Answer
$ 0 \lt x \lt 2$
Work Step by Step
Write $\dfrac{2}{x} $ as $\dfrac{2}{1-[-(x-1)]}$
The series will converge when $|f(x)|=|\dfrac{2}{1-[-(x-1)]}| \lt 1$
Re-write as a power series:
$\Sigma_{n=0}^{\infty} 2 [-(x-1)]^n =\Sigma_{n=0}^{\infty} 2(-1)^n (x-1)^n$
Since this series is a geometric series with ratio $-(x-1)$, it will converge for $|-(x-1)|$ or, $0 \lt x \lt 2$
