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: 48

Answer

Converges

Work Step by Step

Let $u_n=\Sigma_{n=1}^\infty\dfrac{\sec^{-1} n}{n^{(1.3)}}$ We compare with: $\Sigma_{n=1}^\infty\dfrac{\sec^{-1} n}{n^{(1.3)}} \lt \Sigma_{n=1}^\infty \dfrac{{\dfrac{\pi}{2}}}{n^{(1.3)}}$ Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^{(1.3)}}$ shows a convergent $p$-series with $p=1.3 \gt 1$ Thus, the series converges by the direct comparison test.
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