Nested Radical of Ramanujan

Evaluate :

$\sqrt{1+2 \sqrt{1+3 \sqrt{1+\cdots\infty}}}$


Solution :

Note that $(x+1)^2=1+x(x+2)$ , now taking the principal square root of both sides and re-substituting the resulting expression for   $x+1$  multiple times gives us that:

$(x+1)=\sqrt{1+x(x+2)}$

$(x+1)=\sqrt{1+x \sqrt{1+(x+1)(x+3)}}$

$(x+1)=\sqrt{1+x \sqrt{1+(x+1) \sqrt{1+(x+2)(x+4)}}}$


$(x+1)=\sqrt{1+x \sqrt{1+(x+1) \sqrt{1+(x+2) \sqrt{1+(x+3) \sqrt{1+(x+4) \sqrt\cdots}}}}}$


Putting $x=2$ in the above, we get

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