\left(1+x A(x)\right)^{n+\beta+1} \\ &=\sum_{m=0}^n {n+\beta+1 \choose m} [x^{n-m}] B(x) A(x)^m. \end{aligned}$$

The tail of the sum in eq.  over $m$ turns out to be asymptotically negligible. However, in contrast to the preceding arguments, the sum cannot be truncated at a fixed value of $m$ independent of $n$. A cutoff that grows slowly with $n$ has to be introduced. More specifically,

\[lmm:chainrule\_truncation\] If $f,g,A,B$ as above, then there is a constant $C\in {\mathbb{R}}$ such that $$\begin{aligned}
 \label{eqn:chainrule_trancated1} [x^n] f(g^{-1}(x)) &= \sum_{m=0}^{s(n)} {n+\beta+1 \choose m} [x^{n-m}] B(x) A(x)^m + {\mathcal{O}}(\alpha^n \Gamma(n+\beta+2-R)) &&\forall R \in {\mathbb{N}}_0, \end{aligned}$$ where $s(n)= {\lceil 4(R \log n +C)\rceil}$.

It needs to be proven that the partial sum over $m \in (s(n),\infty)$ in eq.  is in ${\mathcal{O}}(\alpha^n \Gamma(n+\beta+2-R))$ for an appropriate function $s(n)$. Recall that $A,B \in {{\mathbb{R}}[[x]]^{\alpha}_{\beta+2}}$. Lemma \[lmm:prod\_hard\_estimate\] guarantees the existence of a $C\in {\mathbb{R}}$ such that $$\begin{gathered}
 \sum_{m=s(n)}^n {n+\beta+1 \choose m} | [x^{n-m}] B(x) A(x)^m | \leq \sum_{m=s(n)}^n {n+\beta+1 \choose m} C^{m+1} \alpha^n \Gamma(n-m+\beta+2), \intertext{for all $n \geq s(n) \geq 0$. The binomial can be expressed in terms of $\Gamma$-functions ${ n+\beta+1 \choose m} = \frac{\Gamma(n+\beta+2)}{\Gamma(n-m+\beta+2) m!}$. Subsequently, an elementary inequality, which quantifies the fast convergence of the Taylor expansion of the exponential, proven in 