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Math Help - Prove

  1. #1
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    Prove

    Prove that:
    \<br />
\frac{1}{2}.\frac{3}{4}...\frac{{2n - 1}}{{2n}} < \frac{1}{{\sqrt {2n + 1} }},n > 1<br />
\<br />
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Ortega View Post
    Prove that:
    \<br />
\frac{1}{2}.\frac{3}{4}...\frac{{2n - 1}}{{2n}} < \frac{1}{{\sqrt {2n + 1} }},n > 1<br />
\<br />
    use induction.

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    use induction.

    RonL
    yes i know to solve this problem using induction but i try another method,
    another method without induction.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Ortega View Post
    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
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  5. #5
    Senior Member JaneBennet's Avatar
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    \begin{array}{rcl}1\cdot3=(2-1)(2+1)=2^2-1&<&2^2\\<br />
3\cdot5=(4-1)(4+1)=4^2-1&<&4^2\\<br />
5\cdot7=(6-1)(6+1)=6^2-1&<&6^2\\<br />
{}&\vdots&{}\\<br />
(2n-3)(2n-1)&<&(2n-2)^2\\<br />
(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}}
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  6. #6
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    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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