Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Review - Exercises - Page 785: 21



Work Step by Step

Since there is a factor $(-1)^{n-1}$ in the summation, it is an Alternating Series. We can use the Alternating Series Test by checking that the two following conditions are true: 1. $|a_{n}|\geq |a_{n+1}|$ for all $n$ 2.$ \lim\limits_{n \to \infty}|a_{n}|=0$ The first condition means that the absolute value of the terms have to get smaller, and the second condition tells us how small should they get: $0!$ Let $a_{n}=(-1)^{n-1}\frac{\sqrt n}{n+1}$, then $|a_{n}|= \frac{\sqrt n}{n+1}$, (1) $|a_{1}|=0.5,|a_{2}|=0.4714,|a_{3}|=0.4330,....$ As we can see, the absolute value of the terms are indeed getting smaller. (2) $ \lim\limits_{n \to \infty}|a_{n}|=\lim\limits_{n \to \infty}\frac{\sqrt n}{n+1}$ $=\lim\limits_{n \to \infty}\frac{\sqrt n}{n}$ $=\lim\limits_{n \to \infty}\frac{1}{\sqrt n}$ $=\frac{1}{\infty}$ $=0$ Thus, both conditions are verified. Hence, the series is convergent.
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.