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Math Help - Series

  1. #1
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    Series

    How would I calculate the sum of the following series:

    \sum _{n=1}^{\infty }{\frac { \left( -2 \right) ^{n-1}}{{6}^{n}}}

    Of course I can use the calculator, but how would I do it manually?
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  2. #2
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    Have you tried to simplify the summation?

    \frac{(-2)^{n-1}}{6^n}= \frac{(-2)^{n}(-2)^{-1}}{(-3\times -2)^n}  = \frac{(-2)^{n}(-2)^{-1}}{(-3)^n\times (-2)^n} = \frac{(-2)^{-1}}{(-3)^n}

    or even noticed the first couple of terms

    \frac{1}{6}-\frac{1}{18}+\frac{1}{54}-\frac{1}{162}+\dots
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  3. #3
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    \sum _{n=1}^{\infty }{\frac { \left( -2 \right) ^{n-1}}{{6}^{n}}}=\sum _{n=1}^{\infty }{\frac { \left( -2 \right) ^{n-1}}{{6}^{n-1}6}}

    =\sum _{n=1}^{\infty }{\frac{1}{6}\left(\frac{-2}{6}\right)^{n-1}}

    =\sum _{n=1}^{\infty }{\frac{1}{6}\left(\frac{-1}{3}\right)^{n-1}}

    This is a geometric series with a=\frac{1}{6}, r=-\frac{1}{3}

    =\dfrac{\frac{1}{6}}{1-\left(-\frac{1}{3}\right)}

    =\dfrac{\frac{1}{6}}{1+\frac{1}{3}}

    =\dfrac{\frac{1}{6}}{\frac{4}{3}}

    =\dfrac{1}{8}
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  4. #4
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    EDIT: Sorry double post
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  5. #5
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    Thanks so much! This makes perfect sense once you showed how this is a geometric series!
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by RU2010 View Post
    How would I calculate the sum of the following series:

    \sum _{n=1}^{\infty }{\frac { \left( -2 \right) ^{n-1}}{{6}^{n}}}

    Of course I can use the calculator, but how would I do it manually?
    \displaystyle \sum _{n=1}^{\infty }\frac { ( -2 ) ^{n-1}}{6^{n}}=-\frac{1}{2} \sum _{n=1}^{\infty }\left(-\;\frac { 1 }{3}\right)^n

    and the sum on the right is a geometric series.

    CB
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