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

Prove that (1/2) ∙ (3/4) ∙ (5/6) ∙ …∙ ((2005)/(2006)) ∙ ((2007)/(2008)) < (1/(√(2009)))

$${Prove}\:\:{that}\:\: \\ $$ $$\frac{\mathrm{1}}{\mathrm{2}}\:\centerdot\:\frac{\mathrm{3}}{\mathrm{4}}\:\centerdot\:\frac{\mathrm{5}}{\mathrm{6}}\:\centerdot\:\ldots\centerdot\:\frac{\mathrm{2005}}{\mathrm{2006}}\:\centerdot\:\frac{\mathrm{2007}}{\mathrm{2008}}\:\:<\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{2009}}} \\ $$

Commented byJDamian last updated on 19/Jul/20

I guess ((2007)/(2009)) should actually be ((2007)/(2008))

$${I}\:{guess}\:\frac{\mathrm{2007}}{\mathrm{2009}}\:{should}\:{actually}\:{be}\:\frac{\mathrm{2007}}{\mathrm{2008}} \\ $$

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

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

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

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