$\newcommand{\nfrac}[2]{\frac{\displaystyle{#1}}{\displaystyle{#2}}}$
Exercises
Ex. 7.03
denote $a_n = \nfrac{f^n(0)}{n!}$.
Observe \begin{align*} f^0(x) &= \nfrac{1}{1-2x} &f^0(0) = 1 \\ f’(x) &= 1 \cdot 2 \cdot (1-2x)^{-2} &f’(0) = 1 \cdot 2 \\ f’’(x) &= 1 \cdot 2 \cdot 2^2 \cdot (1 - 2x)^{-3} &f’’(0) = 1 \cdot 2 \cdot 2^2 \\ \end{align*} Generally $f^n(x) = n! \cdot 2^n (1 - 2x)^{-(n+1)}$ and $f^n(0) = n! \cdot 2^n$. It follows: \begin{align*} \nfrac{1}{1-2x} &= 1 + 2^1 x^1 + 2^2 x^2 + \dots \\ r &= 2x \\ \nfrac{1}{1-2x} &= (2x)^0 + (2x)^1 + (2x)^2 + \dots + (2x)^n \\ &= 1 + 2^1 x + 2^2 x^2 + \dots + 2^n x^n \end{align*} Hence it is valid for \begin{align*} |r| < 1 \\ |2x| < 1 \\ |x| < \nfrac{1}{2} \end{align*}