University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.4 - Comparison Tests - Exercises - Page 509: 22



Work Step by Step

Consider $a_n=\dfrac{n+1}{n^2 \sqrt n}$ or, $a_n=\dfrac{n+1}{n^2 \sqrt n}=\dfrac{n}{n^{5/2}}+\dfrac{1}{n^{5/2}}$ Now, $\Sigma_{n=1} ^\infty \dfrac{1}{n^{5/2}}$ is convergent due to the p-series with $p=\dfrac{5}{2} \gt 1$ and $\Sigma_{n=1} ^\infty \dfrac{1}{n^{3/2}}$ is convergent due to the p-series with $p=\dfrac{3}{2} \gt 1$ Hence, the the sum of the series converges due to the convergent p-series.
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