Com10atorics wrote: Prove that for any integer $n\geq 2$ it holds that $\dbinom {2n}{n}>\frac {4^n}{2n}$. Posted before. See here. Please use the search function.

Notice that $$\binom{2n}{n}=\sum_{i=1}^n \binom{n}{i}^2\geq \frac{(\sum_{i=1}^n \binom{n}{i})^2}{n}=\frac{4^n}{n}>\frac{4^n}{2n}$$