# Compute $\lim\limits_{n\to \infty}\frac{\prod\limits_{k=1}^{n}a_k}{2^n}$

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.

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: $$\lim_{n\to +\infty}\frac{1}{2^n}\prod_{k=1}^{n}a_k = \color{red}{\frac{2}{\pi}}=0.636619772\ldots \tag{4}$$
• can you explain why $$\prod_{k=1}^{n}\cos\left(\frac{\pi}{2^{k+1}}\right)=\frac{1}{2^n\sin\left({\pi}/{2^{n+1}}\right)}\tag{3}$$ – Chris May 4 '15 at 15:57
• @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 $\sin(\pi/2)=1$. – Jack D'Aurizio May 4 '15 at 16:02