Determining the convergence or divergence with the given nth term

  • #1
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[answered]

I want to know why this particular approach is wrong so I can learn from my mistakes.

Homework Statement


[tex]a_n = \frac{ln(n^3)}{2n}[/tex]


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:

Answers and Replies

  • #2
294
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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?
 
  • #3
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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?
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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