Answer
Converges
Work Step by Step
We have: $a_k=\dfrac{\sqrt[k] k}{k^3}$ and $b_k=\dfrac{1}{k^3}$
Now, we need to apply the limit comparison test.
$L=\lim\limits_{k \to \infty}\dfrac{a_k}{b_k}\\=\lim\limits_{k \to \infty}\dfrac{\dfrac{\sqrt[k] k}{k^3}}{1/k^{3}}\\=\lim\limits_{k \to \infty} \sqrt[k] k\\=1$
We see that $0 \lt \lim\limits_{k \to \infty}\dfrac{a_k}{b_k} \lt \infty$
and the series $\Sigma b_k$ convergent p-series with $p=3$.Therefore, the given series converges by the limit comparison test.
