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Determining the convergence or divergence with the given nth term

  1. Apr 13, 2012 #1
    [answered]

    I want to know why this particular approach is wrong so I can learn from my mistakes.
    1. The problem statement, all variables and given/known data
    [tex]a_n = \frac{ln(n^3)}{2n}[/tex]


    3. The attempt at a solution
    For the sake of being time efficient, I will skip writing things like the limit as n approaches infinity etc.

    [tex]a_n = \frac{ln(n^3)}{2n}[/tex]
    I let n^3 = u
    [tex]a_n = \frac{ln(u)}{2n}[/tex]
    Using L'hopital's rule,
    [tex]\frac{3n^2}{2n^3}[/tex]
    Applying it once more,
    [tex]\frac{6n}{6n^2}[/tex]
    Which simplifys to
    [tex]\frac{6}{12n}[/tex]

    This shows that the function diverges. Which is the wrong answer.
     
    Last edited: Apr 13, 2012
  2. jcsd
  3. Apr 13, 2012 #2
    I'm confused, [itex]\frac{6}{12n}[/itex] converges to zero as n goes to infinity. Why do you interpret that as saying the function diverges?
     
  4. Apr 13, 2012 #3
    I'm shocked at my carelessness. I know that a number divided by infinity will equal 0. It is one of basic properties of infinite limits that I always use. So then how could I have done this careless mistake? It baffles me.
     
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