The generating function of Fibonacci numbers

1. Multiply $F_{n+2}=F_{n+1}+F_n$ by $x^{n+2}$ and sum over $n\ge0$:
$$\begin{aligned} G(x)-F_0-F_1x&= \sum_{n\geq 2} F_nx^n = \sum_{n\geq 0} F_{n+2}x^{n+2} \\ &= \sum_{n\geq 0} F_{n+1}x^{n+2} + \sum_{n\geq 0} F_{n}x^{n+2} \\ &= x\bigl(G(x)-F_0\bigr)+x^2G(x), \end{aligned}$$
so $G(x)-x=xG(x)+x^2G(x)$ and $G(x)=\dfrac x{1-x-x^2}$.

2. The singularities of $G$ are the roots of $1-x-x^2$, namely $-\varphi$ and $-\psi=\frac1\varphi\approx0.618$. The smaller in absolute value is $\frac1\varphi$, so $R=\frac1\varphi=\frac{\sqrt5-1}2$. (This agrees with $\frac{F_n}{F_{n+1}}\to\frac1\varphi$.)

3. Since $1-x-x^2=(1-\varphi x)(1-\psi x)$ and $\varphi-\psi=\sqrt5$,
$$\frac x{(1-\varphi x)(1-\psi x)}=\frac1{\sqrt5}\left(\frac1{1-\varphi x}-\frac1{1-\psi x}\right)=\frac1{\sqrt5}\sum_{n\ge0}\left(\varphi^n-\psi^n\right)x^n.$$
Identifying coefficients: $F_n=\dfrac{\varphi^n-\psi^n}{\sqrt5}$.

Check. Expanding $\frac x{1-x-x^2}=x+x^2+2x^3+3x^4+5x^5+8x^6+\cdots$ ✓.