# Ramanujan’s puzzling problem [duplicate]

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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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 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
• I find this problem highly intriguing and a solution to problem is essential since we can’t rely on theories which have no proofs. – user517784 Feb 10 '18 at 14:17
• @kingW3 for your kind information I am not asking for a solution here.i am questioning the foundation of this discovery.Before declaring possible duplicates please read the understand what I am asking from the users of this community. – user517784 Feb 10 '18 at 14:32
• Ramanujan was literally a ducking genius! 🐥 – Jaideep Khare Feb 10 '18 at 14:32
• Yes that is without a doubt.But some things need to be challenged to get a complete understanding of their theories. – user517784 Feb 10 '18 at 14:34