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

Answer

Converges

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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.