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 522: 60



Work Step by Step

We are given that $\dfrac{s_n+s_{n+1}}{2}=s_n+\dfrac{1}{2} (-1)^{n+2} a_{n+1}$ This gives that $s_n=\dfrac{s_{n+1}}{2}-\dfrac{1}{2} (-1)^{n+2} a_{n+1}$ Thus, $s_{20}= 1-\dfrac{1}{2}+\dfrac{1}{3}-......-\dfrac{1}{20}$ So, $s_{20} \approx 0.66 87714032$ Now, $s_{20}+(\dfrac{1}{2}) (\dfrac{1}{21}) = 0.66 87714032+\dfrac{1}{42}=0.692580927$
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