Infinitely nested radical formulas for $\pi$

Question

Has anyone come by the infinitely nested radical formula: $$\Bigg\{\pi=\frac{12}{5}\cdot \lim\limits_{n \to \infty} 2^{n}\cdot\frac{1}{2^{\frac{2^{n}+1}{2^n}}} \sqrt{2^{\frac{2^{n-1}+1}{2^{n-1}}} -\cdot\cdot\cdot\cdot\sqrt{ 2^{\frac{3}{2}} +\sqrt{ 2^2 + (\sqrt{ 6}- \sqrt{ 2})}}}\Bigg\}$$ And can anyone prove it?

Answer 1

From here:

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If there exists an infinite iterative sequence $A=\{a_1, a_2,\cdots a_n\}$, and a sequence $B=\{c_1, c_2,\cdots c_n\}$ obtainable from $A$ such that for all $b, a_n =\cos b$, $a_{n+1} =\cos b/2, c_{n+1}=\sin b/2$. Then $$a_{n+1} =\sqrt{\frac{1+a_n}{2}} \text{ and } c_n =\sqrt{\frac{1-a_n}{2}}$$

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If $a_1 =\frac{\sqrt{6}-\sqrt{2}}{4}$, then $$c_2 = \frac{1}{2^{3/2}}\sqrt{2^2-(\sqrt{6}-\sqrt{2})}, c_3 = \frac{1}{2^{5/4}}\sqrt{2^{3/2}-\sqrt{2^2-(\sqrt{6}-\sqrt{2})}}$$ And thus $$c_n = \frac{1}{2^{\frac{2^n+1}{2^n}}} \sqrt{2^{\frac{2^{n-1}+1}{2^{n-1}}}-\cdots \sqrt{2^{5/4}+\sqrt{2^{3/2}+\sqrt{2^2 +(\sqrt{6}-\sqrt{2}})}}}$$ If $k = \frac{12}{5}\lim_{n \to \infty} 2^{n-1} c_n$, we get the required nested radical. Hope it helps.