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Consider the uncountable-dimensional vector space $V$ consisting of finite linear combinations of infinite sequences of the form $(1,z,z^2,z^3,\dots)$ with $z \neq 1$ in $\mathbb{C}$. Since the sequences $(1,z,z^2,z^3,\dots)$ are linearly independent, there is a linear function $\sigma$ from $V$ to $\mathbb{C}$ that sends the sequence $(1,z,z^2,z^3,\dots)$ to the number $1/(1-z)$; for example, it sends $(2^0+(\frac12)^0,2^1+(\frac12)^1,2^2+(\frac12)^2,\dots)$ to $(1/(1-2))+(1/(1-\frac12)) = -1 + 2 = 1$; thus, in $V$, we could say that $2+\frac{5}{2}+\frac{17}{4}+\dots=1$. (This example sequence was designed to thwart certain approaches to regularizing divergent series, by way of combining two series that converge with respect to different Ostrowski valuations.)

$V$ would be a natural setting in which to do certain kinds of Euler-ish calculations with divergent series, but unfortunately $V$ lacks some desirable closure properties; for instance, it doesn't contain convolutions like $(1,w+z,w^2+wz+z^2,w^3+w^2z+wz^2+z^3,\dots$) which arise when you multiply two geometric series.

Is there an extension of $V$ to a larger space $V'$ that is closed under both addition and convolution, and an extension of $\sigma$ to a linear map from $V'$ to $\mathbb{C}$? (Is it described in the existing literature on divergent series?)

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  • $\begingroup$ Any conditions you want to put on this extension? Linear maps can always be extended from one vector space to another, so you can just take $V'=\mathbb C^{\mathbb N}$. $\endgroup$
    – Wojowu
    Commented Mar 27, 2022 at 19:09
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    $\begingroup$ I think that the $V'$ will be the sequence of all sequences whose generating functions are just rational functions, and $\sigma$ will just evaluate these rational functions at the point $1$. Alternatively, $V'$ will be the collection of all sums $\mathbf{x}+\mathbf{y}$ where $\mathbf{x}$ satisfies some linear recurrence relation, and $\mathbf{y}$ is zero for all but finitely many terms. Rational generating functions are talked about in Richard Stanley's Enumerative Combinatorics text. $\endgroup$
    – Joseph Van Name
    Commented Mar 27, 2022 at 19:48
  • $\begingroup$ @Wojowu : Hadn't thought of that; thanks for pointing it out. I don't want answers that rely on the Axiom of Choice. $\endgroup$
    – James Propp
    Commented Mar 28, 2022 at 17:48
  • $\begingroup$ @Joseph Van Name : I like your answer (though it makes me wish I'd thought harder before posting my question; I've worked with this very same vector space before and forgot I'd done so). Do you want to post it as an official answer, or do you want me to do it? $\endgroup$
    – James Propp
    Commented Mar 28, 2022 at 17:51
  • $\begingroup$ You may post the official answer to your own question. $\endgroup$
    – Joseph Van Name
    Commented Mar 31, 2022 at 2:07

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Joseph Van Name gave a good answer in the Comments:

I think that the $V'$ will be the sequence of all sequences whose generating functions are just rational functions, and $\sigma$ will just evaluate these rational functions at the point 1. Alternatively, $V'$ will be the collection of all sums 𝐱+𝐲 where 𝐱 satisfies some linear recurrence relation, and 𝐲 is zero for all but finitely many terms. Rational generating functions are talked about in Richard Stanley's Enumerative Combinatorics text.

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    $\begingroup$ Certainly $V'$ should consist of sequences with rational generating functions which don't have poles at $1$ if we're going to evaluate them at $1$. $\endgroup$
    – Sam Hopkins
    Commented Apr 1, 2022 at 14:47

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