# 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.