# Determining the convergence or divergence with the given nth term

1. Apr 13, 2012

### Nano-Passion

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

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.

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

Last edited: Apr 13, 2012
2. Apr 13, 2012

### 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?

3. Apr 13, 2012

### Nano-Passion

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.