Proving the limit of a recursive sequence Given $$a_0=0 \quad \&\quad a_n=\sqrt {\frac {1+a_{n-1}}2}$$ for $n>0$.
How do we prove or disprove $$\lim_{n \to \infty}4^n \left(1-a_{n}\right) = \frac{\pi^2}{8}. $$
 A: $$
\sqrt{2}a_{n}=\sqrt{1+a_{n-1}} \Rightarrow a_{n}\ge0 \\
\sqrt{2}a_{n}=\sqrt{1+a_{n-1}} \Rightarrow 2a_{n}^2=1+a_{n-1} \Rightarrow a_{n-1}-1=2a_{n}^2-2=2(a_{n}-1)(a_{n}+1) \\
\Rightarrow (a_{n-1}-1)/(a_{n}-1)=2(a_{n}+1)\gt0 \Rightarrow \text{sgn}(a_{n-1}-1)=\text{sgn}(a_{n}-1) \\
\Rightarrow a_{n-1},a_{n}\gt1 \space\text{or}\space a_{n-1},a_{n}\lt1 \Rightarrow a_{n}\lt1 \quad\{a_0=0,a_1=1/2\} \\[8mm]
$$
$$
0\le a_n \lt1,\quad \text{Let}\space a_n = \cos(x_n) \Rightarrow 0\lt x_n \le{\pi/2} \\
a_0=\cos(x_0)=0 \Rightarrow x_0={\pi/2},\quad \sqrt{2}a_n=\sqrt{1+a_{n-1}} \Rightarrow \\
\sqrt{2}\cos(x_n)=\sqrt{1+\cos(x_{n-1})}=\sqrt{1+2\cos^2(x_{n-1}/2)-1}=\sqrt{2}\cos(x_{n-1}/2) \\
\Rightarrow x_n=2^{-1}x_{n-1}=2^{-2}x_{n-2}=\cdots=2^{-n}x_{n-n}=2^{-n}x_0 \Rightarrow \color{red}{x_n=2^{-n}\pi/2} \\[8mm]
$$
$$
L=\lim_{n\rightarrow\infty}\left[4^n\left(1-a_n\right)\right]=\lim_{n\rightarrow\infty}\left[4^n\left(1-\cos(x_n)\right)\right]=\lim_{n\rightarrow\infty}\left[4^n\left(1-\cos(2^{-n}\pi/2)\right)\right] \Rightarrow \\
L=\lim_{n\rightarrow\infty}\frac{1-\cos(2^{-n}\pi/2)}{2^{-2n}} = \frac{0}{0} \space\{\text{L'Hopital's rule}\}\space =\lim_{n\rightarrow\infty}\frac{\frac{d}{dn}\left(1-\cos(2^{-n}\pi/2)\right)}{\frac{d}{dn}\left(2^{-2n}\right)} \Rightarrow \\
L=\lim_{n\rightarrow\infty}\frac{(-2^{-n}\pi/2)(\log2)\sin(2^{-n}\pi/2)}{-2^{-2n}(2\log2)}=\frac{\pi^2}{8}\lim_{n\rightarrow\infty}\frac{\sin(2^{-n}\pi/2)}{2^{-n}\pi/2} \Rightarrow \color{red}{L=\frac{\pi^2}{8}}
$$
