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Divergence/Convergence of the series (ln(n))^3/n^3

  1. Oct 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Does the following series converge or diverge? give reasons for your answers.
    [tex]\Sigma[/tex](ln(n))^3/n^3

    2. Relevant equations
    Is the direct comparison test appropriate for this question? limit comparison? I have issues with the direct comparison when attempting to compare to the harmonic series, and limit comparison doesn't seem to yield a very simple solution either.


    3. The attempt at a solution
    I attempted to show that the series above is greater than the series ln(n)/n, which is greater than 1/n, the harmonic series. this is apparently not the case, however, as made evident by my graphing calculator on window -1<x<10 & -2<y<2.
     
  2. jcsd
  3. Oct 19, 2009 #2

    Office_Shredder

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    Compare ln(n) and n1/3
     
  4. Oct 19, 2009 #3
    could you be a little more specific...? im not sure how comparing ((ln(n))^3)/(n^3) to ln(n) or n^1/3 would help...you saying use DCT? LCT? thanks for the reply.
     
  5. Oct 19, 2009 #4

    Office_Shredder

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    Which grows faster? ln(n) or n1/3?
     
  6. Oct 19, 2009 #5
    ln(n) grows faster as n->inf...i'm still not seeing how this is pertinent to my problem however. Again, thanks.
     
  7. Oct 19, 2009 #6

    Office_Shredder

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    You might want to double check that with a calculator. You probably went over in class how the exponential function grows faster than any power function, and the logarithm function grows slower; you might want to review your notes and check (I don't see how this problem is possible without making this observation)
     
  8. Oct 19, 2009 #7
    I stand corrected. n^1/3 does indeed grow faster. How do I relate that fact to the sum of the sequence ((ln(n))^3)/n^3? Ohh....wait, so if i set up the inequality to use in the DCT, 1/n<(ln(n)/n)^3, then cube-root everything, multiply the n from the right side, I end up with n^(2/3)<ln(n)...Is that where you were going with this? in that case, it is a direct comparison with 1/n, proving that (ln(n))^3/n^3 diverges. Correct?
     
  9. Oct 19, 2009 #8

    Office_Shredder

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    No, because your end result n2/3<ln(n) isn't true, so you can't conclude 1/n< ln(n)3/n3

    If you know that ln(n)<n1/3 for big n, then ln(n)3<n.

    So what is ln(n)3/n3 smaller than? Think of susbtituting something for ln(n)3
     
  10. Oct 19, 2009 #9
    so since ln(n)^3 is less than n, that means that ln(n)^3/n^3 is less than n/n^3, so ln(n)^3/n^3 is less than 1/n^2, which is a convergent series...?
     
  11. Oct 20, 2009 #10

    Office_Shredder

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    That's right. Keep in mind when you're wording your answer to be precise:

    For large n, ln(n)3<n, so ln(n)3/n3<1/n2, so since convergence is not affected by the leading portions of the series (it doesn't matter what the values of ln(n)3/n3 when n is smaller) by the comparison test the series converges
     
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