Answer
Converges to $\dfrac{6}{3-x}$ for $ -1 \lt \dfrac{x-1}{2} \lt 1$ or, $ -1 \lt x \lt 3$
Work Step by Step
The sum of a geometric series can be found as:
$S=\dfrac{a}{1-r}$
The given series shows a convergent geometric series with first term, $a=3$ and common ratio $r =\dfrac{x-1}{2}$
$S=\dfrac{3}{1-(\dfrac{x-1}{2})}=\dfrac{6}{3-x}$
Hence, the given series converges to $\dfrac{6}{3-x}$ for $ -1 \lt \dfrac{x-1}{2} \lt 1$ or, $ -1 \lt x \lt 3$
