Determining the convergence or divergence with the given nth term

[Nano-Passion]

[answered]

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

Homework Statement

a_n = \frac{ln(n^3)}{2n}

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.

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

This shows that the function diverges. Which is the wrong answer.

 

[Poopsilon]

I'm confused, \frac{6}{12n} converges to zero as n goes to infinity. Why do you interpret that as saying the function diverges?

 

[Nano-Passion]

I'm confused, \frac{6}{12n} 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.