## University Calculus: Early Transcendentals (3rd Edition)

Since, we have $\Sigma_{n=1}^\infty \dfrac{-8}{n}=-8\Sigma_{n=1}^\infty \dfrac{1}{n}$ Here, the given series is a harmonic series with partial sums not bounded. and, since $\Sigma_{n=1}^\infty \dfrac{1}{n}$ diverges, so does our sum. Hence, the given series diverges.