Calculus: Early Transcendentals 8th Edition

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

Chapter 11 - Section 11.5 - Alternating Series - 11.5 Exercises: 16



Work Step by Step

Alternating series test: Suppose that we have a series $\Sigma a_n$, such that $a_{n}=(-1)^{n}b_n$ or $a_{n}=(-1)^{n+1}b_n$, where $b_n\geq 0$ for all $n$. Then if the following two conditions are satisfied, the series is convergent. 1. $\lim\limits_{n \to \infty}b_{n}=0$ 2. $b_{n}$ is a decreasing sequence. In the given problem, $b_{n}=\frac{n}{2^n}$ which satisfies both conditions of the Alternating Series Test as follows: 1. $b_{n}=\frac{n}{2^n}$ is decreasing because the denominator is increasing. 2. $\lim\limits_{n \to \infty}b_{n}=\lim\limits_{n \to \infty}\frac{n}{2^n}$ Since, the limit is in the form of $\frac{\infty}{\infty}$, we can use L-hospital's rule. $=\lim\limits_{n \to \infty}\frac{1}{2^nln2}$ $=\frac{1}{\infty}$ $=0$ Hence, the given series is convergent by the Alternating Series Test.
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.