University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Practice Exercises - Page 553: 52


$\ln (\dfrac{5}{3}) \approx 0.510825624$

Work Step by Step

Since, we have the form of the given series as: $\ln (1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-....+(-1)^{n-1}\dfrac{x^n}{n}$ Here, in the given problem $x=\dfrac{2}{3}$ Thus, the sum of the series is: $\ln (1+\dfrac{2}{3})=\ln (\dfrac{3+2}{3})=\ln (\dfrac{5}{3}) \approx 0.510825624$
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