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: 47


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

Work Step by Step

The nth partial sums are: $ 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)})]$ Now, $\lim\limits_{n \to \infty} s_n=\lim\limits_{n \to \infty} [-\dfrac{1}{\ln 2} +\dfrac{1}{ \ln (n+2)})]=\lim\limits_{n \to \infty} [-\dfrac{1}{\ln 2}] +\lim\limits_{n \to \infty} \dfrac{1}{ \ln (n+2)})]$ or, $=-(\dfrac{1}{\ln 2})+0$ or, $=-\dfrac{1}{\ln 2}$
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