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Let ${A}_{0}=\frac{3}{4}$, and ${A}_{n+1}=\frac{1+\sqrt{{A}_{n}}}{2}$ for all $n\ge 0$
How to find the value of $\prod _{n=1}^{\mathrm{\infty }}{A}_{n}$ ?
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cefflid6y
If we set ${A}_{n}={\mathrm{cos}}^{2}\left({\theta }_{n}\right)$, we get:
${A}_{n+1}=\frac{1+\mathrm{cos}\left({\theta }_{n}\right)}{2}={\mathrm{cos}}^{2}\left(\frac{{\theta }_{n}}{2}\right),$
and since ${\theta }_{0}=\frac{\pi }{6}$, induction gives:
${A}_{n}={\mathrm{cos}}^{2}\left(\frac{\pi }{6\cdot {2}^{n}}\right)={\left(\frac{\mathrm{sin}\left(\frac{\pi }{6\cdot {2}^{n-1}}\right)}{2\mathrm{sin}\left(\frac{\pi }{6\cdot {2}^{n}}\right)}\right)}^{2}$
from which it follows that:
$\prod _{n=1}^{+\mathrm{\infty }}{A}_{n}={\left(\frac{3}{\pi }\right)}^{2}=\frac{9}{{\pi }^{2}}.$