{{\mathbb{R}}[[x]]^{\alpha}_{\beta}}$. Moreover, ${\mathcal{A}}^\alpha_\beta$ is a derivation, that means it fulfills the Leibniz rule $$\begin{aligned}
 \label{eqn:leipniz} ( {\mathcal{A}}^\alpha_\beta ( f \cdot g ))(x) &= f(x) ({\mathcal{A}}^\alpha_\beta g)(x) + g(x) ({\mathcal{A}}^\alpha_\beta f)(x) . \end{aligned}$$

Set $h(x)=( f \cdot g )(x) =f(x) g(x)$. The coefficients $h_n$ are given as a sum by the Cauchy product formula. This sum can be written suggestively as $$\begin{aligned}
 \label{eqn:product_sum_decomp} h_n = \sum_{m=0}^n f_{n-m} g_m &= \sum_{m=0}^{R-1} f_{n-m} g_m + \sum_{m=0}^{R-1} f_m g_{n-m} + \sum_{m=R}^{n-R} f_m g_{n-m} &&\forall n \geq 2R. \end{aligned}$$ Definition \[def:basic\_asymp\_definition\] guarantees that the first two sums have sound asymptotic expansions for large $n$. Together they constitute an asymptotic expansion of $h_n$ up to order $R-1$. We verify this exemplary on the first sum, where the asymptotic expansion from eq.  up to order $R-m-1$ of $f_{n-m}$ can be substituted: $$\begin{aligned}
 \begin{split} \sum_{m=0}^{R-1} f_{n-m} g_m &=\sum_{m=0}^{R-1} g_m \sum _{k=0}^{R-m-1} \alpha^{n-m+\beta-k} \Gamma(n-m+\beta-k) [x^k]({\mathcal{A}}^\alpha_\beta f)(x) + {\mathcal{O}}\left(\alpha^{n}\Gamma(n+\beta-R)\right) \\ &= \sum _{k=0}^{R-1} \alpha^{n+\beta-k} \Gamma(n+\beta-k) \sum_{m=0}^{k}g_m [x^{k-m}]({\mathcal{A}}^\alpha_\beta f)(x) + {\mathcal{O}}\left(\alpha^{n}\Gamma(n+\beta-R)\right). \end{split} \end{aligned}$$ The inner sum $\sum_{m=0}^{k}g_m [x^{k-m}]({\mathcal{A}}^\alpha_\beta f)(x)$ is the $k$-th coefficient of the product $g(x) ({\mathcal{A}}^\alpha_\beta f)(x)$. It remains to be shown that the last sum in eq.  is negligible. Because $f_n,g_n \in {\mathcal{O}}( \alpha^n \Gamma(n+\beta))$, there is a 