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Question Number 104092 by naka3546 last updated on 19/Jul/20

$$Provethat$$ $$\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \dots \cdot \frac{2005}{2006}\cdot \frac{2007}{2008}<\frac{1}{\sqrt{2009}}$$

Commented byJDamian last updated on 19/Jul/20

$$Iguess\frac{2007}{2009}shouldactuallybe\frac{2007}{2008}$$

Answered by 1549442205PVT last updated on 19/Jul/20

$$\mathrm{It}\mathrm{is}\mathrm{easy}\mathrm{to}\mathrm{see}\mathrm{that}\mathrm{if}\frac{\mathrm{a}}{\mathrm{b}}<1\mathrm{then}\mathrm{a}<\mathrm{b}$$ $$\Rightarrow \mathrm{ab}+\mathrm{a}<\mathrm{ab}+\mathrm{b}\Rightarrow \mathrm{a}(\mathrm{b}+1)<\mathrm{b}(\mathrm{a}+1)\Rightarrow \frac{\mathrm{a}}{\mathrm{b}}<\frac{\mathrm{a}+1}{\mathrm{b}+1}.$$ $$\mathrm{Hence},$$ $$\mathrm{Putting}\mathrm{A}=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{2005}{2006}.\frac{2007}{2009}.\mathrm{We}\mathrm{have}:$$ $$\frac{1}{2}<\frac{2}{3},\frac{3}{4}<\frac{4}{5},\frac{5}{6}<\frac{6}{7}...,\frac{2005}{2006}<\frac{2006}{2007},\frac{2007}{2009}<\frac{2009}{2010}$$ $$\mathrm{Multiplying}1004\mathrm{inequlities}\mathrm{side}\mathrm{by}$$ $$\mathrm{side}\mathrm{we}\mathrm{getA}<\frac{2}{3}.\frac{4}{5}.\frac{6}{7}....\frac{2006}{2007}.\frac{2009}{2010}=\mathrm{B}$$ $$\Rightarrow {\mathrm{A}}^2<\mathrm{AB}=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}....\frac{2005}{2006}.\frac{2006}{2007}.\frac{2007}{2009}.\frac{2009}{2010}$$ $$=\frac{1}{2010}\Rightarrow \mathrm{A}<\frac{1}{\sqrt{2010}}<\frac{1}{\sqrt{2009}}(\mathrm{q}.\mathrm{e}.\mathrm{d})$$ $$$$

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