RAMANUJAN'S INFINITE SEQUENCE OF SQUARE ROOT (NESTED RADICAL).

         INFINITE SEQUENCE OF SQUARE ROOT       


INFINITE SEQUENCE:

                An infinite sequence is a list or string of discrete objects, usually numbers, that can be paired off one-to-one with the set of positive integer s {1, 2, 3, ...}. Examples of infinite sequences are N = (0, 1, 2, 3, ...) and S = (1, 1/2, 1/4, 1/8, ..., 1/2 n , ...). 

INFINITE SEQUENCE OF SQUARE ROOT:

Consider, 
               y = x + n 
              (x +n)² = n² + x(x+2n)
              (x + n) = √[n² +x(x+2n)]
              (x + 2n) = √[n² +x(x+2n)(x+3n)]
              (x + 3n) = √[n² +x(x+2n)(x+3n)(x+4n)]
              ...
              ...
              ...
              ...
              ...
              x+n = √[n² +x√n²+(n+x)√n² +(n+2x)....]

 Remarkably the answer is exactly 3. 

 

We have verified the pattern holds up to 4, with the next term under the radical being 6 = √62. Let’s justify why the pattern will continue. Assume the pattern holds where n – 2 is the coefficient in front of the last radical, which is equal to n = √n2. We will then show the pattern continues. We can expand the last radical so that n – 1 is the next coefficient in front of another radical equal to to n + 1 = √(n + 1)2.

Now we can re-apply the pattern to the term (n + 1) and generate the infinite nested radical.
Now you might wonder how anyone could think of such an idea. Seeing numerical patterns was the genius of Ramanujan, and even to this day we are not sure how he came up with some of his formulas.
The pattern can also be derived by squaring a binomial and using substitution, as explained on mathforum.
 The formal proof involves demonstrating convergence which is quite a bit more complicated.








              


           









 

 

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