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 - Page 638: 16

Answer

Diverges

Work Step by Step

$\lim _{k\rightarrow \infty}\dfrac {\sqrt {k^{2}+1}}{k}=\lim _{k\rightarrow \infty }\sqrt {1+\dfrac {1}{k^{2}}}=1\neq 0$ then the series diverges
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