Finding the Laurent Series of $\frac{1}{z^2(z+1)(z-2)} $ in the region $ 0 < |z| < 1 $.
I started to find the Partial Fraction of $ \frac{1}{z^2(z+1)(z-2)} $ which I got to be $ \frac{1}{3z^2}\left(\frac{1}{z-2} - \frac{1}{z+1}\right) $ then I expanded the two fraction terms to their series representations. Which gave me $$ \frac{1}{3z^2}\left(-\frac{1}{2}\sum_{n=0}^{\infty}{\left(\frac{z}{2}\right)^n} - \sum_{n=0}^{\infty}{\left(-z\right)^n} \right). $$ Combining the sums together, I got the following $$ \frac{1}{3z^2}\sum_{n=0}^{\infty}{\left(-\frac{1}{2}\left(\frac{z}{2}\right)^n - \left( -z \right)^n\right)} = \frac{1}{3z^2} \sum_{n=0}^{\infty}{\left(-\left(\frac{1}{2}\right)^{n+1} - \left(-1\right)^n\right)z^n } $$
After the distributing the $ z^2 $ in the denominator and changing the starting index I got the following
$$ \frac{1}{3} \sum_{n=-2}^{\infty}{ \left[\left( -\left( \frac{1}{2} \right)^{n+3} - \left(-1\right)^{n+2} \right)\cdot z^n\right] } .$$
Is there anything I did incorrectly that I need to address or does this look right?