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.