Ramanujam's Problem

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

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

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

$\begin{aligned} 2+1+0 & = \sqrt{0+1^2+2\sqrt{0+1^2+(2+1)\sqrt{0+1^2+(2+2)\sqrt \cdots}}} \\ \implies \boxed{3} & = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt \cdots}}} \end{aligned}$

This one is rather a controversial solution:

Hope anyone can provide another solution.

Here, I'm simply expanding the numbers in a particular order. $3= \sqrt{9} = \sqrt{1+8} = \sqrt{1+(2 \times 4)} = \sqrt{1+2 \sqrt{16}}$ Now, 16 can be written as 15+1

$\Rightarrow 3 = \sqrt{1+2 \sqrt{1+15}} = \sqrt{1+2 \sqrt{1+(3 \times 5)}} = \sqrt{1+2 \sqrt{1+3 \sqrt{25}}}$

Again, we can write 25 as 1+24 and 24 can be split as $4 \times 6$ and then again 6 can be written as $\sqrt{36}$ and so on!