# Solution to "Ramanujan's Puzzle"

Looking at this puzzle, it was not obvious to me that this expression converged to anything meaningful, so I tried an approximate evaluation to see if anything turned up. To do this, I had to break the indefinite sequence of square roots, replace the unwanted portion by a number and evaluate for decreasing n.
I decided that there would be N square roots, each called Tn and that the N+1 th square root would be replaced by an unknown number Q.
This gave the following recurrence relation:-
TN = Q
Tn = sq(1 + n * Tn+1) 1<= n < N
T1 will be an approximation to the given expression
I evaluated this with a short program and was surprised to find that it converged rapidly to a simple value which is independent of N and Q.
When I printed out values of the Tns it became evident that if Q = N + 1 , the whole thing collapses and
Tn = n+1 for all 0 < n < N
But why should Tn converge to n + 1 ?
Suppose that Q is > 0 but < N + 1 i.e. Q = N + 1 - d
Then

 TN-1 = sq(1 + (N - 1)*(N + 1 - d)) = sq( N2 - d*(N-1)) = N * sq( 1 - d/N + 1/N2)

Write x = d/N - 1/N2. Since we chose Q < N + 1 then 0 < x < 1
Expanding the square root by the binomial theorem gives
TN-1 = N * (1 - x/2 + (terms in higher powers of x ))
Thus at each stage the difference between Tn-1 and n almost halves so that convergence is quite rapid.
Now of course, the binomial expansion converges for values of x greater than one.
However for any given Q it is possible to choose N sufficiently greater than Q to ensure that T1 converges to 2 to any required degree of accuracy.

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