<s>[INST] Show that: $\prod_{n=1}^{\infty}\frac{\sqrt{5}F_{2n+2}+1}{\sqrt{5}F_{2n+2}-1}=\phi$ How to show that:
$$P=\prod_{n=1}^{\infty}\frac{\sqrt{5}F_{2n+2}+1}{\sqrt{5}F_{2n+2}-1}=\phi$$
Where $\phi=\frac{1+\sqrt{5}}{2}$
$$P=\frac{\sqrt{5}F_{4}+1}{\sqrt{5}F_{4}-1}\cdot \frac{\sqrt{5}F_{6}+1}{\sqrt{5}F_{6}-1}\cdot \frac{\sqrt{5}F_{8}+1}{\sqrt{5}F_{8}-1}\cdots$$
$$P=\frac{3\sqrt{5}+1}{3\sqrt{5}-1}\cdot \frac{8\sqrt{5}+1}{8\sqrt{5}-1}\cdot \frac{21\sqrt{5}+1}{21\sqrt{5}-1}\cdots$$
I thought maybe I try and alter its form to makes it easy...but it is not apparently
$$P=\frac{2\phi+(3-1)\sqrt{5}}{2\cdot3\phi-(3+1)}\cdot\frac{2\phi+(8-1)\sqrt{5}}{2\cdot8\phi-(8+1)}\cdot\frac{2\phi+(21-1)\sqrt{5}}{2\cdot21\phi-(21+1)}\cdots$$
 [/INST] Let $F_n$ denote the Fibonacci sequence and $L_n$ denote the Lucas sequence. Note the following identities
\begin{eqnarray*}
5F_{2n}^2-1&=&L_{2n-1}L_{2n+1} \\ 
2F_{2n+2}-F_{2n+1}&=&L_{2n+1} \\
5F_{2n+1}F_{2n+2}-2&=&L_{2n+1}L_{2n+2} \\
5F_{2n+2}-L_{2n+2}&=&2L_{2n+1} \\
L_{2n+2}F_{2n+2}-1&=&L_{2n+1}F_{2n+3}.
\end{eqnarray*}
Start by noting that
\begin{eqnarray*}
\frac{3 \sqrt{5}+1}{3 \sqrt{5}-1}=  \left( \frac{1+\sqrt{5}}{2}\right) \left(\frac{-2+5 \sqrt{5}}{11} \right)
\end{eqnarray*}
Now we proceed inductively
\begin{eqnarray*}
P_m=\prod_{n=1}^{m}\frac{\sqrt{5}F_{2n+2}+1}{\sqrt{5}F_{2n+2}-1} = \phi 
 \left( \frac{-2+F_{2m+3}\sqrt{5}}{L_{2m+3}} \right).
\end{eqnarray*}
This follows by multiplying out and using the above formulea.
It remains only to observe that 
\begin{eqnarray*}
\lim_{n \rightarrow \infty} \frac{L_n}{F_{n}} = \sqrt{5}
\end{eqnarray*}
and we are done.
</s>