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Infinite series problem

  1. Oct 26, 2013 #1
    1. The problem statement, all variables and given/known data

    Determine convergence or divergence using any method covered so far*:

    Ʃ(1/(n*ln(n)^2 - n)) from n = 1 to infinity


    *The methods are the following:

    - Dichotomy for positive series (if the partial sums are bounded above and the series is positive, the series converges)

    - Integral Test

    - Convergence of p-Series

    - Comparison test

    - Limit Comparison Test

    2. Relevant equations

    n/a

    3. The attempt at a solution

    The series is negative for n = 1 and 2, so I am left with Comparison test but I am having trouble determining what sequence to compare to. I am not "seeing" it.
     
  2. jcsd
  3. Oct 26, 2013 #2

    LCKurtz

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    Is that denominator ##n\ln^2(n) -n## or ##n\ln (n^2)-n##?
    It doesn't matter if the first few terms of the series are negative. It is the tail end of the series that determines convergence or divergence. So don't rule out the other methods.
     
  4. Oct 26, 2013 #3
    The denominator is the first one you posted, n*(ln(n))^2 - n.

    In my book for the Limit Comparison Test it states "Let a[n] and b[n] be positive sequences." I assumed that "positive sequences" meant positive for all n [1, infinity). So it is really saying that if the sequence converges to a positive value or diverges "positively" than one can use the Limit Comp test on these sequences? Or am I still off base
     
  5. Oct 26, 2013 #4

    LCKurtz

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    If ##\sum_{n=3}^\infty a_n## converges or diverges, adding ##a_1+a_2## to it won't change the convergence or divergence. So analyze the sum from 3. I would suggest a combination of comparison and integral tests.
     
  6. Oct 27, 2013 #5
    Ahh I see thanks for clarifying. I will reattempt the problem.

    EDIT: Got it by simply using integral test from n=3 to infinity. The integral converges so the series must as well. Thanks LCKurtz.
     
    Last edited: Oct 27, 2013
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