Prove

**Ortega** · August 12th 2008, 05:02 AM

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

**CaptainBlack** · August 12th 2008, 05:52 AM

Originally Posted by Ortega

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

use induction.

RonL

**Ortega** · August 12th 2008, 05:54 AM

Originally Posted by CaptainBlack

use induction.

RonL

yes i know to solve this problem using induction but i try another method,
another method without induction.

**CaptainBlack** · August 12th 2008, 09:07 AM

Originally Posted by Ortega

yes i know to solve this problem using induction but i try another method,
another method without induction.

Why do you not want to use induction on this. Also if it is a requirement on the problem, perhaps you should have mentioned it in the original post.

RonL

**JaneBennet** · August 12th 2008, 09:33 AM

$\begin{array}{rcl}1\cdot3=(2-1)(2+1)=2^2-1&<&2^2\\ 3\cdot5=(4-1)(4+1)=4^2-1&<&4^2\\ 5\cdot7=(6-1)(6+1)=6^2-1&<&6^2\\ {}&\vdots&{}\\ (2n-3)(2n-1)&<&(2n-2)^2\\ (2n-1)(2n+1)&<&(2n)^2\end{array}$

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

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

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

**Ortega** · August 12th 2008, 10:41 AM

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.

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