Answer
The series $\mathop \sum \limits_{n = 2}^\infty \dfrac{1}{{{{\ln }^3}n}}$ diverges.
Work Step by Step
From the information prior to Exercise 31 in Section 11.3, we know that for $n$ sufficiently large, the following inequality is valid:
$\ln n \le {n^a}$, ${\ \ \ \ \ \ \ }$ for all $a \gt 0$
Let $a = \dfrac{1}{3}$. So, $\ln n \le {n^{1/3}}$.
Thus, for large $n$, we have
$\dfrac{1}{{{{\ln }^3}n}} \gt \dfrac{1}{n}$
We know that the harmonic series $\mathop \sum \limits_{n = 2}^\infty \dfrac{1}{n}$ diverges. By the Direct Comparison Test, the larger series $\mathop \sum \limits_{n = 2}^\infty \dfrac{1}{{{{\ln }^3}n}}$ also diverges.
