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Let $(a_n) \subset \mathbb{R}$ be a sequence such that the series $\sum_{n=1}^\infty a_n$ converges, but does not converge absolutely. Then there is a partition of the symmetric group ${\rm Sym}(\mathbb{N})$ into

  • sets $S_x \ (x \in \mathbb{R} \cup \{\pm \infty\})$ of all $\sigma \in {\rm Sym}(\mathbb{N})$ for which the series $\sum_{n=1}^\infty a_{n^\sigma}$ converges to $x$, and

  • a set $S_{\rm NaN}$ of all $\sigma \in {\rm Sym}(\mathbb{N})$ for which the series $\sum_{n=1}^\infty a_{n^\sigma}$ does not converge.

Question: Has there any work been done on properties of this partition for particular series $\sum_{n=1}^\infty a_n$?

Basic observations are:

  • all sets in the partition have the same cardinality, and

  • all sets in the partition are closed under multiplication from the right by permutations $\sigma$ which satisfy the condition $\forall n \in \mathbb{N} \ |n - n^\sigma| \leq C$ for some constant $C$.

Added on Oct 11, 2014: Let the spectrum $\mathcal{S}_{\sum a_n}(G)$ of a group $G < {\rm Sym}(\mathbb{N})$ with respect to the series $\sum_{n=1}^\infty a_n$ be the set of all $x \in \mathbb{R} \cup \{-\infty,+\infty,{\rm NaN}\}$ such that $G \cap S_x \neq \emptyset$. --

  • Is it true that if $G$ has full spectrum with respect to a series $\sum_{n=1}^\infty a_n$ (i.e. $\mathcal{S}_{\sum a_n}(G) = \mathbb{R} \cup \{-\infty,+\infty,{\rm NaN}\}$), then $G$ has full spectrum also with respect to any other conditionally convergent series $\sum_{n=1}^\infty b_n$?

  • Are there countable groups whose spectrum with respect to a particular series is dense in $\mathbb{R}$? -- And if yes, does the choice of the series actually matter here?

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    $\begingroup$ All sets in the partition indeed have the same cardinality, but don't we also know that $|S_x|=|{\rm Sym}(\mathbb{N})|=|\mathbb R|$ for all $x\in\mathbb R\cup\{\infty,-\infty,{\rm NaN}\}$? $\endgroup$
    – Joonas Ilmavirta
    Oct 8, 2014 at 11:55
  • $\begingroup$ @JoonasIlmavirta: Yes, of course. $\endgroup$
    – Stefan Kohl ♦
    Oct 8, 2014 at 13:01
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    $\begingroup$ Have you see this paper? "Rearrangement of conditionally convergent series on a small set", by Rafał Filipów, and Piotr Szuca. J. Math. Anal. Appl. 362 (2010) 64–71. It is not exactly the same question, but some of the information there appears relevant. $\endgroup$
    – Andrés E. Caicedo
    Oct 11, 2014 at 17:59
  • $\begingroup$ Is $S_{\sum a_n}$ a subgroup? If we generalize the definition of spectrum to apply to arbitrary subsets of $\rm{Sym}(\mathbb{N})$, then a more general question is: Is it true that for any conditionally convergent $\sum b_n$, $\left|\mathcal{S}_{\sum b_n}\left(S_{\sum a_n}\right)\right| = 1$? $\endgroup$
    – Amit Kumar Gupta
    Oct 11, 2014 at 18:18
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    $\begingroup$ @StefanKohl in my second question, first I consider generalizing the notion of spectrum to apply to any subset of the symmetric group, not just subgroups, i.e. for any $X \subset \rm{Sym}(\mathbb{N})$, consider $\mathcal{S}_{\sum a_n}(X)$. I ask, if $X$ has spectrum of size 1 with respect to one conditionally convergent series, does it have spectrum of size 1 with respect to all such series? $\endgroup$
    – Amit Kumar Gupta
    Oct 12, 2014 at 0:33

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Is it true that if $G$ has full spectrum with respect to a series $\sum_{n=1}^\infty a_n$ then $G$ has full spectrum also with respect to any other conditionally convergent series $\sum_{n=1}^\infty b_n$?

If I understand the definitions correctly, clearly no. Let $G$ be the symmetric group of $2\mathbb{N}$ and consider conditionally convergent series supported on the evens or odds.

Are there countable groups whose spectrum with respect to a particular series is dense in $\mathbb{R}$?

Sure. Let $\sum_{n=1}^\infty a_n$ be a conditionally convergent series. For a countable dense set of $x \in \mathbb{R}$, pick $\sigma_x \in \operatorname{Sym}(\mathbb{N})$ so that the rearrangement by $\sigma_x$ is $x$. Let $G$ be the group generated by the $\sigma_x$.

And if yes, does the choice of the series actually matter here?

Another evens/odds example comes to mind...

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