Find the Taylor expansion around z = 0 and the radius of convergence of $\displaystyle \displaystyle \frac {z}{z^2+z-2}$.

I started out by simplify the expression. $\displaystyle \displaystyle \frac {z}{(z-1)(z+2)}$ Furthermore in separated fashion, I don't know the mathematical term. $\displaystyle \displaystyle \frac {1}{3} \frac{1}{(z-1)} + \frac {2}{3} \frac{1}{(z+2)}$

$\displaystyle \displaystyle - \frac {1}{3} \frac{1}{(1-z)} + \frac {2}{6} \frac{1}{(1+\frac{z}{2})}$ = $\displaystyle \displaystyle \frac {1}{3} \left( \frac{1}{(1+\frac{z}{2})} - \frac{1}{(1-z)} \right) $ = Expansion = $\displaystyle \displaystyle \frac {1}{3} \left( {(1-\frac{z}{2} + \frac{z^2}{4} - \frac{z^3}{8} + O (z^4))} - {(1+z+z^2+z^3 + O (z^4))} \right) $ = $\displaystyle \displaystyle \frac {1}{3} \left( 1-\frac{z}{2} + \frac{z^2}{4} - \frac{z^3}{8} - 1-z-z^2-z^3 + O (z^4)\right) $ = $\displaystyle \displaystyle - \frac {z}{2} -\frac{z^2}{4} - \frac{3z^3}{8} + O (z^4)$

Questions

I couldn't find any formula giving the expansion like for instance $\displaystyle \frac{z^n}{n!}$. How would I go about finding it? Is there a golden rule telling me how far I should develop if it isn't stated? I just happened to chose the power of 4 because my table of formulas does. Is the convergence radius = 1, (z-1) i the closest singularity to z = 0, right?