Ramanujan’s puzzling problem [duplicate]

Question:

$$f_1 (x) = \sqrt {1+\sqrt {x} }$$ $$f_2 (x) = \sqrt{1+ \sqrt {1+2 \sqrt {x} } }$$ $$f_3 (x) = \sqrt {1+ \sqrt {1+2 \sqrt {1+3 \sqrt {x} } } }$$ ... and so on to $$f_n (x) = \sqrt {1+\sqrt{1+2 \sqrt {1+3 \sqrt {\ldots \sqrt {1+n \sqrt {x} } } } } }$$ Evaluate this function as n tends to infinity.

Or logically: Find $\displaystyle{\lim_{n \to \infty}} f_n (x)$ .

MyProblem: Ramanujan discovered $$x+n+a=\sqrt{ax + (n+a)^2 +x \sqrt{a(x+n)+(n+a)^2 +(x+n) \sqrt{\ldots}}}$$ which gives the special cases $$x+1=\sqrt{1+x \sqrt{1 + (x+1) \sqrt{1 + (x+2) \sqrt{1 + (x+2) \sqrt{\ldots}}}}}$$ for x=2 , n=1 and a=0 $$3= \sqrt{1+2 \sqrt{1+3 \sqrt{1+ 4 \sqrt{1+\cdots}}}}$$

Is there any proof of this discovery?Without proof I don’t think that the problem should be solved in this way.And if there is a proof please share it and also tell your approach to solve it.

marked as duplicate by kingW3, rtybase, Macavity, Math Lover, Hans LundmarkFeb 10 '18 at 18:20

• Formally you have $f_{n+1}(x) = f_n(1+(n+1)\sqrt{x})$. Anyway I don't see any obvious solution. – Nathanael Skrepek Feb 10 '18 at 14:17