François Viète

Let $a_0 = \sqrt{\dfrac 12}$ , $a_1 = \sqrt{\dfrac 12 + \dfrac 12 \sqrt{\dfrac 12}}$ , $a_2 = \sqrt{\dfrac 12+\dfrac 12\sqrt{\dfrac 12 + \dfrac 12 \sqrt{\dfrac 12}}}$ and so on. We note that

$\begin{aligned} a_n & = \sqrt{\frac 12 + \frac {a_{n-1}}2} & \small \color{#3D99F6} \text{Squaring both sides} \\ a_n^2 & = \frac 12 + \frac {a_{n-1}}2 \\ 2a_n^2 - 1 & = a_{n-1} & \small \color{#3D99F6} \text{Let }a_n = \cos \theta_n \\ \cos 2\theta_n & = \cos \theta_{n-1} \\ \implies \theta_n & = \frac {\theta_{n-1}}2 = \frac {\color{#3D99F6}\theta_0}{2^n} \end{aligned}$

Then the express is:

$\begin{aligned} \lim_{n \to \infty} \frac 2{\prod_{k=0}^n a_k} & = \lim_{n \to \infty} \frac 2{\prod_{k=0}^n \cos \theta_k} \\ & = \lim_{n \to \infty} \frac 2{\color{#3D99F6}\prod_{k=0}^n \cos \frac {\theta_0}{2^n}} & \small \color{#3D99F6} \text{See note.} \\ & = \lim_{n \to \infty} \frac 2{\color{#3D99F6}\frac {\sin 2 \theta_0}{2^{n+1}\sin \frac {\theta_0}{2^n}}} & \small \color{#3D99F6} \text{Note that }a_0 = \cos \theta_0 = \frac 1{\sqrt 2} \implies \theta_0 = \frac \pi 4 \\ & = \lim_{n \to \infty} \frac 2{\color{#3D99F6}\frac 1{2^{n+1}\sin \frac \pi{2^{n+2}}}} \\ & = \lim_{n \to \infty} 2^{n+2} \sin \frac \pi{2^{n+2}} \\ & = \lim_{n \to \infty} \pi \times \frac {\sin \frac \pi{2^{n+2}}}{\frac \pi{2^{n+2}}} \\ & = \pi \approx \boxed{3.142} \end{aligned}$

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Note: Prove by induction that $\displaystyle P_n = \prod_{k=0}^n \cos \frac \theta{2^k} = \frac {\sin 2\theta}{2^{n+1} \sin \frac \theta{2^n}}$ for all $n \ge 0$ .

For $n=0$ , $\displaystyle P_0 = \prod_{k=0}^0 \cos \frac \theta{2^k} = \cos \theta = \frac {\sin 2\theta}{\sin \theta}$ . Therefore, the claim is true for $n = 0$ . Now assuming that the claim is true for $n$ , then $\displaystyle P_{n+1} = \prod_{k=0}^{n+1} \cos \frac \theta{2^k} = \left(\prod_{k=0}^n \cos \frac \theta{2^k}\right)\cdot \cos \frac \theta{2^{n+1}} = \frac {\sin 2\theta}{2^{n+1} \sin \frac \theta{2^n}} \cdot \cos \frac \theta{2^{n+1}} = \frac {\sin 2\theta}{2^{n+2} \sin \frac \theta{2^{n+1}}}$ . Therefore, the claim is true for $n+1$ and hence true for all $n \ge 0$ .

Let $C$ be a circle of radius $r > 0$ and $(A_n)_{n \ge 3}$ the sequence of areas of the regular polygons inscribed in $C$ with $n$ sides, then: $A_n = \frac{1}{2} n r^2 \sin(\frac{2 \pi}{n}) = n r^2 \sin(\frac{\pi}{n}) \cos(\frac{\pi}{n}) \quad \text{ and } \quad A_{2n} = \frac{1}{2} 2n r^2 \sin(\frac{2 \pi}{2n}) = \frac{1}{2} 2n r^2 \sin(\frac{\pi}{n}) \Rightarrow$ $\displaystyle \frac{A_n}{A_{2n}} = \cos(\frac{\pi}{n}) \Rightarrow A_n = \lim_{k \to \infty} \frac{A_n}{A_{2n}} \cdot \frac{A_{2n}}{A_{2^2 n}} \cdot \frac{A_{2^2n}}{A_{2^3n}} \cdot \ldots \cdot \frac{A_{2^{k - 1}n}}{A_{2^{k}n}} \cdot A_{2^k n} = \cos(\frac{\pi}{n}) \cdot \cos(\frac{\pi}{2n}) \cdot \cos(\frac{\pi}{4n}) \cdot \ldots \cdot \cos(\frac{\pi}{2^k n}) \cdot \ldots \cdot \pi r^2$ because $\lim_{k \to \infty} A_{2^k n} = \pi r^2 = \text{ Area of the circle}$ , (This is known like Archimedes's process) and this implies that $\pi = \frac{\frac{1}{2} n \sin(\frac{2 \pi}{n})}{\cos(\frac{\pi}{n}) \cdot \cos(\frac{\pi}{2n}) \cdot \cos(\frac{\pi}{4n}) \cdot \ldots \cdot \cos(\frac{\pi}{2^k n}) \cdot \ldots \cdot}$ Substituing $n = 4$ , like Viète did, and knowing that $\cos(\frac{\pi}{4}) = \sqrt{\frac{1}{2}}$ and that for angle $\cos(\frac{\alpha}{2}) = \sqrt{\frac{1}{2} + \frac{1}{2} \cos(\alpha)}$ we get the result.