Answer
$x \lt -2$ or $x>2$
$f(x)=\displaystyle \frac{x}{x-2}, \quad x>2$ or $x \lt -2$
Work Step by Step
We aim for terms of Th.9.6 (Convergence of a Geometric Series).
A geometric series with ratio $r$ diverges when $|r| \geq 1$.
If $0 \lt |r| \lt 1$, then the series converges to the sum
$\displaystyle \sum_{n=0}^{\infty}ar^{n}=\frac{a}{1-r},\quad 0 \lt |r| \lt 1$.
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$\displaystyle \sum_{n=0}^{\infty}(\frac{2}{x})^{n}$
The series is geometric, r=$\displaystyle \frac{2}{x}$, converges for
$|\displaystyle \frac{2}{x}| \lt 1 \Rightarrow |x|>2 \Rightarrow x \lt -2$ or $x>2$
$f(x)=\displaystyle \sum_{n=0}^{\infty}(\frac{2}{x})^{n}=\frac{1}{1-\frac{2}{x}}=\frac{1}{\frac{x-2}{x}}$
$=\displaystyle \frac{x}{x-2}, \quad x>2$ or $x \lt -2$
