why generating function $A(z) = 1 + z + z^2 + \cdots$ can be denoted as $\frac{1}{1-z}$ It is easy to see that $1 + z + z^2 + \cdots$ is equal to $\frac{1}{1-z}$ when $1 > z > 0$ and for $z >= 1$, they are not equivalent. So I have thought $\frac{1}{1-z}$ is just a short for the more complex formula $1 + z + z^2 + \cdots$ and it has nothing to do with the value of the function. But later when I look at some materials about generating function (e.g. analytic combinatorics), I find that people also do some algebra operations on the generating function. For example, for the problem of how many trees are there with $N$ nodes, the generating function is $G(z) = z(1 + G(z) + G(z)^2 + \cdots)$ and thus $G(z)=\frac{z}{1-G(z)}$. I regard $\frac{1}{1-G(z)}$ is just a short of $1 + G(z) + G(z)^2 + \cdots$ here and all things look good. To get the closed form of $G(z)$, however, they transform $G(z)=\frac{z}{1-G(z)}$ to $G(z)-G(z)^2=z$, which is an algebra operation. We know that $1 + G(z) + G(z)^2 + \cdots$ and $\frac{1}{1-G(z)}$ are not necessarily the same. So is the solution after we solving $G(z)-G(z)^2=z$ exactly the $G(z)$ we expect?
 A: You are dealing with formal power series with, say, real coefficients; actually the coefficients could be in any commutative ring, let's assume for simplicity it's a field or, if you prefer, the real numbers. 
The formal power series form a ring which contains the ring of polynomials in one variable as a subring. Addition is defined componentwise, while multiplication is defined with the convolution (or Cauchy product):
$$
\biggl(\,\sum_{n\ge0}a_nz^n\biggr)\biggl(\,\sum_{n\ge0}b_nz^n\biggr)=\sum_{n\ge0}c_nz^n
$$
where
$$
c_n=\sum_{k=0}^{n}a_kb_{n-k}\qquad(n\ge 0)
$$
One can check that all ring axioms are satisfied. Now, considering $1-z$ as the formal power series in which all coefficients of $z^n$ are zero from $2$ onwards,
$$
(1-z)\sum_{n\ge0}z^n=
\sum_{n\ge0}z^n-\sum_{n\ge1}z^n=1
$$
so $1-z$ is the inverse of $\sum_{n\ge0}z^n$ in the ring of formal power series, which justifies writing
$$
\sum_{n\ge0}z^n=\frac{1}{1-z}
$$
Convergence is not taken into consideration.
The reasoning with
$$
G(z)=z(1+G(z)+G(z)^2+\dotsb)
$$
can be justified by considering rings of formal power series with coefficients in the ring of formal power series. A bit complicated, but not so difficult either.
A: Generating functions are formal power series, which essentially means that we are not concerned with analytic issues of convergence.  (The symbol $z$ is not meant to stand for a complex number, or any other type of number, for that matter.)
Formal power series do support most of the algebraic operations that you are familiar with, including manipulating closed form "short-hand" such as
$$
1 + z + z^2 + \cdots = \frac{1}{1 - z}. 
$$
