Find product limit of this recursively-defined sequence? 
Problem： if $a_1=3$, $a_n=2a_{n-1}^2-1$, $n\ge2$, find the limit of this expression:
  $$\lim\limits_{n \to ∞} \prod\limits_{k=1}^{n-1} (1+\frac {1}{a_k})$$

The original problem asks to find this:
$$\lim\limits_{n \to ∞} \frac {a_n}{2^na_1a_2.....a_{n-1}}$$ 
The solution uses relation $a_n^2-1=2a_{n-1}^2×2(a_{n-1}^2-1)$ recursively and find out that the limit is $\sqrt2$;
  while I believe there got to be some other ways to do this. 
So I start by finding a simpler form of relation from expression $a_n=2a_{n-1}^2-1$ which is $a_n-1=2(a_{n-1}+1)(a_{n-1}-1)$, and finally I come to the product expression and obviously got stuck, I just don't have enough weaponry to deal with problems like this, trust me I always stumble on this type...
How should one correctly use the recursive relation to gradually approach the result? Is it mostly intuition or experience? And what are the first thing you tried when you first see this problem? 

Nice question, this following is my answer,(if I can't some wrong)
Let $a_{1}=\dfrac{1}{2}(x+\dfrac{1}{x})(x>1)\Longrightarrow x=3+2\sqrt{2}$
since
$$a_{2}=2a^2_{1}-1=\dfrac{1}{2}\left(x^2+\dfrac{1}{x^2}\right)$$
$$a_{3}=2a^2_{2}-1=\dfrac{1}{2}\left(x^4+\dfrac{1}{x^4}\right)$$
by indution you have
$$a_{n}=\dfrac{1}{2}\left(x^{2^{n-1}}+\dfrac{1}{x^{2^{n-1}}}\right)$$
and 
$$\prod_{k=1}^{n}\left(1+\dfrac{1}{a_{k}}\right)=\prod_{k=1}^{n}\dfrac{a_{k}+1}{a_{k}}=\prod_{k=1}^{n}\dfrac{2a^2_{k-1}}{a_{k}}=\dfrac{2^{n-1}(a_{1}+1)}{a_{n}}\prod_{k=1}^{n-1}a_{k}$$
since
\begin{align*}\prod_{k=1}^{n-1}a_{k}&=\dfrac{1}{2^{n-1}}\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{1}{2^{n-1}}\dfrac{x^{2^n}-\dfrac{1}{x^{2^n}}}{x-\dfrac{1}{x}}
\end{align*}
so
$$\prod_{k=1}^{n}\left(1+\dfrac{1}{a_{k}}\right)=\dfrac{x+1/x+2}{x-1/x}\cdot\dfrac{x^{2^n}-\dfrac{1}{x^{2^n}}}{x^{2^{n-1}}+\dfrac{1}{x^{2^{n-1}}}}\to\dfrac{(x+1)}{x-1}=\sqrt{2},n\to+\infty$$
becasuse
$$\lim_{n\to\infty}\dfrac{x^{2^n}-\dfrac{1}{x^{2^n}}}{x^{2^{n-1}}+\dfrac{1}{x^{2^{n-1}}}}=1,x>1$$
