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Math Help - Alternating Harmonic Series

  1. #1
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    Alternating Harmonic Series

    What is the exact value to which the infinite series...converges?
    Last edited by stones44; January 11th 2010 at 05:04 PM.
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    MHF Contributor Bruno J.'s Avatar
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    Hint : integrate \frac{1}{1-x}=1+x+x^2+x^3+... to get an expression for \log (1-x) and then set x=-1.

    (Note : this is an informal proof!)
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    take the maclaurin series for ln(1+x) which is: x - (x^2)/2 + (x^2)/3 +...(-1)^(k+1) (x^k)/k.

    now plug in x=1 and notice that the right side becomes the alternating harmonic series 1 - 1/2 + 1/3 - 1/4 +... so therefore the sum of the alternating harmonic series is ln(2).
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  4. #4
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    of course as a conditional convergent series, we can make it to converge to any value we want.
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    Quote Originally Posted by Krizalid View Post
    of course as a conditional convergent series, we can make it to converge to any value we want.
    By rearranging the series, yes- but this was given in a specific arrangement that converges to ln(2).
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  6. #6
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    Quote Originally Posted by Bruno J. View Post
    Hint : integrate \frac{1}{1-x}=1+x+x^2+x^3+... to get an expression for \log (1-x) and then set x=-1.

    (Note : this is an informal proof!)
    isn't the integral of 1/(1-x) = -ln(1-x) ? so wouldn't it be -ln2
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  7. #7
    MHF Contributor Bruno J.'s Avatar
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    Yes, that's what the integral is. Multiply both sides by -1, and set x=-1.
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