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 10.2 - Infinite Series - Exercises 10.2 - Page 579: 37



Work Step by Step

The nth partial sums are: $ s_n=(\ln \sqrt 2 -\ln \sqrt 1)+(\ln \sqrt 3 -\ln \sqrt 2)+(\ln \sqrt 4 -\ln \sqrt 3)+.....(\ln \sqrt {n+1} -\ln \sqrt n)=\ln \sqrt {(n+1)}$ Now, apply limits, we get $\lim\limits_{n \to \infty} s_n=\lim\limits_{n \to \infty} \ln (\sqrt {(n+1)})=\infty$ Thus, when $s_n \to \infty$ as $n \to \infty$, thus, the series diverges.
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