The Root of the Problem: The Sequel

$\begin{aligned}Let,f(n)=n(n+2)\\ f(n)&=n\sqrt{(n+2)^2}\\ &=n\sqrt{1+\left((n+2)^2-1\right)}\\ &=n\sqrt{1+(n+1)\cdot(n+3)}\\ &=n\sqrt{1+f(n+1)}\\ &=n\sqrt{1+(n+1)\sqrt{1+f(n+2)}}\\ f(n)=n(n+2)&=n\sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+(n+3)\sqrt{1+(n+4)\sqrt{1+\cdots}}}}}\\ \text{putting } n=1 \text{ we get }x,\\ x=f(1)=1(1+2)&=1\sqrt{1+(1+1)\sqrt{1+(1+2)\sqrt{1+(1+3)\sqrt{1+(1+4)\sqrt{1+\cdots}}}}}\\ 3&=1\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\cdots}}}}}\end{aligned}$

Ramanujan's formula (eqn. 26) on nested radical:

$u+n+a = \sqrt{au+(n+a)^2 + u\sqrt{a(u+n)+(n+a)^2 + (u+n)\sqrt{a(u+2n)+(n+a)^2 + (u+2n)\sqrt \cdots}}}$

Putting $u=2$ , $n =1$ and $a=0$ , we have:

$\begin{aligned} 2+1+0 & = \sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + 4 \sqrt{ 1+5\sqrt{1+ \cdots}}}}} \\ \implies x & = \boxed{3} \end{aligned}$