Consider a sequence of functions as follows:-
……and so on to
Evaluate this function as n tends to infinity.
Or logically:
Find
Solution
Ramanujan discovered
which gives the special cases
for 
Proof of Ramanujan’s nested radicals equation
Then keep replacing the the last part term by writing
in the form of
to give
Which is another form of our problem and is referred as Ramanujan’s Nested Radical. Comparing these two expressions & assuming
=
=
=
=
=
-
Thank you for your precious comments and suggestions.
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.
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!
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
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.
really nice.