Answer
The series converges by direct comparison.
Work Step by Step
$\Sigma_{n=2}^{\infty}\frac{1}{n^3-8}$
Let's compare this series with the series
$\Sigma_{n=2}^{\infty}\frac{1}{n^3}$, which converges by p-series, where $p=3$ and $3\gt1$:
If $\frac{1}{n^3-8}\leq\frac{1}{n^3}$, then the original series also converges. Therefore, the original series converges by direct comparison.
