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Math Help - Determinate the sums

  1. #1
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    Determinate the sums

    Hi all i need to determinate the following sums:
    \sum_{n=1}^{\infty} (-1)^n\frac{1}{n(n+1)}
    and
    \sum_{n=1}^{\infty} \frac{1}{n(n+1)}

    How do i approach this? Thanx in advance
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  2. #2
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    Quote Originally Posted by Zaph View Post
    Hi all i need to determinate the following sums:
    \sum_{n=1}^{\infty} (-1)^n\frac{1}{n(n+1)}
    and
    \sum_{n=1}^{\infty} \frac{1}{n(n+1)}

    How do i approach this? Thanx in advance
    Start by substituting the partial fraction decomposition of \frac{1}{n(n+1)}.
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    Quote Originally Posted by mr fantastic View Post
    Start by substituting the partial fraction decomposition of \frac{1}{n(n+1)}.
    Well that gives me \frac{1}{n}  \frac{1}{(n+1)}

    And then what?
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  4. #4
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    Quote Originally Posted by Zaph View Post
    Well that gives me \frac{1}{n} \frac{1}{(n+1)}

    And then what?
    Do you understand what is meant by partial fraction decomposition? \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}. I suggest you now write out the first few terms using this and see what happens. It also might help to review your classnotes or textbook. Telescoping series would be a good place to start.
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    Quote Originally Posted by mr fantastic View Post
    Do you understand what is meant by partial fraction decomposition? \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}. I suggest you now write out the first few terms using this and see what happens. It also might help to review your classnotes or textbook. Telescoping series would be a good place to start.
    Okay..
    I think i found a pattern in the first few terms, as i get

    1 - \frac{1}{2} + \frac{1}{2}-\frac{1}{3} +... so that means 1 - \frac{1}{2} + \frac{1}{2}.. cancels eachother out meaning the sum is 1.

    Then for the alternating series, it gets timed by -1, and by the same idea it sums to -1?
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    Quote Originally Posted by Zaph View Post
    Okay..
    I think i found a pattern in the first few terms, as i get

    1 - \frac{1}{2} + \frac{1}{2}-\frac{1}{3} +... so that means 1 - \frac{1}{2} + \frac{1}{2}.. cancels eachother out meaning the sum is 1.

    Then for the alternating series, it gets timed by -1, and by the same idea it sums to -1?
    EDIT: The alternating series is wrong, gonna calc some more..
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  7. #7
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    Quote Originally Posted by Zaph View Post
    Okay..
    I think i found a pattern in the first few terms, as i get

    1 - \frac{1}{2} + \frac{1}{2}-\frac{1}{3} +... so that means 1 - \frac{1}{2} + \frac{1}{2}.. cancels eachother out meaning the sum is 1.

    Then for the alternating series, it gets timed by -1, and by the same idea it sums to -1?
    Not all terms are multiplied by -1, only where n is odd... You will have an alternating series that is NOT telescopic.


    Have a look at the series

    \sum_{n = 1}^{\infty}\left[(-1)^n\left(\frac{1}{n} - \frac{1}{n - 1}\right)\right] = -\left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) - \left(\frac{1}{3} - \frac{1}{4}\right)
     + \left(\frac{1}{4} - \frac{1}{5}\right) - \left(\frac{1}{5} - \frac{1}{6}\right) + \left(\frac{1}{6} - \frac{1}{7}\right) - \dots + \dots

     = -1 + \frac{2}{2} - \frac{2}{3} + \frac{2}{4} - \frac{2}{5} + \frac{2}{6} - \frac{2}{7} + \dots - \dots

     = -1 + \frac{2}{1} - \frac{2}{1} + \frac{2}{2} - \frac{2}{3} + \frac{2}{4} - \frac{2}{5} + \frac{2}{6} -  \frac{2}{7} + \dots - \dots

     = 1 - 2\sum_{n = 1}^{\infty}\left[\frac{(-1)^{n-1}}{n}\right]


    This infinite sum is the alternating harmonic series, which is known to converge to \ln{2}.


    So we finally have

    \sum_{n = 1}^{\infty}\left[(-1)^n\left(\frac{1}{n} - \frac{1}{n -  1}\right)\right] = 1 - 2\ln{2}.
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