Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.6 - Alternating Series and Conditional Convergence - Exercises 10.6 - Page 603: 54


$$n \geq 999$$

Work Step by Step

Consider: $S_n=a_1-a_2+......+(-1)^{n+1}a_n$ and $$|S-S_n| \leq |a_{n+1}| \\ \implies |a_{n+1} | \lt 0.001 =|\dfrac{n+1}{(n+1)^2+1}| \lt 0.001 \\ \implies (n+1)^2+1 \gt 1000(n+1) \\ \implies n \geq 999$$
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