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Math Help - more sequence/series

  1. #1
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    more sequence/series

    from n=1 to infinity of


    ((-1)^n * n)/3^n
    does it converge or diverge, and what is the sum?
    good luck!!!
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  2. #2
    Member disclaimer's Avatar
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    It's convergent by the alternating series test.

    \sum_{n=1}^{\infty}(-1)^n\frac{n}{3^n}

    \lim_{n\rightarrow{\infty}}\frac{n}{3^n}=0 while a_{n+1}\leq{a_n}.
    Last edited by disclaimer; April 24th 2008 at 09:14 AM.
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  3. #3
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    Hello, alex_nicole!

    \sum^{\infty}_{n=1} (\text{-}1)^n\,\frac{n}{3^n}

    We have: . . . . S \;=\;-\frac{1}{3} + \frac{2}{3^2} - \frac{3}{3^3} + \frac{4}{3^4} + \cdots

    Multiply by \frac{1}{3}\!:\;\;\frac{1}{3}S \;=\qquad-\frac{1}{3^2} + \frac{2}{3^3} - \frac{3}{3^4} + \cdots

    Add:. . . . . . . \frac{4}{3}S \;=\;-\frac{1}{3} + \frac{1}{3^2} - \frac{1}{3^3} + \frac{1}{3^4} + \cdots


    \text{We have: }\qquad\quad \frac{4}{3}S \;=\;-\frac{1}{3}\underbrace{\left(1 - \frac{1}{3} + \frac{1}{3^2} - \frac{1}{3^2} + \cdots\right)}_{\text{gemetric seres}}

    . . The series has the sum: . \frac{1}{1-\left(-\frac{1}{3}\right)} \:=\:\frac{3}{4}

    And we have: . \frac{4}{3}S \:=\:-\frac{1}{3}\left(\frac{3}{4}\right) \:=\:-\frac{1}{4}

    Therefore: . S \;=\;\frac{3}{4}\left(-\frac{1}{4}\right) \;=\;\boxed{-\frac{3}{16}}

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