Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.6 Alternating Series - 8.6 Exercises - Page 657: 13

Answer

Diverges

Work Step by Step

For alternating series $\Sigma_{k=1}^\infty(-1)^k a_k$ (let us consider), to be convergent , we must follow the two conditions such as: a) The magnitude of terms must form a non-increasing sequence. b) $\lim\limits_{k \to \infty} a_k=0$ We are given that $a_k=\dfrac{k}{3k+2}$ 1) In the given sequence, $a_k=\dfrac{k}{3k+2}$; the magnitude of terms are a non-increasing sequence. 2) $\lim\limits_{k \to \infty} a_k=\lim\limits_{k \to \infty} \dfrac{k}{3k+2}=\dfrac{1}{3} \ne 0$ This means that the given sequence diverges.
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