The generating function of the Fibonacci sequence as a rational function 
Prove that if $$f(x)=\sum _1 ^{\infty} a_n \cdot x^n,$$ where $(a_n)$ is the Fibonacci Sequence, and if $|x|<1/2$, then $$(x^2 + x - 1) \cdot f(x) = -x.$$

So far, as an earlier part of this question, I've shown that $a_{k+1}/a_k \le 2$ and that $f(x)$ converges for all $|x|<1/2$.
Although I'm aware that this $f$ converges, I have no method for finding what it converges to! 
This question implies that $f(x) = -x/(x^2 +x -1)$ which, for $-1/2 < x < 1/2$ is inside $(-4/10,2)$. I've also tried factoring the $(x^2 + x -1)$ into my series for $f$, but just don't know where to go from here.

Since the series converges, we are allowed to write
$$(x^2+x-1)f(x)=\sum_{n=1} a_n x^{n+2}+\sum_{n=1}a_n x^{n+1}-\sum_{n=1} a_n x^n$$
Now for the last series
$$\sum_{n=1} a_n x^n=a_1 x+a_2 x^2+\sum_{n=3}a_n x^n=a_1 x+a_2 x^2+\sum_{n=3}(a_{n-2}+a_{n-1})x^n$$
so putting all together you get
$$(x^2+x-1)f(x)=a_1 x^2-a_1 x-a_2 x^2$$
Assuming you are defining the Fibonacci numbers starting with $a_1=a_2=1$, you have the result.
