${a_n}$ series of Fibonacci numbers. $f(x)=\sum_{0}^{\infty}a_nx^n$, show that in the convergence radius: $f(x)= \frac{1}{1-x-x^2}$

I'd really like your help with this following problem: Let ${a_n}$ be a Fibonacci series $a_1=a_0=1$ and $a_{n+2}=a_n+a_{n+1}$ for every $n \geq 0$.

Let $f(x)=\sum_{0}^{\infty}a_nx^n$, I need to find the radius of convergence and to prove that in the range of this radius $f(x)= \frac{1}{1-x-x^2}$.

we know that the convergence radius $R=\lim_{n\to \infty} |\frac {a_n}{a_{n+1}}| $, How can I apply it in this case? Any direction to prove that $f(x)$ is as requested?

Thanks alot!

As for the question about radius of convergence, I'll write the comments by Javaman:

As $\dfrac{a_{n+1}}{a_n} \to \phi$, where $\phi=\dfrac{1+\sqrt 5}{2}$, we see that the limit superior coincides with the limit.

$$\limsup \dfrac{a_{n+1}}{a_n}=\lim \dfrac{a_{n+1}}{a_n}=\phi$$ which gives you the radius of convergence.

Loosely, this is how generating functions work:

Note that we have, $a_n=a_{n-1}+a_{n-2},~~ n\ge2$

$$\begin{align*} f(x) & = \sum _{n=0}^ \infty {a_n x^n} \&=1+a_1x+\sum_{n=2}^\infty(a_{n-1}+a_{n-2})x^n\\&=1+a_1x+\sum_{n=2}^\infty{a_{n-1}x^n+\sum_{n=2}^\infty}a_{n-2}x^n\\&=1+a_1x+x\sum_{n-1=1}^\infty{a_{n-1}x^{n-1}}+x^2\sum_{n-2=0}^\infty{a_{n-2}x^{n-2}}\\&=1+x+x(f(x)-1)+x^2f(x) ~~~\text{as $a_1=1$}\\&=1+xf(x)+x^2f(x)\end{align*}$$

But, as for rigour, I am looking forward to enlightening comments!

You can find the radius of convergence by noting the well-known identity:

$$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \varphi, $$ where $\varphi = \frac{\sqrt{5}+1}{2}$ is the golden ratio.

Once you know the radius of convergence, to find the partial sum $S_N$, just write:

$$\begin{align} S_N := \sum_{n=0}^N a_n x^n &= 1 +x + \sum_{n =2}^N a_n x^n \\ &= 1+x + \sum_{n=2}^N(a_{n-2} + a_{n-1})x^n \\ &= 1+ x + x^2\sum_{n=2}^N a_{n-2}x^{n-2} + x\sum_{n=2}^Na_{n-1} x^{n-1} \\ &= 1 + x + x^2 \sum_{n=0}^{N-2} a_n x^n + x \cdot \sum_{n=1}^{N-1} a_n x^n \\ &= 1 + x + x^2 S_{N-2} + x\sum_{n=0}^{N-1} a_n x^n - x \\ &= 1 + x^2 S_{N-2} + x S_{N-1}. \end{align}$$ Since $S_N = a_Nx^N + S_{N-1} = a_Nx^N + a_{N-1}x^{N-1} + S_{N-2}$, we arrive at $$ S_N = 1 + x^2 (S_N - a_N x^N - a_{N-1}x^{N-1}) + x(S_N - a_Nx^N) $$ Solving for $S_N$ gives $$ S_N = \frac{1 - a_Nx^{N+2} - a_{N-1}x^{N+1} - a_Nx^{N+1}}{1 - x - x^2}. $$ Finally, taking the limit $\lim_{N \to \infty} S_N$ gives the answer as the terms with powers of $x$ in the numerator go to zero (since the original sum under question converges).