sequence of positive numbers satisfying $a_{n+1}=\frac{2}{a_n+a_{n-1}}$, prove it converges

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, $ab=1$.

@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 $1$.