Find limit of this recursive sequence $$
a_0=0,\ a_1=2,\ a_{n+1}=\sqrt{2 - \frac{a_{n-1}}{a_n}} \\
\lim_{n\to\infty}2^na_n\ =\ ?
$$
 A: Let's prove by induction that
$$
a_n=2\sin\left(\frac{\pi}{2^n}\right)
$$
Let $\theta_n=\pi/2^n$. Clearly, $2\sin(\pi)=0=a_0$ and $2\sin(\frac{\pi}{2})=2=a_1$. Furthermore,
$$
a_{n+1}=\sqrt{2-\frac{a_{n-1}}{a_n}}=2\sqrt{\frac{1-\frac{a_{n-1}}{2a_n}}{2}}
$$
But
$$
\frac{a_{n-1}}{2a_n}=\frac{\sin \theta_{n-1}}{2\sin\theta_n}=
\frac{\sin 2\theta_n}{2\sin\theta_n}=
\frac{2\sin \theta_n\cos\theta_n}{2\sin\theta_n}=\cos\theta_n
$$
and
$$
\cos2\alpha=1-2\sin^2\alpha
$$
$$
\sin\alpha=\sqrt{\frac{1-\cos 2\alpha}{2}}
$$
Hence,
$$
a_{n+1}=2\sqrt{\frac{1-\cos\theta_n}{2}}=
2\sin\frac{\theta_n}{2}=2\sin\theta_{n+1}
$$
Using Taylor series expansion,
$$
a_n=2\theta_n+O\left(\theta_n^3\right)=
\frac{\pi}{2^{n-1}}+O\left(8^{-n}\right)
$$
so
$$
2^n a_n=2\pi+O\left(4^{-n}\right)\to 2\pi
$$
A: This is only a partial answer.  I'll complete it if I can, or someone else can take over.
A little fiddling around shows that
$$\begin{align}
a_2&=\sqrt2\\
a_3&=\sqrt{2-\sqrt2}\\
a_4&=\sqrt{2-\sqrt{2+\sqrt2}}\\
a_5&=\sqrt{2-\sqrt{2+\sqrt{2+\sqrt2}}}\\
\end{align}$$
at which point the pattern should be clear, so the question boils down to how quickly does
$$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}$$
approach its limit, which is easily shown to be $2$.
