5
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This question already has an answer here:

$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}$

Apparently, the answer is 3.

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marked as duplicate by r9m, graydad, Najib Idrissi, user147263, Lord_Farin Jan 27 '15 at 18:13

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.

4
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Hint: Observe $$(x+1)^2 - 1 = x(x+2),$$ or $$x+1 = \sqrt{1 + x(x+2)}.$$ Now recursively substitute: $$\begin{align*} x+1 &= \sqrt{1 + x\sqrt{1 + (x+1)(x+3)}} , \\ &= \sqrt{1 + x \sqrt{1 + (x+1)\sqrt{1 + (x+2)(x+4)}}}, \\ &= \sqrt{1 + x \sqrt{1 + (x+1)\sqrt{1 + (x+2)\sqrt{1 + (x+3)(x+5)}}}}, \ldots \end{align*}$$ Of course, you need to do something much more rigorous to complete the proof. I leave that to you.

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