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Series proof

  1. Jun 26, 2010 #1
    Hey,

    I am wondering if anyone could algebraically prove that [tex]\sum{n=0=>inf} (n / 2^n)[/tex] = 2. There is some simple trick to it but I am stumped. :/

    Here is a clearer picture of my summation:
    http://img638.imageshack.us/img638/2114/texer1.png [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 26, 2010 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF. What is the context of your question? What is the application?
     
    Last edited by a moderator: May 4, 2017
  4. Jun 27, 2010 #3

    Redbelly98

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    Staff Emeritus
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    Homework Helper

    I, too, wish to know this. morgul, could you respond?

    If this is a homework problem, or even a self-study problem from a book, we have rules about getting help on a problem like this. Just click the "Rules" link at the top of this page, and then scroll down to the section titled Homework Help at the rules page.
     
    Last edited: Jun 27, 2010
  5. Jun 28, 2010 #4
    Ah, thank you for the replies! This is not a homework problem. This was from an old notebook of mine and the solution is incomplete. It was just bothering me.
     
  6. Jun 29, 2010 #5
    Define

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

    Take a derivative:

    [tex] - 1/(x-1)^2 = f'(x) = - \sum_{n=0}^{\infty} n x^{-n-1} = - (1/x) \sum_{n=1}^{\infty} n x^{-n}.[/tex]

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

    Set x=2.
     
  7. Jun 29, 2010 #6
    Define

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

    Take a derivative:

    [tex] - 1/(x-1)^2 = f'(x) = - \sum_{n=0}^{\infty} n x^{-n-1} = - (1/x) \sum_{n=1}^{\infty} n x^{-n}.[/tex]

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

    Set x=2.
     
  8. Jun 29, 2010 #7
    Alternatively, summation by parts yields partial sum
    [tex]
    \sum_{n=1}^N \frac{n}{2^n}= -\sum_{n=1}^{N-1} 2\left[(\frac{1}{2})^n-1\right]+2N(1-2^{-N})=...
    [/tex]
     
    Last edited: Jun 29, 2010
  9. Jun 29, 2010 #8
    Alternatively, summation by parts yields partial sum
    [tex]
    \sum_{n=1}^N \frac{n}{2^n}= \sum_{n=1}^{N-1} 2\left[(\frac{1}{2})^n-1\right]+2N(1-2^{-N})=...
    [/tex]
    Sorry for double post.
     
    Last edited: Jun 29, 2010
  10. Jun 29, 2010 #9

    Mute

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    Homework Helper

    You've made a slight mistake.

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

    So,

    [tex]\sum_{n=0}^\infty \frac{n}{x^n} = -x\frac{d}{dx}\left(\frac{x}{x-1}\right) = \frac{x^2}{(x-1)^2} - \frac{x}{x-1}[/tex]
     
    Last edited: Jun 29, 2010
  11. Jun 29, 2010 #10
    Yes, thank you. The web site glitched on me before I could see the renderings of my formulas.
     
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