University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.9 - Convergence of Taylor Series - Exercises - Page 542: 30


$x+\dfrac{1}{2}x^2+\dfrac{5}{6}x^3+\dfrac{7}{12} x^4+...$

Work Step by Step

We know that the Taylor Series for $ \ln (1+x)$ and $\dfrac{1}{1-x}$ is defined as: $ \ln (1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-...$ and $\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n =1+x+x^2+....+x^n$ Now, $(\ln (1+x)) \cdot (\dfrac{1}{1-x})=(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-..)\cdot (1+x+x^2+....+x^n)$ or, $=x+\dfrac{1}{2}x^2+\dfrac{5}{6}x^3+\dfrac{7}{12} x^4+...$
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