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I would like to show that $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}$ holds for all natural numbers. I got stuck here:

$\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\cdot\frac{2n+1}{2n+2}\leq \frac{1}{\sqrt{3n+1}}\frac{2n+1}{2n+2}.$

I would appreciate your help.

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    Related: http://math.stackexchange.com/questions/1899857/mathematical-problem-induction-frac12-cdot-frac34-cdots-frac2n-12n-frac/.2017-02-09

1 Answers 1

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You want to show that $ \frac1{\sqrt{3n+1}}\;\frac{2n+1}{2n+2}\leq\;\frac1{\sqrt{3n+4}}. $ Since everything is positive, this inequality is the same as $ (3n+4)(2n+1)^2\leq (3n+1)(2n+2)^2. $ After expanding and cancelling the $n^3$ terms we get $ 12n^2+19n\leq 24n^2+20n. $ This inequality holds trivially for any $n\in\mathbb{N}$, and now you can retrace the steps back to your desired inequality.