Determine if the series $\frac{((\ln(n))^3}{n}$ is convergent or divergent. 
Determine if the series $\frac{((\ln(n))^3}{n}$ is convergent or divergent.

What I Tried :- If I were to use the comparison test would I end up with $(\ln(n))^3 > 1/n^2 > 0$. So
$\frac{1}{n^2}$ is convergent by $p$-test as $(p=2>1)$.
Therefore the original series is convergent by comparison test.
Can anyone help me understand If I am heading in the right direction?

That does not work.
The comparison test states that if $\sum b_{k}$ converges and $0\leq a_{n} \leq b_{n}$ for all natural numbers $n$ greater than some fixed natural number $N$, then $\sum a_{k}$ converges.
On the other hand if $\sum a_{k}$ diverges and $0\leq a_{m} \leq b_{m}$ for all natural numbers $m$ greater than some fixed natural number $M$, then $\sum b_{k}$ diverges.
Hint: What happens if you compare $\frac{(\ln n)^{3}}{n}$ with $\frac{1}{n}$
