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Limit/Direct Comparison for Series Question

  • Thread starter Mad Season
  • Start date
  • #1

Homework Statement


[tex]\sum_{n=1}^{\infty} \frac{7n}{6n^2 ln(n)+2}[/tex]

Determine whether the series converges or diverges.


Homework Equations


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


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:

Answers and Replies

  • #2
jbunniii
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Does

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

converge or diverge?
 
  • #3
106
2
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:
  • #4
106
2
Does

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

converge or diverge?
This can be easily calculated with the condensation test :D
 
  • #5
Does

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

converge or diverge?
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?
 
  • #6
106
2
Yes it diverges.
 
  • #7
Thanks for the help, appreciate it.
 

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