Answer
Converges
Work Step by Step
Let $u_n=\dfrac{(\ln n)^2}{n^3}$ and $v_n=\dfrac{1}{ n^2}$
Now, $\lim\limits_{n \to \infty}\dfrac{u_n}{v_n} =\lim\limits_{n \to \infty}\dfrac{(\dfrac{\ln n}{n^3})^2}{1/ n^2}$
But $\lim\limits_{n \to \infty} \dfrac{(\ln n)^2 }{n}=\dfrac{\infty}{\infty}$
We will have to apply L-Hospital's rule.
This implies that $ \lim\limits_{n \to \infty} \dfrac{2 (\ln n)(\dfrac{1}{n})}{1}=0$
Thus, the series converges by the comparison test.
