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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.

rtybase
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  • 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. –  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. –  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. –  Feb 10 '18 at 14:34
  • @CalculusProgrammer The first answer there provides the info I think you're looking for, consider looking at the linked question(answer). – kingW3 Feb 10 '18 at 14:36
  • Ok I went through it . –  Feb 10 '18 at 14:38
  • @CalculusProgrammer Does it answer what you're looking for? – kingW3 Feb 10 '18 at 14:43
  • No since his solution is incomplete and further discussion about this problem is written in an article.When I clicked on the link I selected to read the article for free and was asked to register an account to read it.So actually my question is still unanswered. –  Feb 10 '18 at 14:47
  • Yes I got it now.Click this link to read the supportive article on this proof :-https://gauravtiwari.org/nested/ –  Feb 10 '18 at 14:58

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