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?