Answer
Diverges
Work Step by Step
Here, we have the given series as $\Sigma_{n=1}^{\infty} \dfrac{1}{ \sqrt[3] n-1}$
Let us consider that $a_n=\Sigma_{n=1}^{\infty} \dfrac{1}{ \sqrt[3] n-1} $ and $b_n=\Sigma_{n=1}^{\infty} \dfrac{4}{ \sqrt[3] n}$
We can see that $a_n \leq b_n$ and $b_n$ shows a p-series with $p=\dfrac{1}{3}$ and $p=\dfrac{1}{3} \lt 1$
This implies that the series $b_n$ by the p-series test.
Hence, the given series also diverges by the direct comparison test.
