Let the first series be $\sum_{k=0}^{\infty} u_{k}(x-a)^{k}$.
Let the second series be $\sum_{k=0}^{\infty} v_{k}(x-a)^{k}$
Their convergence areas are $R_{1}$ and $R_{2}$.
I don't know how to prove that $f(x)*g(x)=\sum_{k=0}^{\infty} w_{k}(x-a)^{k}$, where $w_{k}=\sum_{k} u_{k} v_{k-n}$ and that its convergence area is $R\geq\min{R_{1},R_{2}}$.
I thought about the convergence area so that if the smallest convergence area is 0, then R must be greater or equal than it. But I don't know how that helps me or how to prove anything.
If $w_k=\sum_{n=0}^ku_nv_{k-n}$, then\begin{align}w_k(x-a)^k&=\sum_{n=0}^ku_nv_{k-n}(x-a)^k\\&=\sum_{n=0}^ku_n(x-a)^nv_{k-n}(x-a)^{k-n}.\end{align}So, $w_k(x-a)^k$ is the $k$th term of the Cauchy product of the series $\sum_{k=0}^\infty u_k(x-a)^k$ and $\sum_{k=0}^\infty v_k(x-a)^k$. So, if $|x-a|<R_1$ and $|x-a|<R_2$, the series$$\sum_{k=0}^\infty w_k(x-a)^k$$converges, since it is the Cauchy product of two absolutely convergent series. Therefore, the radius of convergence is at least $\min\{R_1,R_2\}$.