Why is generating function proof of Fibonacci formula correct? The proof goes as follows:-
Let $F = 1 + x + 2x^2 + 3x^3 + 5x^4 + 8x^5 + ...$
Then
$$\begin{align} 1 + Fx + Fx^2 &= 1 + (x + x^2 + 2x^3 + 3x^4 + ...) + (x^2 + x^3 + 2x^4 + 3x^5) \\
 1 + Fx + Fx^2 &= 1 + x + (x^2+x^2 + 2x^3+x^3 + 3x^4+2x^4 + ...) \\ 
 1 + Fx + Fx^2 &= F \\
  \frac{1}{1-x-x^2} &= F  \end{align} $$
We rearrange the terms and we get this result which can be then manipulated further to find formula for nth Fibonacci term.
I want to focus on the rearrangement of terms in an infinite series. This is no different from results like $ 1 + 2 + 4 + 8 + ... = -1$.
The usual answer given is that we cannot treat an infinite summation like this like real numbers and apply normal addition and subtraction rules. We must apply sophisticated techniques like limits to evaluate these sums. So $ 1 + r + r^2 + r^3 + ... $ only makes sense when $|r| < 1$.
So going back to the Fibonacci proof above. We are applying normal addition rules for an infinite summation and also the result $\displaystyle  F = \frac{1}{1-x-x^2} $ doesn't make sense for any value of $x$. But we still use this result to complete the proof. Why does it work?
Isn't this the same as using the result $ 1 + 2 + 4 + 8 + ... = -1 $ to prove other results? It would be absurd to use it as basis for other proofs.
Please shed some light on this. I am really confused.

Denote by $F_n$ the $n$-th Fibonacci number
$$
F_0=0,\qquad F_1=1,\qquad F_{n+2}=F_{n+1}+F_n
$$
and consider the series
$$
F=\sum_{n\ge0}F_nx^n
$$
in the ring of formal power series with coefficients in $\mathbb{R}$. This ring has the property that any series
$$
\sum_{n\ge0}a_nx^n
$$
with $a_0\ne0$ is invertible. Your computation takes places in this algebraic structure, where convergence is of no concern.
We can operate formally and write
\begin{align}
1+xF+x^2F
&=1+\sum_{n\ge0}F_nx^{n+1}+\sum_{n\ge0}F_nx^{n+2} \\[6px]
&=1+\sum_{n\ge1}F_{n-1}x^n+\sum_{n\ge2}F_{n-2}x^n \\[6px]
&=1+x+\sum_{n\ge2}(F_{n-1}+F_{n-2})x^n \\[6px]
&=1+x+\sum_{n\ge2}F_nx^n \\[6px]
&=F
\end{align}
so
$$
1=(1-x-x^2)F
$$
The formal power series $1-x-x^2$ (with zero coefficient terms for $x^n$ with $n\ge3$) is invertible, so
$$
F=(1-x-x^2)^{-1}
$$
The rest is manipulation of elements in the ring of power series, using the fact that
$$
(1-\alpha x)^{-1}=\sum_{n\ge0}\alpha^nx^n
$$
No limit and no convergence, just algebra.
