Ta có:
$\dfrac{C_{2n}^n}{2^{2n}}=\dfrac{(2n)!}{2^{2n}n!n!}=\dfrac{1.3.5.\ldots.(2n-1)}{2.4.6.\ldots.(2n)}=\prod_{i=1}^n\dfrac{2i-1}{2i}$
$\dfrac{2i-1}{2i}<\dfrac{2i-1}{\sqrt{4i^2-1}}=\dfrac{\sqrt{2i-1}}{\sqrt{2i+1}}$
Từ đó, suy ra:
$\dfrac{C_{2n}^n}{2^{2n}}<\prod_{i=1}^n\dfrac{\sqrt{2i-1}}{\sqrt{2i+1}}=\dfrac{1}{\sqrt{2n+1}}$