Weierstrass had drawn attention to the fact that there exist functions which are continuous for every value of $x$ but do not possess a derivative for any value. We now consider the celebrated function given by Weierstrass to show this fact. It will be shown that if $f(x)= # Solving Ramanujan’s Puzzling Problem Consider a sequence of functions as follows:-$ 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

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