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.

  • 5
    $\begingroup$ A good start might be to note that $F_{n+1}^2-F_{n-1}^2 = F_{2n}$. (Which you can see by plugging in Binet's formula and a bit of algebraic simplification). $\endgroup$ – hmakholm left over Monica Sep 4 '15 at 20:27

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 $\phi^4=\phi^2+2\phi+1=3\phi+2$)

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  • $\begingroup$ I was getting close to this, but you were faster. (+1) $\endgroup$ – robjohn Sep 4 '15 at 22:15
  • $\begingroup$ I posted mine anyway, then decided it was too close to yours (same idea with only a bit of different presentation), so I deleted it. $\endgroup$ – robjohn Sep 4 '15 at 23:41
  • $\begingroup$ Interesting that $a_{2n} = a_nb_n$. This helps the evaluation of the partial products. $\endgroup$ – marty cohen Sep 5 '15 at 0:49
  • $\begingroup$ @martycohen: I don't see that that is used in this proof. Is it? $\endgroup$ – robjohn Sep 5 '15 at 3:47

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