Lemma \ref{lmm:estimate_exp}, can be applied with $s(n)= {\lceil 4(R \log n +C)\rceil}$: } = \sum_{m=s(n)}^n \frac{C^{m+1}}{m!} \alpha^n \Gamma(n+\beta+2) \leq C\alpha^n \frac{\Gamma(n+\beta+2)}{n^{R}} \in {\mathcal{O}}(\alpha^n \Gamma(n+\beta+2-R)). \end{gathered}$$

\[lmm:chainrule\_asymp\_expansion\_first\] If $f,g,A,B$ as above, then $f \circ g^{-1} \in {{\mathbb{R}}[[x]]^{\alpha}_{\beta+2}}$ and $$\begin{aligned}
 [x^k] \left( {\mathcal{A}}^\alpha_{\beta+2} f \circ g^{-1} \right)(x) &= [x^{k}] \left( {\mathcal{A}}^\alpha_{\beta+2} B(x) (1+xA(x))^{k-1} e^{\frac{A(x)}{\alpha} }\right)(x) && \forall k\in {\mathbb{N}}_0. \end{aligned}$$

The proof proceeds by substitution of the asymptotic expansion up to order $R-1$ of $B(x) A(x)^m$ into eq. . This can be done, because $n-m$ is large. The result is $[x^n]f(g^{-1}(x))=$ $$\begin{gathered}
 \sum_{m=0}^{s(n)} {n+\beta+1 \choose m} \sum_{k=0}^{R-1} \alpha^{n+\beta+2-k-m} \Gamma(n+\beta+2-k-m) [x^k] \left( {\mathcal{A}}^\alpha_{\beta+2} B(x) A(x)^m \right)(x), \intertext{where the rest term, ${\mathcal{O}}(\alpha^n \Gamma(n+\beta+2-R))$, was omitted. An elementary variant of the Chu-Vandermonde identity (Lemma \ref{lmm:identity}) can be used to expand the product of the binomial and $\Gamma$-function. This expansion allows us to perform the summation over $m$ behind the coefficient extraction:} = \sum_{k=0}^{R-1} \sum_{m=0}^{s(n)} \sum_{l=0}^m \alpha^{n+\beta+2-k} { l+ k -1 \choose l} \frac{ \Gamma(n+\beta+2-k-l) }{ (m-l)! } [x^k] \left( {\mathcal{A}}^\alpha_{\beta+2} B(x) \left(\frac{A(x)}{\alpha}\right)^m \right) (x) \\ = \sum_{k=0}^{R-1} \sum_{l=0}^{s(n)} \alpha^{n+\beta+2-k-l} { l+ k -1 \choose l}\Gamma(n+\beta+2-k-l) [x^{k}] \left( {\mathcal{A}}^\alpha_{\beta+2} B(x) A(x)^l \sum_{m=0}^{s(n)-l} \frac{\left(\frac{A(x)}{\alpha }\right)^{m}}{m!} \right) (x) \intertext{The sum over $l$ can be truncated at order $R-1-k$, 