calculus - Could $4+2+4+2+4+2+cdots = -1$?

In physics classes, on this StackExchange and even in blogs the sum $1 + 2 + 3 + 4 + \cdots = - \frac{1}{12}$ has been under the microscope.

As a consequence of the trapezoid rule, the error to the Riemann sum is bounded by the second derivative.

$$\int_0^N f(x) \; dx = \frac{1}{2}f(0) + f(1) + \dots + f(N-1) + \frac{1}{2}f(N) + O\,(N \| f \|_{\dot{C}^2} )$$

Letting $f(x) = (1-x/N)_+$ we get $\sum_{i=0}^{N-1} 1 = -\frac{1}{2} + \int_0^N 1 \, dx + O(1)$.

I am wondering what happen if we use the Simpson rule:

$$\int_0^N f(x) \; dx = \frac{1}{3} \big( f(0) + 4f(1) + 2f(2) + \dots + 4f(N-1) + f(N) \big) + O\,(N || f ||_{\dot{C}^4} )$$

Now we plug in $f(x) = (1-x)_+$ and get

$$\frac{1}{3} (4 + 2 + 4 + 2 + \cdots ) = -\frac{1}{3} + \int_0^N f(x) \; dx + O(N^{-3})$$

Is that still consistent with the other types of sums you get from Euler-Macularin type summation methods?