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

ramanujan - Solving Ramanujan's Puzzling Problem=$ 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.

Updated: December 21st, 2017

Gaurav Tiwari

A designer by profession, a mathematician by education but a Blogger by hobby. Loves reading and writing. Just that.

This Post Has 15 Comments

  1. I’m not entirely sure, but using C++ (with n = 1,000,000) I numerically evaluated it to the function f(x) = 2. But as I said, not quite sure!

  2. There are two slightly different versions of this nested radical, so you need to be careful.

    The version posed by Ramanujan was
    sqrt(1 + 2*sqrt(1 + 3*sqrt(1 + 4*sqrt(1 + … = 3

    Your version is almost the same:
    sqrt(1 + 1*sqrt(1 + 2*sqrt(1 + 3*sqrt(1 + … = sqrt(1 + 3) = 2.

  3. Hi there,

    Just a small typo — I think you meant to write the limit as n tends to infinity. On all of the limits you wrote in that article, you unfortunately said that x goes to infinity.

    x_x

  4. I think you might be one of the best bloggers in India today. We are having a TEDx conference, and it would be great to have you as a Speaker. I am sure you can come up with a very interesting talk. Let me know however I can contact you.

Leave a Reply

Your email address will not be published. Required fields are marked *