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 10.9 - Convergence of Taylor Series - Exercises 10.9 - Page 625: 30



Work Step by Step

Consider the Taylor Series for $ \ln (1+x)$ and $\dfrac{1}{1-x}$ as follows: $ \ln (1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-...$; $\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n =1+x+x^2+....+x^n$ This implies that $\ln (1+x) \cdot (\dfrac{1}{1-x})=(x-\dfrac{1}{2}x^2+\dfrac{1}{3}x^3-..)\cdot (1+x+x^2+....+x^n)$ Hence, $\ln (1+x) \cdot (\dfrac{1}{1-x})=x+\dfrac{x^2}{2}+\dfrac{5x^3}{6}+\dfrac{7x^4}{12}+...$
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