Htykuut

Let $pi(z) = sum_{k=0}^{infty} pi_k z^k$. We assume that $pi(z) neq 0$ for all $|z| leq 1$ and $sum_{k=0}^{infty} k^s |pi_k| < infty$ for some $s geq 1$. I have two questions:

Question 1: In a paper that I am reading, the author defines $alpha(z) = pi(z)^{-1}$ and says that $alpha(z)$ has the expansion $alpha(z) = 1 - sum_{k=1}^{infty} alpha_k z^k$. I am wondering why does $alpha(z)$ admit such expansion? Is it because that we can prove $pi(z)$ to be holomorphic and nonzero on some open set $U$ containing the unit disk, and so $alpha(z)$ is holomorphic on $U$ as well? I can only see that the convergence radius is 1 and do not know how it can be generalized to $1+varepsilon$.

Question 2: The author also claims that $sum_{k=1}^{infty} k^s |alpha_k| < infty$. Is there any theorem entailing this convergence?

I’m feeling too lazy and overstuffed to answer in detail. But maybe the following start will help. My strategy is to work formally, and keep track of convergence data as I go.

Writing $Pi(z)=sum_0^inftypi_kz^k$, the hypotheses imply that $Pi(0)=pi_0ne0$. Thus $pi_0^{-1}Pi(z)=1+sum_1^inftytau_kz^k=1+T(z)$, with $T$ having same domain of convergence as $Pi$. Now the reciprocal of $1+T$ is $1+sum_1^infty(-T)^k$, and I’m going to have to leave all the rest, especially the computation of the domain of convergence, to you.