Your idea matters-15(Fahim's sequence)

Let $\rho$ be the golden ratio , and $\overline{\rho} = 1-\rho$ be its conjugate. Indexing the Fibonacci sequence so that $F_1 = F_2 = 1,$ the Binet formula says that $F_n = \frac1{\sqrt{5}}\left(\rho^n - {\overline{\rho}}^n\right).$ Call Fahim's sequence $G_n$ ; then the same methods as Binet's formula lead to the formula $G_n = \rho^n + {\overline{\rho}}^n.$ (You can prove this by induction, even if you don't see how to derive it.)

So $\begin{aligned} \lim_{n\to\infty} \frac{G_n}{F_n} &= \sqrt{5} \lim_{n\to\infty} \frac{\rho^n + {\overline{\rho}}^n}{\rho^n - {\overline{\rho}}^n} \\ &= \sqrt{5} \lim_{n\to\infty} \frac{1 + ({\overline{\rho}}/\rho)^n}{1 - ({\overline{\rho}}/\rho)^n} \\ &= \sqrt{5}, \end{aligned}$ since $|{\overline{\rho}}/\rho| < 1.$

In general, a sequence $\{ a_n \} _{n \in \mathbb N}$ defined by

$a_n = \begin{cases} a & \text{for }n=1 \\ b & \text{for }n=2 \\ a_{n-1}+a_{n-2} & \text{else} \end{cases}$

can be calculated by the explicit formula

$a_n = c_1 \phi^n + c_2 \overline\phi^n$

where $\phi$ and $\overline\phi$ are the golden ratio and its conjugate respectively.

The constants $c_1$ and $c_2$ depend on the starting values $a$ and $b$ . We can set up two equations and solve for $c_1$ and $c_2$ .

$\begin{aligned} a = a_1 &= c_1 \phi + c_2 \overline\phi \\ b = a_2 &= c_1 \phi^2 + c_2 \overline\phi^2 \end{aligned}$

$\begin{aligned} c_1 = \frac {-a_1+a_1\phi+a_2} {\phi+2} \\ c_2 = \frac {a_1+a_1\phi-a_2\phi} {1-2\phi} \end{aligned}$

With this, we can now also solve for the general ratio $r$ of two such sequences $f_n = c_1\phi^n+c_2\overline\phi^n$ and $g_n = c_3\phi^n+c_4\overline\phi^n$ .

$\begin{aligned} r &= \lim_{n \to \infty} \frac {g_n}{f_n} \\ &= \lim_{n \to \infty} \frac {c_3 \phi^n + c_4 \overline\phi^n} {c_1 \phi^n + c_2 \overline\phi^n} \\ &= \lim_{n \to \infty} \frac {c_3+ c_4 \left( \frac{\overline\phi}{\phi} \right)^n} {c_1 + c_2 \left( \frac{\overline\phi}{\phi} \right)^n} \\ &= \boxed{\frac {c_3} {c_1}} && \left|\overline\phi / \phi \right| <1 \\ &= \frac {2g_2-g_1+g_1\sqrt{5}} {2f_2-f_1+f_1\sqrt{5}} \end{aligned}$

In our specific case, we have $f_1=1; f_2=1; g_1=1; g_2=3$ which results in

$\boxed{r = \frac {5+\sqrt{5}}{1+\sqrt{5}} = \sqrt{5}}$

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$\begin{aligned} & \frac {5+\sqrt{5}}{1+\sqrt{5}} \\ =& \frac {\sqrt{5} \cdot \left( \sqrt{5}+1 \right)} {\sqrt{5}+1} \\ =& \sqrt{5} \end{aligned}$