f(N) = floor(N)
= 1 + 1 + 1 + ... + 1, whenever N's an integer
= 1^0 + 2^0 + 3^0 + ... + N^0.
Then f(N) ~ N - 1/2 is f's simplest unbiased polynomial estimator. It's a succinct description of f and its asymptotic behavior.Next, let's do something like integration, and increase those zeroth powers to one, ie. consider the function
g(N) = 1^1 + 2^1 + 3^1 + ... + (N - 1)^1
= 1 + 2 + 3 + ... + (N - 1).
In this case, the estimator ends up being g(N) ~ N^2/2 - N/2 - 1/12
~ integral(N - 1/2) - 1/12,
where I've written this estimate as the difference of a divergent term and a constant term.Here's the point: the divergent term is really just left over (integrated up or 'induced') from our asymptotic description of the degree 0 sum. In a sense, it's degenerate and doesn't provide any sort of 'new' information. So in the same sense, the interesting semantic content of 1 + 2 + 3 + ... is just the constant term: -1/12.
Ps. This way of thinking is related to trace formula truncation.
We wish to use the same familiar notation with the infinite that we do with the finite, but we must keep in mind that while they do share similarities, they are not the same operation. The way certain ambiguities are resolved in order to extend the finite to the infinite can lead to counterintuitive results and absurdities that may not even be apparent at first.
For me, seeing how one approach to generalizing infinite sums yields the -1/12 result, and how that result actually has some relevance in physics is quite profound and insightful and I am happy that you reminded me of it.
Although it relies on a particular interpretation of the left side of the equation, it's used in some areas of physics.