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: 32



Work Step by Step

$\Sigma_{k=1}^\infty2k^\frac{-3}{2}=2\Sigma_{k=1}^\infty\frac{1}{k^\frac{3}{2}}$ $p=\frac{3}{2}\gt1$, therefore $\Sigma_{k=1}^\infty\frac{1}{k^\frac{3}{2}}$ converges as does $2\Sigma_{k=1}^\infty\frac{1}{k^\frac{3}{2}}$
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