Answer
$n=11$
Work Step by Step
Recall the formula that $\text{Error}=|R_n| \leq a_{n+1}$
Here, we have
$R_n \leq \dfrac{1}{(2n+1)^3} \lt 10^{-4}$
and $ (2n+1)^3 \gt 10^{4}$
or, $2n+1 \gt \sqrt[3] {10^4}$
or, $n \gt 10.25$
or, $n=11$
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
Published by
Pearson
ISBN 10:
0321947347
ISBN 13:
978-0-32194-734-5
Chapter 8 - Sequences and Infinite Series - 8.6 Alternating Series - 8.6 Exercises - Page 657: 34
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