University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.2 - Infinite Series - Exercises - Page 498: 42



Work Step by Step

We have: $a_n=\dfrac{6}{(2n-1)(2n+1)}=3(\dfrac{1}{2n-1}-\dfrac{1}{2n+1})$ Here, $ s_n=3(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}+.....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1})=3(1-\dfrac{1}{2n+1})$ Thus, we have the sum: $\lim\limits_{n \to \infty} s_n=3(1-\dfrac{1}{2n+1})=3$
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