Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.3 - The Integral Test - Exercises 10.3 - Page 586: 16



Work Step by Step

Consider the series $\Sigma_{n=1}^\infty \dfrac{-2}{n\sqrt n}$ It can be re-written as: $\Sigma_{n=1}^\infty \dfrac{-2}{n\sqrt n}=-2\Sigma_{n=1}^\infty \dfrac{1}{n^{3/2}} $ This shows that the series is a $p$-series with $p=\dfrac{3}{2} \geq 1$ Thus, the series is convergent.
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