Expand f(z) into partial fractions:
[tex]\dfrac1{z(z-2)} = \dfrac12 \left(\dfrac1{z-2} - \dfrac1z\right)[/tex]
Recall that for |z| < 1, we have the power series
[tex]\displaystyle \frac1{1-z} = \sum_{n=0}^\infty z^n[/tex]
Then for |z| > 2, or |1/(z/2)| = |2/z| < 1, we have
[tex]\displaystyle \frac1{z-2} = \frac1z \frac1{1 - \frac2z} = \frac1z \sum_{n=0}^\infty \left(\frac 2z\right)^n = \sum_{n=0}^\infty \frac{2^n}{z^{n+1}}[/tex]
So the series expansion of f(z) for |z| > 2 is
[tex]\displaystyle f(z) = \frac12 \left(\sum_{n=0}^\infty \frac{2^n}{z^{n+1}} - \frac1z\right)[/tex]
[tex]\displaystyle f(z) = \frac12 \sum_{n=1}^\infty \frac{2^n}{z^{n+1}}[/tex]
[tex]\displaystyle f(z) = \sum_{n=1}^\infty \frac{2^{n-1}}{z^{n+1}}[/tex]
[tex]\displaystyle \boxed{f(z) = \frac14 \sum_{n=2}^\infty \frac{2^n}{z^n} = \frac1{z^2} + \frac2{z^3} + \frac4{z^4} + \cdots}[/tex]