Answer
$0.00144$
Work Step by Step
$s_{3}=\Sigma_{n=1}^{3}\frac{1}{(2n+1)^{6}}$
$=\frac{1}{(2(1)+1)^{6}}+\frac{1}{(2(2)+1)^{6}}+\frac{1}{(2(3)+1)^{6}}$
$=\frac{1}{3^{6}}+\frac{1}{5^{6}}+\frac{1}{7^{6}}$
$=0.00144$
Calculus 8th Edition
by Stewart, James
Published by
Cengage
ISBN 10:
1285740629
ISBN 13:
978-1-28574-062-1
Chapter 11 - Infinite Sequences and Series - 11.3 The Integral Test and Estimates of Sums - 11.3 Exercises - Page 766: 39
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