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This question already has an answer here:

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.

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marked as duplicate by kingW3, rtybase, Macavity, Math Lover, Hans Lundmark Feb 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.

  • $\begingroup$ Formally you have $f_{n+1}(x) = f_n(1+(n+1)\sqrt{x})$. Anyway I don't see any obvious solution. $\endgroup$ – Nathanael Skrepek Feb 10 '18 at 14:17
  • $\begingroup$ I find this problem highly intriguing and a solution to problem is essential since we can’t rely on theories which have no proofs. $\endgroup$ – user517784 Feb 10 '18 at 14:17
  • $\begingroup$ @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. $\endgroup$ – user517784 Feb 10 '18 at 14:32
  • $\begingroup$ Ramanujan was literally a ducking genius! 🐥 $\endgroup$ – Jaideep Khare Feb 10 '18 at 14:32
  • $\begingroup$ Yes that is without a doubt.But some things need to be challenged to get a complete understanding of their theories. $\endgroup$ – user517784 Feb 10 '18 at 14:34