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!
Best Answer
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
$\lim\limits_{n\to\infty}4^n\big(1-a_n\big)=\lim\limits_{n\to\infty}4^n\left[1-\cos\left(\dfrac\theta{2^{n-1}}\!\right)\right]=$
$=\lim\limits_{n\to\infty}4^n\left(\dfrac\theta{2^{n-1}}\!\right)^{\!2}\left[\dfrac{1-\cos\left(\frac\theta{2^{n-1}}\!\right)}{\left(\frac\theta{2^{n-1}}\!\right)^2}\right]=$
$=\lim\limits_{n\to\infty}4\theta^2\left[\dfrac{1-\cos\left(\frac\theta{2^{n-1}}\!\right)}{\left(\frac\theta{2^{n-1}}\!\right)^2}\right]=4\theta^2\cdot\dfrac12=2\theta^2=$
$=2\big(\!\arccos a_1\!\big)^2\,.$