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Math Help - Show that the sum of the series is greater than 24

  1. #1
    Junior Member Kaloda's Avatar
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    Show that the sum of the series is greater than 24

    Show that 1/(√1+√3)+1/(√5+√7)+1/(√9+√11) . . . + 1/(√9997+√9999) > 24.

    I've no idea how to solve this. Please help. Ty.
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  2. #2
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    Re: Show that the sum of the series is greater than 24

    Hey Kaloda.

    Are you familiar with mathematical induction?
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  3. #3
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    Re: Show that the sum of the series is greater than 24

    Try taking advantage of conjugates. For example, the first term can be rewritten \frac{1}{\sqrt{3}+\sqrt{1}}\cdot \frac{\sqrt{3}-\sqrt{1}}{\sqrt{3}-\sqrt{1}} = \frac{\sqrt{3}-\sqrt{1}}{2}.

    In each case, the denominator is 2, so we can factor out a half.

    The series eventually becomes \frac{1}{2}\sum^{2499}_{k=0}[\sqrt{4k+3} - \sqrt{4k+1}].

    Does this look nicer? Good luck.
    -AB
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  4. #4
    Junior Member Kaloda's Avatar
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    Re: Show that the sum of the series is greater than 24

    @chiro:

    No. Mathematical Induction is supposed to be our next topic.
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  5. #5
    Junior Member Kaloda's Avatar
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    Re: Show that the sum of the series is greater than 24

    Quote Originally Posted by abender View Post
    Try taking advantage of conjugates. For example, the first term can be rewritten \frac{1}{\sqrt{3}+\sqrt{1}}\cdot \frac{\sqrt{3}-\sqrt{1}}{\sqrt{3}-\sqrt{1}} = \frac{\sqrt{3}-\sqrt{1}}{2}.

    In each case, the denominator is 2, so we can factor out a half.

    The series eventually becomes \frac{1}{2}\sum^{2499}_{k=0}[\sqrt{4k+3} - \sqrt{4k+1}].

    Does this look nicer? Good luck.
    -AB

    Any additional steps? I seem stucked.
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  6. #6
    Senior Member jakncoke's Avatar
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    Re: Show that the sum of the series is greater than 24

    I've been racking my head all day to do this without induction, i'm curious if anyone has an answer. I took a look at trying to do this with continued fractions but i had no idea what i was doing, if only i was ramanujan.
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  7. #7
    Junior Member Kaloda's Avatar
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    Re: Show that the sum of the series is greater than 24

    Can you show me the solution using Mathematical Induction?
    I know the steps but I don't know how to apply it in cases that utilize summations.
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  8. #8
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    Re: Show that the sum of the series is greater than 24

    Quote Originally Posted by Kaloda View Post
    Any additional steps? I seem stucked.
    Try writing out the first few terms of that sum. You'll obtain what is known as a telescoping sum.
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  9. #9
    MHF Contributor MarkFL's Avatar
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    Re: Show that the sum of the series is greater than 24

    If I were going to use induction, I would state the hypothesis:

    \frac{1}{2}\sum_{k=0}^n\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\frac{1}{2}\sqrt{n+1}-1

    or equivalently:

    \sum_{k=0}^n\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\sqrt{n+1}-2 where n\in\mathbb{N}_0

    base case P_0:

    \sum_{k=0}^0\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\sqrt{0+1}-2

    \sqrt{3}-\sqrt{1}>1-2 true.

    Consider:

    6n+9=2\left(3n+\frac{9}{2} \right)

    6n+9=2\left(2(2n+3)-\frac{2n+3}{2} \right)

    6n+9=2\left(\sqrt{4(2n+3)^2}-\sqrt{\frac{(2n+3)^2}{4}} \right)

    6n+9>2\left(\sqrt{4(2n+3)^2-1}-\sqrt{\frac{(2n+3)^2+4n+3}{4}} \right)

    6n+9>2\left(\sqrt{(4n+5)(4n+7)}-\sqrt{(n+1)(n+3)} \right)

    8n+12-2\sqrt{(4n+5)(4n+7)}>2n+3-2\sqrt{(n+1)(n+3)}

    (4n+7)-2\sqrt{(4n+5)(4n+7)}+(4n+5)>(n+2)-2\sqrt{(n+1)(n+3)}+(n+1)

    (\sqrt{4n+7}-\sqrt{4n+5})^2>(\sqrt{n+2}-\sqrt{n+1})^2

    \sqrt{4n+7}-\sqrt{4n+5}>\sqrt{n+2}-\sqrt{n+1}

    \sqrt{4(n+1)+3}-\sqrt{4(n+1)+1}>\sqrt{(n+1)+1}-\sqrt{n+1}

    Now, adding this to the hypothesis, we have:

    \sum_{k=0}^n\left(\sqrt{4k+3}-\sqrt{4k+1} \right)+\sqrt{4(n+1)+3}-\sqrt{4(n+1)+1}>\sqrt{n+1}-2+\sqrt{(n+1)+1}-\sqrt{n+1}

    \sum_{k=0}^{n+1}\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\sqrt{(n+1)+1}-2

    \frac{1}{2}\sum_{k=0}^{n+1}\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\frac{1}{2}\sqrt{(n+1)+1}-1

    We have derived P_{n+1} from P_n, thereby completing the proof by induction, and we may now state:

    \frac{1}{2}\sum_{k=0}^{2499}\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\frac{1}{2}\sqrt{2500}-1=24
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  10. #10
    Senior Member jakncoke's Avatar
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    Re: Show that the sum of the series is greater than 24

    I don't know why i didn't think of this earlier.

    But if you have  \sum_{k=0}^{4998} \frac{1}{\sqrt{2k+3} + \sqrt{2k+1}} then
    take  f(k) = \frac{1}{\sqrt{2k+3} + \sqrt{2k+1}}
    then  \int_0^{4999} f(k) dk \leq \sum_{k=0}^{4998} \frac{1}{\sqrt{2k+3} + \sqrt{2k+1}} \leq \int_0^{4998} f(k) dk + f(0)
    So after computing the integral i got  \frac{1}{6}(-(2k+1)^{3/2} + 2k\sqrt(2k+3) + 3\sqrt(2k+3)) , and after evaluating i got
    so  49.99 \leq \sum_{k=0}^{4998} \frac{1}{\sqrt{2k+3} + \sqrt{2k+1}} \leq 50.11
    Last edited by jakncoke; December 15th 2012 at 09:34 PM.
    Thanks from Kaloda
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