Find the value of the infinite product $\sqrt\frac12\cdot\sqrt{\frac12+\sqrt\frac12}\cdot\sqrt{\frac12+\sqrt{\frac12+\sqrt\frac12}}\cdots$ Find $\sqrt{\frac{1}{2}}.\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}}.\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}}}....\infty$

Let $x=\sqrt{\frac{1}{2}}.\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}}.\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}}}....\infty$
$\log x=\frac{1}{2}\log(\frac{1}{2})+\frac{1}{2}\log(\frac{1}{2}+\sqrt\frac{1}{2})+\frac{1}{2}\log(\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}})+....$
I do not know how to solve it further.

Let $$x = \prod_{n = 1}^{\infty} y_n \qquad \text{and} \qquad y_n = \sqrt {\frac {1} {2} + \sqrt {\frac {1} {2} + \cdots + \sqrt {\frac {1} {2}}}},$$ and $$y = \lim_{n \to \infty} y_n.$$ We have $$y_{n + 1} = \sqrt {\frac {1} {2} + y_n},$$ which, passing to limit, gives $$y = \sqrt {\frac {1} {2} + y}.$$ We have $y = \frac {\sqrt {3}} {2} + \frac {1} {2}$, which is $> 1$, and that's why $x$ diverges to $+ \infty$.
