Let $F(z)=\displaystyle \sum\_{k=0}^\infty a\_kz^k,\;|z|<R $ and $F(R)=\displaystyle \sum\_{k=0}^\infty a\_kR^k$ (the series converges).


Assume that $F(\alpha\_j)=0,\;j=1,2,\dots ,m$, where all $|\alpha\_j|<R$, Then $$F(z)=(z-\alpha\_1)\dots (z-\alpha\_m)\cdot \displaystyle \sum\_{k=0}^\infty b\_kz^k,\;|z|<R. $$
This is obvious, because $F(z)/[(z-\alpha\_1)\dots (z-\alpha\_m)]$ is analytic in $|z|<R.$


**My question**. Does $\sum\_{k=0}^\infty b\_kR^k$ converge?


The question is related to the research on the $q$-Bernstein polynomials. The relevant information is contained, for example, in [S. Ostrovska, The $q$-Versions of the Bernstein Operator: From Mere Analogies to Further Developments](http://link.springer.com/article/10.1007/s00025-016-0530-2), Results in Mathematics, 69(3), 275-295, Section 4.2.



Convergence of a series
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 $S\_N$.


