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Math Help - Integrating a series and finding its sum

  1. #1
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    Integrating a series and finding its sum

    Given:
    \sum^{\infty}_{n=1}\frac{n}{2^{n-1}}x^{n-1}

    Radius of convergence is 2.

    Defined as a function:
    f(x)=\sum^{\infty}_{n=1}\frac{n}{2^{n-1}}x^{n-1}
    x\in]-2,2[

    Show that for |x|<2 the following is true:
    \int_{0}^{x}f(t)dt=\frac{2x}{2-x}

    My attempt:
    I first start by changing the definition under the sum from n=1 to n=0:
    f(x)=\sum^{\infty}_{n=1}\frac{n}{2^{n-1}}x^{n-1}=\sum^{\infty}_{n=0}\frac{n+1}{2^{(n+1)-1}}x^{(n+1)-1}=\sum^{\infty}_{n=0}\frac{n+1}{2^n}x^n

    To integrate a series, I use the following formula:
    \int_{0}^{b}\sum^{\infty}_{n=0}c_nx^ndx=\sum^{\inf  ty}_{n=0}\frac{c_n}{n+1}b^{n+1}

    Using that formula I get:
    \int_{0}^{x}\sum^{\infty}_{n=0}\frac{n+1}{2^n}t^nd  t=\sum^{\infty}_{n=0}\frac{\frac{n+1}{2^n}}{n+1}x^  {n+1}=\sum^{\infty}_{n=0}\frac{n+1}{(n+1)2^n}x^{n+  1}=\sum^{\infty}_{n=0}2^{-n}x^{n+1}

    But now I have no idea what to do next.
    I know that the sum for a power series is:
    \sum^{\infty}_{n=0}x^n=\frac{1}{1-x}
    But the problem with this formula is that it only counts for |x|<1 whereas I have |x|<2

    I also don't know how to express my series as x^n.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Integrating a series and finding its sum

    Note that
    \sum_{n=0}^{\infty} 2^{-n}x^{n+1} = \sum_{n=0}^{\infty} \frac{x^n x}{2^n} = x \sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^n
    This geometric serie converges if \left|\frac{x}{2}\right|<1 \Rightarrow |x|<2 with sum
    \frac{1}{1-\frac{x}{2}} = \frac{2}{2-x}
    therefore (with the factor x) the sum is \frac{2x}{2-x}
    Thanks from MathIsOhSoHard
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