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Thread: 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
    $\displaystyle
    \sum,(\frac{1}{n(n+2)})
    $
    n=1

    And I know that you use partial fractions and get

    $\displaystyle 1=A(n+2)+B(n)$
    or
    infinity
    $\displaystyle
    \sum,(\frac{1}{2n}-\frac{1}{2n+4})
    $
    n=1

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

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

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

    $\displaystyle \frac{1}{2}$

    and the answer is

    $\displaystyle \frac{3}{4}$

    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
    $\displaystyle
    \sum,(\frac{1}{n(n+2)})
    $
    n=1

    And I know that you use partial fractions and get

    $\displaystyle 1=A(n+2)+B(n)$
    or
    infinity
    $\displaystyle
    \sum,(\frac{1}{2n}-\frac{1}{2n+4})
    $
    n=1

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

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

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

    $\displaystyle \frac{1}{2}$

    and the answer is

    $\displaystyle \frac{3}{4}$

    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 $\displaystyle \frac 12 \left( \frac 1n - \frac 1{n + 2} \right)$. 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
    $\displaystyle \frac{1}{2}-\frac{1}{n+1}$

    it was suppose to be

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

    so now I get

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

    which is

    $\displaystyle \frac{1}{4}$

    but if I added them they would be

    $\displaystyle \frac{3}{4}$

    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
    Feb 2007
    From
    New York, USA
    Posts
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    5
    Quote Originally Posted by xpack View Post
    No I messed up with

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

    it was suppose to be

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

    so now I get

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

    which is

    $\displaystyle \frac{1}{4}$

    but if I added them they would be

    $\displaystyle \frac{3}{4}$

    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 $\displaystyle \frac 1{2n} - \frac 1{2n + 4}$, 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, $\displaystyle \lim_{n \to \infty} \left( \frac 12 - \frac 1{2n + 4} \right) \ne \frac 12 - \frac 14$. That is 1/2. But that term is wrong anyway, that is not the expanded form.
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