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Calculus 2 - Infinite Series

  1. Nov 6, 2011 #1
    Determine if the following series converges or diverges:

    Ʃ[k=1,inf] ( k^2-1 )/( k^3+4 )

    I don't see how to solve this problem

    the divergence test is inconclusive

    the ratio test is inconclusive

    the root test is inconclusive

    the integral test... not sure how to integrate this...

    the comparison test... I thought about comparing it to k^2/k^3=1/k but because 1/k > ( k^2-1 )/( k^3+4 ) and Ʃ 1/k is the divergent harmonic series so we cannot conclude anything

    the limit comparison test... Not sure what to use for the other series but 1/k and
    lim k->inf (k(k^2-1))/(k^3+4) = 0 but because 1/k is divergent harmonic series I don't know how to apply the comparison test in this case... we could conclude that the original series converges if Ʃ1/k converged to but sense it doesn't I don't think I can conclude anything from this... there's only 3 cases from the limit comparison test that which we can conclude something from, what do we do in a situation like this were it's not one of those cases???

    this is not a p-series

    this is not a geometric series

    I'm lost as to what to do. Thanks for any help.
     
  2. jcsd
  3. Nov 6, 2011 #2

    SammyS

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    Compare to [itex]\displaystyle \sum_{k=1}^{\infty}\frac{1}{k}\,.[/itex]
     
    Last edited: Nov 6, 2011
  4. Nov 6, 2011 #3
    do you mean

    Ʃ-1/k^2 ?

    is that prime suppose to be there?
     
  5. Nov 6, 2011 #4

    Mark44

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    The dominant term in the numerator is k2, and the dominant term in the denominator is k3, so for large k, (k2 - 1)/(k3 + 4) ≈ k2/k3 = 1/k.
     
  6. Nov 6, 2011 #5
    Ʃ[k=1,inf]-1/k^2 < Ʃ[k=1,inf] ( k^2-1 )/( k^3+4 )

    Ʃ[k=1,inf]-1/k^2 is a convergent p series

    so we cannot conclude anything because the smaller one is convergent not the bigger one?
     
  7. Nov 6, 2011 #6
    but sense
    1/k > ( k^2-1 )/( k^3+4 )
    and 1/k diverges I thought we couldn't conclude anything sense the larger one diverges not the smaller one?
     
  8. Nov 6, 2011 #7

    Mark44

    Staff: Mentor

    There's a subtlety to the comparison test that you're not getting.

    Suppose we're investigating Ʃan, a series of positive terms. Also, suppose that Ʃcn is a series that is known to converge and Ʃdn is a series that is known to diverge.

    If an ≤ cn, then Ʃan converges.
    If an ≥ dn, then Ʃan diverges.

    Notice that there are two cases that aren't covered here:
    when an ≥ cn,
    and when an ≤ dn.

    We can't conclude anything about the series Ʃan in those two cases.
     
  9. Nov 6, 2011 #8

    Mark44

    Staff: Mentor

    Do you know this, or do you just think it's true?

    Do you know the limit comparison test?
     
  10. Nov 6, 2011 #9
    Ya I know about this. Wouldn't this be one of the cases that aren't covered were

    a_n > b_n

    and a_n is the divergent harmonic series that we are comparing to Ʃ1/k and b_n is the original series ( k^2-1 )/( k^3+4 )

    and ya it seems to be true for large values of k

    and ya i know about the limit comparison test
     
  11. Nov 6, 2011 #10

    SammyS

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    Yes, you are right about this.

    Compare your series to [itex]\displaystyle \sum_{k=1}^{\infty}\frac{1}{k+2}\,[/itex] term by term.

    For k>2, [itex]\displaystyle \frac{ k^2-1 }{ k^3+4}>\frac{1}{k+2}\,.[/itex]

    You can show this by subtracting one from the other.

    [itex]\displaystyle \frac{ k^2-1 }{ k^3+4}-\frac{1}{k+2}=\frac{(k-2) (2 k+3)}{(k+2) (k^3+4)}[/itex]
     
  12. Nov 6, 2011 #11
    ah ya I was able to conclude that it diverges by using the limit comparison test with 1/k

    the above also works because the series would diverge by the integral test so the larger one must as well by the comparison test.

    Thanks for the help.
     
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