Answer
Diverges
Work Step by Step
Consider $a_n=\dfrac{1}{n \sqrt[n] n}$ and $b_n=\dfrac{1}{ n}$
Now, $\lim\limits_{n \to \infty}\dfrac{a_n}{b_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{1}{n \sqrt[n] n}}{\dfrac{1}{ n}}$
or, $ =\lim\limits_{n \to \infty} \dfrac{1}{\sqrt[n] n}$
and $\lim\limits_{n \to \infty} \dfrac{1}{n^{1/n}}=\dfrac{1}{1}=1 \ne 0 \ne \infty $
Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^2}$ is a divergent series due to the p-series test with $p \leq 1$
Thus, the series diverges by the limit comparison test.