Fun with 2018 #13 (Archimedes and the beauty of circle)

$\begin{aligned} L & = \lim_{n \to \infty} \frac {2018 \sin \left(\frac {2\pi}{2018} \right)}{\prod_{k=0}^n \cos \left(\frac \pi{2^k\cdot 2018}\right)} & \small \color{#3D99F6} \text{Let } \theta = \frac \pi {2018} \\ & = \lim_{n \to \infty} \frac {2018 \sin \left( 2\theta \right)}{\color{#3D99F6} \prod_{k=0}^n \cos \left(\frac \theta {2^k}\right)} & \small \color{#3D99F6} \text{See note.} \\ & = \lim_{n \to \infty} \frac {2018 \sin \left( 2\theta \right)}{\color{#3D99F6}\frac {\sin (2\theta)}{2^{n+1}\sin \frac \theta{2^n}}} \\ & = \lim_{n \to \infty} 2018\cdot 2^{n+1} \sin \frac \theta {2^n} \\ & = \lim_{n \to \infty} 4036 \cdot \frac {\theta \color{#3D99F6}\sin \frac \theta {2^n}}{\color{#3D99F6}\frac \theta{2^n}} & \small \color{#3D99F6} \text{Note that }\lim_{x \to 0} \frac {\sin x}x = 1 \\ & = 4036 \theta = 4036 \cdot \frac \pi{2018} \\ & = 2\pi \approx \boxed{6.283} \end{aligned}$

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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 $n \ge 0$ .