The short answer is that if you take the partial sums of the first n terms, you get the sequence 1, 1, 2, 3, 4, … which settles down to have its nth element being n−1 and thus is a representative sequence for the number ω−1. I think you’ll be able to follow the maths in the paper quite well, especially if trying a few examples of things you’d like to sum or integrate for yourself on paper.
(There is some tricky stuff to do with ultrafilters, but that mainly comes up as a way of settling the matter for which of two sequences represents the higher number when they keep trading the lead infinitely many times.)
Thanks! I’m glad this has 1 + 0 + 1 + 1 = ω − 1, but I’m going to need to go read more to understand why ;)
The short answer is that if you take the partial sums of the first n terms, you get the sequence 1, 1, 2, 3, 4, … which settles down to have its nth element being n−1 and thus is a representative sequence for the number ω−1. I think you’ll be able to follow the maths in the paper quite well, especially if trying a few examples of things you’d like to sum or integrate for yourself on paper.
(There is some tricky stuff to do with ultrafilters, but that mainly comes up as a way of settling the matter for which of two sequences represents the higher number when they keep trading the lead infinitely many times.)