Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Review - Exercises - Page 785: 29

Answer

$$\frac{\pi}{4}$$

Work Step by Step

$$\Sigma_{n=1}^{\infty}[tan^{-1}(n+1)-tan^{-1}n]=\lim\limits_{k \to \infty}\Sigma_{n=1}^{k}[tan^{-1}(n+1)-tan^{-1}n]$$$$=\lim\limits_{k \to \infty}tan^{-1}(k+1)-tan^{-1}1$$$$=\frac{\pi}{2}-\frac{\pi}{4}$$$$=\frac{\pi}{4}$$
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