limit of : $a_{n+2} =\frac{1}{a_n} + \frac{1}{a_{n+1}}$

It is easy to show that the sequence is bounded. If $\sqrt{2}/M \le a_n,a_{n+1} \le \sqrt{2}M$, then also $\sqrt{2}/M \le a_{n+2} \le \sqrt{2}M$.

In fact, if we let $b_n=\max(|\log (a_n/\sqrt{2})|,|\log (a_{n+1}/\sqrt{2})|)$, then $b_{n+1}\le b_n$. Therefore (since $b_n$ is non-negative) $b_n$ converges to its infimum. Let us denote $\lim_{n\to\infty} b_n = B$. All that is left is to do is prove that $B=0$. Suppose that $B>0$, then for some $N$, we have $B-\epsilon<b_n<B+\epsilon$ for all $n>N$. So $|\log (a_{N}/\sqrt{2})|<B+\epsilon$ and $|\log (a_{N+1}/\sqrt{2})|<B+\epsilon$. If both $a_N$ and $a_{N+1}$ are on different sides of $\sqrt{2}$, then $|\log (a_{N+2}/\sqrt{2})|<B-2\epsilon$ (with $C$ some constant) and the same for $a_{N+3}$. If both are on the same side then $a_{N+2}$ is on the other side, and the argument is the same. In any case we have that $b_{N+3}<B-2\epsilon$ and so a contradiction.

If $a_n$ converges towards $a$, then the limit satisfies $a = \frac 1 a + \frac 1 a$.

Maybe prove this. Suppose $$ {\sqrt{2}}(1+\varepsilon)>a_{n+1}>\frac{\sqrt{2}}{1+\varepsilon},\qquad {\sqrt{2}}(1+\varepsilon)>a_{n}>\frac{\sqrt{2}}{1+\varepsilon}. $$ If $a_{n+1},a_n$ are on the same side of $\sqrt{2}$, then $a_{n+2}$ is on the opposite side and $$ {\sqrt{2}}(1+\varepsilon)>a_{n+2}>\frac{\sqrt{2}}{1+\varepsilon}. $$ If $a_{n+1},a_n$ are opposite sides of $\sqrt{2}$, then $$ {\sqrt{2}}\left(1+\frac{\varepsilon}{2}\right)>a_{n+2}>\frac{\sqrt{2}}{\displaystyle 1+\frac{\varepsilon}{2}}. $$