Calculus: Early Transcendentals (2nd Edition)

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

Chapter 8 - Sequences and Infinite Series - Review Exercises - Page 658: 9

Answer

Divergent

Work Step by Step

Let us consider that $\lim\limits_{k \to \infty}a_k=\lim\limits_{k \to \infty} \dfrac{(-1)^n}{0.9^n})\\=\lim\limits_{k \to \infty}(- \dfrac{1}{0.9})^n\\=\lim\limits_{k \to \infty} (\dfrac{-10}{9})^n$ This shows a divergent geometric sequence.
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