${a_n}$ 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!

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).
