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Math Help - Sum of a series

  1. #1
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    Sum of a series & maclin series

    So I have written down some notes, and I have forgotten why i wrote what i wrote down.


    Find the sum of the series
    \sum_{n=1}^\infty nx^n

    I wrote down these steps

    \sum_{n=1}^\infty nx^n= x \sum_{n=1}^\infty nx^{n-1}=x* \frac{1}{(1-x)^2}= \frac{x}{(1-x)^2} for abs(x)<1

    I don't know how they come about...
    Last edited by bfpri; April 25th 2010 at 08:35 PM.
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  2. #2
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    Use \sum nx^{n-1} = \frac{\mathrm{d}}{\mathrm{d}x} \sum x^n.
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  3. #3
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    Quote Originally Posted by maddas View Post
    Use \sum nx^{n-1} = \frac{\mathrm{d}}{\mathrm{d}x} \sum x^n.
    Edit: Never mind...
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  4. #4
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    Quote Originally Posted by Prove It View Post
    Edit: Never mind...
    :[
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    Quote Originally Posted by maddas View Post
    Use \sum nx^{n-1} = \frac{\mathrm{d}}{\mathrm{d}x} \sum x^n.
    Maddas is right. Than take the derivative of \frac{1}{1-x}. Since \frac{1}{1-x}=\sum x^n for |x|<1.
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  6. #6
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    Quote Originally Posted by maddas View Post
    :[
    OK, I'll fill in the steps then :P


    \sum_{n = 1}^{\infty}nx^n = x\sum_{n =1}^{\infty}nx^{n - 1}.


    Notice that \frac{d}{dx}\left(\sum_{n = 1}^{\infty}x^n\right) = \sum_{n = 1}^{\infty}nx^{n - 1}.

    Also notice that \sum_{n = 1}^{\infty}x^n is a geometric series with a = x and r = x.

    So \sum_{n = 1}^{\infty}x^n = \frac{x}{1 - x} for |x| < 1.


    Therefore \frac{d}{dx}\left(\sum_{n =1}^{\infty}x^n\right) = \frac{d}{dx}\left(\frac{x}{1 - x}\right)

     = \frac{1}{(1- x)^2}.

    So this means

    \sum_{n = 1}^{\infty}nx^{n - 1} = \frac{1}{(1-x)^2}

    x\sum_{n = 1}^{\infty}nx^{n -1} = \frac{x}{(1-x)^2}

    \sum_{n = 1}^{\infty}nx^n = \frac{x}{(1-x)^2}, provided of course that |x| < 1.
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  7. #7
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    :D
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  8. #8
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    they're using power series as functions..got it

    thanks for the help there!

    I have another question though if you don't mind...Don't want to make another topic

    For the maclin series of \sqrt{1+x}, after the limit ratio test is

     \frac{abs(x)}{2}  \lim_{n\to\infty}\frac{2n-1}{n+1} = abs(x)<1. Hence 1 is the radius of convergence. I understand how most of everything comes about, but what i don't understand is why 2n-1 is there. since the original was 1*3*5...(2n-3), shouldn't that be (2n-1)*(2n-2)?
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