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Limit/Direct Comparison for Series Question

  1. Sep 11, 2012 #1
    1. The problem statement, all variables and given/known data
    [tex]\sum_{n=1}^{\infty} \frac{7n}{6n^2 ln(n)+2}[/tex]

    Determine whether the series converges or diverges.


    2. Relevant equations
    Denominator is growing faster, so the limit as n --> to infinity should equal zerio


    3. The attempt at a solution

    I tried isolating the highest power of the both the numerator and denominator. Which is:
    [tex]\frac{7n}{6n^2 ln(n)}[/tex] = [tex]\frac{7}{6n ln(n)}[/tex]

    What would I do next? Would I compare the simplified bn to an for a limit comparison test?

    I also tried a direct comparison through: [tex]\frac{1}{6n^2+2}[/tex]
    But I can't tell if that would work. Would the an be less than bn?


    Any feedback and help appreciated.
     
    Last edited: Sep 11, 2012
  2. jcsd
  3. Sep 11, 2012 #2

    jbunniii

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    Does

    [tex]\sum \frac{1}{n \ln n}[/tex]

    converge or diverge?
     
  4. Sep 11, 2012 #3
    This might not be the only way but its pretty cool. Since the series is positive and decreasing you can test the series [tex]\sum 2^n a_{2^n}[/tex] where [tex] a_{n}[/tex] is your sequence after dividing the first factor of n. Your series converges if and only if the above series converges.

    http://en.wikipedia.org/wiki/Cauchy_condensation_test
     
    Last edited: Sep 11, 2012
  5. Sep 11, 2012 #4
    This can be easily calculated with the condensation test :D
     
  6. Sep 11, 2012 #5
    I just did the integral test for this series, and it diverges. So basically, I can uses the limit comparison test with the bn:

    [tex]\frac{7}{6n ln(n)}[/tex]

    So ultimately, the series diverges then. Right?
     
  7. Sep 11, 2012 #6
    Yes it diverges.
     
  8. Sep 11, 2012 #7
    Thanks for the help, appreciate it.
     
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