Evaluate $A_0=\dfrac{3}{4}$, and $A_{n+1}=\dfrac{1+\sqrt{A_n}}{2}$

Question

Let $A_0=\dfrac{3}{4}$, and $A_{n+1}=\dfrac{1+\sqrt{A_n}}{2}$ for all $n\geq0$.

How to find the value of $\displaystyle\prod_{n=1}^\infty A_n$ ?

I don't have any idea.

Thank you.

Answer 1

If we set $A_n = \cos^2(\theta_n)$, we get: $$ A_{n+1} = \frac{1+\cos(\theta_n)}{2} = \cos^2\left(\frac{\theta_n}{2}\right),$$ and since $\theta_0=\frac{\pi}{6}$, induction gives: $$ A_n = \cos^2\left(\frac{\pi}{6\cdot 2^n}\right)=\left(\frac{\sin\left(\frac{\pi}{6\cdot 2^{n-1}}\right)}{2\sin\left(\frac{\pi}{6\cdot 2^n}\right)}\right)^2 $$ from which it follows that: $$ \prod_{n=1}^{+\infty}A_n = \left(\frac{3}{\pi}\right)^2 = \color{red}{\frac{9}{\pi^2}}.$$