What is the value of the following limit: $\lim\limits_{n\to \infty} 4^n(1-a_n)$ where $a_{n+1} = {\sqrt{\frac{1+a_n}{2}}}$? I have the following recursive $(a_n)_{n>0}$ sequence defined with $a_{n+1} = {\sqrt{\frac{1+a_n}{2}}}$ and where $-1 \leqslant a_n \leqslant 1$.
And I must calculate the following:
$$\lim\limits_{n\to \infty} 4^n\left(1-a_n\right)$$
I tried figuring it out with proving that the sequence is decreasing, but I could not prove it with any way. The limit calculated is clearly $\,1\,$, but it just give the case of $\,\infty\cdot0\,$ which is not very helpful. I can think of the Cesaro-Stolz theorem maybe as a starting point, but could not work it out.
Thank you in advance for helping out!
 A: Since $a_n \in [-1,1]$ for all $n\ge 1$, there exists a sequence $x_n \in [0,\pi]$ such that $a_n=\cos x_n$ for all $n \ge 1$. It follows that
$$
\cos x_{n+1}=a_{n+1}=\sqrt{\frac{1+\cos x_n}{2}}=\sqrt{\cos^2\left(\frac{x_n}{2}\right)}=\cos\left(\frac{x_n}{2}\right)
$$
By induction, we have
$$
a_n=\cos\left(\frac{x_1}{2^{n-1}}\right) \quad \forall n\ge 1.
$$
Now
$$
\lim_n4^n(1-a_n)=\lim_n2^{2n+1}\sin^2\left(\frac{x_1}{2^n}\right)=\lim_n2^{2n+1}\left(\frac{x_1}{2^n}\right)^2=2x_1^2=2[\cos^{-1}(a_1)]^2
$$
A: As Christophe pointed out, we can set $\;\theta=\arccos a_1\in[0,\pi]\;.$
First of all, we will prove by induction that
$a_n=\cos\left(\dfrac\theta{2^{n-1}}\!\right)\;\;$ for any $\;n\in\mathbb N\,.\quad\color{blue}{(*)}$
Base case :
if $\;n=1\,,\;$ we get that $\;a_1=\cos\theta=\cos\left(\dfrac\theta{2^0}\!\right)\;,$
hence $\,(*)\,$ is true for $\,n=1\,.$
Induction step :
let $\,k\in\Bbb N\,$ be given and suppose $\,(*)\,$ is true for $\,n=k\,.$
$a_{k+1}=\sqrt{\dfrac{1+a_k}2}=\sqrt{\dfrac{1+\cos\big(\frac\theta{2^{k-1}}\big)}2}=\cos\left(\dfrac\theta{2^k}\!\right)\,.$
Thus, $\,(*)\,$ holds for $\,n=k+1\,$ and the proof of the induction step is complete.
By the principle of induction, $\,(*)\,$ is true for all $\,n\in\Bbb N\,.$

Now, we will calculate the limit
$\lim\limits_{n\to\infty}4^n\big(1-a_n\big)=\lim\limits_{n\to\infty}4^n\left[1-\cos\left(\dfrac\theta{2^{n-1}}\!\right)\right]=$
$=\lim\limits_{n\to\infty}4^n\left(\dfrac\theta{2^{n-1}}\!\right)^{\!2}\left[\dfrac{1-\cos\left(\frac\theta{2^{n-1}}\!\right)}{\left(\frac\theta{2^{n-1}}\!\right)^2}\right]=$
$=\lim\limits_{n\to\infty}4\theta^2\left[\dfrac{1-\cos\left(\frac\theta{2^{n-1}}\!\right)}{\left(\frac\theta{2^{n-1}}\!\right)^2}\right]=4\theta^2\cdot\dfrac12=2\theta^2=$
$=2\big(\!\arccos a_1\!\big)^2\,.$
A: Several commenters have pointed out that you should consider the double angle cosine equation, $\cos(2\theta) = 2\cos^{2}(\theta) - 1$. This rearranges to
$$
\cos^{2}(\theta) = \frac{1+\cos(\theta)}{2}
$$
which is analogous to an equivalent form of the general equation for your series $a_{n}$
$$
a_{n+1} = \sqrt\frac{1+a_{n}}{2}
$$
which would be
$$
a^{2}_{n+1} = \sqrt\frac{1+a_{n}}{2}
$$
As you can see, if you set $a_{1}$ to $\cos(\theta)$, then $a_{2}=\cos(\frac{\theta}{2})$, and it should be clear $a_{n}=\cos(\frac{\theta}{2^{n-1}})$.
Then
$$
\begin{align}
4^{n}(1-a_{n}) &= 4^{n}\left(1-\cos\left(\frac{\theta}{2^{n-1}}\right)\right) \\
4^{n}(1-a_{n}) &= 4^{n}\left( 1 - \left( 1 - \frac{1}{2}\frac{\theta^{2}}{(2^{n-1})^{2}}+\ldots\right)\right) \\
4^{n}(1-a_{n}) &= 4^{n}\left(  \frac{1}{2}\frac{\theta^{2}}{4^{n-1}}-\ldots\right) \\
4^{n}(1-a_{n}) &= 2\theta^{2}-\ldots
\end{align}
$$
Where it should be clear that the Taylor series of the cosine is truncated in the right place (i.e., where the other terms will limit out with increasing $n$).
The last result says that you can replace the thing you were asked about (on the left hand side), with the expression on the right, which you know. Just replace $\theta = \cos^{-1}(a_{1})$.
