Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.7 Strategy for Testing Series - 11.7 Exercises - Page 786: 27



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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