An Unexpected Number 2

We note that $p_0 = F_2$ and $p_1 = F_3$ , where $F_n$ is the $n$ th Fibonacci number. Then $p_2 = F_2+F_3 = F_4$ , $p_3 = F_5$ , $p_4 = F_6$ , ..., implying that $p_n = F_{n+2}$ . Similarly, $q_n = F_{n+1}$ . Then we have:

$\begin{aligned} L & = \lim_{n \to \infty} \frac {p_n}{q_n} = \lim_{n \to \infty} \frac {F_{n+2}}{F_{n+1}} & \small \color{#3D99F6} \text{Since }F_n = \frac {\varphi^n - (-\varphi)^{-n}}{\sqrt 5} \text{, where} \\ & = \lim_{n \to \infty} \frac {\varphi^{n+2} - (-\varphi)^{-n-2}}{\varphi^{n+1} - (-\varphi)^{-n-1}} & \small \color{#3D99F6} \varphi = \frac {1+\sqrt 5}2 \text{ is the golden ratio.} \\ & = \lim_{n \to \infty} \frac {\varphi - (-\varphi)^{-2n-3}}{1 - (-\varphi)^{-2n-2}} & \small \color{#3D99F6} \text{Divide up and down by } \varphi^{n+1} \\ & = \varphi \approx \boxed{1.618} \end{aligned}$