Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 9 - Infinite Series - 9.6 Alternating Series; Absolute And Conditional Convergence - Exercises Set 9.6 - Page 646: 16

Answer

Converges absolutely

Work Step by Step

Apply the ratio test. Therefore, $ L=\lim\limits_{k \to \infty} |\dfrac{a_{k+1}}{a_k}|=\lim\limits_{k \to \infty} \dfrac{1}{(k+1)!} \times \dfrac{k!}{1}\\=\lim\limits_{k \to \infty} \dfrac{k !}{(k+1)!} \\=\lim\limits_{k \to \infty} \dfrac{1}{k+1} \\=0 \lt 1$ So, we can conclude that the given series converges absolutely by the ratio test.
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