Given a recursion $a_{n+ 1}= a_{n}^{2}- 2$ with $a_{0}= 3.$ Find the asymptotic analysis of $$\frac{a_{n+ 1}^{2}}{\prod_{0}^{n+ 1}a_{i}}- \sqrt{5}$$ Maybe this helps - https://artofproblemsolving.com/community/c4h2308107p18323223 -
And this is my way of thinking, I use two inequalities $$\sqrt{5}\leq \frac{a_{n+ 1}^{2}}{\prod_{0}^{n+ 1}a_{i}}\leq\sqrt{5}+ \frac{2}{3}\cdot\left ( \varphi- 1 \right)^{-2^{n+ 1}}$$ with $\varphi= \frac{3+ \sqrt{5}}{2}.$ From the recurrence sequence $a_{n+ 1}= a_{n}^{2}- 2\Rightarrow a_{n+ 1}= \varphi^{2^{n}}+ \varphi^{-2^{n}},$ we have $$\frac{a_{n+ 1}^{2}}{\prod_{i= 0}^{n+ 1}a_{i}}= \frac{\varphi^{2^{n+ 1}}+ \varphi^{-2^{n+ 1}}}{\prod_{i= 1}^{n}\left ( \varphi^{2^{i}}+ \frac{1}{\varphi^{2^{i}}} \right )}= (\varphi- \frac{1}{\varphi})\left ( 1+ \frac{2\varphi^{-2^{n+ 2}}}{1- \varphi^{-2^{n+ 2}}} \right )> \varphi- \frac{1}{\varphi}= \sqrt{5}$$ Because I'm busy now.. OK, I'll get it done by tonight, I haven't posted the solution for the rightside inequality, and how can we use it to find the $o\left ( 1 \right ).$ I need to the help, thanks a real lot !
Edit. The rightside inequality is just single variable of $f\left ( n \right )\leq g\left ( n \right )$ with $$f\left ( n \right )= \sqrt{5}+ \frac{2}{3}\cdot\varphi^{-2^{n+ 1}}, g\left ( n \right )= \sqrt{5}\left ( 1+ \frac{2\varphi^{-2^{n+ 2}}}{1- \varphi^{-2^{n+ 2}}} \right ), 0< \varphi^{-2^{n+ 1}}\leq\frac{7-3\sqrt{5}}{2}$$
First of all I think it's strange your product goes up to $i=n+1$, as that factor cancels with the numerator.
Assuming that is what you mean, notice that
$$\prod_{i=0}^{n+1} a_i = F_{2^{n+3}},$$
where $F_k$ is the $k$th Fibonacci number. That alongside with the formula
$$ a_n = \left(\frac{2}{\sqrt{5}+3}\right)^{2^n}+\left(\frac{1}{2} \left(\sqrt{5}+3\right)\right)^{2^n} $$
should do the trick.