Prove that:

$\displaystyle \

\frac{1}{2}.\frac{3}{4}...\frac{{2n - 1}}{{2n}} < \frac{1}{{\sqrt {2n + 1} }},n > 1

\

$

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- Aug 12th 2008, 05:02 AMOrtegaProve
Prove that:

$\displaystyle \

\frac{1}{2}.\frac{3}{4}...\frac{{2n - 1}}{{2n}} < \frac{1}{{\sqrt {2n + 1} }},n > 1

\

$ - Aug 12th 2008, 05:52 AMCaptainBlack
- Aug 12th 2008, 05:54 AMOrtega
- Aug 12th 2008, 09:07 AMCaptainBlack
- Aug 12th 2008, 09:33 AMJaneBennet
$\displaystyle \begin{array}{rcl}1\cdot3=(2-1)(2+1)=2^2-1&<&2^2\\

3\cdot5=(4-1)(4+1)=4^2-1&<&4^2\\

5\cdot7=(6-1)(6+1)=6^2-1&<&6^2\\

{}&\vdots&{}\\

(2n-3)(2n-1)&<&(2n-2)^2\\

(2n-1)(2n+1)&<&(2n)^2\end{array}$

$\displaystyle \therefore\ 1\cdot3^2\cdot5^2\cdots(2n-1)^2(2n+1)\ <\ 2^2\cdot4^2\cdot6^2\cdots(2n)^2$

$\displaystyle \Rightarrow\ 1\cdot3\cdot5\cdots(2n-1)\sqrt{2n+1}\ <\ 2\cdot4\cdot6\cdots2n$

$\displaystyle \Rightarrow\ \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\ <\ \frac{1}{\sqrt{2n+1}}$ - Aug 12th 2008, 10:41 AMOrtega
Thks Jane,this is what i look for.i saw the solution for the problem into the book,and they solve it complicated,frankly i didnt understand anything of their result.Your solution is very good and simple to understand it.

BRAVO.

*i hope that you understand me.