Find limit of this recursive sequence

$$ a_0=0,\ a_1=2,\ a_{n+1}=\sqrt{2 - \frac{a_{n-1}}{a_n}} \\ \lim_{n\to\infty}2^na_n\ =\ ? $$

Solution 1:

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 $$ 2^n a_n=2\pi+O\left(4^{-n}\right)\to 2\pi $$

Solution 2:

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 $2$.