University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.5 - Absolute Convergence; The Ratio and Root Tests - Exercises - Page 516: 31



Work Step by Step

Consider $a_n=\dfrac{\ln n}{n}$ Since, we know that $\ln n \geq 1$ for the all value of $n$. Thus, $\Sigma_{n=1}^\infty \dfrac{\ln n}{n} \geq \Sigma_{n=1}^\infty \dfrac{1}{n}$ we see that the series $\Sigma_{n=1}^\infty \dfrac{1}{n}$ is a divergent harmonic series. Hence, the given series Diverges by the direct comparison test.
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