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Quick question

  1. Mar 2, 2005 #1
    Hey. This has been bugging me for a long time:
    why does summation from n=1 to infinity of (-1)^n or i^n or 1/n or -1/n not converge, because summation from n=1 to infinity of 1/n^2 conveges. Don't the terms in (-1)^n or i^n or 1/n or -1/n(-1)^n tend to 0?
     
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  3. Mar 2, 2005 #2

    t!m

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    While tending towards 0 is a requirement for convergence, it does not guarantee convergence. The most classic example, which you cited, would be the harmonic series, 1/n which does have a limit of 0 but still diverges. Though a limit of anything other than 0 would guarantee divergence.
     
  4. Mar 3, 2005 #3

    Galileo

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    [itex](-1)^n[/itex] and [itex] \quad i^n[/itex] do not tend to zero, so the associate series is not convergent.
     
  5. Mar 3, 2005 #4

    dextercioby

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    Well,the trick is with the series
    [tex] \sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k} [/tex].
    IIRC it converges to
    [tex]\ln 2 [/tex].

    Daniel.
     
  6. Mar 3, 2005 #5

    saltydog

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    Hum, interesting. Think I might try to show that. From your post the other day, seems Euler Gamma functions may be involved.
     
  7. Mar 3, 2005 #6
    I think it's just a matter of using the Taylor series for the logarithm.
     
  8. Mar 3, 2005 #7

    dextercioby

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    You can start from this series
    [tex] \frac{1}{1+x}=1-x+x^{2}-x^{3}+x^{4}-... [/tex]

    and integrate it term by term and you'll have to put x=1 in the new series to find the formula which i've hinted.

    Daniel.
     
  9. Mar 3, 2005 #8

    saltydog

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    Thanks, I understand now (should know that I know). Actually I'd like to understand this one:

    [tex] \sum_{k=1}^{n}\frac{1}{2k}=\frac{1}{2}\psi(n+1)+\frac{1}{2}\gamma [/tex]

    I'll do some work on it and if I can't figure it out, I'll make a separate post in the general math and well, ask for help.
     
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