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Math Help - An Interesting Infinite Geometric Series

  1. #1
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    An Interesting Infinite Geometric Series

    Express the infinite series 1 + (1/2)x + (1/3)x^2 + (1/4)x^3 + (1/5)x^4 + ... as an elementary function of x.
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  2. #2
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    Re: An Interesting Infinite Geometric Series

    Quote Originally Posted by icemanfan View Post
    Express the infinite series 1 + (1/2)x + (1/3)x^2 + (1/4)x^3 + (1/5)x^4 + ... as an elementary function of x.
    Without worrying too much about convergence:

    Let

    f(x)=1+\frac{x}{2}+\frac{x^2}{3}+...

    then:

    \frac{d}{dx} xf(x)=1+x+x^2+... = \frac{1}{1-x}

    CB

    (Also, you should have asked permission to post this in the maths challenge area of MHF)
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  3. #3
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    Re: An Interesting Infinite Geometric Series

    Quote Originally Posted by CaptainBlack View Post
    Without worrying too much about convergence:

    Let

    f(x)=1+\frac{x}{2}+\frac{x^2}{3}+...

    then:

    \frac{d}{dx} xf(x)=1+x+x^2+... = \frac{1}{1-x}

    CB
    It did occur to me to solve the problem that way, but then how do you determine what f(x) is?

    \frac{d}{dx} xf(x) = xf'(x) + f(x) = 1+x+x^2+... = \frac{1}{1-x}

    xf'(x) + f(x) = \frac{1}{1-x}

    Not sure where you go from there. My solution is entirely different.
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    Re: An Interesting Infinite Geometric Series

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  5. #5
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    Re: An Interesting Infinite Geometric Series

    Thanks, uniquesailor. That was easier than I made it. Actually, I realized my original solution was wrong, but I'm not sure why. This is what I did:

    e^x + \frac{e^x}{2x} + \frac{e^x}{3x^2} + \frac{e^x}{4x^3} + ... = e^xg(x)

    \frac{d}{dx} e^x + \frac{e^x}{2x} + \frac{e^x}{3x^2} + \frac{e^x}{4x^3} =

    e^x + \frac{e^x}{2x} - \frac{e^x}{x^2} + \frac{e^x}{3x^2} - \frac{e^x}{x^3} + \frac{e^x}{4x^3} - \frac{e^x}{x^4} ... =

    e^x + \frac{e^x}{2x} + \frac{e^x}{3x^2} + \frac{e^x}{4x^3} + ... - \frac{e^x}{x^2} - \frac{e^x}{x^3} - \frac{e^x}{x^4} - ... =

    e^x g(x) - \frac{e^x}{x^2}(1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + ...)

    Hence

    \frac{d}{dx} e^x g(x) = e^x g(x) - \frac{e^x}{x^2} \cdot \frac{x}{x - 1}

    e^x g(x) + e^x g'(x) = e^x g(x) - e^x \cdot \frac{1}{x(x - 1)}

    e^x g'(x) = -e^x \cdot \frac{1}{x(x - 1)}

     g'(x) = -\frac{1}{x(x - 1)}

     g(x) = \ln (x) - \ln (x - 1) + C

     f(x) = g(\frac{1}{x}) = \ln (\frac{1}{x}) - \ln (\frac{1 - x}{x}) + C

     f(x) = -\ln (1 - x) + C

    Somehow I lost a factor of x in there...
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