<s>[INST] Showing the sequence converges $a_{1}=\frac{1}{2}$, $a_{n+1}=\frac{1}{2+a_{n}}$ Showing the sequence converges $a_{1}=\frac{1}{2}$,  $a_{n+1}=\frac{1}{2+a_{n}}$.
I already know that if $(a_{n})$ converges then it does to  $\sqrt{2}-1$.But i dont't know how to prove that this sequence cenverges.
EDIT
I think that the subsequence $(a_{2n+1})$ is monotonic decreasing and the subsequence $(a_{2n})$ is monotonic increasing.
 [/INST] Here is another proof.
Say $$a_n = \frac{p_n}{q_n}$$
then from the identity we have that
$$p_{n+1} = q_n, q_{n+1} = 2q_n + p_n = 2q_n + q_{n-1}$$
$q_n$ are of the form
$$ A (\sqrt{2} + 1)^n + B (1 - \sqrt{2})^n $$
($\sqrt{2} + 1$ and $1 - \sqrt{2}$ are roots of $x^2 = 2x + 1$)
Thus $$a_n = \frac{q_{n-1}}{q_n}  = \frac{A (\sqrt{2} + 1)^{n-1} + B (1 - \sqrt{2})^{n-1}}{A (\sqrt{2} + 1)^n + B (1 - \sqrt{2})^n}\to \frac{1}{1 + \sqrt{2}} = \sqrt{2} -1 $$
(Divide both numerator and denominator by $(\sqrt{2} + 1)^n$)
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