Polynomial for Fibonacci $II$

$\begin{aligned} P_n(x) & = 1 + x + 2x^2 + 4x^3 + 8x^4 + \cdots & \small \color{#3D99F6} F_n = 2^{n-2} \text{ for all }n \ge 2 \text{ (see proof)} \\ & = 1 + x \color{#3D99F6} \left(1 + 2x+(2x)^2 + (2x)^3 + \cdots \right) & \small \color{#3D99F6} \text{An infinite geometric progression} \\ & = 1 + \frac x{1-2x} & \small \color{#3D99F6} \text{for }|2x| < 1 \end{aligned}$

$\implies P_n \left(\frac 13\right) = 1 + \frac {\frac 13}{1-\frac 23} = \boxed 2$

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Proof by induction for the claim $F_n = 2^{n-2}$ for all $n\ge 2$ :

For $n=2$ , $F_2 = 2^{2-2} = 1$ . Therefore, the claim is true for $n=2$ . Assuming the claim is true for $n$ then:

$\begin{aligned} F_{n+1} & = F_n + \color{#3D99F6} F_{n-1} + F_{n-2} + F_{n-3} + \cdots + F_1 \\ & = F_n + \color{#3D99F6} F_n \\ & = 2F_n = 2 (2^{n-2} = 2^{n-1} \end{aligned}$

Therefore the claim is also true for $n+1$ and hence true for all $n \ge 2$ .

$\lim\limits_{n \to \infty}P_{n}(x) = \displaystyle\sum_{i=1}^{\infty} F_{i}x^{i-1} = \frac{1}{1-(x+x^2+x^3+\dots +x^{n-1})}$

$\frac{1}{1-x}=1+x+x^2+x^3+\dots+x^{n}$ where $n \to \infty$

$\Rightarrow \frac{x}{1-x}=x+x^2+x^3+\dots+x^{n+1}$ $\displaystyle\sum_{i=1}^{\infty} F_{i}x^{i-1} = \frac{1}{1-(\frac{x}{1-x})}$ $\Rightarrow \lim\limits_{n \to \infty}P_{n}(x) =\frac{1-x}{1-2x}$ $\Rightarrow \lim\limits_{n \to \infty}P_{n}(\frac{1}{3}) = \boxed{2}$

For the proof of generating functions click here