Problem
I first came across this statement $\sum_{n=1}^\infty n = -\frac{1}{12}$ a couple of years ago.
Why does an integral test for convergance not disprove this. That is, with an integral of $$\int_0^\infty x dx$$
I see the integral test requires a monotonically decreasing function and this will be why you can't use it for the infinte sum of positive integers. I fail to understand why this function requires a decreasing function. Perhaps this is beyond the scope of what I'm trying to understand.
Context
I saw a video challenging the viewer to solve the Balancing bricks problem and I proceeded to solve it to the point of the Sum of the harmonic series which I looked up information.
I saw the integral test as a way to show this series also had an infinite sum as demonstrated in the Harmonic Series Wikipedia Article. This immediately reminded me about that ol' friend $\sum_{n=1}^\infty n = -\frac{1}{12}$ and I was lead to wonder why that integral test couldn't be applied. Cue much internet searching and now this question.
Let me take an easier example.
By definition of a series, with $A_N= \sum_{n=0}^N 2^n = 2^{N+1}-1$ : $$\sum_{n=0}^\infty 2^n = \lim_{N \to \infty} A_N = \lim_{N \to \infty}2^{N+1}-1 = +\infty$$
Now you can regularize this divergent series : $$\text{only for } |z| < 1, \qquad \sum_{n=0}^\infty z^n = \frac{1}{1-z}, \qquad \overset{\color{red}{(\mathcal{P})}}{\sum_{n=0}^\infty} 2^n =\left.\frac{1}{1-z}\right|_{z = 2} = -1$$ where $\displaystyle\overset{\color{red}{(\mathcal{P})}}{\sum} $ means we are using the power series regularization, completely changing the meaning of the symbol $\sum$
Here what we are talking about is $$\overset{\color{red}{(\mathcal{Z})}}{\sum_{n=1}^\infty} n =\frac{-1}{12}$$ where $\displaystyle \overset{\color{red}{(\mathcal{Z})}}{\sum}$ is the zeta regularization.