If the series $f(z)=\sum^{\infty}_{n=0}a_n(z-i)^n$ and $g(z)=\sum^{\infty}_{n=0}b_n(z+i)^n$ both have radius of convergence $\frac{3}{2}$, how about the series of $h(z)=f(z)+g(z)$ centered at $0$?
I was thinking like say the function $f(z)=\frac{1}{1-z}$. If the center is at $i$ then the radius of convergence is $\sqrt{2}$. If the center is $0$, then the radius of convergence is 1. So we can not find some area that $overlap$ then find the radius. Really have little ideas, any help?
Since $D(0,1/2)\subset D(i,3/2)\cap D(i,3/2),$ $f+g$ is analytic in $D(0,1/2).$ Hence the ROC of the power series of $f+g$ at $0$ is at least $1/2.$ To show the ROC can equal $1/2,$ let
$$f(z) = \frac{1}{z+i/2},\,\, g(z) = \frac{1}{z-i/2}.$$
At the other extreme, we can let
$$f(z) = \frac{1}{z+i5/2} + \frac{1}{z-i5/2},\,\, g(z) = -f(z).$$
Then $f+g=0$ in $D(0,1/2),$ so the ROC here is $\infty.$
See if you can find examples to show any $R\in (1/2,\infty)$ can be the ROC.