Com10atorics 07.06.2021 20:48 Prove that for any integer $n\geq 2$ it holds that $\dbinom {2n}{n}>\frac {4^n}{2n}$.
Tintarn 07.06.2021 20:51 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.
Ru83n05 19.08.2021 03:16 Notice that $$\binom{2n}{n}=\sum_{i=1}^n \binom{n}{i}^2\geq \frac{(\sum_{i=1}^n \binom{n}{i})^2}{n}=\frac{4^n}{n}>\frac{4^n}{2n}$$