**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, 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?

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