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 T_{n} and that
the N+1 th square root would be replaced by an unknown number Q.

This gave the following recurrence relation:-

T_{N} = Q

T_{n} = sq(1 + n * T_{n+1}) 1<= n < N

T_{1} 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 T_{n}s it became evident that if Q = N + 1 ,
the whole thing collapses and

T_{n} = n+1 for all 0 < n < N

But why should T_{n} converge to n + 1 ?

Suppose that Q is > 0 but < N + 1 i.e. Q = N + 1 - d

Then

T_{N-1} | = | sq(1 + (N - 1)*(N + 1 - d)) |

= | sq( N^{2} - d*(N-1)) | |

= | N * sq( 1 - d/N + 1/N^{2}) |

Write x = d/N - 1/N^{2}. Since we chose Q < N + 1 then 0 < x < 1

Expanding the square root by the binomial theorem gives

T_{N-1} = N * (1 - x/2 + (terms in higher powers of x ))

Thus at each stage the difference between T_{n-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 T_{1} converges to 2
to any required degree of accuracy.

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