Fan of Ramanujan? Fan of Factorials? Desserts?

Using the Ramanujan's formula for nested radical (26) :

$x+n+a = \sqrt{ax + (n+a)^2 + x\sqrt{a(x+n) + (n+a)^2 + (x+n)\sqrt{a(x+2n) + (n+a)^2 + (x+2n)\sqrt{\cdots}}}}$

Setting $x=2(0.5!)$ , $n=0$ and $a=0$ , we have:

$\begin{aligned} 2(0.5!) & = \sqrt{2(0.5!)\sqrt{2(0.5!)\sqrt{2(0.5!)\sqrt{\cdots}}}} \\ (2(0.5!))^2 & = 2(0.5!) \sqrt{2(0.5!)\sqrt{2(0.5!)\sqrt{2(0.5!)\sqrt{\cdots}}}} \end{aligned}$

Note that $0.5!) = \Gamma (0.5+1) = \Gamma \left(\frac 32 \right) = \frac 12 \Gamma \left(\frac 12 \right) = \frac 12 \sqrt \pi$ , where $\Gamma (\cdot)$ denotes that gamma function . Therefore $(2(0.5!))^2 = \pi \approx \boxed{3.14}$ in three significant figures.

Because the pattern goes on forever, we can substitute it into itself:

$x = (2)(0.5!)\sqrt{x}$

$\sqrt{x} = (2)(0.5!)$

$x = 4(0.5!)^2$

$x = \pi$

$\frac{1}{2}! = \Gamma(\frac{3}{2}) = \frac{1}{2}\Gamma(\frac{1}{2})$

Using Euler's Reflection formula,

$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)} \quad \text{ for all non integer values of } z \text{ and taking } z = \frac{1}{2} \implies \Gamma(\frac{1}{2}) = \sqrt{\pi} \text{ then,}$

$\frac{1}{2}\Gamma(\frac{1}{2}) = \frac{1}{2}\sqrt{\pi} \implies 2\cdot (\frac{1}{2})! = \sqrt{\pi}$

Let $x = \sqrt{\pi}\sqrt{\sqrt{\pi}\sqrt{\cdots}} \text{ then } \, x^2 = \pi x \, \therefore \, x(x-\pi) = 0 \implies x = \pi \approx \boxed{3.14}$