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Math Help - How do you find the value of this series?

  1. #1
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    How do you find the value of this series?

    summation notion beginning at n=1...

    [(-1)^n]/n


    No calculator use please. Any way to figure where it converges?
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  2. #2
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    The power series for \ln(1+x) is x-\tfrac12x^2 + \tfrac13x^3-\ldots+(-1)^{n-1}\tfrac1nx^n+\ldots. It converges when |x|<1 but also when x=1 (though this is harder to prove).
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  3. #3
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    Uh... what? I don't understand. My series isn't even a power series. I'm not asking whether it converges, but where it converges.
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  4. #4
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    Quote Originally Posted by Kaitosan View Post
    Uh... what? I don't understand. My series isn't even a power series. I'm not asking whether it converges, but where it converges.
    He's actually telling you how to find the sum (which is usually difficult to do in closed form) Try substituting x = 1 .
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  5. #5
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    Gotcha. Thanks guys.
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  6. #6
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    Quote Originally Posted by Opalg View Post
    The power series for \ln(1+x) is x-\tfrac12x^2 + \tfrac13x^3-\ldots+(-1)^{n-1}\tfrac1nx^n+\ldots. It converges when |x|<1 but also when x=1 (though this is harder to prove).
    As I thought about that, I actually think you meant...


    -ln(1+x)


    Since my series start at n=1. Also, I think I can use the expression ln[1/(1+3x)] since it barely converges to 1/3 lol.

    Right?
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  7. #7
    Super Member redsoxfan325's Avatar
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    Yes, it is -\ln(1+x). Here's why:

    Start with the alternating geometric series: \frac{1}{1+x}=1-x+x^2-x^3+...

    Now make each side negative: -\frac{1}{1+x}=-1+x-x^2+x^3-...

    Integrate both sides: -\ln(1+x)=-x+\frac{x^2}{2}-\frac{x^3}{3}+\frac{x^4}{4}

    So -\ln(1+x) = \sum_{n=1}^{\infty}\frac{(-1)^n}{n}x^n

    Plugging in x=1 gives you \boxed{\sum_{n=1}^{\infty}\frac{(-1)^n}{n}=-\ln(2)}
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  8. #8
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    Heh yeah. Thank you redsoxfan! This information is kinda new to me.
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