Existence of limit in a recurrence equation: $\alpha_{n+2}=\frac{\alpha_{n}+(1/\alpha_{n+1})}{2}$ Let be $\boldsymbol{\alpha_{n+2}=\frac{\alpha_{n}+(1/\alpha_{n+1})}{2}}$ a recurrence equation with known $\alpha_0$ and $\alpha_1$. How do you prove that $\lim_{n\to\infty}\alpha_n$ exists? Note that no conditions are to be assumed about $\alpha_0$ and $\alpha_1$.
I tried to solve this problem with the usual techniques for the classical sequence $\alpha_{n+2}=\frac{\alpha_{n+1}+1/\alpha_{n+1}}{2}$, but I did not get any important. My main problem is the unknown signs of the initial conditions, so the sequence can oscillate around 0, but obviously the limit should be 1 or -1.
On the other hand, if this sequence can be divergent, then what conditions about $\alpha_0$ and $\alpha_1$ should be necessary and sufficient in order to ensure that it is convergent?
 A: If the sequence is convergent, it must tend to $1$ or to $-1$, depending on the initial conditions. It follows that it cannot be convergent for all initial conditions. In fact it is even not defined for all initial conditions. The set of
initial conditions for which it is defined is complicated.
Let $a_n=(x_{n+1},x_n)$. Then $a_{n+1}=\left((x_{n+1}+1/x_n)/2,x_{n+1}\right)$,
so $a_{n+1}=F(a_n)$, where 
$$F(x,y)=((x+1/y)/2,x)$$
The limit must be a fixed point of this $F$, $F(a)=a$. Solving this we find
$x=y=\pm1$. Linearizing the map near a fixed point we find that the eigenvalues of the Jacobi matrix 
$$\lambda_{1,2}=\frac{1\pm i\sqrt{7}}{4},$$
so $|\lambda_i|<1$, both fixed points are attracting so each of them has a basin of attraction (the set of initial conditions for which the sequence converges to this fixed point).
But the problem is that the set of initial conditions for which your sequence is defined is complicated. Indeed, when $y=0$, $F$ is undefined. Therefore
the second iterate is undefined when $x=0$, the third iterate is undefined when $xy=-1$, and so on. We obtain a countable set of algebraic curves for which the sequence is undefined. Now from the general considerations, there is also a complicated set where it is defined but has no limit.
