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Homework Help: Summation differentiation geometric series

  1. Feb 2, 2010 #1
    1. The problem statement, all variables and given/known data
    find the sum for

    [tex]\sum_{k=1}^{\infty} kx^{k} [/tex]


    2. Relevant equations

    [tex]\sum_{k=0}^{\infty} x^{k} = \frac{1}{1-x}; -1 < x < 1 [/tex]

    3. The attempt at a solution

    [tex] \sum_{k=1}^{\infty} kx^{k} = \sum_{n=0}^{\infty}(n+1)x^{n+1} = x\sum_{n=0}^{\infty} (n+1)x^{n} = x \frac{d}{dx} \sum_{n=0}^{\infty}x^{n+1}[/tex]

    I'm not sure how to proceed; thoughts?
     
  2. jcsd
  3. Feb 2, 2010 #2

    rock.freak667

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    I think from here

    [tex]\sum_{k=0}^{\infty} x^{k} = \frac{1}{1-x}[/tex]


    You should just differentiate both sides w.r.t. x and then change the index.
     
  4. Feb 2, 2010 #3
    does it matter that it's xn+1 and not xn?
     
  5. Feb 2, 2010 #4

    tiny-tim

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    Hi Somefantastik! :smile:

    (ooh, you just changed it! :biggrin:)

    That's fine … now you need to get that ∑xn+1 to be part of a ∑xn starting at 0 (so you may need to subtract something :wink:).
     
  6. Feb 2, 2010 #5
    How about

    [tex] x \frac{d}{dx} \sum_{n=0}^{\infty}x^{n+1} = x \frac{d}{dx}\left[x\left(\sum_{n=0}^{\infty}x^{n}\right)\right] = x \frac{d}{dx}\left[x\frac{1}{1-x}\right][/tex]
     
  7. Feb 2, 2010 #6

    tiny-tim

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    Yes, that's fine also. :smile:

    ( x/(1-x) = 1/(1-x) - 1 )

    (though in the form 1/(1-x) - 1, it's easier to differentiate! :wink:)
     
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