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Thread: Integrating a series and finding its sum

  1. #1
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    Integrating a series and finding its sum

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

    Radius of convergence is 2.

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

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

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

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

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

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

    I also don't know how to express my series as $\displaystyle x^n$.
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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
    $\displaystyle \sum_{n=0}^{\infty} 2^{-n}x^{n+1} = \sum_{n=0}^{\infty} \frac{x^n x}{2^n} = x \sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^n$
    This geometric serie converges if $\displaystyle \left|\frac{x}{2}\right|<1 \Rightarrow |x|<2$ with sum
    $\displaystyle \frac{1}{1-\frac{x}{2}} = \frac{2}{2-x}$
    therefore (with the factor $\displaystyle x$) the sum is $\displaystyle \frac{2x}{2-x}$
    Thanks from MathIsOhSoHard
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