Series of iterates of a power series Let $f = \sum_{n \ge 2} a_{n} x^{n}$ be a formal power series of order higher/equal two. By $$ f^{\circ n} := \underbrace{f \circ \dots \circ f}_{\text{ n times}}$$ we denote the $n^{th}$ iterate of $f$ and set $$f^{0} = x. $$ It is easily seen that the order of $f^{\circ n}$ is higher/equal $2^{n}$, so we may define the formal series  $$S(z):= \sum_{n\ge 0} f^{\circ n}.$$
If we assume that $f$ is a convergent power series, it is also not hard to show that $\sum_{n = 0}^{N} f^{\circ n}$ converges locally uniformly to $S$, especially $S$ is again a convergent power series.
Note however that $S \neq (x + f)^{ \circ -1}$ - it is in general not the composite inverse of (x+f).
I don't have a particular reason for studying this series, but it seems to be interesting to ask - what is $S$?. In the sense that whether $S$ relates to $f$ in any "familiar" way, i.e. is the value of some better known operator evaluated at $f$ or does it have any interesting properties?
Maye someone already had a thought on it,thanks for all answers.
EDIT: Here is the proof the $S(z)$ is analytic if $f$ is of order $\ge 2$ and convergent:
Since $f(0)=0$ you can find a neighborhood in which the modulus of $f$ is strictly smaller then one, so there exist $R>0$ and $M \in (0,1)$ so that $$\vert a_{n} \vert \le \frac{M}{R^{n}}.$$ And you obtain $$\vert f(x) \vert \le \vert \frac{x}{R} \vert ^{2} \frac{ M}{1 - \vert \frac{x}{R}\vert }$$
Since $0<M<1$ there is an $s>0$ so that $\frac{ M}{1 - \vert \frac{x}{R}\vert } \le 2$ for all $x \in D_{s}:= \{ \vert x \vert \le s\}$ and we choose $s$ so small that $\vert \frac{s}{R} \vert \le \frac{1}{2}$.
Now I claim that on $D_{s}$ we have $$\vert f^{\circ n }(x) \vert \le \left\vert\frac{x}{R}\right\vert^{n} 2^{n},$$ which is proven by induction:
Let $n=1$. Then $\vert f(x) \vert \le \vert\frac{x}{R} \vert^{2} \frac{ M}{1 - \vert \frac{x}{R}\vert } \le \vert\frac{x}{R} \vert \vert\frac{x}{R} \vert 2 \le \vert\frac{x}{R} \vert$.
Now the step $n\to n+1$:
$$ \vert f(f^{\circ n}) \vert \le \vert \frac xR \vert^{2 n} 2^{2n -2}
\frac{ M}{1 - \underbrace{\vert (\frac xR^{n}) 2^n\vert}_{\le \vert \frac{x}{R}\vert }} \le \vert \frac{x}{R} \vert^{2 n} 2^{2n -2} \cdot 2 \le 2^{n} \vert \frac{x}{R} \vert^{n+1}$$
 A: I've done one example.
Consider the function $f(x)=2 x^2-x^3+1/4 x^4 $ The top left segment of the Carlemanmatrix, let's call it $F$ is $$ F_{8 \times 8}=\small \begin{bmatrix} 
 1 & . & . & . & . & . & . & . \\
 0 & 0 & . & . & . & . & . & . \\
 0 & 2 & 0 & . & . & . & . & . \\
 0 & -1 & 0 & 0 & . & . & . & . \\
 0 & 1/4 & 4 & 0 & 0 & . & . & . \\
 0 & 0 & -4 & 0 & 0 & 0 & . & . \\
 0 & 0 & 2 & 8 & 0 & 0 & 0 & . \\
 0 & 0 & -1/2 & -12 & 0 & 0 & 0 & 0
 \end{bmatrix} $$ such that the coefficients for $f(x)$ are in the second column (let's define the columnindex $c=1$ and rows and column-indexes always begin at 0) In the next column there are the coefficients for $f(x)^2$ and in the first column that (trivial one) for $f(x)^0 = 1$.  (Note: my convention here is the transpose of the example in the wikipedia for some historical reasons) 
The coefficients for the h'th iterate of $f(x)$ occur in the second column of the h'th power of $F$.      
Then the coefficients for the sum of all iterates occur in the second column of the sum of all powers of $F$. But that means, that we could define the matrix $A$ as result of the geometric series of $F$, which is known as Neumann series and would be computed by
$$  \sum_{k=0}^\infty F^k = (I - F)^{-1} $$   
Unfortunately, the top left element of that geometric series $(I-F)$ is zero, so it is not invertible. But for any finite truncation we can use the submatrix without the first column and row and then this is invertible:
$$  A =  ((I - F)^*)^{-1} $$ 
The top left segment of $A$ is then
$$ A=\small \begin{bmatrix}
 1 & . & . & . & . & . & . & . \\
 2 & 1 & . & . & . & . & . & . \\
 -1 & 0 & 1 & . & . & . & . & . \\
 33/4 & 4 & 0 & 1 & . & . & . & . \\
 -8 & -4 & 0 & 0 & 1 & . & . & . \\
 -4 & 2 & 8 & 0 & 0 & 1 & . & . \\
 11 & -1/2 & -12 & 0 & 0 & 0 & 1 & . \\
 985/8 & 1025/16 & 9 & 16 & 0 & 0 & 0 & 1
 \end{bmatrix} $$
The coefficients in the first column in $A$ allow now to compute the sum of any contiguous segment of the iterates of $f°^h(x)$ just by the function 
$$g(x_0,x_m) = \sum_{h=0}^{m-1} f°^h(x_0) = \sum_{k=1}^\infty (x_0^k-x_m^k) \cdot A_{k-1,0} $$
where $x_m = f°^m(x_0)$    
This is a pretty general scheme and the current example is just to get an intuition. For other polynomials and even power series $f(x)$ one might need some finetuning and exceptions, but I think it is pretty obvious what's going on in general.    
[update] I should add, that the current example gives a pretty small range of convergence, and perhaps one should take a better one. We can only use such $x_0$ , where $|f(x_0)|<|x_0|$, and I tried with $x=0.01$ and smaller... [/update]
A: If $f(z),g(z)$ are the generating functions for combinatorial classes $\mathcal{F},\mathcal{G}$, then
the composition $f(g(z))$ is the generating function for the class of objects of $\mathcal{F}$ whose atoms have been replaced with elements of $\mathcal{G}$.
So, for example, if $f(z)=z^2(1-z)^{-1}$ is considered to be the g.f. for rooted trees of depth 1 with at least two leaf nodes, then $f^{\circ n}$ is the g.f. for balanced trees of depth $n$ in which each internal node has at least two child nodes, and $S$ is the g.f. for all such balanced trees, satisfying $S(z) = z + S(f(z))$.
A: Another interpretation: Let $T_{f}(g) := f \circ g$ denote the right-composition operator associated with $T_{f}$, let $H := id + T_{f}$. Then $H^{-1} = \sum_{n=0}^{\infty} T_{f}^{n}$. Note that $T_{f}^{n}(f) =f^{ \circ (n+1)}$. Thus $$\sum_{k=0}^{\infty} f^{\circ n} = x + H^{-1}(f).$$
