Suppose that $\sum_0^\infty a_nz^n$ has radius of convergence $1$ and suppose that $|z_0|=r<R$. Let $g(z)=\sum_0^\infty a_n (z-z_0)^n$.
Problem: Prove that $g(z)$ has radius of convergence at least $R-r$.
I saw this question in Beals and couldn't figure it out! I started expanding binomially, but I had trouble writing the coefficients in the form $g(z)=\sum_0^\infty b_nz^n$.
Any suggestions? Note: Not a homework problem. I am studying for a test on series and this was recommended for studying.
Edit: Hmm...maybe there is a typo. Though I'm not quite sure how an offcenter power series would still have radius of convergence $R$. To me, it seems somewhat intuitive that the radius of $g(z)$ would still be $R-r$.