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prove the inequality if you can: $\frac{1}{2}\cdot\frac{2}{3}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}$

Thanks.

Martin Sleziak
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user145717
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2 Answers2

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I assume that $n$ is a integer positive number. Then:

$$\frac{1}{2}\cdot\frac{2}{3}\cdots\frac{2n-1}{2n} = \frac{(2n-1)!}{(2n)!} = \frac{1}{2n} < \frac{1}{\sqrt{2n+1}}$$

$$\sqrt{2n+1}<2n$$ $$2n+1 < 4n^2$$ $$4n^2-2n-1>0$$

Real solutions of $4n^2-2n-1=0$ are $n=\frac{1 \pm \sqrt{5}}{4}$ ($n=0.809...$, $n=-0.309...$). This parabola is always positive for every number greater than $n=0.809$ and hence it is satisfied for every integer $n\geq1$.

the_candyman
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    Another way to see that the inequality always hold is to note that $2n+1 < 4n^2$ for $n=1$, and that the right-hand-side increases much faster than the left-hand side. – Fredrik Meyer May 01 '14 at 07:57
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$$\frac{1}{2}<\frac{2}{3},$$ $$\frac{3}{4}<\frac{4}{5},$$ $$.........................$$ $$\frac{2n-1}{2n}<\frac{2n}{2n+1},$$ $$\frac{1}{2}\cdot\frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{2}{3}\cdot\frac{4}{5}\cdots\frac{2n}{2n+1}/\cdot \left(\frac{1}{2}\cdot\frac{3}{4}\cdots\frac{2n-1}{2n}\right)$$ $$\left(\frac{1}{2}\cdot\frac{3}{4}\cdots\frac{2n-1}{2n}\right)^2<\frac{1}{2}\cdot\frac{3}{4}\cdots\frac{2n-1}{2n}\cdot \frac{2}{3}\cdot\frac{4}{5}\cdots\frac{2n}{2n+1}$$ $$\left(\frac{1}{2}\cdot\frac{3}{4}\cdots\frac{2n-1}{2n}\right)^2<\frac{1}{2n+1}$$ $$\frac{1}{2}\cdot\frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}$$

bof
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Madrit Zhaku
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    The factor $\frac23$ on the left is a typo for $\frac34$, isn't it? The left side of the inequality is supposed to be $\Pi_{k=1}^n\frac{2k-1}{2k}$, I guess? – bof May 01 '14 at 08:20
  • for me as I do not understand very well inequalities seems understandable with the above solution, thanks sir – user145717 May 01 '14 at 17:40