Finding $\omega$ with an arithmetic series

Question

I just watched a really interesting video by Vsauce about counting past infinity here, and from which I have now learned that the next number after infinity, is $\omega$, then $\omega+1$, and so on.

From another video I have seen, showing how every positive number adds to -1/12, allow me to pose a question:

If every (positive) number added together equals -1/12, then we can determine the equation is:

$\sum_{i=0}^{\infty}i=-\frac{1}{12}$

From the first video, we know that $\omega$ is equal to $\infty+1$, so by that logic, would the following equation be correct?

$\sum_{i=0}^{\omega}i=\frac{11}{12}$

Answer 1

You are making a huge mess.

1. $\omega$ is not a number according to the common conception, it is an ordinal. The arithmetics of ordinal numbers is pretty different from the arithmetics over $\mathbb{Z}$: for instance, $1+\omega\neq \omega+1$. Additionally, $\mathbb{Z}$ is a set while $\mathbb{ON}$ is a proper class by the Burali-Forti paradox;
2. The series $\sum_{n\geq 1}n$ is simply divergent. Of course, for any $s>1$ we have $$\zeta(s)=\sum_{n\geq 1}\frac{1}{n^s}$$ and the Riemann $\zeta$ function has an analytic continuation to the complex plane, allowing to state $\zeta(-1)=-\frac{1}{12}$, but this is pretty different from stating that $\sum_{n\geq 1}n = -\frac{1}{12}$;
3. By 1. and 2., $\sum_{n=1}^{\omega}n$ is doubly meaningless.