Holy Fibonacci!

Using Binet's Formula $F_n \; = \; \tfrac{1}{\sqrt{5}}\big[\phi^n - (-\phi^{-1})^{n}\big]$ where $\phi = \tfrac12(1 + \sqrt{5})$ , we can show that $F_{n+1}^2 - F_{n-1}^2 \; = \; F_{2n} \;,$ so we are calculating $\prod_{n=2}^\infty \frac{F_{2n}-1}{F_{2n}+1} \;.$ Using Binet's Formula again, $F_{2n} \pm 1 \; = \; \tfrac{1}{\sqrt{5}}\big[\phi^{2n} - \phi^{-2n}\big] \pm 1 \; = \; \tfrac{1}{\sqrt{5}}\phi^{-2n}\big[\phi^{4n} \pm \sqrt{5}\phi^{2n} - 1\big]$ Since $\phi^2 - \phi^{-2} = \sqrt{5}$ , we deduce that $\begin{array}{rcl} F_{2n} \pm 1 & = &\displaystyle \tfrac{1}{\sqrt{5}}\phi^{-2n}\big[\phi^{4n} \pm \phi^{2n +2} \mp \phi^{2n-2} - 1\big] \\ & = & \displaystyle\tfrac{1}{\sqrt{5}}\phi^{-2n}\big[(\phi^{2n} \pm \phi^2)(\phi^{2n} \mp\phi^{-2})\big] \; = \; \tfrac{1}{\sqrt{5}}\phi^{-2n}\big[(\phi^{2n-2} \pm 1)(\phi^{2n+2} \mp 1)\big] \\ & = & \displaystyle \tfrac{1}{\sqrt{5}}\phi^{-2n}a_{\pm}(n-1) a_{\mp}(n+1) \end{array}$ where $a_{\pm}(n) \; =\; \phi^{2n} \pm 1 \;.$ Thus $\frac{F_{2n}-1}{F_{2n}+1} \; = \; \frac{a_{-}(n-1) a_{+}(n+1)}{a_{+}(n-1)a_{-}(n+1)}$ and hence $\begin{array}{rcl} \displaystyle\prod_{n=2}^N \frac{F_{2n}-1}{F_{2n}+1} & = & \displaystyle \prod_{n=2}^N \frac{a_{-}(n-1) a_{+}(n+1)}{a_{+}(n-1)a_{-}(n+1)} \; =\; \prod_{n=2}^N \frac{a_{-}(n-1)}{a_{-}(n+1)} \times \prod_{n=2}^N \frac{a_{+}(n+1)}{a_{+}(n-1)} \\ &= & \displaystyle \frac{a_{-}(1)a_{-}(2)}{a_{-}(N)a_{-}(N+1)} \times \frac{a_{+}(N)a_{+}(N+1)}{a_{+}(1)a_{+}(2)} \end{array}$ since the last two products telescope neatly. Thus $\prod_{n=2}^N \frac{F_{2n}-1}{F_{2n}+1} \; = \; \frac{a_{-}(1)a_{-}(2)}{a_{+}(1),a_{+}(2)} \times \frac{a_{+}(N)}{a_{-}(N)} \times \frac{a_{+}(N+1)}{a_{-}(N+1)} \; = \; \tfrac13\frac{a_{+}(N)}{a_{-}(N)} \times \frac{a_{+}(N+1)}{a_{-}(N+1)}\;.$ Letting $N \to \infty$ , the fact that $F_{2N} \to \infty$ implies that $\frac{a_{+}(N)}{a_{-}(N)} \to 1$ , and hence $\prod_{n=2}^\infty \frac{F_{2n}-1}{F_{2n}+1} \; = \; \tfrac13 \;.$