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: 3

Answer

Divergent

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. Given: $-\frac{2}{5}+\frac{4}{6}-\frac{6}{7}+\frac{8}{8}-\frac{10}{9}+....$ The terms are becoming larger, $\frac{2}{5}\lt \frac{4}{6} \lt \frac{6}{7} \lt \frac{8}{8} \lt\frac{10}{9}+....$ Thus, Alternating Series Test does not apply. We can come up with the formula that generates the terms: General Term $\Sigma _{n=1}^{\infty}(-1)^{n}\frac{2n}{n+4}$ Evaluate as $n$ approaches infinity. Thus, $-\frac{2}{5}+\frac{4}{6}-\frac{6}{7}+\frac{8}{8}-\frac{10}{9}+....=\Sigma_{n=1}^{\infty}(-1)^{n}\frac{2n}{n+4}$ $=\Sigma_{n=1}^{\infty}(-1)^{n}\frac{2}{1+\frac{4}{n}}$ $=\Sigma_{n=1}^{\infty}(-1)^{n}(2)= DNE$ which means that the limit does not exist, so the series diverges by the Test of Divergence.
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