# ask the Limit of a sequence

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

I have no idea how to compute the limit of a sequence, please give some hints.

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I might try rewriting the sequence in terms of $b_n = 2^n a_n$. –  hardmath May 15 '14 at 0:57
Take the limit of both sides of the recursion and solve for the $a_n$ limit assuming it exists. –  RRL May 15 '14 at 1:15
@RRL: I'd like to see how you would do that! –  user21820 May 15 '14 at 1:58

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

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There's a minor error. $a_n = 2 \sin(\frac{\pi}{2^n})$ so that $a_0 = 2 \sin(\pi) = 0$ and $a_1 = 2 \sin(\frac{\pi}{2}) = 2$. So the limit would be $2 \pi$ instead. –  user21820 May 15 '14 at 2:39
I've edited away the error. Anyway it's a nice solution! And by the way it is one of the formulae at mathworld.wolfram.com/PiFormulas.html. –  user21820 May 15 '14 at 6:32
@user21820 Oops, you're right. Thanks for fixing it! –  Marcin Łoś May 15 '14 at 7:01

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 $2$.

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oops I'm silly, had an algebra error on my paper! –  DanZimm May 15 '14 at 1:52
It's $2\cos\left(\pi/2^n\right)$, as can be shown by half-angle formula starting with e.g. $\pi$. You beat me to it ;) I hope now you can complete the answer. Hint: first two terms of Taylor series are enough. –  Marcin Łoś May 15 '14 at 1:53
@MarcinŁoś, do you want to post a complete solution (or hint) as a separate answer? You should get credit for it. –  Barry Cipra May 15 '14 at 2:24