How to evaluate this infinite product ? (Fibonacci number) Let $F_n$ be Fibonacci numbers.
How to evaluate
$$\prod_{n=2}^\infty \left(1-\frac{2}{F_{n+1}^2-F_{n-1}^2+1}\right)\text{ ?}$$
It seem like that
$$\prod_{n=2}^\infty \left(1-\frac{2}{F_{n+1}^2-F_{n-1}^2+1}\right)=\frac{1}{3}$$
But how to prove it?
Thank in advances.
 A: The product is indeed equal to $\frac{1}{3}$.
We can use Binet's formula, which states that
$$F_n=\frac{1}{\sqrt{5}}\left(\phi^n - \left(-\frac{1}{\phi}\right)^n\right)$$
for all $n$ where $\phi=\frac{1+\sqrt{5}}{2}$
We can plug this directly into the product which we want to evaluate, but our life becomes slightly easier if we first simplify $F_{n+1}^2-F_{n-1}^2$ using other means.
We note that the addition formula for Fibonacci numbers tells us that $$F_{m+n}=F_{m+1}F_n + F_{m}F_{n-1}$$
and that in particular
$$F_{2n}=F_n(F_{n+1}+F_{n-1})$$
We can then write $F_{n+1}^2-F_{n-1}^2$ as
$$F_{n+1}^2-F_{n-1}^2=(F_{n+1}-F_{n-1})(F_{n+1}+F_{n-1})=F_n(F_{n+1}+F_{n-1})=F_{2n}$$
as noted by Henning Makholm in the comments on the question.
The product then becomes
$$\prod_{k=2}^\infty \frac{F_{2n}-1}{F_{2n}+1}$$
Using Binet's formula, the terms in the product become
$$\frac{F_{2n}-1}{F_{2n}+1}=\frac{\frac{1}{\sqrt{5}}\left(\phi^{2n}-\frac{1}{\phi^{2n}}\right)-1}{\frac{1}{\sqrt{5}}\left(\phi^{2n}-\frac{1}{\phi^{2n}}\right)+1}=\frac{\phi^{4n}-\sqrt{5}\phi^{2n}-1}{\phi^{4n}+\sqrt{5}\phi^{2n}-1}$$
Noting that $\phi^2-\frac{1}{\phi^2}=\sqrt{5}$, we see that this can be factorised as
$$\frac{\left(\phi^{2n}-\phi^2\right)\left(\phi^{2n}+\phi^{-2}\right)}{\left(\phi^{2n}-\phi^{-2}\right)\left(\phi^{2n}+\phi^2\right)} = \frac{\left(\phi^{2n-2}-1\right)\left(\phi^{2n+2}+1\right)}{\left(\phi^{2n+2}-1\right)\left(\phi^{2n-2}+1\right)}=\frac{a_{n-1}b_{n+1}}{a_{n+1}b_{n-1}}$$
where we define
$$a_n=\phi^{2n}-1$$
and
$$b_n=\phi^{2n}+1$$
The product then becomes
$$\prod_{k=2}^\infty \frac{a_{k-1}b_{k+1}}{a_{k+1}b_{k-1}}$$
which telescopes, and we see that the product is equal to
$$\lim_{n\to\infty} \frac{a_1 a_2 b_n b_{n+1}}{a_n a_{n+1} b_1 b_2}$$
It is straightforward to see that
$$\lim_{n\to\infty} \frac{b_n}{a_n} = 1$$
so that the product is then equal to the value
$$\frac{a_1 a_2}{b_1 b_2}=\frac{(\phi^2-1)(\phi^4-1)}{(\phi^2+1)(\phi^4+1)}=\frac{(\phi^2-1)^2}{\phi^4+1}=\frac{\phi^2}{3\phi+2+1}=\frac{1}{3}$$
as desired.
(Since $\phi^2=\phi+1$, and so $\phi^4=\phi^2+2\phi+1=3\phi+2$)
