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Math Help - Simplifying series

  1. #1
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    Simplifying series

    Note: Sorry for all these questions, got my finals in a few days.

    The question
    Starting with the geometric series \frac{1}{1 - z} = 1 + z + z^2 + ... + z^n + ... for |z| < 1, simplify the following series for |z| < 1, (remember you can integrate or differentiate a power series term by term within its disc of convergence):
    \sum_{n = 1}^{\infty}{nz^n}

    At first I thought I could pull the n out, and write it as \frac{n}{1 - z}, but it doesn't really make sense. Perhaps I'm just sleep deprived, but I'm having trouble getting this solved.

    Any suggestions?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Simplifying series

    Quote Originally Posted by Glitch View Post
    Note: Sorry for all these questions, got my finals in a few days.

    The question
    Starting with the geometric series \frac{1}{1 - z} = 1 + z + z^2 + ... + z^n + ... for |z| < 1, simplify the following series for |z| < 1, (remember you can integrate or differentiate a power series term by term within its disc of convergence):
    \sum_{n = 1}^{\infty}{nz^n}

    At first I thought I could pull the n out, and write it as \frac{n}{1 - z}, but it doesn't really make sense. Perhaps I'm just sleep deprived, but I'm having trouble getting this solved.

    Any suggestions?
    Hint:
    \sum_{n = 1}^{\infty}{nz^n}=z\sum_{n = 1}^{\infty}{nz^{n-1}}


    f(z)=\sum_{n = 1}^{\infty}{z^n}=

    Then:


    f'(z)=(\sum_{n = 1}^{\infty}{z^n})'=\sum_{n = 1}^{\infty}{nz^{n-1}}=(\frac{1}{1-z})'

     \sum_{n = 1}^{\infty}{nz^n}=z(\frac{1}{1-z})'
    Last edited by Also sprach Zarathustra; June 17th 2011 at 07:20 AM.
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  3. #3
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    Re: Simplifying series

    \displaystyle \begin{align*}\sum_{n = 1}^{\infty}nz^n &= \sum_{n = 1}^{\infty}(n + 1)z^n - \sum_{n = 1}^{\infty}z^n \\ &= \frac{d}{dz}\left(\sum_{n = 1}^{\infty}z^{n + 1}\right) - \sum_{n = 1}^{\infty}z^n\end{align*}
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  4. #4
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    Re: Simplifying series

    Thanks guys, Also sprach Zarathustra your post really helped.
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