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: 44



Work Step by Step

We have: $a_n=\dfrac{2n+1}{n^2(n+1)^2}=\dfrac{1}{n^2}-\dfrac{1}{(n+1)^2}$ Here, $ s_n=(1-\dfrac{1}{4})+(\dfrac{1}{4}-\dfrac{1}{9})+...[\dfrac{1}{n^2}-\dfrac{1}{(n+1)^2}]=1-\dfrac{1}{(n+1)^2}$ Thus, we have the sum: $\lim\limits_{n \to \infty} s_n=\lim\limits_{n \to \infty} [1-\dfrac{1}{(n+1)^2}]=1$
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