TITLE: Evaluate $A_0=\dfrac{3}{4}$, and $A_{n+1}=\dfrac{1+\sqrt{A_n}}{2}$
QUESTION [8 upvotes]: 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.

REPLY [21 votes]: 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:
$$ \prod_{n=1}^{+\infty}A_n = \left(\frac{3}{\pi}\right)^2 = \color{red}{\frac{9}{\pi^2}}.$$