Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - 10.6 Polar Equations of Conics - Exercises 10.6 - Page 604: 60

Answer

$$0.692580927$$

Work Step by Step

Given: $\dfrac{s_n+s_{n+1}}{2}=s_n+\dfrac{1}{2} (-1)^{n+2} a_{n+1}$ After simplification, we have: $s_n=\dfrac{s_{n+1}}{2}-\dfrac{(-1)^{n+2}}{2} a_{n+1}$ This implies that $s_{20}= 1-\dfrac{1}{2}+\dfrac{1}{3}-......-\dfrac{1}{20} \approx 0.66 87714032$ and $s_{20}+(\dfrac{1}{2}) (\dfrac{1}{21}) = 0.66 87714032+\dfrac{1}{42}$ or, $s_{20}+(\dfrac{1}{2}) (\dfrac{1}{21}) =0.692580927$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.