Mathematical Equality

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

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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.

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I think there are some nontrivial issues of a) convergence of fourier series here and b) interchanging summation and integration.

most troubling is the fact that the fourier series of this function (to see this it's probably easiest to write it as f(x) = x + {x-1/2}, where { } denotes the greatest integer function) will exhibit the Gibbs Phenomenon.