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Sum of a Geometric series

  1. May 2, 2006 #1
    Hi Folks,

    I have this here geometric series which I'm supposed to find the sum of:

    Given

    [tex]\sum_{n=0} ^{\infty} \frac{2n+1}{2^n}[/tex]

    I the sum into sub-sums

    [tex]\sum_{n=0} ^{\infty} 2^{-n} + \sum_{n=0} ^{\infty} \frac{1}{2}^{n-1} [/tex]

    taking [tex]2^{-n}[/tex]

    Since [tex]x^n[/tex] converges towards 1/1+x therefore I differentiate on both sides

    [tex]1/(1+x)^2 = \sum_{n=0} ^{\infty} n \cdot 2^{-n} x^{n-1}[/tex]

    I multiply with x on both sides and obtain

    [tex]x/(1+x)^2 = \sum_{n=0} ^{\infty} n * 2^{-n} x^n[/tex]

    if I set x = 1 on both sides I get

    [tex](1/4) ? = \sum_{n=1} ^{\infty} n*2^{-n} = 2[/tex]

    My teacher says that the expression has to give 2 on the left side, and not (1/4).

    What am I doing wrong? Any surgestions?

    Best Regards
    Fred
     
    Last edited: May 2, 2006
  2. jcsd
  3. May 2, 2006 #2

    AKG

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    Your first line is wrong. When you break your sum up into two sums, you don't do it right. Also, when you say xn converges towards 1/1+x, I suppose you mean [itex]\sum x^n[/itex] converges to 1/1+x. However, this is wrong, it converges to 1/1-x. After that, I have no idea what you're doing. I don't know where you're getting the "=" sign from. Sorry, I think just about everything in your post makes no sense.
     
  4. May 3, 2006 #3

    HallsofIvy

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    [tex]\sum_{n=0} ^{\infty} \frac{2n+1}{2^n}[/tex]

    is not a geometric series!
     
  5. May 3, 2006 #4

    Curious3141

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    Firstly, you separated the sum wrongly. Here's the correct version :

    [tex]\sum_{n=0} ^{\infty} \frac{2n+1}{2^n} = \sum_{n=0} ^{\infty}(n.2^{1-n})+\sum_{n=0} ^{\infty}(2^{-n})[/tex]

    Hint : For the first part-sum on the RHS, try dividing by 2 then subtracting away from the original part-sum.
     
    Last edited: May 3, 2006
  6. May 3, 2006 #5

    Curious3141

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    Actually, it works out to be one ! :biggrin:
     
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