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Series convergence

  1. Jun 25, 2012 #1
    Hi, while reading some artificial intelligence book, i came upon the following sum. How can I evaluate it analytically, so not guess it by computing many terms? It's easy to see by ratio test that it converges (intuitively too, since its a linear vs exponential function).

    [itex]\sum_{i=1}^\infty \frac{i}{2^i}[/itex]
     
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  3. Jun 25, 2012 #2

    micromass

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    Do you know the sum of

    [tex]\sum_{n=1}^{+\infty} x^n[/tex]

    for |x|<1 ??

    Now take derivatives.
     
  4. Jun 25, 2012 #3
    I know the sum, it evaluates to:
    [tex]\frac{1}{1-x}-1[/tex]

    But, how will the derivatives help? The i-th derivative with respect to x gives me:

    [tex]\sum_{n=1}^{\infty} \frac{n!}{(n-i)!}x^{n-i}[/tex]
     
    Last edited: Jun 25, 2012
  5. Jun 25, 2012 #4


    Why would you take the i-th derivative if you were said "the derivative? I find that i as index pretty confussing and annoying, so I'll change it:
    [tex]\frac{1}{1-x}-1=\sum_{n=1}^\infty x^n\Longrightarrow \frac{1}{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}[/tex]

    and now just choose a convenient x within the convergence radius...

    DonAntonio
     
  6. Jun 26, 2012 #5
    I offer another way (not very rigorous, but still intuitive.)
    The nth term of the sum can be written as
    [tex]\left(\frac{1}{2^n}+\frac{1}{2^n}...+\frac{1}{2^n}\right)[/tex]
    where we have the fraction repeated n times. Now look at this structure:

    1/2
    1/4+1/4
    1/8+1/8+1/8
    .............

    Now sum columns instead of the rows!
     
  7. Jun 26, 2012 #6
    But how do you choose that x? Shouldn't

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

    But then x will be a term dependent on n...
     
  8. Jun 26, 2012 #7

    ...and thus [itex]\,\displaystyle{ x^{n-1}=\frac{1}{2}\left(\frac{1}{2}\right)^{n-1}}[/itex]...! And of course, taking off a factor of 0.5, x is a constant.

    DonAntonio
     
  9. Jun 26, 2012 #8
    Ah right lol, thank you very much.
     
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