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 10.4 - Comparison Tests - Exercises 10.4 - Page 591: 22



Work Step by Step

Let $u_n=\dfrac{n+1}{n^2 \sqrt n}$ This can be re-written as: $u_n=\dfrac{n+1}{n^2 \sqrt n}=\dfrac{n}{(n)^{5/2}}+\dfrac{1}{(n)^{5/2}}$ Here, the first series $\Sigma_{n=1} ^\infty \dfrac{1}{(n)^{5/2}}$ is a convergent p-series with $p=\dfrac{5}{2} \gt 1$ Also, the second series $\Sigma_{n=1} ^\infty \dfrac{1}{(n)^{3/2}}$ is convergent p-series with $p=\dfrac{3}{2} \gt 1$ Thus, the given series converges by the p-series.
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