Answer
Interval of Convergence: $0 \lt x \lt 16$ and $\Sigma_{n=0}^{\infty} (\dfrac{\sqrt{x}}{2}-1)^n=\dfrac{2}{4-\sqrt x}$
Work Step by Step
Remember that the series will converge when
$|f(x)|=|\dfrac{\sqrt{x}}{2}-1| \lt 1$
This implies that $-1 \lt \dfrac{\sqrt x}{2}-1 \lt 1$
and $0 \lt \dfrac{\sqrt x}{2} \lt 2$
or, $0 \lt x \lt 16$
This means that the interval of convergence is : $0 \lt x \lt 16$
Next, we will find the sum of the given geometric series
$S=\dfrac{a}{1-r}=\dfrac{1}{1-(\dfrac{\sqrt{x}}{2}-1)}=\dfrac{2}{4-\sqrt x}$
and $\Sigma_{n=0}^{\infty} (\dfrac{\sqrt{x}}{2}-1)^n=\dfrac{2}{4-\sqrt x}$