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Math Help - Sum of series

  1. #1
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    Sum of series

    By expressing the rth term of the series \sum^n_{r=1} \frac{(-1)^{r+1}(2r+1)}{r(r+1)} in partial fractions, or otherwise, find the sum of the series.
    If this sum is denoted by S_n prove that for all odd values of n, S_nS_{n+1}=1

    I found the partial fractions to be:
    \sum^n_{n=1}(\frac{-1}{r}+\frac{1}{r+1})
    and the sum of the series:
    \frac{1}{n+1}-1
    but the sum is given as 1+\frac{(-1)^{n+1}}{n+1}
    Thanks!
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by arze View Post
    By expressing the rth term of the series \sum^n_{r=1} \frac{(-1)^{r+1}(2r+1)}{r(r+1)} in partial fractions, or otherwise, find the sum of the series.
    If this sum is denoted by S_n prove that for all odd values of n, S_nS_{n+1}=1

    I found the partial fractions to be:
    \sum^n_{n=1}(\frac{-1}{r}+\frac{1}{r+1})
    and the sum of the series:
    \frac{1}{n+1}-1
    but the sum is given as 1+\frac{(-1)^{n+1}}{n+1}
    Thanks!
    Try your partial fractions again.
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  3. #3
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    Ok this is the way I worked it out:
    For the part \frac{}{r}...\frac{}{r+1}
    When r=0, the other part of becomes \frac{(-1)^{1}(1)}{1}
    hence \frac{-1}{r}
    then the next part, r=-1, and the part becomes \frac{(-1)^{0}(-1)}{-1}
    hence \frac{1}{r+1}
    hence we get the partial fraction -\frac{1}{r}+\frac{1}{r+1}
    Is this correct?
    Thanks!
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  4. #4
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    No, the partial fraction decomposition is

    [LaTeX ERROR: Convert failed]

    Also here's how to get it

    [LaTeX ERROR: Convert failed]

    So you have [LaTeX ERROR: Convert failed] is valid for all r, so let

    r = 0, you get A = 1
    r = -1/2, you get A/2-B/2 = 0 which implies A = B

    Thus you get your result.

    Then just multiply the result by (-1)^{r+1} and then just do your sum.
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  5. #5
    MHF Contributor chisigma's Avatar
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    Even more simple...

    \displaystyle \frac{2r+1}{r\ (r+1)} = \frac{2}{r} - \frac{1}{r\ (r+1)}= \frac{1}{r} + \frac{1}{r+1} (1)

    ... and now...

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

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

    At this point You have to sum (2) and (3) toghether...

    Kind regards

    \chi \sigma
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  6. #6
    Math Engineering Student
    Krizalid's Avatar
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    \dfrac{2r+1}{r(r+1)}=\dfrac{r+r+1}{r(r+1)}=\dfrac{  1}{r+1}+\dfrac{1}{r}.

    that's actually even easier.
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