Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.7 - Strategy for Testing Series - 11.7 Exercises - Page 746: 27

Answer

Convergent

Work Step by Step

$\Sigma_{k=1}^{\infty}\frac{k lnk }{(k+1)^{3}}$ We note that $\sqrt k \gt lnk$ Then $\frac{\sqrt k}{k^{2}}=\frac{1}{k^{3/2}}$ As a series $\Sigma_{k=1}^{\infty}\frac{1}{k^{3/2}}$ converges because it is a p-series with $p=\frac{3}{2}\gt 1$
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