How to write this series as a ratio of polynomials?

Question

I have the series:

$g(z) = \sum^{\infty}_{0} b_{n}z^{n}$ where $z\in$$\mathbb{C}$.

$b_{n}$ is the $n^{th}$ number in the sequence, i.e. $(b_{0},b_{1},b_{2},b_{3},...)$ = (1,1,2,3,...)

How can I write this expression as a ratio of polynomials?

I don't really understand what I am even being asked or why.

I've tried rewriting as:

$g(z) = 1 + z + 2z^{2} + ... + (a_{n-1}+a_{n-2})z^{n} + ...$

I could try writing out z in expanded form $x+iy$ and observing what happens with alternate powers, but I do not see how that gets me closer to a polynomial form.

Answer 1

$$g(z) = \sum^{\infty}_{0} F_{n}z^{n}=\frac{z}{1-z-z^2}$$

Answer 2

$g(z) - zg(z) = \sum_\limits{n=1}^\infty b_nz^n - \sum_\limits{n=1}^\infty b_nz^{n+1}\\ g(z)(1 - z) = \sum_\limits{n=1}^\infty b_nz^n - \sum_\limits{n=2}^\infty b_{n-1}z^{n} \\ g(z)(1 - z) = b_1z + \sum_\limits{n=2}^\infty (b_n-b_{n-1})z^n$

By the definition of the Fibonacci sequence.

$b_{n} = b_{n-1} + b_{n-2}\\ b_{n} - b_{n-1} = b_{n-2}$

$g(z)(1 - z) = z + \sum_\limits{n=2}^\infty b_{n-2}z^n\\ g(z)(1 - z) = z + z^2 g(z)\\ g(z) = \frac {z}{1-z-z^2}$