TITLE: Recurrent sequence limit
QUESTION [9 upvotes]: 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?

REPLY [6 votes]: Dedicated to @Ant who asked "How did you come up with that?" and to all those that might be interested by a (somewhat) different, possibly more "natural" answer, i.e., @Almagest @David  @Eugen Covaci  @Turambar.
An alternate change of variable is possible, based on hyperbolic trigonometry. The treatment I am going to give is in fact very parallel to the excellent solution of @Geromty. Setting:
$$a_1=2 \cosh(y)=3$$
Due to relationship $\cosh(2\alpha)=2 \cosh^2(\alpha)-1$, by an immediate recurrence:
$$a_k=2 \cosh(2^{k-1}y)$$
Thus:
$$\Pi_{k=1}^{n}a_{k}=2^n\Pi_{k=0}^{n-1} \cosh(2^ky)=\dfrac{\sinh(2^ny)}{\sinh y} \ \ \ (1)$$
The last closed form formula is a classical result (see Appendix).
Thus $$\dfrac{a_{n+1}}{a_{1}a_{2}\cdots a_{n}}= 2\cosh(2^ny)\dfrac{\sinh(y)}{\sinh(2^ny)}=2\sinh(y) \dfrac{\cosh(2^ny)}{\sinh(2^ny)}$$
which tends (from above), when $n \rightarrow \infty$, to $$2\sinh(y)=2\sqrt{\cosh^2y-1}=2\sqrt{\dfrac{9}{4}-1}=\sqrt{5}.$$
Appendix: Proof of the second equality in (1).
Let us first recall the formula:
$$\sinh(2 \alpha)=2 \sinh(\alpha)\cosh(\alpha) \ \ \ \ (2)$$
Let  $P_n$ be the product in (1). Let us multiply $P_n$ by $\sinh(y)$. Then, using (2), one gets 
$$2^{n-1} \sinh(2y)\cosh(2y)\cosh(4y)\cosh(8y)...\cosh(2^{n-1}y).$$
We can anew apply formula (2), and again in cascade, reducing finally in this way the initial product ($\sinh(y)P_n$) to $\sinh(2^ny)$.
Remark: there is, evidently, a similar result in circular trigonometry $$2^n\Pi_{k=0}^{n-1} \cos(2^ky)=\dfrac{\sin(2^ny)}{\sin y}.$$