Mathematically correct way of using the generator functions Let $(F_n)_{n \in \mathbb{N}}$ be the Fibonacci sequence: $F_0=0$, $F_1=1$ and $\forall n \in \mathbb{N}: F_{n+2}=F_{n+1}+F_{n}$. Now let
$$f(x):=\sum\limits_{n \in \mathbb{N}} F_n x^n$$
So we have that
$$f(x)=x+\sum\limits_{n > 1} F_n x^n$$
$$f(x)=x+\sum\limits_{n > 1} F_{n-1} x^n+\sum\limits_{n > 1} F_{n-2} x^n$$
$$f(x)=x+\sum\limits_{n > 1} F_{n-1} x^{n-1} x+\sum\limits_{n > 1} F_{n-2} x^{n-2} x^2$$
$$f(x)=x+x\sum\limits_{n \in \mathbb{N}} F_{n} x^n+x^2\sum\limits_{n \in \mathbb{N}} F_{n} x^n$$
So
$$f(x)=x+xf(x)+x^2f(x)$$
$$f(x)=\frac{x}{1-x-x^2}$$
And if we go on, we can find a closed form for $F_n$.  
Q: Can we do this in general to get a closed form for the generating function $g(x)=\sum\limits_{n \in \mathbb{N}} a_n x^n$ without knowing anything about it's convergence? And if the answer is no, then how could we make it mathematically correct?
 A: You certainly can in
the case of a linear recurrence
$a_n
=\sum_{k=1}^m c_k a_{n-k}
$.
If you assume a solution
of the form
$a_n = r^n$
then you get
$r^n
=\sum_{k=1}^m c_k r^{n-k}
$.
Multiplying by $r_{m-n}$
we get
$r^{n+m-n}
=\sum_{k=1}^m c_k r^{n-k+m-n}
$
or
$r^{m}
=\sum_{k=1}^m c_k r^{m-k}
=\sum_{k=0}^{m-1} c_{m-k} r^{k}
$.
This is called the
characteristic polynomial
of the recurrence
and its roots determine
the growth of $a_n$
depending on the initial conditions.
You can also get
the generating function
of the recurrence.
Let
$A(x)
=\sum_{n=0}^{\infty} a_nx^n
$.
Then,
naming some things along the way, 
$\begin{array}\\
A(x)
&=\sum_{n=0}^{\infty} a_nx^n\\
&=\sum_{n=0}^{m-1} a_nx^n+\sum_{n=m}^{\infty} a_nx^n\\
&=B(x)+\sum_{n=m}^{\infty} \sum_{k=1}^m c_k a_{n-k}x^n\\
&=B(x)+\sum_{n=m}^{\infty} \sum_{k=1}^m x^kc_k a_{n-k}x^{n-k}\\
&=B(x)+\sum_{k=1}^m x^kc_k\sum_{n=m}^{\infty}  a_{n-k}x^{n-k}\\
&=B(x)+\sum_{k=1}^m x^kc_k\sum_{n=m-k}^{\infty}  a_{n}x^{n}\\
&=B(x)+\sum_{k=1}^m x^kc_k\left(\sum_{n=0}^{\infty}  a_{n}x^{n}-\sum_{n=0}^{m-k-1}  a_{n}x^{n}\right)\\
&=B(x)+\sum_{k=1}^m x^kc_k\left(A(x)-A_k(x)\right)\\
&=B(x)+A(x)\sum_{k=1}^m x^kc_k-\sum_{k=1}^m x^kc_kA_k(x)\\
\end{array}
$
so
$A(x)(1-\sum_{k=1}^m x^kc_k)
=B(x)-\sum_{k=1}^m x^kc_kA_k(x)
$.
You can work this further
to get explicit forms
for the polynomial on the right,
but I'll stop here.
