Answer
$0 \lt x \lt 2$
$f(x)=\displaystyle \frac{x-1}{2-x},\ \quad 0 \lt x \lt 2$
Work Step by Step
We aim for terms of Th.9.6 (Convergence of a Geometric Series).
A geometric series with ratio $r$ diverges when $|r| \geq 1$.
If $0 \lt |r| \lt 1$, then the series converges to the sum
$\displaystyle \sum_{n=0}^{\infty}ar^{n}=\frac{a}{1-r},\quad 0 \lt |r| \lt 1$.
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Adjusting for the subscript,
$\displaystyle \sum_{n=1}^{\infty}(x-1)^{n}=(x-1)^{1}+(x-1)^{2}+(x-1)^{3}+...$
$=(x-1)[1+(x-1)^{1}+(x-1)^{2}+...]$
$=(x-1)\displaystyle \sum_{n=0}^{\infty}(x-1)^{n}$
The series converges for $|x-1| \lt 1 \Rightarrow$
$-1 \lt x-1 \lt 1\qquad/+1$
$0 \lt x \lt 2$
$f(x)=(x-1)\displaystyle \sum_{n=0}^{\infty}(x-1)^{n}$
$=(x-1)\displaystyle \frac{1}{1-(x-1)}=\frac{x-1}{2-x},\ 0 \lt x \lt 2$
