Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.5 Alternating Series - 11.5 Exercises - Page 776: 16



Work Step by Step

Alternating series test: Suppose that we have 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 condition 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 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 the form of $\frac{\infty}{\infty}$, we can use L-hospital rule. $=\lim\limits_{n \to \infty}\frac{1}{2^nln2}$ $=\frac{1}{\infty}$ $=0$ Hence, the given series is convergent by Alternating Series Test.
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