Solution of problem T7/363 in M&Y

Let $(x_n)$ be a sequence given by $x_1=5,x_{n+1}=x_n^2-2(n\geq 1)$. Calculate $\lim_{n\to\infty}\dfrac{x_{n+1}}{x_1x_2\cdots x_n}$.

My solution.

Assume that $x_n=2u_n\forall n$, then $u_1=\dfrac{5}{2},u_{n+1}=2u_n^2-1\forall n$.

Now, write $u_1=\dfrac{1}{2}(a+\dfrac{1}{a})(a>1)$. By induction on $n$ we have $u_n=\dfrac{1}{2}\left(a^{2^{n-1}}+a^{-2^{n-1}}\right)\forall n\geq 1$. Therefore $x_n=\left(a^{2^{n-1}}+a^{-2^{n-1}}\right)\forall n\geq 1$.

Final, $x_1x_2\cdots x_n=\prod_{i=1}^n(a^{2^{i-1}}+a^{-2^{i-1}})=\prod_{i=1}^n\dfrac{ a^{2^i}-a^{-2^i}}{ a^{2^{i-1}}-a^{-2^{i-1}}}=\\ \dfrac{a^{2^n}-a^{-2^n}}{a^1-a^{-1}}\forall n\geq 1$, so that limit is equal to $\lim_{n\to\infty} (a^1-a^{-1})\cdot \dfrac{a^{2^n}+a^{-2^n}}{a^{2^n}-a^{-2^n}}=a^1-a^{-1}.\Box$