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Sequence Problem

  1. Mar 6, 2005 #1
    This seems like a simple problem....almost embarassed about it, but I can't figure it out:

    Determine if the following sequence converges or diverges (2n-1)/(3n^2 +1), n= 1,2,3,..... If the sequence converges, find its limit.

    I think the series converges to 0...............am i right? :confused:
  2. jcsd
  3. Mar 6, 2005 #2


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    You mean the sequence converges to 0. Well, for one, why don't you just take the limit as n approaches infinity? Or, if you're not allowed to do that, show that the sequence only has positive numbers, and that it is decreasing (decreasing for all n > 2, in this case). Therefore it is bounded and monotone, hence convergent. Then assume that it converges to some small positive number e. Choose a natural number n > 2/(3e) and show that the sequence for this n is less than e. This gives a contradiction, showing that the sequence does not converge to any positive number, hence it must converge to 0.
  4. Mar 7, 2005 #3
    sequence....opps yea...................i did find the limit to get my answer using L' Hopital's rule........i was thinking this was the "nth term test" which cannot show convergence.....only divergence....but then i realized this is a sequence and that test was for a series.
    Last edited: Mar 7, 2005
  5. Mar 7, 2005 #4


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    L'Hopital's rule seems like over-kill here. Standard method for dealing with fractions where n goes to infinity: Divide both numerator and denominator by the highest power of n (here n2) so that every term involves (1/n) to a power. As n goes to infinity, that goes to 0.
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