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Math Help - Series

  1. #1
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    Series

    ((-4)^(n-1))/5^n
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  2. #2
    o_O
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    Are you trying to determine convergence?

    \sum \frac{(-4)^{n-1}}{5^n} \ = \ \sum \frac{(-1)^{n-1} 4^{n-1}}{5^n} \cdot {\color{red}\frac{4}{4}} \ = \ \frac{1}{4}\sum (-1)^{n-1}\left(\frac{4}{5}\right)^n

    Use the alternating series test.
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  3. #3
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    oops sorry i meant to ask what was the sum of that equation.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by amiv4 View Post
    oops sorry i meant to ask what was the sum of that \color{red}\text{series}.
    \forall{x}\backepsilon|x|<1~\sum_{n=0}^{\infty}x^n  =\frac{1}{1-x}

    So you have \frac{-1}{4}\sum_{n=0}^{\infty}\left(\frac{4}{5}\right)^n which fits the bill, so the sum is \frac{-1}{4}\cdot\frac{1}{1+\frac{4}{5}}
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