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First note that \[ \dfrac{1}{z+1} = \begin{cases} \displaystyle\dfrac{1}{1 - (-z)} = \sum_{n=0}^\infty (-z)^n & |z| \l 1 \\ \dfrac{1}{z} \dfrac{1}{1 - (-\frac{1}{z})} = \dfrac{1}{z} \displaystyle\sum_{n=0}^\infty \left( -\dfrac{1}{z} \right)^n & |z| > 1 \end{cases} \] and that \[ \dfrac{1}{z-3} = \begin{cases} -\dfrac{1}{3}\dfrac{1}{1 - \frac{z}{3}} = \displaystyle -\dfrac{1}{3} \sum_{n=0}^\infty \left( \dfrac{z}{3} \right)^n & |z| \l 3 \\ \dfrac{1}{z} \dfrac{1}{1 - \frac{3}{z}} = \dfrac{1}{z} \displaystyle\sum_{n=0}^\infty \left( \dfrac{3}{z} \right)^n & |z| > 3 \end{cases} \] using geometric series.
$\displaystyle \sum_{n=0}^\infty (-z)^n - \dfrac{1}{3} \sum_{n=0}^\infty \left( \dfrac{z}{3} \right)^n$
$\displaystyle \dfrac{1}{z} \sum_{n=0}^\infty \left( -\dfrac{1}{z} \right)^n - \dfrac{1}{3} \sum_{n=0}^\infty \left( \dfrac{z}{3} \right)^n$
$\displaystyle \dfrac{1}{z} \sum_{n=0}^\infty \left( -\dfrac{1}{z} \right)^n + \dfrac{1}{z} \displaystyle\sum_{n=0}^\infty \left( \dfrac{3}{z} \right)^n$