For the ODE $$3z^2u''+8zu'+(z-2)u=0$$ construct a series solution of the form $$\sum_{k=0}^{\infty} A_kz^{k+r}$$ that is bounded as $z\rightarrow 0$. Take $A_0=1$ and compute explicitly the terms up to and including the one with $k=2$.
Now, I have determined that $$(3r(r-1)+8r-2)A_0z^r+\sum_{k=1}^{\infty} (3(k+r)(k+r-1)+8(k+r)-2)A_k+A_{k-1})z^{k+r}=0$$
Hence the indicial equation is $3r^2+5r-2=0$, with roots $r=\frac{1}{3}, -2$. The recurrence relation is $$A_k=-\frac{A_{k-1}}{3(k+r)(k+r-1)+8(k+r)-2}, \ \ k\geq 1$$ But how do I know which value for $r$ gives an unbounded solution or not?