1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Absolute Convergence

  1. Dec 2, 2007 #1
    [tex]\sum_{n=2}^{\infty}\frac{1}{(lnn)^{n}}[/tex]

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

    [tex]\sum_{n=2}^{\infty}(\frac{1}{ln3})^3[/tex]

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

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

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

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

    When n=3, r equals

    [tex]|\frac{1}{ln3}|<1[/tex]

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

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Absolute Convergence
  1. Absolute convergence (Replies: 1)

  2. Absolute Convergence (Replies: 6)

  3. Absolute Convergence (Replies: 1)

  4. Absolute Convergence (Replies: 7)

  5. Absolute Convergence (Replies: 4)

Loading...