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Series diverge or converge?

  1. Jun 12, 2012 #1
    1. The problem statement, all variables and given/known data
    series for (n^2+1) / (n^3+1)


    2. Relevant equations



    3. The attempt at a solution
    I was under the impression that if the denominator is increasing at a larger rate than the numerator, then it will become smaller and smaller approaching zero? So it would converge

    But I did the working and the comparison test dictates that it must diverge

    Which one is right?
     
  2. jcsd
  3. Jun 12, 2012 #2

    vela

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    Just because the terms of the series approach 0 doesn't mean the series will converge.

    The converse, however, is true: If the series converges, the terms approach 0. The n-th term test is the contrapositive of this statement: If the terms don't approach 0, the series diverges. Note it says nothing about what happens when the terms do go to 0.
     
    Last edited: Jun 12, 2012
  4. Jun 12, 2012 #3
    Well, 1 + 1/2 + 1/3 + 1/4 + ... diverges but the terms approach 0 (see "Harmonic Series" in wikipedia)
     
  5. Jun 12, 2012 #4
    The comparison test is right.
     
  6. Jun 12, 2012 #5

    Curious3141

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    Expressing the general term as [itex]\frac{1}{n - \frac{n-1}{n^2+1}}[/itex] and comparing to the harmonic series gives a quick answer.
     
  7. Jun 12, 2012 #6
    Dividing top and bottom by n2 (or even n3) might do what you intended a little nicer, me thinks. But like Curious said, it gives you a quick answer, and for instance is not a proof, that I know of.
     
  8. Jun 13, 2012 #7
    Use a comparison test, with n^2<n^2+1, and 1/(n^3+n^2)<1/(n^3+1), combining this will gives 1/(n+1)<Sn
     
  9. Jun 13, 2012 #8
    I think you're confusing series and sequences
     
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