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Don's question at Yahoo! Answers (Taylor series)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello Don,

Denoting $t=x+3$ and using the geometric series:

$$\frac{1}{x}=\frac{1}{t-3}=-\frac{1}{3}\cdot\frac{1}{1-\frac{t}{3}}=-\frac{1}{3}\sum_{n=0}^{\infty} \left(\frac{t}{3}\right)^n \qquad \left(\;\left|\frac{t}{3}\right|<1\;\right)$$ Hence, $f(x)=-\displaystyle\sum_{n=0}^{\infty}\frac{(x+3)^n}{3^{n+1}}$, valid expasion for $|x+3|<3$, or equivalently for $x\in (-6,0).$