# Determining the convergence or divergence with the given nth term

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.

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