We can normalise $R=1\ $. By induction the question boils down to this: Let $$ F(z)=(z-a)\sum_{n=0}^\infty b_nz^n=-ab_0+\sum_{n=1}^\infty (b_{n-1}-ab_n)z^n. $$ If this series converges for $z=1$, we have to show that $\sum_{n=0}^\infty b_n$ converges.
Let $S_N=\sum_{n=1}^Nb_{n-1}$ and let $T_N=S_N-aS_{N+1}\ $. Then we assume that $T_N$ converges. As $|a|<1$, the series $\displaystyle\sum_{j=0}^\infty a^j$ converges and hence the series $\displaystyle\sum_{j=0}^\infty a^jT_{N+j}\ $ converges. Now $\displaystyle\sum_{j=0}^Ma^jT_{N+j}=S_N-a^{M+1}\ S_{N+M+1}\ \ $. This means that $a^jS_{N+j}\ $ converges for $j\to\infty$. Let $A_N$ denote the limit, then $\displaystyle F_N=\sum_{j=0}^\infty a^jT_{N+j}=S_N-A_N$. Now $$ A_N=\lim_ja^{j+k}\ S_{N+k+j}\ =a^k\lim_ja^j\ S_{N+k+j}\ =a^kA_{N+k}\ \ , $$ so that $A_N=a^{-N}A$ for some $A\in\mathbb C$. We show that $A=0$. For this write $\displaystyle S_N(z)=\sum_{j=1}^Nb_{n-1}\ z^{n-1}$. Then $S_N(a)$ converges to $\displaystyle S(a)=\sum_{n=1}^\infty n_{n-1}\ a^{n-1}$. We have $A=\lim_N a^NS_N$ and $$ a^jS_j-aS_j(a)=\sum_{k=1}^{j-1}b_{n-1}\ (a^j-a^k)=a^jS_{j-1}-aS_{j-1}\ (a). $$ The left hand side converges to $A-aS(a)$ and the right hand side to $aA-aS(a)$. We conclude that $A=aA$ and hence $A=0$.
Therefore $F_N=S_N$. We claim that $F_N$ is a Cauchy sequence. For this consider $$ F_{N+k}-F_N=\sum_{j=0}^\infty a^j\left(T_{N+j}-T_{N+k+j}\ \ \right). $$ As $T_N$ is a Cauchy-sequence, the right hand side becomes arbitrarily small as $N$ increases. Therefore $F_N$ is Cauchy, hence convergent and so is $S_N$.