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.

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

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.

4. 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.

5. 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.

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

7. 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.

8. Ramanujan always the best

9. 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.

10. really nice.