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Math Help - Sum of the series

  1. #1
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    Sum of the series

    How do I find the sum of the series: (sum. n=1, infinity) [2^(n+1)]/pi^n
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  2. #2
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    Quote Originally Posted by schnek View Post
    How do I find the sum of the series: (sum. n=1, infinity) [2^(n+1)]/pi^n

    This looks to be a geometric series, and you are trying to find the sum as it goes to infinite. So lets break it down

    \sum_{n=1}^{\infty}\frac{2^{n+1}}{\pi^n}

    Now If you can remember properties of exponents we can transform the numerator to help us

    \sum_{n=1}^{\infty}\frac{2^{n}\cdot2^1}{\pi^n}

    We can pull that out as a constant multiplicative of this geometric series

    \sum_{n=1}^{\infty}2\cdot\frac{2^{n}}{\pi^n}


    \sum_{n=1}^{\infty}2\cdot\frac{2^{n}}{\pi^n}

    We can use another property of exponents here


    \sum_{n=1}^{\infty}2\cdot(\frac{2}{\pi})^n

    Now the rest is easy, Our constant multiple lets call it a which is the first term in the series, is  a = 2\cdot\frac{2}{\pi}
    and your common ration r is r=\frac{2}{\pi}, now if we recall the formula for a sum of the geometric series as it goes to infinity

    \frac{a}{1-r}

    You should be able to get the rest
    Last edited by RockHard; November 23rd 2009 at 06:58 PM.
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  3. #3
    Super Member redsoxfan325's Avatar
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    One small caveat...the sum starts at 1, so a should be 2\cdot\frac{2}{\pi}.
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  4. #4
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    Hello, schnek!

    Find the sum of the series: . S \;=\;\sum^{\infty}_{n=1} \frac{2^{n+1}}{\pi^n}
    We have: . S \;=\;\frac{2^2}{\pi} + \frac{2^3}{\pi^2} + \frac{2^4}{\pi^3} + \hdots

    This is a geometric series with first term a = \frac{4}{\pi} and common ratio r = \frac{2}{\pi}

    The sum is: . S \;=\;\frac{a}{1-r} \;=\;\frac{\frac{4}{\pi}}{1 - \frac{2}{\pi}} \;=\;\frac{4}{\pi-2}

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  5. #5
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    Aw, I knew I was gonna make a careless mistake, that the numerator is the first term in the series. , Thanks Sox, learned from this mistake
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