Ok, here's my attempt at an elementary explanation. Consider the function f(N) = floor(N) = 1 + 1 + 1 + ... + 1, whenever N's an integer = 1^0

Ok, here's my attempt at an elementary explanation. Consider the function 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.