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
The series will converge when $|f(x)|=|\dfrac{\sqrt{x}}{2}-1| \lt 1$
So, $-1 \lt \dfrac{\sqrt x}{2}-1 \lt 1$
$ \implies 0 \lt \dfrac{\sqrt x}{2} \lt 2$
$ \implies 0 \lt x \lt 16$
Now, the sum of the given geometric series is:
$S=\dfrac{a}{1-r}=\dfrac{1}{1-(\dfrac{\sqrt{x}}{2}-1)}=\dfrac{2}{4-\sqrt x}$
Thus, the Interval of Convergence is:
$0 \lt x \lt 16$
and $\Sigma_{n=0}^{\infty} (\dfrac{\sqrt{x}}{2}-1)^n=\dfrac{2}{4-\sqrt x}$