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 - Section 10.2 - Infinite Series - Exercises 10.2 - Page 580: 42

Answer

$3$

Work Step by Step

Consider $a_n=\dfrac{6}{(2n-1)(2n+1)}=3(\dfrac{1}{2n-1}-\dfrac{1}{2n+1})$ The nth partial sums are: $ 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})$ Now, apply limits to find the sum That is, $\lim\limits_{n \to \infty} s_n=3(1-\dfrac{1}{2n+1})=3$
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