Answer
$0 \lt x \lt 2$
Work Step by Step
Firstly, we will write $\dfrac{2}{x} $ as $\dfrac{2}{1-[-(x-1)]}$
Remember that the series will converge when
$|f(x)|=|\dfrac{2}{1-[-(x-1)]}| \lt 1$
Re-write as a power series:
$\Sigma_{n=0}^{\infty} (2) [-(x-1)]^n =(2) \Sigma_{n=0}^{\infty} (-1)^n (x-1)^n$
This series is a geometric series with a common ratio $-(x-1)$, and it will converge for $|-(x-1)|\lt 1$
This implies that $0 \lt x \lt 2$
