University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.3 - The Integral Test - Exercises - Page 504: 18



Work Step by Step

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.
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