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.
 A: Show that 
$$
a_2\le a_4\le \cdots\le a_{2n}\le a_{2n-1}\cdots\le a_3\le a_1,
$$
i.e., show that $\{a_{2n}\}$ is increasing, while $\{a_{2n-1}\}$ is decreasing.
And also $a_{2n}\le a_{2n-1}$.
This can be done inductively:
$$
a_{2k} \le a_{2k+2}\,\,\Rightarrow\,\, \sqrt{2-a_{2k}} \ge \sqrt{2-a_{2k+2}}
\,\,\Rightarrow\,\, \sqrt{2-\sqrt{2-a_{2k}}} \le \sqrt{2-\sqrt{2-a_{2k+2}}}.
$$
So $\{a_{2n}\}$, $\{a_{2n-1}\}$ converge. 
A: Hint: near the fixed point $x = 1$ the function $x\to\sqrt{2-x}$ is contractive.
A: This iteration $x_{n+1} = f(x_n)$ with $f(x) = \sqrt{2-x}$ has a nice attractive fixed point at $(1,1)$
You can fiddle with the starting value here: GeoGebra interactive worksheet.

We have
$$
f'(x) = -\frac{1}{2\sqrt{2-x}}
$$
and
$$
\lvert f'(1) \rvert = 1/2 < 1
$$
so $x^* = 1$ is attractive in a neighbourhood.
A: Hint: write $a_n=1+b_n$ or $a_n=1-b_n$, whichever makes $b_n$ positive. How does $b_n$ behave?
Elaboration: we have $a_n=1+b_n$ for odd $n$ and $a_n=1-b_n$ for even $n$ (why so?). So, for example, for even $n$ we can write $a_{n+1}=\sqrt{2-a_n}$ as $1+b_{n+1}=\sqrt{2-(1-b_n)}=\sqrt{1+b_n}$. Now you can compare $b_{n+1}$ and $b_n$. Proceed similarly for odd $n$.
A: 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
$$
(c_n+2)(c_n-1)\Bigl(c_n-\frac{1+\sqrt{5}}{2}\Bigr)\Bigl(c_n-\frac{1-\sqrt{5}}{2}\Bigr)\ge0
$$
A: Convergence speed:
For $x=1+t$ close to $1$,
$$1+t_{n+1}=\sqrt{1-t_n}\approx1-\frac{t_n}2$$
and 
$$t_{n+k}\approx t_n\left(-\frac12\right)^k.$$
The convergence is linear (one more exact bit per iteration). This is well illustrated by a logarithmic plot of the residue:

A: Potentially interesting alternative approach: if $a_{n+1}=\sqrt{2-a_n}$, let's take $a_n=2\cos\theta_n$ and hammer through some algebra to see that $2\cos\theta_{n+1}=2\sin\frac{\theta_n}{2}=2\cos\frac{\pi-\theta_n}{2}\implies\theta_{n+1}=\frac{\pi-\theta_n}{2}$. 
Noting that $\frac{\pi}{3}=\frac{\pi-\frac{\pi}{3}}{2}$, we can see that $(\theta_{n+1}-\frac{\pi}{3})=\frac{-1}{2}(\theta_n-\frac{\pi}{3})$, showing that 


*

*$\theta_n-\frac{\pi}{3}\to0$

*$\theta_n\to\frac{\pi}{3}$

*$a_n\to 2\cos\frac{\pi}{3}=1$


With a bit more work, you can push this a bit further to find an exact form for $a_n$, but that is arguably overkill for your question.
A: Here is a relatively simple argument that shows the convergence of the sequence $(a_n)_{n\geq 1}$.
First, we show by induction that  $1/3<a_n<5/3$ for $n\geq 1$.  This is clearly the case for $n=1$, since $a_1=\sqrt{2}$.  Assuming the inductive hypothesis, we have: 
\begin{eqnarray*}
1/3&<&a_n<5/3 \\
5/3&>&2-a_n>1/3 \\
5/3>\sqrt{5/3}&>&a_{n+1}>\sqrt{1/3}>1/3\,.
\end{eqnarray*}
QED
Next, use the recurrence to get
$$a_{n+1}^2-1=(2-a_n)-1=-(a_n-1)\,.$$ Hence 
$|a_{n}-1|=|a_{n+1}+1|\,|a_{n+1}-1|>4/3\,|a_{n+1}-1|$ and $|a_{n+1}-1|<3/4 \, |a_{n}-1|$ (for $n\geq 0$).  Consequently, 
$$|a_{n}-1|<(3/4)^n$$
(for $n\geq 1$).  It follows that $\lim_{n\to \infty}a_n=1$.
