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 - Practice Exercises - Page 636: 27


conditionally convergent.

Work Step by Step

The alternating series test states that if the following conditions are met then the series is convergent. 1. $\lim\limits_{ n\to \infty} b_n=0$ 2.$b_n$ is a decreasing sequence. A $p$-series has the form of $\Sigma_{n=k}^{\infty}\dfrac{1}{n^p}$.It is convergent iff $p \gt 1$ otherwise diverges. From the given problem, we have $\Sigma_{n=1}^{\infty}|\dfrac{(-1)^n}{\sqrt n}|$ The given series can be re-written as: $a_n=\Sigma_{n=1}^{\infty}\dfrac{1}{ n^{1/2}}$ Here, $p=1/2=0.5 \lt 1$ Thus, the given series is divergent. So, it can not be absolutely convergent. From the given series, we get 1. $\lim\limits_{ n\to \infty} b_n=\lim\limits_{ n\to \infty} \dfrac{1}{2}\dfrac{1}{ n^{1/2}}=0$ 2.$b_n=\dfrac{1}{\sqrt n}$is a decreasing sequence. Therefore, the given series is conditionally 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.