Ramanujan would like it

Ramanujan discovered that:

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

When $x = 2$ , $n=1$ and $a=0$ , then we have:

$2+1+0 = \sqrt{1+ 2\sqrt{1 + 3\sqrt{1 + 4\sqrt{1+...}}}} = \boxed{3}$