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: 47


$-\dfrac{1}{\ln 2}$

Work Step by Step

Here, we have: $ s_n=(\dfrac{1}{\ln 3}-\dfrac{1}{\ln 2})+(\dfrac{1}{\ln 4}-\dfrac{1}{\ln 3})+.....+(\dfrac{1}{\ln (n+2)}-\dfrac{1}{\ln (n+1)})=-\dfrac{1}{\ln 2} +\dfrac{1}{ \ln (n+2)})$ Thus, we have the sum: $\lim\limits_{n \to \infty} s_n=\lim\limits_{n \to \infty} [-\dfrac{1}{\ln 2} +\dfrac{1}{ \ln (n+2)})]=\lim\limits_{n \to \infty} [-\dfrac{1}{\ln 2}] +\lim\limits_{n \to \infty} \dfrac{1}{ \ln (n+2)})]=-\dfrac{1}{\ln 2}+0=-\dfrac{1}{\ln 2}$
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