})}{(n^n e^{-n} \sqrt{2\pi n})^2} = \frac{4^n (n)^{2n} e^{-2n} \sqrt{4\pi n} (1+\frac{1}{16n})}{n^{2n} e^{-2n}* 2\pi n} = \frac{4^n (1+\frac{1}{16n})}{\sqrt{\pi n}} \leq \frac{4^n}{\sqrt{2n}}$.
