Answer
Divergent
Work Step by Step
The given series can be re-written as:
$\Sigma_{n=1}^\infty \dfrac{-8}{n}=(-8)\Sigma_{n=1}^\infty (\dfrac{1}{n})$
This shows a harmonic series with unbounded partial sums.
Therefore, we have the series $\Sigma_{n=1}^\infty \dfrac{1}{n}$ diverges, so its sum does as well.
Thus, the series is divergent.