Answer
Converges
Work Step by Step
Let $u_n=\dfrac{1}{\sqrt{n^3+2}}$ and $v_n=\dfrac{1}{ \sqrt{n^3}}$
Now, $\lim\limits_{n \to \infty}\dfrac{u_n}{v_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{1}{\sqrt{n^3+2}}}{1/\sqrt{n^3}}$
$ \implies \lim\limits_{n \to \infty} \sqrt{\dfrac{n^3}{(n^3+2)}}=1$
Here, $v_n=\Sigma_{n=1}^\infty \dfrac{1}{ \sqrt{n^3}}=\dfrac{1}{ n^{3/2}}$ is a convergent p-series with $p=\dfrac{3}{2}$
Thus, the given series converges by the limit comparison test.