Answer
The series converges if $x ϵ (-\infty, -2)$ U $(2,\infty)$ and the sum is $\frac{x}{x-2}$
Work Step by Step
$\Sigma^{\infty}_{n=0} \frac{2^{n}}{x^{n}} = \Sigma^{\infty}_{n=0}(\frac{2}{n})^{n}$
$a=1$ and $r=\frac{2}{x}$
The series converges when
$|r| \lt 1$
$|\frac{2}{x}| \lt 1$
$-1 \lt \frac{2}{x} \lt 1$
$-1 \lt \frac{2}{x}$ or $\frac{2}{x} \lt 1$
$x \lt -2$ or $x \gt 2$
The series converges if $x ϵ (-\infty, -2)$ U $(2,\infty)$
$\Sigma^{\infty}_{n=0} \frac{2^{n}}{x^{n}} = \frac{a}{1-r}$
$=\frac{1}{1-\frac{2}{x}}$
$=\frac{x}{x-2}$
