sequence of positive numbers satisfying $a_{n+1}=\frac{2}{a_n+a_{n-1}}$, prove it converges Assume that $(a_n)$ is a sequence of positive real numbers satisfying $a_{n+1}=\frac{2}{a_n+a_{n-1}}$ for $n=2,3,\dots$.
Prove that $(a_n)$ is convergent and find the limit.

I have no idea how to prove convergence.
I tried to prove that $(a_n)$ is bounded and monotone, but both trials failed.
If convergence is proven, the limit is easy: if $g=\lim a_n$, then $g\ge 0$ because all $a_n$'s are positive, the equivalent relation $a_{n+1}(a_n+a_{n-1})=2$ gives $g(g+g)=2$, so finally $g=1$.
 A: I believe the following will help. 
Let $\overline{\lim}a_n=a$ and $\underline{\lim}a_n=b$.
It's obvious that $a\geq b$.
Now, let $$\lim\limits_{k\rightarrow\infty}{a_{n_k}}=a$$ and $$\lim\limits_{m\rightarrow\infty}{a_{n_m}}=b.$$
We know that
$$a_{n_k}=\frac{2}{a_{n_k-1}+a_{n_k-2}}.$$
Consider $\{k_l\}$ such that $$\lim\limits_{l\rightarrow\infty}a_{n_{k_l}-1}=c$$ and $$\lim\limits_{l\rightarrow\infty}a_{n_{k_l}-2}=d.$$
Thus,$$a=\frac{2}{c+d}\leq\frac{2}{b+b}$$ and we obtain $$ab\leq1.$$
By the same way we'll get $$b\geq \frac{2}{2a}$$ and $$ab\geq1.$$
Id est, $ab=1$.
A: @Michael Rozenberg's result is true. But, there should be proof of boundedness of the sequence. This is in fact easy induction. 
Let $a_0, a_1>0$. Find $\alpha>0$ such that $\alpha<\min(a_0, a_1)\leq \max(a_0,a_1)<\frac1{\alpha}$. 
Suppose $\alpha<\min(a_{n-1},a_n)\leq \max(a_{n-1},a_n)<\frac1{\alpha}$. Then by the recurrence, we have
$$
a_{n+1}=\frac2{a_n+a_{n-1}}< \frac 2{2\alpha} = \frac1{\alpha}
$$ 
and $$a_{n+1}=\frac2{a_n+a_{n-1}}> \frac2{2/\alpha}=\alpha.$$
Thus, it follows by induction that $\alpha<a_n<\frac1{\alpha}$ for all $n\geq 0$. 
Then as @Michael Rozenberg did, it follows that $\limsup a_n=a>0$, $\liminf a_n=b>0$ satisfy $ab=1$. Along with $b\leq a$, we obtain $b\leq 1\leq a$. 
It seems that one application of the recurrence is not enough for proving the convergence. Thus, we try applying the recurrence twice. 
$$
a_{n+1}=\frac2{a_n+a_{n-1}}=\frac2{\frac2{a_{n-1}+a_{n-2}}+a_{n-1}}.
$$
Let $\{a_{n_k}\}$ be a subsequence of $\{a_n\}$ such that $a_{n_k}\rightarrow a$.  Then 
$$
a_{n_k}=\frac2{\frac2{a_{n_k-2}+a_{n_k-3}}+a_{n_k-2}}.
$$
Take further subsequence $n_{k_l}$ of $n_k$ such that $a_{n_{k_l}-2}$ and $a_{n_{k_l}-3}$ both converge, to the limits $\beta$, $\gamma$ respectively. Then taking $l\rightarrow\infty$ in the recurrence $$
a_{n_{k_l}}=\frac2{\frac2{a_{n_{k_l}-2}+a_{n_{k_l}-3}}+a_{n_{k_l}-2}},
$$
we obtain
$$
a=\frac2{\frac2{\beta+\gamma}+\beta}.
$$
Now, replacing $\gamma$ by $a$ makes the RHS larger. Thus, 
$$
a\leq \frac2{\frac2{\beta+a}+\beta} .
$$
This yields
$$
a\left(\frac2{\beta+a}+\beta\right) \leq 2,
$$
$$
a\left(2+ \beta(\beta+a)\right)\leq 2(\beta+a),
$$
$$
a\beta(\beta+a)\leq 2\beta,
$$
$$
a(\beta+a)\leq 2.
$$
Replacing $\beta$ by $b$, we have
$$
a(b+a)\leq 2. 
$$
Since $ab=1$, we have $a^2\leq 1$ which yields $a\leq 1$. 
Therefore, $1\leq a \leq 1$, so $a=1$. By $ab=1$, also we have $b=1$. This prove the convergence of $\{a_n\}$ and the limit is $1$. 
