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Math Help - Telescoping Sum

  1. #1
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    Telescoping Sum

    I have a problem similar to this this thread

    http://www.mathhelpforum.com/math-he...um-series.html

    But my problem is
    infinity
    <br />
\sum,(\frac{1}{n(n+2)})<br />
    n=1

    And I know that you use partial fractions and get

    1=A(n+2)+B(n)
    or
    infinity
    <br />
\sum,(\frac{1}{2n}-\frac{1}{2n+4})<br />
    n=1

    and expanded I have
    \frac{1}{2}-\frac{1}{n+1}

    And I have to find the sum of the series but when I do I get it is equal to

    lim \sum,(\frac{1}{2}-\frac{1}{n+1})
    n->infinity

    \frac{1}{2}

    and the answer is

    \frac{3}{4}

    What am I doing wrong?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
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    New York, USA
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    Quote Originally Posted by xpack View Post
    I have a problem similar to this this thread

    http://www.mathhelpforum.com/math-he...um-series.html

    But my problem is
    infinity
    <br />
\sum,(\frac{1}{n(n+2)})<br />
    n=1

    And I know that you use partial fractions and get

    1=A(n+2)+B(n)
    or
    infinity
    <br />
\sum,(\frac{1}{2n}-\frac{1}{2n+4})<br />
    n=1

    and expanded I have
    \frac{1}{2}-\frac{1}{n+1}

    And I have to find the sum of the series but when I do I get it is equal to

    lim \sum,(\frac{1}{2}-\frac{1}{n+1})
    n->infinity

    \frac{1}{2}

    and the answer is

    \frac{3}{4}

    What am I doing wrong?
    you almost surely canceled too many terms. Check your simplification again. Yup, that's what happened. I did it and it works fine. You partial fraction decomposition is correct and everything. Though I used the form \frac 12 \left( \frac 1n - \frac 1{n + 2}  \right). Just a bit easier for the arithmetic, and I got the right answer.
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  3. #3
    Newbie
    Joined
    Sep 2009
    Posts
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    No I messed up with

    The expanded form
    \frac{1}{2}-\frac{1}{n+1}

    it was suppose to be

    \frac{1}{2}-\frac{1}{2n+4}

    so now I get

    \frac{1}{2}-\frac{1}{4}

    which is

    \frac{1}{4}

    but if I added them they would be

    \frac{3}{4}

    So i dont know why i'm getting a minus though
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
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    From
    New York, USA
    Posts
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    Quote Originally Posted by xpack View Post
    No I messed up with

    The expanded form
    \frac{1}{2}-\frac{1}{n+1}

    it was suppose to be

    \frac{1}{2}-\frac{1}{2n+4}

    so now I get

    \frac{1}{2}-\frac{1}{4}

    which is

    \frac{1}{4}

    but if I added them they would be

    \frac{3}{4}

    So i dont know why i'm getting a minus though
    having read your post the first time, I came away with your partial fraction decomp to be \frac 1{2n} - \frac 1{2n + 4}, that is the correct one. when you expand that and cancel terms, you would not end up with what you wrote. And by the way, \lim_{n \to \infty} \left( \frac 12 - \frac 1{2n + 4} \right) \ne \frac 12 - \frac 14. That is 1/2. But that term is wrong anyway, that is not the expanded form.
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