Recurrent sequence limit Let $a_n$ be a sequence defined: $a_1=3; a_{n+1}=a_n^2-2$
We must find the limit: $$\lim_{n\to\infty}\frac{a_n}{a_1a_2...a_{n-1}}$$
My attempt
The sequence is increasing and does not have an upper bound. Let $b_n=\frac{a_n}{a_1a_2...a_{n-1}},n\geq2$. This sequence is decreasing(we have $b_{n+1}-b_{n}<0$. How can I find this limit?  

since  take $a_{1}=x+\dfrac{1}{x},x>1$,then 
$$a_{2}=x^2+\dfrac{1}{x^2},a_{3}=x^4+\dfrac{1}{x^4}\cdots,a_{n}=x^{2^{n-1}}+\dfrac{1}{x^{2^{n-1}}}$$
so we have
$$a_{1}a_{2}\cdots a_{n}=\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}\right)\cdots\left(x^{2^{n-1}}+\dfrac{1}{x^{2^{n-1}}}\right)=\dfrac{x^{2^n}-\dfrac{1}{x^{2^n}}}{x-\dfrac{1}{x}}$$
so
$$\dfrac{a_{n+1}}{a_{1}a_{2}\cdots a_{n}}=\left(x-\frac{1}{x}\right)\dfrac{x^{2^n}+\dfrac{1}{x^{2^n}}}{x^{2^n}-\dfrac{1}{x^{2^n}}}\to \dfrac{x^2-1}{x},n\to+\infty$$
and
$$\left(x-\dfrac{1}{x}\right)^2+4=\left(x+\dfrac{1}{x}\right)^2=9\Longrightarrow x-\dfrac{1}{x}=\sqrt{5}$$
 another approach
$$a^2_{n+1}-4=a^4_{n}-4a^2_{n}=a^2_{n}(a^2_{n}-4)=a^2_{n}a^2_{n-1}(a^2_{n-1}-4)=\cdots=a^2_{n}a^2_{n-1}\cdots a^2_{1}(a^2_{1}-4)$$
so we have
$$\dfrac{a_{n+1}}{a_{1}a_{2}\cdots a_{n}}=\sqrt{a^2_{1}-2}\cdot \dfrac{a_{n+1}}{\sqrt{a^2_{n+1}-4}}\to\sqrt{a^2_{1}-4}=\sqrt{5},a_{n+1}\to+\infty$$
since $a_{n+1}\to +\infty,n\to +\infty$
