How did Ramanujan get this result?

We know Ramanujan got this result $$\sqrt{1+2\sqrt{1+3\sqrt{1+\cdots }}}=3$$ and he used the formula $$x+n+a=\sqrt{ax+{{(n+a)}^{2}}+x\sqrt{a(x+n)+{{(n+a)}^{2}}+(x+n)\sqrt{\cdots }}}$$ where $x=2,n=1,a=0$ ,we get the first result, but I don't know how to prove it, can you help me?

$$(x+n+a)^2 = x^2 + n^2 + a^2 + 2an + 2ax + 2nx$$ $$= ax + (n+a)^2 + x(x + a + 2n)$$

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

which you can substitute for $(x+n) + n + a$ again and again to get the sequence of iterated roots.

The convergence is because the sequence is monotone increasing but bounded above by $x+n+a$ for $n > 0$, $a,x \ge 0$ (after enough ($k$) iterations, $x + k*n + a$ will be greater than 1.)

Just use the following formula repeatedly

$$n^2=1+(n-1) (n+1)$$

then we have

$$2=\sqrt{1+1\times 3}$$

$$2=\sqrt{1 +1 \sqrt{1+2\times 4}}$$

$$2=\sqrt{1 +1 \sqrt{1+2 \sqrt{1+3\times 5}}}$$

$$2=\sqrt{1 +1 \sqrt{1+ 2 \sqrt{1+ 3 \sqrt{1+4\times 6}}}}$$

$$2=\sqrt{1+1 \sqrt{1+2 \sqrt{1+ 3 \sqrt{1+4 \sqrt{1+5\times 7}}}}}$$

$$2=\sqrt{1+1 \sqrt{1+2 \sqrt{1+3 \sqrt{1+4 \sqrt{1+5 \sqrt{1+6\times 8}}}}}}$$

$$\cdots$$

By using

$$n+(n-1) n(n+1) =n^3$$

I found the below new results

$$\sqrt[3]{2+1\times2 \sqrt[3]{3+2\times 3 \sqrt[3]{4+3\times 4 \sqrt[3]{5+4\times 5 \sqrt[3]{6+5\times 6 \sqrt[3]{7+...\ }}}}}}=2$$

$$\sqrt[3]{3+2\times 3 \sqrt[3]{4+3\times 4 \sqrt[3]{5+4\times 5 \sqrt[3]{6+5\times 6 \sqrt[3]{7+6\times 7 \sqrt[3]{8+...\ }}}}}}=3$$

$$\sqrt[3]{4+3\times 4 \sqrt[3]{5+4\times 5 \sqrt[3]{6+5\times 6 \sqrt[3]{7+6\times 7 \sqrt[3]{8+7\times 8 \sqrt[3]{9+...\ }}}}}}=4$$

$$\cdots$$

More result can be found below:

http://eslpower.org/Notebook.htm