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Math Help - partial sums

  1. #1
    Super Member Random Variable's Avatar
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    partial sums

    Let  \displaystyle \sum_{n=0}^{\infty} a_{n} x^{n} be a convergent series.

    Let  S_{n} be the nth partial sum of the above series.

    Then \displaystyle \sum^{\infty}_{n=1} S_{n} x^{n} = \frac{1}{1-x} \sum^{\infty}_{n=0} a_{n} x^{n} .

    Should this be obvious? I don't get it.
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  2. #2
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    It might help to note that \displaystyle \frac{1}{1 - x} = \sum_{n = 0}^{\infty}x^n...
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  3. #3
    Super Member Random Variable's Avatar
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     \displaystyle \sum_{n=0}^{\infty} S_{n} x^{n} = \sum^{\infty}_{n=0} \sum^{n}_{k=0} (a_{k}x^{k}) x^{n}

    Reverse the order of summation?
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  4. #4
    Super Member Random Variable's Avatar
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    It makes sense when you write out the terms.

     \displaystyle \frac{1}{1-x} \sum^{\infty}_{n=0} a_{n} x^{n} = \frac{a_{0} + a_{1} x + a_{2}x^{2} + ...}{1-x}

     \displaystyle = (a_{0} + a_{1} x + a_{2}x^{2} + ...)(1 + x + x^{2} + ...) \ , \ \text{for} \ |x| < 1

     \displystyle= a_{0} + (a_{0}+ a_{1}) x + (a_{0}+a_{1}+a_{2}) x^{2} + ...

     \displaystyle = \sum^{\infty}_{n=0} S_{n} x^{n}
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    Let  \displaystyle \sum_{n=0}^{\infty} a_{n} x^{n} be a convergent series.

    Let  S_{n} be the nth partial sum of the above series.

    Then \displaystyle \sum^{\infty}_{n=1} S_{n} x^{n} = \frac{1}{1-x} \sum^{\infty}_{n=0} a_{n} x^{n} .

    Should this be obvious? I don't get it.
    I think you're overthinking this, by definition assuming all the proper conditions are met the Cauchy Product says that \displaystyle \sum_n x^n\sum_n a_n x^n=\sum_n (\sum_{k=0}^n a_k 1)x^n
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