University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.6 - Alternating Series and Conditional Convergence - Exercises - Page 521: 56


$n \gt e^{e^{1000}}-3$

Work Step by Step

We have $S_n=a_1+a_2+......+(-1)^{n+1}a_n$ Now, $|S-S_n| \leq |a_{n+1}|$ So, $|a_{n+1} | \lt 0.001 $ or, $|\dfrac{1}{\ln (\ln (n+3))}| \lt 0.001$ or, $\ln (n+3) \gt e^{1000}$ or, $n \gt e^{e^{1000}}-3$ Thus, we need $n \gt e^{e^{1000}}-3$ terms to estimate the sum of the entire series when an error is less than $0.001$.
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