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Math Help - summation problem; find closed-form

  1. #1
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    summation problem; find closed-form

    Hello all:

    I seen this summation problem and was wondering what insight the learned on this site have:

    Given \frac{2-r}{r(r+1)(r+2)}=\frac{2}{r+2}-\frac{3}{r+1}+\frac{1}{r}, show that \sum_{r=1}^{n}\frac{2-r}{r(r+1)(r+2)}=\frac{2}{n+2}-\frac{1}{n+1}

    I ran this through Maple and got \frac{2}{n+2}-\frac{1}{n+1}

    but how is that derived?. I tried partial sums but hit a wall. It may be easier than I am making it.
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    MHF Contributor red_dog's Avatar
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    \displaystyle \sum_{r=1}^n\frac{2-r}{r(r+1)(r+2)}=\sum_{r=1}^n\left(\frac{2}{r+2}-\frac{3}{r+1}+\frac{1}{r}\right)=
    \displaystyle =\frac{2}{3}-\frac{3}{2}+\frac{1}{1}+
    \displaystyle +\frac{2}{4}-\frac{3}{3}+\frac{1}{2}+
    \displaystyle +\frac{2}{5}-\frac{3}{4}+\frac{1}{3}+
    .................................................. ........................
    \displaystyle +\frac{2}{n}-\frac{3}{n-1}+\frac{1}{n-2}+
    \displaystyle +\frac{2}{n+1}-\frac{3}{n}+\frac{1}{n-1}+
    \displaystyle +\frac{2}{n+2}-\frac{3}{n+1}+\frac{1}{n}=
    \displaystyle =\frac{2}{n+1}+\frac{2}{n+2}-\frac{3}{n+1}-\frac{3}{2}+1+\frac{1}{2}=
    \displaystyle =\frac{2}{n+2}-\frac{1}{n+1}-1+1=
    \displaystyle =\frac{2}{n+2}-\frac{1}{n+1}
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  3. #3
    Eater of Worlds
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    You're a good egg, Red Dog. I was on the same track as you but went off ona tangent. I see.
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    i ham a good egg
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by galactus View Post
    \frac{2-r}{r(r+1)(r+2)}=\frac{2}{r+2}-\frac{3}{r+1}+\frac{1}{r}, show that \sum_{r=1}^{n}\frac{2-r}{r(r+1)(r+2)}=\frac{2}{n+2}-\frac{1}{n+1}
    We have that
    \sum_{r = 1}^n \frac{1}{r + 1} = \sum_{r = 1}^{n} \frac{1}{r} - 1  + \frac{1}{n + 1}

    and
    \sum_{r = 1}^n \frac{1}{r + 2} = \sum_{r = 1}^{n} \frac{1}{r} - 1 - \frac{1}{2} + \frac{1}{n + 1} + \frac{1}{n + 2}

    So....
    \sum_{r=1}^{n}\frac{2-r}{r(r+1)(r+2)}

    = \sum_{r=1}^{n} \left ( <br />
\frac{2}{r+2}-\frac{3}{r+1}+\frac{1}{r} \right )

    = 2 \sum_{r=1}^{n} \frac{1}{r+2} - 3 \sum_{r=1}^{n} \frac{1}{r+1} + \sum_{r=1}^{n} <br />
\frac{1}{r}

    = 2 \left ( \sum_{r = 1}^{n} \frac{1}{r} - 1 - \frac{1}{2} + \frac{1}{n + 1} + \frac{1}{n + 2} \right ) - 3 \left ( \sum_{r = 1}^{n} \frac{1}{r} - 1  + \frac{1}{n + 1} \right ) + \sum_{r=1}^{n} \frac{1}{r}

    = 2 \sum_{r = 1}^{n} \frac{1}{r} - 2 - 1 + \frac{2}{n + 1} + \frac{2}{n + 2} - 3 \sum_{r = 1}^{n} \frac{1}{r} + 3 - \frac{3}{n + 1} + \sum_{r=1}^{n} \frac{1}{r}

    Most of the terms cancel and we are left with
    \sum_{r=1}^{n}\frac{2-r}{r(r+1)(r+2)} =  \frac{2}{n + 2} - \frac{1}{n + 1}

    -Dan
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by galactus View Post
    Hello all:

    I seen this summation problem and was wondering what insight the learned on this site have:

    Given \frac{2-r}{r(r+1)(r+2)}=\frac{2}{r+2}-\frac{3}{r+1}+\frac{1}{r}, show that \sum_{r=1}^{n}\frac{2-r}{r(r+1)(r+2)}=\frac{2}{n+2}-\frac{1}{n+1}

    I ran this through Maple and got \frac{2}{n+2}-\frac{1}{n+1}

    but how is that derived?. I tried partial sums but hit a wall. It may be easier than I am making it.
    I have a nice method that gives the sum in terms of hypergeometric functions

    RonL
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    Quote Originally Posted by CaptainBlank View Post
    I have a nice method that gives the sum in terms of hypergeometric functions
    That does sound cool!

    My question is where are these "hypergeometric series" used? I keep on hearing and hearing their names but not once I used them.
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  8. #8
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    Yes, CB, would you share them with us. I always like learning something new with math. The thing about math, one can always learn something new.
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  9. #9
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    Quote Originally Posted by galactus View Post
    The thing about math, one can always learn something new.
    Except for Gauss. I am sure I mentioned this before. One book I had said Gauss was the last mathematician to know everything.

    It might seem that is a bad thing, but I would rather know everything (I plan to in a few years) rather than learn new things.
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  10. #10
    Grand Panjandrum
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    Define the functions:

    <br />
f_{k,n}(x)=\sum_{r=1}^n \frac{x^{r+k}}{r+k}<br />

    Then the sum in question can be written in terms of f_{0,n}(1), f_{1,n}(1), and f_{2,n}(1).

    But:

    <br />
f'_{k,n}(x)=\sum_{r=1}^n x^{r+k-1}=\frac{x^k(1-x^n)}{1-x}<br />
    so:

    <br />
f_{k,n}(x)=\int_0^x \frac{\zeta^k(1-\zeta^n)}{1-\zeta} d\zeta<br />

    Which can be written in terms of the hypergeometric functions, but my copy of Abramowitz and Stegun is at work
    so I can't check which bits of the formal answer that I have in my note book are valid for the values of k and x=1 that we need, so that is my lot for the weekend at least.

    RonL
    Last edited by CaptainBlack; June 29th 2007 at 03:06 PM.
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