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Compute $$\lim\limits_{n\to\infty}\frac{\prod\limits_{k=1}^{n}a_k}{2^n}$$ where $a_k=\sqrt{2+a_{k-1}}$ and $a_1=\sqrt{2}$.

I proved $\lim\limits_{n\to\infty}a_n$ exists and found it (it's 2), but $\lim\limits_{n\to\infty}\frac{\prod\limits_{k=1}^{n}a_k}{2^n}$ is much more challenging for me.

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Assuming $a_n=2\cos\theta$, we have: $$ a_{n+1}=\sqrt{2+2\cos\theta} = 2\cos\frac{\theta}{2}\tag{1}$$ hence it follows that: $$ a_n = 2 \cos\left(\frac{\pi}{2^{n+1}}\right)\tag{2}$$ and: $$ \frac{1}{2^n}\prod_{k=1}^{n}a_n = \prod_{k=1}^{n}\cos\left(\frac{\pi}{2^{k+1}}\right)=\frac{1}{2^n\sin\left(\frac{\pi}{2^{n+1}}\right)}\tag{3} $$ by the sine duplication formula. That gives: $$ \lim_{n\to +\infty}\frac{1}{2^n}\prod_{k=1}^{n}a_k = \color{red}{\frac{2}{\pi}}=0.636619772\ldots \tag{4}$$

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  • $\begingroup$ can you explain why $$\prod_{k=1}^{n}\cos\left(\frac{\pi}{2^{k+1}}\right)=\frac{1}{2^n\sin\left({\pi}/{2^{n+1}}\right)}\tag{3} $$ $\endgroup$ – Chris May 4 '15 at 15:57
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    $\begingroup$ @Chris: multiply the product in the LHS by $\sin(\pi/2^{n+1})$ and telescope by using $\sin(x)=2\sin(x/2)\cos(x/2)$ and $\sin(\pi/2)=1$. $\endgroup$ – Jack D'Aurizio May 4 '15 at 16:02
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    $\begingroup$ @Chris: you're welcome :) $\endgroup$ – Jack D'Aurizio May 4 '15 at 17:17

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