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I think that the sequence $$R_n=\frac{f_{2n+1}}{f_{2n-1}},$$ (where $f_n$ is the fibonacci sequence) converges to $\varphi^2$ where $\varphi$ is the "golden ratio." A quick check with the calculator gives $$\begin{align}R_6 =\frac{f_{13}}{f_{11}} =\frac{233}{89} &\approx 2.61798\end{align},$$ and $\varphi^2\approx 2.61803$, so it's kinda close already. I'm pretty sure this is a well know result, but after some googleing I couldn't find any proof for it. Does anyone have one? Or perhaps could lead me in the right direction to proving it myself? Thanks!

T. Alexander
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    This follows directly from the ratio of consecutive Fibonacci numbers converging to $\varphi,$, which is fairly well known indeed, see for example here or here. – dxiv Sep 15 '17 at 02:07

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$$\lim_{n\to \infty}R_n=\lim_{n\to \infty}\frac{f_{2n+1}}{f_{2n-1}}=\\ \lim_{n\to \infty}\frac{f_{2n+1}}{f_{2n}}.\frac{f_{2n}}{f_{2n-1}}=\\ \lim_{n\to \infty}\frac{f_{2n+1}}{f_{2n}}\lim_{n\to \infty}\frac{f_{2n}}{f_{2n-1}}\to \phi .\phi$$

Khosrotash
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It is well known that if $F(n)$ is the fibonacci sequence then $$\lim_{n\rightarrow \infty} \frac{F(n+1)}{F(n)} = \phi \approx 1.618$$

Do you know why? Can you derive it? Have a crack at it yourself and I'll update my answer later.

From this result, it easily follows that the ratio of two Fibonacci numbers that are $k$ places apart is $\phi^k$

$$\begin{align} \lim_{n\rightarrow \infty} \frac{F(n+k)}{F(n)} &= \lim_{n\rightarrow \infty} \left({\frac{F(n+k)}{F(n+k-1)}\cdot \frac{F(n+k-1)}{F(n+k-2)}\cdot \cdots\ \cdot \frac{F(n+1)}{F(n)}}\right) \\[3ex] &= \underbrace{\lim_{n\rightarrow \infty} \frac{F(n+k)}{F(n+k-1)}\cdot \lim_{n\rightarrow \infty}\frac{F(n+k-1)}{F(n+k-2)}\cdots\ \cdot \lim_{n\rightarrow \infty}\frac{F(n+1)}{F(n)}}_{k \text{ terms}}\\[2ex] &= \underbrace{\phi \cdot \phi \cdots \phi}_{k \text{ terms}} = \phi^k \end{align}$$

In your particular example the terms $2k+1$ and $2k-1$ are just 2 places apart so their ratio will tend to $\phi^2$

Thanassis
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You also could use Binet's formula $$F_n=\frac{\varphi^n-\psi^n}{\varphi-\psi}\qquad \text{where}\qquad \varphi = \frac{1 + \sqrt{5}}{2} \qquad \text{and}\qquad \psi = \frac{1 - \sqrt{5}}{2} = 1 - \varphi = - {1 \over \varphi} $$ Since $\psi < 1$ and $\varphi>1$, for large values of $n$, then $$F_n\approx \frac{\varphi^n}{\sqrt 5}$$ For illustation purposes $F_{10}=55$ while the above approximation would give $\approx 55.0036$.

With this, you can build a lot of things.

Claude Leibovici
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