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:


$ \displaystyle{\lim_{n \to \infty}} f_n (x) $ .


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=$ 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.

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Gaurav Tiwari
Gaurav Tiwari is a professional graphic & web designer from New Delhi, India. is his personal space where he writes on blogging, digital marketing, content writing, learning and business growth. Gaurav has contributed in developing more than 325 brands worldwide and while you are reading this, he's busy building a couple more.

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

  3. 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.

  4. 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.


  5. wow – wouldnt have a clue where to start!

  6. 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.

  7. Ramanujan always the best

  8. I really love the version that starts 3= because it smells like a magic number but really implicates the architecture of the number system we use.

  9. really nice.

  10. Ramanujan isn’t human, he is a Beast

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

  12. thank you , this is one of my try out question in my school.
    quiet confused since i saw this crazy square root .lol

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