If $x_1=5$, $x_{n+1}=x_n^2-2$, find $\lim x_{n+1}/(x_1\cdots x_n)$ If
$$\left\{x_{n}\right\}\mid x_{1}=5,x_{n+1}=x_{n}^{2}-2,\forall n\geq 1$$
find
$$\lim_{n\to\infty}\frac{x_{n+1}}{x_{1}x_{2}\cdots x_{n}}.$$
If someone could help me out with tags, it'd be lovely. I think this is calculus and real-analysis, but I'm not sure--I had the problem scribbled down on a post-it, and I forget where it's from.
 A: This is not an answer, just too big for a comment. I hope it'll help someone else finish a proof. 
Write
$$
y_n = \frac{x_n}{x_1 \dots x_{n-1}}, \quad n \ge 2.
$$
Then $x_{n+1} = x_n^2 - 2$ implies $y_{n+1} = y_n - \frac{2}{x_1 \dots x_n}$, hence by forming a telescopic sum, one gets
$$
y_n - y_2 = \sum_{i=2}^{n-1} y_{i+1} - y_i = \sum_{i=2}^{n-1} \frac{-2}{x_1 \dots x_i}
$$
and therefore 
$$
y_n = y_2 - \sum_{i=2}^{n-1} \frac{2}{x_1 \dots x_i}= \frac {23}{5} - \frac 25 \sum_{i=2}^{n-1} \frac{1}{x_2 \dots x_i}.
$$
Computing the limit is therefore equivalent to being able to compute
$$
\sum_{i=2}^{n-1} \frac 1{x_2 \dots x_i}.
$$
This sum is quite easily shown to converge (pretty fast in top of that, since if we remove the "$-2$"'s in the recursion definition of the sequence $x_n$, we get something like $\sum \frac 1{5^{2^1}\dots 5^{2^n}}$ which converges REALLY faster than the geometric series). But I have no idea yet how to compute the series. Maybe this is not the way to go.
Hope that helps,
