# 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.**

You can either start a new conversation or continue an existing one.Please don't use this comment form just to build backlinks. If your comment is not good enough and if in some ways you are trying to just build links — your comment will be deleted. Use this form to build a better and cleaner commenting ecosystem. Students are welcome to ask for help, freebies and more. Your email will not be published or used for any purposes.I’m not entirely sure, but using C++ (with n = 1,000,000) I numerically evaluated it to the function f(x) = 2.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!

You’re right..! After using google, I got this Link , which was also saying the same. But I wasn’t satisfied.

Answer is 3 according to formulation 27 at http://mathworld.wolfram.com/NestedRadical.html .

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.

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.

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

x_x Corrected Now. Thanks.

wow – wouldnt have a clue where to start!

wow – wouldnt have a clue where to start!

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.

Ramanujan always the best

second equation is wrong x+1=sq(1+xsq(1+(x+1))sq(1+(x+2)sq(1+(x+3))))

thank you , this is one of my try out question in my school.

quiet confused since i saw this crazy square root .lol

thank you , this is one of my try out question in my school.

quiet confused since i saw this crazy square root .lol