Fermat to infinity

There are many solutions, but the one I found uses properties of Fermat numbers.

First note that $a_n=a_{n-1}^2-2 \Rightarrow a_n+1=(a_{n-1}+1)(a_{n-1}-1) \Rightarrow a_{n-1}-1=\frac{a_n+1}{a_{n-1}+1}$ . We will use this later.

Second, I claim $a_n=\frac{2^{2^{n+1}}+1}{2^{2^{n}}}$ for all $n \geq 1$ , which we prove with mathematical induction. For $n=1$ , clearly $\frac{17}{4}=a_1=\frac{2^{2^{1+1}}+1}{2^{2^1}}$ . Now assume $a_k=\frac{2^{2^{k+1}}+1}{2^{2^{k}}}$ for some $k \geq 1$ , and consider $a_{k+1}$ . By the defining recursion, $a_{k+1}=a_k^2-2=(\frac{2^{2^{k+1}}+1}{2^{2^{k}}})^2-2=\frac{2^{2^{k+2}}+2 \cdot 2^{2^{k+1}}+1}{2^{2^{k+1}}}-\frac{2 \cdot 2^{2^{k+1}}}{2^{2^{k+1}}}=\frac{2^{2^{k+2}}+1}{2^{2^{k+1}}}$ , as desired. Therefore, $a_n=\frac{2^{2^{n+1}}+1}{2^{2^{n}}}$ for all $n \geq 1$ .

Finally, the product $(1-\frac{1}{a_1})(1-\frac{1}{a_2})(1-\frac{1}{a_3})(1-\frac{1}{a_4}) \cdots =(\frac{a_1-1}{a_1})(\frac{a_2-1}{a_2})(\frac{a_3-1}{a_3})(\frac{a_4-1}{a_4}) \cdots = \\ (\frac{a_2+1}{a_1+1}\frac{a_3+1}{a_2+1}\frac{a_4+1}{a_3+1}\frac{a_5+1}{a_4+1} \cdots )(\frac{1}{a_1a_2a_3a_4 \cdots })=\lim_{N\to\infty} (\frac{a_{N+1}+1}{a_1+1})(\frac{2^{2^1}2^{2^2}2^{2^3}2^{2^4} \cdots 2^{2^N}}{(2^{2^2}+1)(2^{2^3}+1)(2^{2^4}+1)(2^{2^5}+1) \cdots (2^{2^{N+1}}+1)})$ .
At this point we use the fact for Fermat numbers that $(2^{2^0}+1)(2^{2^1}+1)(2^{2^2}+1) \cdots (2^{2^{n-1}}+1)=2^{2^n}-1$ . Therefore, we get

$\lim_{N\to\infty} \frac{4}{21}(\frac{2^{2^{N+2}}+1}{2^{2^{N+1}}}+1)(\frac{2^{2^{N+1}-1}}{2^{2^{N+2}}-1})(\frac{15}{2})= \lim_{N\to\infty} \frac{5}{7} (\frac{2^{2^{N+2}}+2^{2^{N+1}}+1}{2^{2^{N+2}}-1})=\frac{5}{7}$ .
So $A+B=5+7=12$ .