$\newcommand{\fib}{\operatorname{fib}}$
I am trying to solve the Fibonacci recurrence using generating functions, but I've run into a bit of a snag. Here's what I've done so far:
$$\begin{align} &\fib(0) = 0\\ &\fib(1) = 2\\ &\fib(n) = \fib(n-1) + \fib(n-2) \end{align} $$
$$\begin{align} A(x) &= \sum_{i=0}^\infty a_ix^i=\sum_{i=1}^{\infty}a_ix^i\\ &=x+\sum_{i=2}^\infty a_ix^i\\ &=x+\sum_{i=2}^\infty (a_{i-1} + a_{i-2})x^i\\ &=x+x\sum_{i=1}^\infty a_ix^i + x^2\sum_{i=0}^{\infty}a_ix^i\\ \\ A(x)&=x+xA(x)+x^2A(x)\\ \\ &=\frac{-x}{x^2+x-1} \end{align} $$
I then need to rearrange $A(x)$ back into the form $A(x)=\sum_{i=0}^\infty f(i)x^i$ to obtain $f(i)$, which should be Binet's closed-form equation for the $i$'th Fibonnacci number, but I don't know any techniques to do so. Is there a method for rearranging arbitrary rationals into infinite sums?
The general technique is to use partial fractions. You may not have had much experience decomposing things into slightly messy partial fractions, so I’ll go into a bit of detail.
I would write it $$A(x)=\frac{x}{1-x-x^2}\;.$$ Now factor the denominator: $1-x-x^2=(1-\alpha x)(1-\beta x)$; equating coefficients, we see that $\alpha\beta=-1$ and $\alpha+\beta=1$, so $\beta=1-\alpha$, and $\alpha(1-\alpha)=-1$, i.e., $\alpha^2-\alpha-1=0$. The quadratic formula then gives us
$$\alpha=\frac{1\pm\sqrt5}2\;,$$
so we can set $\alpha=\frac12(1+\sqrt5)$ and $\beta=\frac12(1-\sqrt5)$. Now a partial fraction decomposition of $A(x)$ is possible:
$$A(x)=\frac{x}{1-x-x^2}=\frac{A}{1-\alpha x}+\frac{B}{1-\beta x}\;,$$
so $A(1-\beta x)+B(1-\alpha x)=x$. Thus, $A+B=0$, and $-\beta A-\alpha B=1$, and you can solve for $A$ and $B$. Finally, you can easily convert each of these rational functions to an infinite series:
$$\frac{A}{1-\alpha x}=A\sum_{n\ge 0}(\alpha x)^n=A\sum_{n\ge 0}\alpha^nx^n\;,$$
for instance.