Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.6 Power Series - Exercises - Page 578: 45


$$\ln (1+x) =\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{n}}{n}$$

Work Step by Step

Given $$\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n} x^{n}=1-x+x^{2}-x^{3}+\cdots$$ By integration, we get \begin{align*} \int \frac{1}{1+x}&=\int \sum_{n=0}^{\infty}(-1)^{n} x^{n} dx\\ \ln (1+x)&=C+ \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{n+1}}{n+1} \end{align*} To find $C$, put $x=0$, then $C= 0$, hence \begin{align*}\ln (1+x) &= \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{n+1}}{n+1} \\ &=\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{n}}{n}\\ &=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots\end{align*}
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