because the summands can be estimated by $C^l P(l) \alpha^n \Gamma(n+\beta+2-k-l)$ with some $C\in {\mathbb{R}}$ and $P \in {\mathbb{R}}[l]$ each depending on $k$. Corollary \ref{crll:exponential_inO} asserts that the truncated part is in ${\mathcal{O}}(\alpha^n \Gamma(n+\beta+2-R))$. The subsequent change in summation variables $k \rightarrow k+l$ gives rise to,} \begin{gathered} = \sum_{k=0}^{R-1} \alpha^{n+\beta+2-k}\Gamma(n+\beta+2-k) [x^{k}] \left( {\mathcal{A}}^\alpha_{\beta+2} B(x) \sum_{l=0}^{k} { k -1 \choose l}(x A(x))^l \sum_{m=0}^{s(n)-l} \frac{\left(\frac{A(x)}{\alpha }\right)^{m}}{m!} \right)(x), \end{gathered} \end{gathered}$$ which results in the statement after noting that the sums over $l$ and $m$ can be completed, because $\lim_{n\rightarrow \infty} s(n) = \infty$.

The rest of the proof is merely an algebraic exercise. We start with the expression from Lemma \[lmm:chainrule\_asymp\_expansion\_first\] for $[x^k] \left( {\mathcal{A}}^\alpha_{\beta+2} f \circ g^{-1} \right)(x)$ and use the chain and product rules from Proposition \[prop:derivation\], Theorem \[thm:chain\_analytic\] as well as $ {\mathcal{A}}^\alpha_\beta x^2\partial_x ={\mathcal{A}}^\alpha_{\beta+2}\partial_x=(\alpha^{-1} - x \beta + x^2 \partial_x){\mathcal{A}}^\alpha_\beta $ from Proposition \[prop:ordinary\_derivative\] to expand it. Accordingly, for all $k\in {\mathbb{N}}_0$: $[x^k] \left( {\mathcal{A}}^\alpha_{\beta+2} f \circ g^{-1} \right)(x)=$ $$\begin{gathered}
 \left[x^k\right] \left( {\mathcal{A}}^\alpha_{\beta+2} B(x) (1+x A(x))^{k-1} e^{\frac{A(x)}{\alpha} } \right)(x) = \left[x^k\right] \left( {\mathcal{A}}^\alpha_{\beta+2} f(x)g'(x)\left(\frac{g(x)}{x}\right)^{\beta-k+1} e^{\frac{\frac{x}{g(x)}-1}{\alpha x} } \right)(x) \\ = \left[x^k\right] e^{\frac{\frac{x}{g(x)}-1}{\alpha x} } \left(\frac{g(x)}{x}\right)^{\beta-k+1} \Bigg( g'(x)({\mathcal{A}}^\alpha_{\beta+2} f)(x) \\ + f(x) g'(x) \frac{1}{g(x)} \left((\beta-k+1) - \alpha^{-1} \frac{1}{g(x)} \right) ({\mathcal{A}}^\alpha_{\beta+2} 