I would like to rewrite the series $$\sum\_{n=0}^\infty \frac{1}{n!}(\Delta^\varepsilon)^n a\_n,$$ where $\Delta^\varepsilon=\sum\_{k=1}^\infty \varepsilon^k b\_k$, as a series in $\varepsilon$ $$\sum\_{n=0}^\infty c\_n \varepsilon^n$$ (i.e., I need to find the $c\_n$).


Obviously, I can do this term-by-term. But the general case seems quite difficult.



Rewriting a series $\sum_{n=0}^\infty \frac{1}{n!}(\Delta^\varepsilon)^n a_n$ in the form $\sum_{n=0}^\infty c_n \varepsilon^n$
We have, using Cauchy products, at least for $\varepsilon$ small enough, that 
$$(\Delta^{\varepsilon})^n=\sum\_{k=1}^{+\infty}\left(\sum\_{j\_1+\dots+j\_n=k}\prod\_{i=1}^nb\_{j\_i}\right)\varepsilon^k,$$
hence, if the coefficients have a good behaviour  

\begin{align}
\sum\_{n=0}^{+\infty}\frac 1{n!}(\Delta^{\varepsilon})^na\_n&=\sum\_{n=0}^{+\infty}\frac 1{n!}a\_n\sum\_{k=1}^{+\infty}\left(\sum\_{j\_1+\dots+j\_n=k}\prod\_{i=1}^nb\_{j\_i}\right)\varepsilon^k\\\
&=\sum\_{k=1}^{+\infty}\sum\_{n=0}^{+\infty}\frac 1{n!}a\_n\left(\sum\_{j\_1+\dots+j\_n=k}\prod\_{i=1}^nb\_{j\_i}\right)\varepsilon^k.\\\
\end{align}
So we can take 
$$c\_k:=\sum\_{n=0}^{+\infty}\frac 1{n!}a\_n\left(\sum\_{j\_1+\dots+j\_n=k}\prod\_{i=1}^nb\_{j\_i}\right)$$
for $k\geq 1$ and $c\_0=0$. 


