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: 45



Work Step by Step

Here, we have: $ s_n=(1-\dfrac{1}{\sqrt 2})+(\dfrac{1}{\sqrt 2}-\dfrac{1}{\sqrt 3})+(\dfrac{1}{\sqrt k}-\dfrac{1}{\sqrt {k +1}})=1-\dfrac{1}{\sqrt {k+1}}$ Thus, we have the sum: $\lim\limits_{n \to \infty} s_n=\lim\limits_{n \to \infty} [1-\dfrac{1}{\sqrt {k+1}}]=\lim\limits_{n \to \infty}1- \lim\limits_{n \to \infty}\dfrac{1}{\sqrt {k+1}} =1$
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