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: 55

Answer

$n \geq 4$

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, $|\dfrac{1}{((n+1)+3 \sqrt {n+1})^3}| \lt 0.001$ or, $((n+1)+3 \sqrt {n+1})^3 \gt 1000$ or, $n \geq 4$ Thus, we need $n \geq 4$ terms to estimate the sum of the entire series when an error is less than $0.001$.
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