Showing that $\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n+1}L_{2^n+1}}\right)=\frac{3}{\phi^2}$

Question

Infinite product

$F_{n}:=[1,1,2,3,5,8,\cdots]$ and

$L_{n}:=[1,3,4,7,\cdots]$

for $n=1,2,3,\cdots$ respectively.

$\frac{1+\sqrt5}{2}=\phi$

Show that,

$$\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n+1}L_{2^n+1}}\right)=\frac{3}{\phi^2}$$

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We took the idea from this site

Expand the product

(1)

$$\left(1+\frac{1}{F_{3}L_{3}}\right)\cdot\left(1+\frac{1}{F_{5}L_{5}}\right)\cdot\left(1+\frac{1}{F_{9}L_{9}}\right)\cdots=\frac{3}{\phi^2}$$

$F_{n}=\frac{\phi^n-(-\phi)^{-n}}{\sqrt5}$

$L_{n}=\phi^n+(-\phi)^{-n}$

$F_{n}L_{n}=\frac{\phi^{2n}-(-\phi)^{-2n}}{\sqrt5}=F_{2n}$

Rewrite (1)

$$\left(1+\frac{1}{F_{6}}\right)\cdot\left(1+\frac{1}{F_{10}}\right)\cdot\left(1+\frac{1}{F_{18}}\right)\cdots=\frac{3}{\phi^2}$$

Still doesn't help much to get from LHS to RHS.

We have $F_{2n}=F^2_{n+1}-F^2_{n-1}$

I have substituted in, but the formula seem too messy and more complicated.

Can anybody please give a hand?

Answer 1

Hint. One may use the following result,

$$ \prod_{n=1}^{\infty}\left(1+\frac{F_b}{F_{2^na+b}}\right)=\frac{1-(-1)^b\phi^{-2a-2b}}{1-\phi^{-2a}} \tag1 $$

which is proved by J. Sondow here (see proposition 1 on page 3), where $\phi=\dfrac{1+\sqrt5}{2}$ and where $F_{n}:=[1,1,2,3,5,8,\cdots]$ are the Fibonacci numbers.

Applying $(1)$ with $a=2$, $b=2$ using $F_{2^n+1}L_{2^n+1}=F_{2^{n+1}+2}$ gives the announced infinite product.

Answer 2

Use $$F_{4k+2}+1=F_{2k+2}L_{2k}\tag{1}$$ with $k=2^{n-1}$. Also note $$\prod_{k=1}^{n}L_{2^{k}}=\prod_{k=1}^{n}\phi^{2^k}+(-\phi)^{-2^{k}}=\frac{\phi^{2^{k+1}}-(-\phi)^{-2^{k+1}}}{\phi^2-(-\phi)^{-2}}=\frac{F_{2^{k+1}}}{F_{2}}\tag{2} $$ From $\text{(1)}$, note that $$\left(1+\frac{1}{F_{6}}\right)\cdot\left(1+\frac{1}{F_{10}}\right)\cdot\left(1+\frac{1}{F_{18}}\right)\cdots \left(1+\frac{1}{F_{2^{k+1}+2}}\right)=\frac{F_{4}L_{2}L_{4} \dots L_{2^k}}{F_{2^{k+1}+2}} $$ Using $\text{(2)}$, we have $$\frac{F_{4}L_{2}L_{4} \dots L_{2^k}}{F_{2^{k+1}+2}}=\frac{3F_{2^{k+1}}}{F_{2^{k+1}+2}}$$ But $$\lim_{k \to \infty}\frac{3F_{2^{k+1}}}{F_{2^{k+1}+2}}=\frac{3}{\phi^2}$$