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; \(\mathbf{\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots\frac{2n-1}{2n}\leq\frac{1}{\sqrt{3n+1}}}\)

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