Proving a sequence is convergent by convergence of odd and even subsequences? I want to use this method to prove convergence of: $a_{n}=\sqrt{2-a_{n-1}}$, $a_{0}=\frac{2}{3}$.
Here is my attempt at proof:
It can be proven inductively that $0< a_{n}<2$ for all $n$.
I want to show that each of the odd and even subsequences are monotone.
$a_{2n+3}-a_{2n+1}=\sqrt{2-a_{2n+2}}-\sqrt{2-a_{2n}}=\frac{a_{2n}-a_{2n+2}}{\sqrt{2-a_{2n+2}}+\sqrt{2-a_{2n}}}=\frac{a_{2n}-\sqrt{2-\sqrt{2-a_{2n+1}}}}{\sqrt{2-a_{2n+2}}+\sqrt{2-a_{2n}}}$
I'm stuck as I'm not able to show that $(a_{2n+1})_{n\geq0}$ is decreasing.
Thank's in advance.

If $0<x<1$ then $\sqrt {1-x}>1-x/(2-x)$ (by squaring both sides and calculating). This deserves more publicity. We will use this in (2).(ii). below.
If $a_n$ converges to $L$ then $0\leq L=\sqrt {2-L}\implies (L^2=2-L\land 0\leq L)\implies L=1.$ So it is natural to let $a_n=1+b_n.$
We have $0<|b_n|<1.$ (Equivalent to $0<a_n<2$).
(1).  If $1>b_n>0$ then $1+b_{n+1}=\sqrt {2-(1+b_n)}=\sqrt {1-b_n}<1$, which implies $b_{n+1}<0.$..... If $0>b_n>-1$ then $1+b_{n+1}=\sqrt {1-b_n}>1$, which implies $b_{n+1}>0.$
(2).(i).  If $b_n<0$ then $b_{n+1}>0$ so $$1+2|b_{n+1}|=1+2b_{n+1}<(1+b_{n+1})^2=1-b_n=1+|b_n|$$ which implies $0<|b_{n+1}|<|b_n|/2.$
(2). (ii). If $b_n>0$ then  $b_{n+1}<0$ so $$1-|b_{n+1}|=1+b_{n+1}=\sqrt {1-b_n}\;> 1-b_n/(2-b_n)$$ which implies $|b_{n+1}|<b_n/(2-b_n)<b_n=|b_n|.$
Applying all this, we have:
(3).(i).  If $b_n<0$ we have $b_{n+1}>0$ so $|b_{n+2}|<|b_{n+1}|<|b_n|/2.$
(3).(ii). If $b_n>0$ we have $b_{n+1}<0$ so $|b_{n+2}|<|b_{n+1}|/2<|b_n|/2.$
