Answer
Converges
Work Step by Step
Let $u_n=\dfrac{(\sqrt[n] n)}{n^{(2)}}$ and $v_n=(\dfrac{1}{ n^2})$
Now, $\lim\limits_{n \to \infty}\dfrac{u_n}{v_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{(\sqrt[n] n)}{n^{(2)}}}{1/n^2}$
$ \implies \lim\limits_{n \to \infty} \sqrt[n] n=\lim\limits_{n \to \infty} n^{(1/n)}=1$
Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^2}$ shows a convergent $p$-series test with $p \gt 1$.
Hence, the series converges by the limit comparison test.