Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.4 The Divergence and Integral Tests - 8.4 Exercises: 15

Answer

Diverges

Work Step by Step

$\lim _{k\rightarrow \infty }\dfrac {\sqrt {k}}{\ln ^{10}k}=\dfrac {\dfrac {\partial }{\partial k}\sqrt {k}}{\dfrac {\partial }{\partial k}\ln ^{10}k}=\dfrac {\dfrac {1}{2\sqrt {k}}}{\ln ^{9}k\times \dfrac {1}{k}}=\dfrac {\sqrt {k}}{2\ln ^{9}k}=\dfrac {\dfrac {\partial }{\partial k}\sqrt {k}}{\dfrac {\partial }{\partial k}\left( 2\ln ^{9}k\right) }...=\dfrac {\sqrt {k}}{2^{10}}=\infty \neq 0$ then the series diverges
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