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

  • $\begingroup$ 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. $\endgroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.