# Convergence of a series

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, Results in Mathematics, 69(3), 275-295, Section 4.2.

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 $S_N$.
• I think there is a small misprint: it should be $S_N-a^{M+1}S_{N+M+1}$ in the formula for the sum $\sum_{j=0}^M a^jT_{N+j}.$ – Deepti Aug 4 '16 at 6:29
• @Anton: Where does your proof use that $F(a)=0$ ? It is clear that the statement does not hold without this assumption. – Alexandre Eremenko Aug 4 '16 at 7:41
• @Alexandre Eremenko: In the definition of $F(z)$. – user1688 Aug 4 '16 at 7:56
• @Anton: Does it matter that the limit $A$ depends on $N$? – Deepti Aug 4 '16 at 10:00