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Converge absolutely or conditionally, or diverge?

  1. Jan 28, 2008 #1
    1. The problem statement, all variables and given/known data

    heres the problem: http://img145.imageshack.us/img145/161/55180818ir8.png

    does this series converge absolutely, conditionally, or does it diverge?

    2. Relevant equations



    3. The attempt at a solution

    whats series to test it with?
    at first glance it looks like an alternating series. reminds me of a failing p series in the denominator.
     
  2. jcsd
  3. Jan 28, 2008 #2

    HallsofIvy

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    The very, very simplest of convergence tests: if a series converges the individual terms must go to 0. Does that happen here? What is
    [tex]lim_{n\rightarrow \infty} \frac{(-1)^n}{10^{1/n}}[/tex]?
     
  4. Jan 28, 2008 #3
    Okay, it appears to be alternating, however it is not decreasing because of the denominator, it is increasing!

    thus diverging by the alternating series test, correct?
     
  5. Jan 28, 2008 #4

    Tom Mattson

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    What he's using there is the nth term test for divergence. It should have been the first test that you learned, no?
     
  6. Jan 28, 2008 #5
    yes, an easy test, how do i take the nth term test to a variable with a n root? the 10^1/n is confusing me. please help.
     
  7. Jan 28, 2008 #6

    HallsofIvy

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    No, what I said was "If the individual terms do not go to 0, then the series must diverge"!

    What is the limit of [tex]\frac{(-1)^n}{10^{(1/n)}}[/tex]? Is it 0?

    If you have trouble with that, what about [itex]10^{1/n}[/itex] itself? Calculate a few values, say n= 10 and higher.
     
  8. Jan 28, 2008 #7
    i definately can see that this does not go to 0.

    However, how do I prove this on paper?

    It fails convergence obviously.

    Can i use direct comparison with harmonic series, and say that the harmonic series is smaller or equal to this series, thus showing if the harmonic series diverges, then the original series diverges.
     
  9. Jan 29, 2008 #8

    Tom Mattson

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    Just take the limit of the summand as n goes to infinity. You don't need to compare the series with anything.
     
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