@DanielFischer I wanna ask, according to Ramanujan, if
$$f(x)=\sqrt{ab+(a+c)^2+b\sqrt{ab+(a+c)^2+(b+c)\sqrt{ab+(a+c)^2+(b+2c)\sqrt{ab+\cdots}}}}$$
then
$$f(x)=a+b+c$$
How does one prove this one?
Also this one, is it possible the following problem has a unique solution
$$x=\sqrt{i+\sqrt{i^2+\sqrt{i^3+\sqrt{i^4+\sqrt{i^5+\sqrt{i^6+\cdots}}}}}}$$
where $i=\sqrt{-1}$.