Let $a_0=5/2$ and $a_k=a^2_{k-1}-2$ for all $k\geq 1$.The question is to compute $$\prod_{k=0}^{\infty} \left(1-\frac{1}{a_k}\right)$$

I tried to calculate few terms.$a_0=5/2$, $a_1=17/4,a_2=257/16$ it seems that $a_k$ is of the form $2^{2^k}+2^{-2^{k}}$ however I am not sure about it.How to proceed without doing much guesswork.Any ideas?


The explicit formula $a_k=2^{2^k}+2^{-2^{k}}$ is straightforward to prove by induction.
We also have $$ a_k+1 = a_{k-1}^2-1 = (a_{k-1}-1)(a_{k-1}+1) $$ hence $$ a_{k-1}-1 = \frac{a_k+1}{a_{k-1}+1} $$ and $$ \prod_{k=1}^{N}\left(a_k-1\right)=\prod_{k=2}^{N+1}\left(a_{k-1}-1\right)=\frac{a_{N+1}+1}{a_1+1}. $$ Since $$ \frac{x^{2^M}-y^{2^M}}{x-y} = \left(x^{2^{M-1}}+y^{2^{M-1}}\right)\cdots(x+y) $$ by picking $M=N+1, x=2$ and $y=\frac{1}{2}$ we get $$a_N\cdot a_{N-1}\cdots a_0 =\frac{2^{2^{N+1}}-2^{-2^{N+1}}}{\frac{3}{2}}$$ and $$ \prod_{k=0}^{N}\frac{a_k-1}{a_k}=\frac{3}{5}\cdot\frac{\prod_{k=1}^{N}(a_k-1)}{\prod_{k=1}^{N}a_k} = \frac{\left(\frac{3}{2}\right)^2}{a_1+1}\left(1+\frac{1}{2^{2^{N+1}}-2^{-2^{N+1}}}\right).$$ By letting $N\to \infty$ it follows that the wanted product equals $\frac{\left(\frac{3}{2}\right)^2}{a_1+1}= \color{red}{\large\frac{3}{7}}.$

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    $\begingroup$ Dang it! I was going to answer it for once! $\endgroup$ – Simply Beautiful Art Dec 3 '17 at 2:21
  • $\begingroup$ Submit it anyway. I feel that independent answers are fine. $\endgroup$ – marty cohen Dec 3 '17 at 4:04
  • $\begingroup$ @martycohen Nah, it would be very close to an exact of Jack's answer. $\endgroup$ – Simply Beautiful Art Dec 3 '17 at 12:05

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