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.
 A: Denote the $n$-th factor by $x_n$. The sequence $(x_n)$  is increasing, and
$$
 x_2 = \sqrt{\frac{1}{2}+\sqrt\frac{1}{2}} > \sqrt{\frac{1}{2}+\frac{1}{2}} = 1
$$
Therefore
$$
 x_1 x_2 \cdots x_n \ge x_1 x_2^{n-1} \to \infty \, .
$$
A: Begin by investigating the sequence
$$x_0:=0,\quad x_{k+1}:=\sqrt{{1\over2}+x_k}\quad(k\geq0)\ .$$
It has a certain limit $\xi$. Knowing $\xi$ draw conclusions about the limit
$$\lim_{n\to\infty}\>\prod_{k=1}^n x_k\ .$$ 
A: 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 $+ \infty$.
A: Hint: Let $a =\dfrac{1}{2}$ and $b$ be the expression you are trying to calculate, then $b^2 = a + b$. You can use quadratic equation to finish.
