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Math Help - Series- Real Analysis

  1. #1
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    Question Series- Real Analysis

    Hi

    I was wondering if we would use the Ratio Test on this series? If not any ideas?

    Sigma: 2.4.6.....(2n) divided by 3.5.7.....(2n+1)
    Last edited by hebby; February 16th 2010 at 08:19 PM.
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  2. #2
    Member mabruka's Avatar
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    Sorry but i dont think anyone will understand that. use /frac{}{} and \sum between tags to make it clearer.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by hebby View Post
    Hi

    I was wondering if we would use the Ratio Test on this series? If not any ideas?

    Sigma: 2.4.6.....(2n) divided by 3.5.7.....(2n+1)
    a_{n}=\frac{2\cdots(2n)}{3\cdots(2n+1)} and so \frac{a_{n+1}}{a_n}=\frac{2\cdots(2n)\cdot(2n+2)}{  3\cdots\cdot(2n+1)\cdot(2n+3)}\cdot\frac{3\cdots(2  n+1)}{2\cdots(2n)}=\frac{2n+2}{2n+3}. So the ratio test doesn't work. But...try showing that it's always greater than something....any guess what?
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  4. #4
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    The Raabe Test, could be used here and also we can compare it with a harmonic series 1/2n+1, which diverges and thus the given series diverges?
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  5. #5
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    Use the inequality

    \frac{1}{n+1} < \frac{2\cdots(2n)}{3\cdots(2n+1)}

    then do direct comparison with the harmonic series.

    Edit: you can prove the inequality by induction on n.

    \frac{1}{2} < \frac{2}{3} proves it true for n=1.

    Now assume it is true for k \leq n.

    Then

    \frac{1}{n+1} < \frac{2\cdots(2n)}{3\cdots(2n+1)}

    \frac{1}{n+1} \cdot \frac{2n+2}{2n+3} < \frac{2\cdots(2n)}{3\cdots(2n+1)} \cdot \frac{2n+2}{2n+3}

    \frac{1}{n+1} \cdot \frac{n+1}{n+2} < \frac{1}{n+1} \cdot \frac{2n+2}{2n+3} <  \frac{2\cdots(2n)}{3\cdots(2n+1)} \cdot \frac{2n+2}{2n+3}

    \frac{1}{n+2} < \frac{2\cdots(2n+2)}{3\cdots(2n+3)}

    and the induction is complete.
    Last edited by icemanfan; February 16th 2010 at 08:48 PM.
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  6. #6
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    can we also use the Raabe test?...an+1/an => 1-a/n
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  7. #7
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    Quote Originally Posted by hebby View Post
    can we also use the Raabe test?...an+1/an => 1-a/n
    I am unfamiliar with that test.
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by hebby View Post
    can we also use the Raabe test?...an+1/an => 1-a/n
    Yes.
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  9. #9
    Member mabruka's Avatar
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    Wow how could you you figure that out!
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