[m_the0ry]

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can't you just factor out N so that this summation becomes a geometric series that converges to 1?

[/ QUOTE ]

This brings us back to the hardest part of the problem, which is mathematically defining the 'nearest integer' function. I think it would be hard to rigorously show that you can factor out the numerator for this function considering 1/N does not necessarily equal [N/1]^-1.

I had to use fourier series to define the nearest integer function as follows:

NearestInt = x + 1/pi * sigma(k=1 => infinity)[sin(2*pi*k*x + 2*pi*k)

From that the expansion of this problem becomes

N = sigma(i=1 | infinity){N/(2^i) + 1/pi * sigma(k=1 | infinity)[sin(pi*k*N/(2^(i-1)) + pi*k)}

The sin series converges to zero both with respect to i and k, because the harmonics cancel themselves out as well as each other, so you're left with the real number series

N = sigma(i=1 | infinity){N/2^(i)}

which clearly converges to N.