Infinite Ordinal Sum When working with ordinal numbers, would it be correct to say that:
$$ \sum_{i=0}^{\infty}1 = \omega$$
Or does this simply not make sense? In the ordinals, does the notation $\sum^\infty_{i=0}$ even make sense or would $\sum_{i=0}^\alpha$, with $\alpha$ being a (potential infinite) ordinal, be the only correct notation?
Thank you very much.
 A: Ordinal summation requires an ordinal index. And $\infty$ is not an ordinal. 
Other than that, the summation does make sense in general. If $I$ is a linearly ordered set, and $x_i$ is a linearly ordered set for each $i\in I$, then $\sum_{i\in I}x_i$ would be the order type obtained by replacing $i$ with $x_i$, and considering the "[somewhat-]lexicographic order" obtained.
If $I$ is an ordinal and each $x_i$ is an ordinal, it turns out that the sum is an ordinal as well. Which is why everything works out.
As far as notation goes, I'd probably go for $\sum_{i<\alpha}$ and not $\sum_{i=1}^\alpha$. Which will also allow you to catch those pesky limit cases.
A: It makes sense (and is true) under the following conventions:


*

*If $(x_n)_n$ is an increasing sequence of ordinals, then $\lim\limits_{n\to\infty} x_n=\bigcup_n x_n$.

*$\sum_{n=1}^\infty x_n=\lim\limits_{n\to\infty} \sum_{i=1}^n x_n$


As with reals, this notation is rather dangerous though.
EDIT: In general, one can define the limit of a sequence $(x_n)_n$ as follows. Work within an ordinal bigger or equal to $\bigcup_nx_n+1$, and give it the order topology. Then $\lim\limits_{n\to\infty} x_n$ will be the topological limit of the sequence. 
