Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.5 - Absolute Convergence; The Ratio and Root Tests - Exercises 10.5 - Page 598: 27



Work Step by Step

Let us consider $a_n=\dfrac{\ln n}{n^3}$ Apply direct comparison test. This implies that $\Sigma_{n=1}^\infty \dfrac{\ln n}{n^3} \leq \Sigma_{n=1}^\infty \dfrac{(n)}{(n^3)}=\Sigma_{n=1}^\infty \dfrac{1}{n^{(2)}}$ Here, the series $\Sigma_{n=1}^\infty \dfrac{1}{n^2}$ shows a convergent $p$-series with $p=2 \gt 1$ Thus, the series converges by the direct comparison test.
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