Generating functions are functions? On this Wikipedia page it says the following: "in fact, the generating function is not actually regarded as a function".
This comment is made only because we don't need convergence in a generating function?  I feel a bit confused, since for example, there are cases in which we must derive the generating function.  I would like to have a more formal idea of ​​this comment. Or it could be the case that Wikipedia is wrong
 A: In full generality a generating function is a formal power series, so can be thought of as just a sequence $a_n$ of, say, complex numbers. We think of this sequence as a "sum" $f(z) = \sum a_n z^n$ in some sense (and this can be made precise using a "$z$-adic metric") but $z$ is not actually a complex number and this sum is not actually being taken in the usual sense of an infinite sum of complex numbers.
Moreover, because $a_n$ is completely arbitrary this sum, if interpreted as an "actual sum" of complex numbers, need not converge for any nonzero value of $z$, so $f(z)$ does not necessarily have any interpretation as an "actual function" of the complex variable $z$. It can have such an interpretation for sufficiently small $z$ if the radius of convergence is positive, but the radius of convergence could be zero. For example, consider the generating function
$$f(z) = \sum n! z^n$$
of the factorials. The radius of convergence is zero, so $f(z)$ as an "actual sum" of complex numbers does not converge for any $z \neq 0$. However this is still a perfectly well-defined generating function and one can still ask and answer interesting questions about it. For example it has a (formal) logarithm
$$\log \sum n! z^n = z + \frac{3}{2} z^2 + \frac{13}{3} z^3 + \frac{71}{4} z^4 + \dots $$
whose coefficients count the number of subgroups of finite index in the free group $F_2$: see this blog post for details.
In practice this distinction doesn't matter too much because most generating functions have positive radius of convergence.
