Answer
The series $\mathop \sum \limits_{n = 2}^\infty \dfrac{1}{{{n^{\ln n}}}}$ converges.
Work Step by Step
We know that $\ln \left( {{{\rm{e}}^2}} \right) = 2$.
Since $\ln n$ is increasing, for large $n$ ($n \gt {{\rm{e}}^2}$), we have $\ln n \gt 2$.
Thus, for large $n$, the following inequality is valid (please see the figure attached):
$\dfrac{1}{{{n^{\ln n}}}} \lt \dfrac{1}{{{n^2}}}$
The larger series $\mathop \sum \limits_{n = 2}^\infty \dfrac{1}{{{n^2}}}$ converges (because $p = 2 \gt 1$ by the Convergence of $p$-Series).
Therefore, by the Direct Comparison Test, the smaller series $\mathop \sum \limits_{n = 2}^\infty \dfrac{1}{{{n^{\ln n}}}}$ also converges.
