Find all values of x... (Infinite Sums)

This is a geometric sequence with the first term $a_1=\frac{1}{x^2}$ and the common ration $q=\frac{2}{x}$. Indeed the $n^{th}$ term of the series is given by
\[a_n=\frac{2^{n-1}}{x^{{n+1}}}=\frac{2^{-2}2^{n+1}}{x^{n+1}}=\frac{1}{4}(\frac{2}{x})^{n+1}.\]

We know that geometric series converge if and only if
\[|q|<1  \Rightarrow  |\frac{2}{x}|<1  \Rightarrow  |x|>2.\]
Also the infinite series will converge to
\[S=\frac{a_1}{1-q}=\frac{\frac{1}{x^2}}{1-\frac{2}{x}}=\frac{\frac{1}{x^2}}{\frac{x-2}{x}}\]
\[=\frac{1}{x(x-2)}=\frac{1}{x^2-2x}.\]