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This question already has an answer here:

show that $$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\cdots}}}}}=3$$

I know $$3=\sqrt{1+8}=\sqrt{1+2\sqrt{16}}=\sqrt{1+2\sqrt{1+15}}=\sqrt{1+2\sqrt{1+3\sqrt{1+4\cdot6}}}=\cdots$$

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marked as duplicate by user21820, Omnomnomnom, Sharkos, Hans Lundmark, Tom Oldfield Jun 5 '14 at 15:32

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.

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For your question answer is given in this http://www.isibang.ac.in/~sury/ramanujanday.pdf

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  • $\begingroup$ Yes there is answer to this problem, for general f(x). $\endgroup$ – user152394 Jun 5 '14 at 14:44

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