Live now! Full branding package giveaway worth $11,895. # Solving Ramanujan’s Puzzling Problem Consider a sequence of functions as follows:-$ 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) $. ### Solution 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 $$3= \sqrt{1+2 \sqrt{1+3 \sqrt{1+ 4 \sqrt{1+\cdots}}}}$$ Comparing these two expressions & assuming =$ X $, we can write the problem as:$ \displaystyle {\lim_{n \to \infty}} f_n (x) $=$ \sqrt {1+X} $=$ \sqrt {1+3} $=$ \sqrt {4} $=$ 2 \$

For further info please refer the comments below. There is also a supportive article on Ramanujan Nested Radicals on this blog.

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