University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Practice Exercises - Page 552: 27


Conditionally convergent.

Work Step by Step

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 $p \gt 1$ and 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}{\sqrt{n}}=0 $ 2.$b_n=\dfrac{1}{\sqrt n}$is a decreasing sequence. Therefore, the given series is conditionally convergent.
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