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Question Number 157630 by Oberon last updated on 25/Oct/21

Commented by ehab last updated on 25/Oct/21

Answered by mindispower last updated on 26/Oct/21

$$u_0=\sqrt{\frac{1}{2}\mathrm{\Pi }},U_1=\sqrt{\frac{1}{2}+\frac{1}{2}{U}_0}.U_2=\sqrt{\frac{1}{2}+\frac{1}{2}{U}_1}$$$$U_{n+1}=\sqrt{\frac{1}{2}+\frac{1}{2}{U}_n}$$$$\Rightarrow \sqrt{2}{U}_{n+1}=\sqrt{1+U_n}$$$$U_n=cos\left(q_n\right)\Rightarrow \sqrt{2}cos\left(q_{n+1}\right)=\sqrt{1+cos\left(q_n\right)}=\sqrt{2}cos\left(\frac{q_n}{2}\right)$$$$\Rightarrow q_{n+1}=\frac{q_n}{2}\Rightarrow q_n=a{\left(\frac{1}{2}\right)}^n$$$$\frac{1}{\sqrt{2}}=cos\left(a\right)\Rightarrow a=\frac{\pi }{4}$$$$\Rightarrow U_n=cos\left(\frac{\pi }{4.2^n}\right)ust$$$$wewant\frac{1}{2}\prod _{n⩾1}.cos\left(\frac{\pi }{{42}^n}\right)$$$$\underset{k=1}{\prod ^n}cos\left(\frac{\pi }{{42}^k}\right)=\frac{sin\left(\frac{\pi }{2}\right)}{2^{n+1}sin\left(\frac{\pi }{{42}^n}\right)}...cos\left(x\right)=\frac{sin\left(2x\right)}{2sin\left(x\right)}$$$$=\underset{n\to \mathrm{\infty }}{\lim }.\frac{1}{2^{n+1}sin\left(\frac{\pi }{{42}^n}\right)}.\frac{1}{2}=\underset{n\to \mathrm{\infty }}{\lim }.\frac{1}{2^{n+2}.\frac{\pi }{4.2^n}}=\frac{1}{\pi }$$$$justusedsin\left(x\right)=x+o\left(x\right),x\to 0$$$$\Rightarrow \frac{1}{\pi }=\frac{1}{2}\sqrt{\frac{1}{2}}.\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}.\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}}......$$

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