<s>[INST] Convergence of $a_{n+1}=\sqrt{2-a_n}$ I'm attempting to prove the convergence (or divergence, though I strongly suspect it converges) of a sequence defined as  $a_{n+1}=\sqrt{2-a_n}$ with $a_1=\sqrt{2}$.
I cannot use the monotonic sequence theorem as the sequence is not monotonically increasing. In fact, the first few values of the sequence are:
$a_1 =\sqrt{2}\approx 1.4142$
$a_2 =\sqrt{2-\sqrt{2}}\approx .7653$
$a_3 =\sqrt{2-\sqrt{2-\sqrt{2}}}\approx 1.1111$
Thus, it seems that $a_{n \to \infty} \to 1$
It seems that the sequence is behaving similarly to $\frac{\sin x}{x}$, leading me to think that the squeeze theorem may be useful. Still, I cannot seem to make any progress besides numerical computation of successive terms.
 [/INST] You have surely proved that $a_n\le 2$ for all $n$.
Consider the sequences $b_n=a_{2n-1}$ and $c_n=a_{2n}$. The recursions are
$$
b_{n+1}=a_{2n+1}=\sqrt{2-a_{2n}}=\sqrt{2-\sqrt{2-a_{2n-1}}}=
\sqrt{2-\sqrt{2-b_n}}
$$
Let's show $(b_n)$ is decreasing:
\begin{gather}
b_{n+1}\le b_n\\
\sqrt{2-\sqrt{2-b_n}}\le b_n\\
2-\sqrt{2-b_n}\le b_n^2\\
2-b_n^2\le \sqrt{2-b_n}\\
4-4b_n^2+b_n^4\le 2-b_n\\
(b_n+2)(b_n-1)\Bigl(b_n-\frac{1+\sqrt{5}}{2}\Bigr)\Bigl(b_n-\frac{1-\sqrt{5}}{2}\Bigr)\le0
\end{gather}
and we just need to show $1\le b_n\le\sqrt{2}$ (work out why).
The basis of the induction is obvious, as $b_1=\sqrt{2}$. Suppose $1\le b_n\le \sqrt{2}$; then
\begin{gather}
1\le b_{n+1}\le\sqrt{2}\\
1\le 2-\sqrt{2-b_n}\le 2\\
0\le\sqrt{2-b_n}\le 1\\
0\le 2-b_n\le 1\\
1\le b_n\le 2
\end{gather}
Since this is true, we are done.
Therefore $(b_n)$ is a decreasing and bounded sequence, so it converges to $L$ such that
$$
L=\sqrt{L-\sqrt{2-L}}
$$
The only nonnegative values for $L$ are $1$ and $(1+\sqrt{5})/2$, which however is greater than $b_1$, so we see $L=1$.
Now do the same in order to prove $(c_n)$ is increasing; actually you have to see that $0<c_n\le 1$, so
$$
(c_n+2)(c_n-1)\Bigl(c_n-\frac{1+\sqrt{5}}{2}\Bigr)\Bigl(c_n-\frac{1-\sqrt{5}}{2}\Bigr)\ge0
$$
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