## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 10: Infinite Sequences and Series - Practice Exercises - Page 636: 27

#### Answer

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.

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.