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Math Help - MacLaurin series/Ratio test.

  1. #1
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    Red face MacLaurin series/Ratio test.

    For f(x) = loge(1-x)
    Clarification: The question isnít loge(1+x), itís loge(1-x)

    a) Find the MacLaurin series: f(0) +f'(0)x + f''(0)/2! + f'''(0)/3!....etc

    My question is, to what extent do I keep applying the series to, since the series goes on forever and there are no constraints specified.

    b) Also, Iím having trouble finding the interval of convergence for the MacLauren series, how do I find the interval of convergence?

    c) According to the question I am given, I was to find an open interval (e,f) from b) since the statement of the ratio test uses a strict inequality.
    And now I have to use an appropriate test of series convergence for the MacLaurin series found at the start at x = e and x = f

    I am clueless as to how to answer these questions, could somebody please show me the steps? I am keen to understand how to go about tackling these questions.

    Thank you very much, I really appreciate your help.
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    Re: MacLaurin series/Ratio test.

    \frac{1}{1-x} = 1 + x + x^2 + x^3 + ... for |x| < 1

    \int \frac{dx}{1-x} = C - x - \frac{x^2}{2} - \frac{x^3}{3} - ...

    \ln(1-x) = C - x - \frac{x^2}{2} - \frac{x^3}{3} - ...

    x = 0 \implies C = 0

    \ln(1-x) = \left(- x - \frac{x^2}{2} - \frac{x^3}{3} - ... \right) = -\sum_{n=1}^{\infty} \frac{x^n}{n}

    interval of convergence is -1 \le x < 1 ,

    now ... why is x = -1 included and x = 1 not ?
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    Re: MacLaurin series/Ratio test.

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