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# Inequality of a product expression (I.S.I. B.Math 2011 Subjective Problem 3)

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

May 8, 2014