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Math Help - [SOLVED] Power Series Proof

  1. #1
    jallison88
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    [SOLVED] Power Series Proof

    How would one prove that the sum, from n=0 to infinite, of the alternating power series (-1)^n/[3^n*(2n+1)] equals Pi/[2(3)^.5]? I'm having some difficulty conceptualizing the process.
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  2. #2
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    Quote Originally Posted by jallison88
    How would one prove that the sum, from n=0 to infinite, of the alternating power series (-1)^n/[3^n*(2n+1)] equals Pi/[2(3)^.5]? I'm having some difficulty conceptualizing the process.
    Quote Originally Posted by jallison88
    How would one prove that the sum, from n=0 to infinite, of the alternating power series (-1)^n/[3^n*(2n+1)] equals Pi/[2(3)^.5]? I'm having some difficulty conceptualizing the process.
    \sum^{\infty}_{k=0}\frac{(-1)^k}{3^k(2k+1)}
    \sum^{\infty}_{k=0}\left(\frac{1}{3}\right)^k\frac  {(-1)^k}{2k+1}.
    I happen to know that,
    \sum^{\infty}_{k=0}\frac{(-1)^k}{2k+1}=\frac{\pi}{4} because it is arctangent series.
    and also know that,
    \sum^{\infty}_{k=0}\left(\frac{1}{3}\right)^k= \frac{3}{2} because it is geometric series. But I do not see how to deal with both of them together.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by jallison88
    How would one prove that the sum, from n=0 to infinite, of the alternating power series (-1)^n/[3^n*(2n+1)] equals Pi/[2(3)^.5]? I'm having some difficulty conceptualizing the process.
    Look at the power series for \arctan(x), then let x=1/ \sqrt{3}, and then Bob's your uncle.

    RonL
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