Prove-that-1-2-3-4-5-6-2005-2006-2007-2008-lt-1-2009-

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

P r o v e t h a t

\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 by JDamian last updated on 19/Jul/20

I g u e s s \frac{2007}{2009} s h o u l d a c t u a l l y b e \frac{2007}{2008}

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

It is easy to see that if \frac{a}{b} < 1 then a < b

\Rightarrow ab + a < ab + b \Rightarrow a (b + 1) < b (a + 1) \Rightarrow \frac{a}{b} < \frac{a + 1}{b + 1} .

Hence,

Putting A = \frac{1}{2} . \frac{3}{4} . \frac{5}{6} \dots . \frac{2005}{2006} . \frac{2007}{2009} . We have :

\frac{1}{2} < \frac{2}{3}, \frac{3}{4} < \frac{4}{5}, \frac{5}{6} < \frac{6}{7} \dots, \frac{2005}{2006} < \frac{2006}{2007}, \frac{2007}{2009} < \frac{2009}{2010}

Multiplying 1004 inequlities side by

side we getA < \frac{2}{3} . \frac{4}{5} . \frac{6}{7} \dots . \frac{2006}{2007} . \frac{2009}{2010} = B

\Rightarrow A^{2} < AB = \frac{1}{2} . \frac{2}{3} . \frac{3}{4} . \frac{4}{5} . \frac{5}{6} \dots . \frac{2005}{2006} . \frac{2006}{2007} . \frac{2007}{2009} . \frac{2009}{2010}

= \frac{1}{2010} \Rightarrow A < \frac{1}{\sqrt{2010}} < \frac{1}{\sqrt{2009}} (q . e . d)