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Math Help - Power Series of -ln(1-x)/x?

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    Question Power Series of -ln(1-x)/x?

    From the list of basic power series: -ln(1-x) = x + \frac{x^2}{2} +\frac{x^3}{3} + \frac{x^4}{4} +...

    How do you write a power series for f(x)= \frac{-ln(1-x)}{x} ?

    Using the result how do you write a series expression for the \int_{0}^{1} \frac{-ln(1-x)}{x}dx ?


    I started out by looking at the theorem for power series expansion and finding the derivatives of f(x), but that turned out to be a mess. Help is much appreciated!
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    Instead of solving for the second power series directly just divide the power series you have by x.
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    Quote Originally Posted by blgonz03 View Post
    From the list of basic power series: -ln(1-x) = x + \frac{x^2}{2} +\frac{x^3}{3} + \frac{x^4}{4} +...

    How do you write a power series for f(x)= \frac{-ln(1-x)}{x} ?

    Using the result how do you write a series expression for the \int_{0}^{1} \frac{-ln(1-x)}{x}dx ?


    I started out by looking at the theorem for power series expansion and finding the derivatives of f(x), but that turned out to be a mess. Help is much appreciated!
    -ln(1-x) = x + \frac{x^2}{2} +\frac{x^3}{3} + \frac{x^4}{4} +...

    \frac{-ln(1-x)}{x} = 1 + \frac{x}{2} +\frac{x^2}{3} + \frac{x^3}{4} +...
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    Quote Originally Posted by skeeter View Post
    -ln(1-x) = x + \frac{x^2}{2} +\frac{x^3}{3} + \frac{x^4}{4} +...

    \frac{-ln(1-x)}{x} = 1 + \frac{x}{2} +\frac{x^2}{3} + \frac{x^3}{4} +...


    but why?

    \displaystyle \int_{0}^{1} \frac{-ln(1-x)}{x}dx  = \int_{0}^{1} ln(1-x) \cdot -\frac{1}{x}dx  = \int_0^{-\infty} u \, du

    ... which diverges.
    But, it doesn't diverg. It equals \frac{-\pi^2}{6}
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    Quote Originally Posted by Drexel28 View Post
    But, it doesn't diverg. It equals \frac{-\pi^2}{6}
    that would be the series ... \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by skeeter View Post
    that would be the series ... \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}
    Yes, and using some analytic methods one can show that \displaystyle \int_0^1\frac{\ln(1-x)}{x}dx=\int_0^1\sum_{n=1}^{\infty}\frac{x^{n-1}}{n}=\sum_{n=1}^{\infty}\int_0^1\frac{x^{n-1}}{n}dx=\sum_{n=1}^{\infty}\frac{1}{n^2}
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  7. #7
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    Quote Originally Posted by Drexel28 View Post
    Yes, and using some analytic methods one can show that \displaystyle \int_0^1\frac{\ln(1-x)}{x}dx=\int_0^1\sum_{n=1}^{\infty}\frac{x^{n-1}}{n}=\sum_{n=1}^{\infty}\int_0^1\frac{x^{n-1}}{n}dx=\sum_{n=1}^{\infty}\frac{1}{n^2}
    learn something new everday ... saw that right after when I went back and integrated the power series.
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