This will not be altogether unrelated to [this earlier question](https://mathoverflow.net/questions/213046/rearrangements-that-never-change-the-value-of-a-sum).


For which classes $C$ of bijections from $\{1,2,3,\ldots\}$ to itself is it the case that for all sequences $\{a\_i\}\_{i=1}^\infty$ of real numbers, if $\displaystyle f\mapsto \lim\_{n \to\infty} \sum\_{i=1}^n a\_{f(i)}$ remains the same for all $f\in C$ then it remains the same for all bijections $f$ from $\{1,2,3,\ldots\}$ to itself?


Thus if you have a conditionally convergent series, then at least one of the rearrangements that change the value of the sum is within $C$.



How many rearrangements must fail to alter the value of a sum before you conclude that none do?
**Update.** A research collaboration growing out of this question and some of its answers has now resulted in the following article, providing an account of the rearrangement number:



> 
> A. Blass, J. Brendle, W. Brian, J. D. Hamkins, M. Hardy, and P. B. Larson, [The rearrangement number](http://jdh.hamkins.org/the-rearrangement-number/), manuscript under review.
> 
> 
> **Abstract.** How many permutations of the natural numbers are needed so that
>  every conditionally convergent series of real numbers can be
>  rearranged to no longer converge to the same sum? We show that the
>  minimum number of permutations needed for this purpose, which we
>  call the rearrangement number, is uncountable, but whether it equals
>  the cardinal of the continuum is independent of the usual axioms of
>  set theory. We compare the rearrangement number with several
>  natural variants, for example one obtained by requiring the
>  rearranged series to still converge but to a new, finite limit. We
>  also compare the rearrangement number with several well-studied
>  cardinal characteristics of the continuum. We present some new
>  forcing constructions designed to add permutations that rearrange
>  series from the ground model in particular ways, thereby obtaining
>  consistency results going beyond those that follow from comparisons
>  with familiar cardinal characteristics. Finally we deal briefly
>  with some variants concerning rearrangements by a special sort of
>  permutations and with rearranging some divergent series to become
>  (conditionally) convergent.
> 
> 
> 




---


**Original answer.** 
$\newcommand\N{\mathbb{N}}\newcommand\P{\mathbb{P}}$This is a great question. Let me focus
on the title question, namely, the question of how many functions
one might need to ensure your rearrangement property. There are a
few things one can say.


**Theorem.** No countable family of functions suffice to ensure
your rearrangement property. Specifically, for any countably many
permutations $f\_n:\N\to\N$, there is a series $\sum\_k a\_k$ of real
numbers, such that $\sum\_k a\_{f\_n(k)}=0$ for every $f\_n$, but
$\sum\_k a\_k=0$ is only conditionally convergent.


**Proof.** Indeed, I claim further that for any given series
$\sum\_k b\_k$, we may create a new series $\sum\_k a\_k$ by padding
the original series with zeros, but maintaining the order of the
nonzero terms, in such a way that for every function $f\_n$, the
nonzero terms of $\sum\_k a\_{f\_n(k)}$ appear in exactly the same
order as the original series $\sum\_k b\_k$, except for at most $n$
values. So all these series have the same sum, even though the
original series may be only conditionally convergent.


So fix any series $\sum\_k b\_k$, and any countably many
permutations $f\_n:\N\to\N$. We may start by defining $a\_0=b\_0$. At
stage $n$, we will have specified finitely many $a\_k$, in such a
way that the nonzero entries specified so far include $b\_0,
b\_1,\ldots,b\_n$, in that order, but we have padded with a possibly
very large number of zeroes in between, and furthermore such that
for every $m<n$, the nonzero values of $a\_{f\_m(k)}$ that have been
specified also agree with $b\_0,\ldots,b\_n$, in that order (except for $m$ values at most), but with the zeroes possibly inserted differently.


At step $n$, consider the functions $f\_m$ for $m\leq n$, and find
some index $k\_n$ that is sufficiently large so that $k\_n$ and
$f\_m(k\_n)$ are larger than any index we have yet used, for all
$m\leq n$. Let $a\_{k\_n}=b\_n$, and pad the sequence with zeros
$a\_k=0$ at all the other indices up to $k\_n$. Since we add the new
term $b\_n$ such a far distance out, the rearranged non-zero terms
$a\_{f\_m(k)}$ maintain the same order of $\sum\_n b\_n$ as far as we
have yet specified them (except possibly for the errors that we present before stage $m$, when the function $f\_m$ began to be considered). 


It follows that all the particular rearrangements $\sum\_k
a\_{f\_n(k)}$ using functions $f\_n$ have the same value as $\sum\_k
a\_k$, which agrees with $\sum\_k b\_k$. But if $\sum\_k b\_k$ is only
conditionally convergent, then of course there is some other
rearrangement with a different sum. **QED**


I take this theorem to suggest a new cardinal characteristic of
the continuum. Specifically, let us define $\kappa$ to be the
size of the smallest family $\cal C$ of permutations that have
your desired rearrangement property. So $\kappa$ is the answer to the title question of "how many?" because fewer than $\kappa$ are insufficient, by definition, but there is a family of $\kappa$ many permutations that work. So far, we've proved that
$\kappa$ is uncountable, and clearly it is at most continuum.


**Corollary.** If the continuum hypothesis holds, then
$\kappa=\frak{c}$ is the continuum.


This conclusion is also consistent with the failure of the
continuum hypothesis.


**Theorem.** It is relatively consistent with ZFC that no family of fewer than continuum many permutations has your rearrangement property, even when the continuum is large. Indeed, Martin's axiom implies $\kappa=\frak{c}$.


**Proof.** Assume that Martin's axiom holds, and that $\cal C$ is
a family of fewer than the continuum many permutations
$f:\N\to\N$. Fix any series $\sum\_k b\_k$, and let $\P$ be the
partial order consisting of pairs $(s,F)$, where $s$ is a finite
sequence of real numbers, whose nonzero values agree with a finite
initial segment of those appearing in $\sum\_k b\_k$, but possibly
padded with extra zeros, and $F$ is a finite subset of $\cal C$.
The order is $(s,F)\leq (s',F')$ just in case $s\supseteq s'$ and
$F\supseteq F'$, so that $(s,F)$ specifies more of the sequence
and $F$ mentions more functions, and furthermore, for any $f\in
F$, the nonzero portion of $s$ beyond $s'$ appears in the same
order in $s$ as it does in the rearrangement of $s$ by $f$. In
other words, once we add a function $f$ to $F$, then we will
insist that all further extensions of $s$ respect $f$.


As a forcing notion, $\P$ has the countable chain condition,
because any two conditions $(s,F)$, $(s,F')$ with the same first
part are compatible, simply by using $(s,F\cup F')$, and there are
only countably many ways to pad a finite initial segment of
$\sum\_n b\_n$ with zeros. So Martin's axiom will apply to this
forcing notion.


For each $m$, the collection of conditions $(s,F)$ for which $s$
includes the first $m$ terms of $\sum\_k b\_k$ is dense, since the
construction in the main theorem above shows how to handle
finitely many functions at once. Also, for any particular
$f\in\cal C$, it is dense for it to be added to the second
coordinate.


Since we have thus specified fewer than continuum many dense sets,
by Martin's axiom there is a filter in $\P$ that meets all these
dense sets. Such a filter amounts to a series $\sum\_k a\_k$, which
agrees fully on its nonzero terms with the series $\sum\_k b\_k$, in
the same order, and such that for every $f\in \cal C$, the
rearrangement $\sum\_k a\_{f(k)}$ also agrees with $\sum\_k b\_k$
except on finitely many terms.


So all these rearrangements have the same sum as $\sum\_k b\_k$,
even if this series is only conditionally convergent. **QED**


It remains to see whether a small family can ever suffice.


**Question.** Is it consistent with ZFC that $\kappa$ is less
than the continuum?


That is, is it consistent with ZFC that a small family can
suffice?


