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Series divergence

  1. Jul 25, 2009 #1
    [tex]\sum (n^2-arctan(n)) / (n^3 + sin(n))[/tex] n=0 to ∞

    I know this series diverges, but how would you use the comparison test to compare it to (n^2 / n^3 meaning the harmonic series 1/n)

    Thank you very much
     
    Last edited: Jul 25, 2009
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  3. Jul 25, 2009 #2

    HallsofIvy

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    [itex]|sin(n)|\le 1|[/itex] so the denominator can be replaced by [itex]n^3+ 1> n[/itex]. [itex]0\le arctan(n)\le 2\pi[/itex] so the numerator can be replaced by [itex]n^2- 2\pi< n^2[/itex].
     
  4. Jul 25, 2009 #3
    Thanks for the reply

    I thought [itex]- \pi/2\le arctan(n)\le \pi/2[/itex] for one

    and two, based on your reasoning, our series will end up being smaller than [itex]n^3/n^2 = n[/itex] which in turn diverges, therefore this test is inconclusive. :confused:

    And the denominator [itex]n^3 + sin(n)[/itex] will make the whole series bigger that a series with a denominator of [itex]n^3+ 1[/itex] but having a numerator that is smaller than the numerator of the second series with the denominator of [itex]n^3+ 1[/itex], will make it hard to decide which series is bigger.

    thank you
     
  5. Jul 25, 2009 #4

    Dick

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    Call the nth term of the series an. Then you know limit n*an as n->infinity is 1, right? Compare the series with 1/(2n) for sufficiently large n.
     
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