For \mathbf{n\in\mathbb{N}} prove that \mathbf{\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots\frac{2n-1}{2n}\leq\frac{1}{\sqrt{2n+1}}}

Discussion

Note that \mathbf{ \frac{2n}{2n+1} \ge \frac{2n-1}{2n} } since simple cross multiplication gives \mathbf{ 4n^2 \ge 4n^2 - 1 } which is true for all \mathbf{n\in\mathbb{N}}

Hence \mathbf{\frac{1}{2} \le \frac{2}{3} , \frac{3}{4} \le \frac{4}{5} } etc.

Let \mathbf{ x = \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots\frac{2n-1}{2n} \implies x \le \frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdots\frac{2n}{2n+1} \implies x^2 \le \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdots\frac{2n-1}{2n}\cdot\frac{2n}{2n+1}}

All the terms cancel cross wise except 2n+1.

Thus \mathbf{ x = \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots\frac{2n-1}{2n} \le \frac{1}{2n+1} }

(proved)

Special Note

It is possible to show (by induction) a much stronger result; \mathbf{\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots\frac{2n-1}{2n}\leq\frac{1}{\sqrt{3n+1}}}