Answer
Converges
Work Step by Step
Consider $a_n=\dfrac{(\ln n)^2}{n^3}$ and $b_n=\dfrac{1}{ n^2}$
Now, $\lim\limits_{n \to \infty}\dfrac{a_n}{b_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{(\ln n)^2}{n^3}}{\dfrac{1}{ n^2}}=\lim\limits_{n \to \infty} \dfrac{(\ln n)^2 }{n}$
and $\lim\limits_{n \to \infty} \dfrac{(\ln n)^2 }{n}=\dfrac{\infty}{\infty}$
Thus, we need to apply L-Hospital's rule.
so, we have $ \lim\limits_{n \to \infty} \dfrac{2 \ln n(1/n)}{1}=0$
Hence, the series converges by the comparison test.