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Absolute Convergence

  1. Dec 2, 2007 #1

    If I treat it as a geometric series, then when n=3, r is smaller than 1


    What matters is that it's ultimately decreasing, would that be the correct approach? Even problem, no answer!

    Last edited: Dec 2, 2007
  2. jcsd
  3. Dec 2, 2007 #2


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    What is r?
  4. Dec 2, 2007 #3
    Well, as for an answer I would like to say that:

    When n=3, r equals


    Which is true, and so, what matters is that it is ultimately decreasing.
  5. Dec 2, 2007 #4


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    In the ratio test r = Limn-->infinity|Ln(n+1)-(n+1)/Ln(n)-n|.

    If r < 1 then the series converges.

    But ultimately the ratio test is about whether you have a decreasing sequence; so I think you can use the ratio test here.

    Also, for n > 1, absolute convergence is the same as convergence because Ln(n) > 0.
    Last edited: Dec 2, 2007
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