Infinite? INFINITE!

Ramanujan discovered in 1911 that:

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

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

$2+1+0 = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\cdots}}}}}$

Therefore, the answer is $\boxed 3$ .

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Proof: Let $f(x) = 1+x$ . Then $f(x) = \sqrt{(1+x)^2} = \sqrt{1+2x + x^2} = \sqrt{1+x(2+x)} = \sqrt{1+xf(x+1)}$ . Similarly, $f(x+1) = \sqrt{1+(x+1)f(x+2)}$ , $f(x+2) = \sqrt{1+(x+2)f(x+3)}$ and so on. Then we have:

$\begin{aligned} x+1 & = \sqrt{1 + x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\sqrt{1+\cdots}}}}} & \small \color{#3D99F6} \text{For }x = 2 \\ \implies 3 & = \sqrt{1 + 2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\cdots}}}}} \end{aligned}$