So originally I needed to turn the function $f(x)=\frac{3}{2-x}$ into a power series. I think I did this successfully and got $$\sum_{n=0}^\infty \frac{3x^{n}}{2^{n+1}} $$ Now I'm struggling to find the radius of convergence. I set up a form of the ratio test to find it: $$\lim_{n\to \infty}|\frac{\frac{3x^{n+1}}{2^{n+2}}}{\frac{3x^{n}}{2^{n+1}}}|$$ and got $|\frac{x}{2}|$. I'm not sure what this means in terms of the radius of convergence
$\begingroup$
$\endgroup$
Why make things mre comples than they are? You obtained this power series from the factorisation $$\frac3{2-x}=\frac32\,\frac1{1-\cfrac x2},$$ and using the power series expansion of $ \dfrac1{1-u}$ with $u=\dfrac x2$, so the radius of convergence is defined by $$\biggl|\frac x2\biggr|<1\iff |x|<2.$$