I know that various methods exist to assign values for non-converging series e.g. Cesàro summation, Hölder summation, Lambert summation, etc.
I was curious if there exists an axiomatic framework, where one looks at the set of partial operators on infinite sequences and choses a subset $\mathcal{S}$ of those satisfying certain axioms, then for each sequence $a$ one could look at the set $\{S(a): S\in\mathcal{S}\}$, which would even be an intervall if the axioms are deliberately chosen so that convexity of the resulting set is ensured.
One such axiom set that comes to mind would be
- Equality to $\Sigma$ (the 'true' sum operator) for absolute converging series (or if this is too strong, at least for finite and non-negative, converging sequences)
- Linearity (in the sense that if $S(a)$ and $S(b)$ are both defined, then $S(\lambda a+\mu b)$ must be defined and equal $\lambda S(a)+\mu S(b)$)
- Shifting the series to the right and filling in zeros from the left should not change the result
- Stability under bounded permutations (i.e. $\pi\in S_\omega$ is bounded if $\forall i>0\in\mathbf{N}:|i-\pi(i)|<k$ for some $k>0$)
Also maybe some kind of continuity, although I don't know which metric would be apt to use on the space of sequences?
[EDIT] Now that I think about it, maybe this might just result in the interval $[\liminf_{n\to\infty} \sum_{i<n} a_i,\limsup_{n\to\infty} \sum_{i<n}a_i]$, which would be boring of course.
Best Answer
EDIT:
My original answer actually defined a trivial operator -- the fixed formalisation is credit to Kenny Lau on Zulip (see the link for discussion regarding non-triviality).
Feel free to play with it yourself. And check out the challenge (proving that there exists a sequence that does not have a sum (see the proof in math here). Actually providing an example (e.g. $1/n$) may be quite hard (the proof in the chat uses generating functions, which should be hard in Lean), but proving that a sequence with linearly independent shifts has no sum is almost done -- you just need to prove that the forced sum is a sum operator.
(Draft)
OLD ANSWER:
Here's something to get you started -- I wrote it in Lean, a formal proof-checker, because these things are tricky and I wanted to be completely sure I was being rigorous. I suppose we also need
sum_con
for convergent sums, but I'm not sure where infinite series are in the Lean math library!