On rearrangements of conditionally convergent series

M. Sc. Thesis of Balázs Gerencsér

Supervisor: Gábor Tusnády

2007

Eötvös Loránd University, Institute of Mathematics

Contents

1 Introduction 2 2 Preliminaries, notations 4 3 Small local deviations 6 4 Incompatibility of series 12 5 Reconstruction of series 22 6 Random permutations 27 7 Final words 35

Chapter 1

Introduction

Given a real series a_{1}, a_{2}, … it is meaningful to examine the “infinite commutativity” of addition. It is a well known fact that in general

∑_{j=1}^{∞} a_{j} ≠ ∑_{j=1}^{∞} a_{π(j)}

may happen for a permutation π of positive integers. In this paper we will focus on these conditionally convergent series where this can happen.

The starting point is Riemann’s theorem about the possibility of obtaining any sum (more precisely any liminf and limsup) by an appropriate rearrangement. One may want to know more than existence, namely some characterization of rearrangements that preserve convergence or even the sum of conditionally convergent series.

A favorite object to analyze is the set of permutations of ℕ which preserve convergence and sum for all conditionally convergent series. We will use a necessary and sufficient condition by R. P. Agnew [1], which states that a universally sum preserving rearrangement can break the intervals [1, n] into only a limited number of intervals. There are a few different characterizations, a famous one is that of F. W. Levi [5], which examines how elements of the permutation jump over an integer n. Although these are exact conditions, they might be hard to check in some situation. Fortunately there are several simply formulated sufficient conditions. A few of them is presented by P. Schaefer, [8]. For example, it is enough that there is an integer B such that π(j) ≤ j + B for all j. Similar results can be found in U. Elias [2], J. R. Stefánsson [10], and U. C. Guha [4].

The original problem is simple enough and there is the possibility of countless modifications, generalizations or special cases.

One orientation of research is to require some extra properties of the permutation, and check whether there exist a permutation which preserves the summability of a given series or not. J. H. Smith proved in [9] existence of a permutation with the cycle structure fixed for any conditionally convergent series, only excluding trivial cases.

In [13], G. Tusnády estabilished connection between the relation of series and relation of their sets of convergence preserving permutations. This will be detailed later.

A possible extension is to change the range of the series from ℝ to something else. This may increase the complexity of the problems because multiple types of convergence can coexist. For example, take real-valued mea-

surable functions on [0, 1]. As proved by Nikishin in [6], if ∑_{k=1}^{∞} f_{k} → F in

measure, and ∑_{k=1}^{∞} f_{k}^{2}(t) < ∞ almost everywhere, then the sum will converge

to the same limit F almost everywhere for an appropriate rearrangement. G. Giorgobiani gave similar statements in [3] when the functions take their value from a normed space.

Starting from the fact that permutations of ℕ form a group, one can meet new questions. Some algebraic problems are covered by G. S. Stoller in [11] and P. A. B. Pleasants in [7]. A detailed examination was done by Q. F. Stout in [12]. In this work, the “closure” of a set of permutation is defined as the set of permutations which preserve the sum of the same subset of series. The closure of a set of series is defined in a similar way. It is hard to produce the closure of a single permutation, but the closure of a single series is proved to be the linear span of the series and all absolutely convergent series.

In the present work, we will examine some structural questions about the set of convergence preserving permutations for a specific series. This will be developed in the next three chapters. After that, we will face the natural problem of reconstruction. The question is whether we can rebuild the series if only the set of convergence preserving permutations are given. In the last chapter we will have an outlook on a stochastic counterpart, where we use random permutations for rearrangement.

Preliminaries, notations

We will work with real valued series that will be identified as the elements of ℝ^{ℕ}. Let us define the subset of series with convergent sums

S = {a = {a_{i}}_{i=1}^{∞} ∈ ℝ^{ℕ} | ∃A ∈ ℝ lim_{n→∞}(A − ∑_{i=1}^{n} a_{i}) = 0} ,

and the subset of absolutely convergent series:

S_{0} = {a = {a_{i}}_{i=1}^{∞} ∈ ℝ^{ℕ} | ∃A ∈ ℝ lim_{n→∞}(A − ∑_{i=1}^{n} |a_{i}|) = 0} .

We would also like to define the set of series which might be convergent after an appropriate rearrangement. For an a ∈ ℝ^{ℕ} \ S_{0} it is clearly necessary that lim_{n→∞} a_{n} = 0 and that both the positive and the negative elements of the series must sum up to ∞. This is also sufficient, as it can be easily seen. We can start a rearrangement by the first positive element, then choose the first few negative elements until the sum goes below 0. Then continue with the first few unused nonnegative elements until the sum reaches 0, and so on. The condition that both the positive and the negative elements have infinite sums ensure that this procedure will never get stuck, and the condition about elements converging to 0 implies that the deviation from 0 will tend to zero, hence this rearrangement forces the partial sums to go to 0. Now we define:

S̃ = S_{0} ∪ {a = {a_{i}}_{i=1}^{∞} ∈ ℝ^{ℕ} | lim_{i→∞} a_{i} = 0, ∑_{i=1}^{∞} |a_{i}|^{+} = ∞, ∑_{i=1}^{∞} |a_{i}|^{−} = ∞} .

When investigating a series, we will often work with the subseries of nonnegative elements so let us introduce:

S^{+} = {a = {a_{i}}_{i=1}^{∞} ∈ ℝ^{ℕ} | ∀i ∈ ℕ a_{i} ≥ 0, lim_{i→∞} a_{i} = 0} .

S_{0}^{+} = {a = {a_{i}}_{i=1}^{∞} ∈ ℝ^{ℕ} | ∀i ∈ ℕ a_{i} ≥ 0, ∑_{i=1}^{∞} a_{i} < ∞} .

Later we will need series with almost nonnegative elements, which means that we allow a subseries of negative elements with convergent sum.

S̃^{+} = S^{+} − S_{0}^{+}.

When we are interested in possible rearrangements, multiplying a series with a constant or adding an absolutely convergent series doesn’t make sense. So we call two conditionally convergent series a, b ∈ S \S_{0} equivalent, a ≡ b, if a − λb ∈ S_{0} for some λ ∈ ℝ. Obviously λ ≠ 0. This equivalency is also meaningful for elements of S̃ \ S_{0}.

Let us move on to permutations. Let P be the set of all permutations of ℕ:

P = {π : ℕ → ℕ | π is a bijection} .

A permutation describes a rearrangement of a series

πa = {a_{π_{i}}}_{i=1}^{∞}.

Note that for any π, ρ ∈ P this definition has the property

π(ρa) = (ρπ)a.

We introduce the notation:

s(a, π) = lim_{n→∞} ∑_{i=1}^{n} a_{π_{i}}.

This might be finite or infinite, or even might not exist. It is natural to use |s(a, π)| < ∞ to indicate that a permutation assigns a convergent sum to a. When we omit π, then we use the identity permutation.

Given a series a, we are interested which permutations lead to a convergent sum. We denote the set of these permutations by K(a) and call it the convergence set of a:

K(a) = {π ∈ P | |s(a, π)| < ∞} .

It is obvious that if a ∈ S_{0}, then K(a) = P, if a ∉ S̃, then K(a) = ∅. There are some permutations which do not ruin the convergence of any conditionally convergent series, these form the set P_{0}, that is,

P_{0} = ⋂_{a∈S} K(a).

Small local deviations

Given a series a, let us define the partial sums s_{n} = ∑_{i=1}^{n}a_{i}. A necessary con-

dition for s_{n} to converge is that locally (whatever that means) the deviations should be small. An easy case to test this when several adjacent elements of a have the same sign. In this case one occurring deviation is the absolute sum of these adjacent elements.

To exploit this idea, we restrict the permutations so that the order within nonnegative and negative elements is fixed. This means that we split the series a into b and c, where b contains the nonnegative and c contains the negative elements. Now a legal permutation looks like

b_{1}, b_{2}, …, b_{n_{1}}, c_{1}, c_{2}, …, c_{m_{1}}, b_{n_{1}+1}, …, b_{n_{2}}, c_{m_{1}+1}, …, c_{m_{2}}.

The necessary condition described above can be formulated as

lim_{k→∞} ∑_{i=n_{k}+1}^{n_{k+1}} b_{i} = 0  (3.1)

and the same for c. Moreover, if this is true for b and a ∈ S̃ \ S_{0}, then one can build a convergent permutation where the elements of b are segmented this. Simply after each block, insert a few (maybe 0) elements of c until the partial sum decreases below 0. This method provides the lim inf is 0 using lim_{i→∞} c_{i} = 0, and (3.1) forces the lim sup to be 0 by the same reason.

For a fixed b, we try to find some structure of the sets {n_{1}, n_{2}, …} = N ⊂ ℕ which satisfy (3.1). Let H_{b} be the system of these sets. There are some trivial cases, we would like to avoid these.

If ∑_{i=1}^{∞} b_{i} < ∞, this means ∑_{k=1}^{∞} ( ∑_{i=n_{k}+1}^{n_{k+1}} b_{i}) < ∞, but this involves (3.1), so

H_{b} = 2^{ℕ}. If lim sup_{i→∞} b_{i} > 0, then lim sup_{k→∞} ∑_{i=n_{k}+1}^{n_{k+1}} b_{i} > lim sup_{i→∞} b_{i} > 0, so H_{b} = ∅.

To avoid these cases, we will assume that a ∈ S̃ \ S_{0} or equivalently b, −c ∈ S^{+} \ S_{0}^{+} through the whole chapter. Obviously this means that every element of H_{b} must be infinite. These sets are somehow related to the speed of convergence to 0. It is clear that H_{b^{1}} ⊂ H_{b^{2}} if λb^{1} > b^{2} element-wise for some positive λ. A similar statement is true for c.

We start the examination of the structure of H_{b} by showing a method to build a third element if two others are given.

Definition 1 Let N, K ∈ H_{b}, N = {n_{1} < n_{2} < …}, K = {k_{1} < k_{2} < …} Let us define H = N ∧ K = {h_{1} < h_{2} < …}, where

h_{1} = max(n_{1}, k_{1}),
h_{i+1} = max(min(N ∩ [h_{i} + 1, ∞)), min(K ∩ [h_{i} + 1, ∞))), i ≥ 1.

We should note that N ∧ K will also have infinite elements. This is true because N and K are infinite, thus we will take the minimum of non-empty sets, so the algorithm will not stop after finite number of steps.

Proposition 1 We have H ∈ H_{b}.

Proof: We want to show that an interval [h_{i} +1, h_{i+1}] can be covered with at most two intervals of type [n_{i′} + 1, n_{i′+1}] or [k_{i′′} + 1, k_{i′′+1}]. Without the loss of generality we may suppose h_{i} = n_{i′}. If h_{i+1} = n_{i′+1}, then

∑_{j=h_{i}+1}^{h_{i+1}} b_{j} = ∑_{j=n_{i′}+1}^{n_{i′+1}} b_{j}.

If h_{i+1} = k_{i′′+1}, then

∑_{j=h_{i}+1}^{h_{i+1}} b_{j} ≤ ∑_{j=n_{i′}+1}^{n_{i′+1}} b_{j} + ∑_{j=k_{i′′}+1}^{k_{i′′+1}} b_{j}.

Now for any ε > 0, let T be large enough such that for n_{i′+1} > T

∑_{j=n_{i′}+1}^{n_{i′+1}} b_{j} < ε,

and the same be true for K, then choose h_{i} > T. From the previous observation,

∑_{j=h_{i}+1}^{h_{i+1}} b_{j} < 2ε.

This means (3.1) for H, so H ∈ H_{b}.

□

If N ⊂ K ⊂ ℕ, then K gives a finer slicing than N, and it is clear that N ∈ H_{b} implies K ∈ H_{b}. We use this latter property to define a partial ordering on 2^{ℕ}:

Definition 2 Let H, K ⊂ ℕ. We say H ≤ K if for any b ∈ S^{+}\S_{0}^{+} H ∈ H_{b} implies K ∈ H_{b}. We say H ≅ K if H ≤ K and K ≤ H. H < K means H ≤ K but H ≇ K.

Proposition 2 H ≤ K if and only if there exists C ∈ ℕ such that

∀i ≥ 1 |[k_{i}, k_{i+1}] ∩ H| ≤ C.

Proof: First assume the inequality holds for all i, and choose b such that H ∈ H_{b}. We shall prove K ∈ H_{b}.

Fix ε > 0. H ∈ H_{b} means that there exists a T such that for h_{j+1} > T

∑_{l=h_{j}+1}^{h_{j+1}} b_{l} < ε.

In the case k_{i} > T, choose the maximal j′ and the minimal j′′ for [k_{i}, k_{i+1}] ⊂ [h_{j′}, h_{j′′}] to hold. Our condition ensures j′′ − j′ ≤ C + 1. Now

∑_{l=k_{i}+1}^{k_{i+1}} b_{l} ≤ ∑_{l=h_{j′}+1}^{h_{j′′}} b_{l} = ∑_{l=h′_{j}+1}^{h_{j′+1}} b_{l} + ∑_{l=h_{j′+1}+1}^{h_{j′+2}} b_{l} + … + ∑_{l=h_{j′′−1}+1}^{h_{j′′}} b_{l} < (C + 1)ε.

So in the end, for (C + 1)ε, T will be an adequate bound.

Conversely, suppose the inequality does not hold. This means that there exist a series {i_{n}}_{n=1}^{∞} such that for

A_{n} := [k_{i_{n}}, k_{i_{n}+1}] ∩ H, |A_{n}| > n.

Let us define the series b as follows:

b_{j+1} = \{ \begin{matrix}\frac{1}{n}, & if j ∈ A_{n} \\ 0, & otherwise \end{matrix} .

Clearly ∑_{j=h_{i}+1}^{h_{i+1}} a_{j} is \frac{1}{n}, if h_{i} ∈ A_{n}, 0 otherwise, and using A_{n} is bounded,

(3.1) holds. On the other hand, ∑_{j=k_{in}+1}^{k_{in+1}} a_{i} > \frac{n − 1}{n}, so K ∉ H_{b}. This shows

H ≰ K.

□

Proposition 3 Let N, K ⊂ ℕ, H = N ∧ K. In this case, H ≤ N and H ≤ K.

Proof: It is enough to prove the first, we will do this using Proposition 2. Let us choose n_{i} ∈ N. [n_{i}, n_{i+1}] ∩ H might be empty, but if not, let h_{l} be the first element of [n_{i}, n_{i+1}] ∩ H. Looking at the definition of ∧ it follows that h_{l+1} ≥ n_{i+1}, so using C = 2 the condition of Proposition 2 will hold.

□

By Proposition 1 we may notice the operation ∧ produces an element of H_{b} which is smaller than the elements we started with.

It is also true that H_{b} does not have a minimal element.

Proposition 4 Suppose K ∈ H_{b} for some b ∈ S^{+} \ S_{0}^{+}. Then there exists H < K, H ∈ H_{b}.

Proof: We will delete some elements from K such that the condition of Proposition 2 does not hold. It is enough if we drop n consecutive elements of K somewhere for every n. Let us choose i_{1} = 1, and then recursively choose i_{n} ≥ i_{n−1} + n to ensure

∑_{j=k_{i}+1}^{k_{i+1}} b_{j} < \frac{1}{n^{2}}

for every i ≥ i_{n}. Now delete the elements k_{i_{n}+j} for all 1 ≥ n, 0 ≤ j ≤

n − 1 to get H. H ≤ K is obvious, so is K ≰ H. The series ∑_{j=h_{i}+1}^{h_{i+1}} b_{j} is

the combination of a subseries of ∑_{j=k_{i}+1}^{k_{i+1}} b_{j} and of a series less than \frac{1}{i}, this

occurs when we pass deleted elements of K. Both subseries converge to 0, consequently (3.1) holds, H ∈ H_{b}.

□

We will show the previously defined partial ordering form a lattice structure on 2^{ℕ}/ ≅, with ∪ and ∧ as the lattice operations. To see this, we need the following two propositions.

Proposition 5 Let N, K, H ⊂ ℕ. If H ≤ N and H ≤ K then H ≤ N ∧ K. If H ≥ N and H ≥ K then H ≥ N ∪ K.

Proof: Let us start with the first statement. Suppose H ∈ H_{b} for some b ∈ S^{+} \ S_{0}^{+}, then our condition provides N, K ∈ H_{b}. By Proposition 1 this implies N ∧ K ∈ H_{b}. What we wrote here is the definition of H ≤ N ∧ K. To prove the second statement, we will use Proposition 2. H ≥ N means that there is an appropriate C_{N}, and similarly a C_{K}, thus

|[h_{i}, h_{i+1}] ∩ (N ∪ K)| ≤ |[h_{i}, h_{i+1}] ∩ N| + |[h_{i}, h_{i+1}] ∩ K| ≤ C_{N} + C_{K}.

This is true for every i, therefore H ≥ N ∪ K.

□

Proposition 6 For any N, K ⊂ ℕ, (N ∪K)∧N = N and (N ∧K)∪N ≅ N.

Proof: The first statement is trivial from the definition of ∧. Note that the max will always choose the element of the coarser partitioning. Concerning the second one, (N ∧ K) ∪ N ⊃ N, so (N ∧ K) ∪ N ≥ N. The opposite direction is a consequence of the previous Proposition, N ∧ K ≤ N, N ≤ N, so (N ∧ K) ∪ N ≤ N, and we are ready.

□

Now we can sum up what we found until now about any H_{b}:

Theorem 1 For any b ∈ S^{+} \ S_{0}^{+}, H_{b} is a filter of 2^{ℕ}/ ≅ with no minimal elements.

By the way, we can exclude some trivial cases. ∅ ∈ H_{b} would mean 2^{ℕ} = H_{b} but this is impossible for b ∈ S^{+} \ S_{0}^{+}. Clearly the same is true if H_{b} = 2^{ℕ} \ [∅].

This is a well formed necessary condition. An open question is how far this is from being sufficient. We still have some work to do, as shown by the following statement.

Proposition 7 Take N ⊂ ℕ, and H = {H ⊂ ℕ | N < H}. This H will be a filter with no minimal elements, but H ≠ H_{b} for any b ∈ S^{+} \ S_{0}^{+}.

Proof: It is clear that H is a filter. To prove it has no minimal elements, it is enough to show that for any N < K, one can find N < H < K. Using Proposition 2, N < K implies we can find disjoint intervals I_{j} = [n_{i_{j}}, n_{i_{j+1}}] such that |I_{j} ∩ K| ≥ j. Define

H = N ∪ ⋃_{j=1}^{∞}(I_{2j} ∩ K).

Hereby the intervals I_{2j} ensure N < H and I_{2j+1} ensure H < K.

Suppose H = H_{b}. N ∉ H_{b} means that there are infinitely many I_{j} = [n_{i_{j}} + 1, n_{i_{j+1}}] for which ∑_{k∈I_{j}} b_{k} > ε for some ε > 0. The fact that lim_{k→∞} b_{k} = 0

causes these intervals to become longer and longer, so by thinning we may suppose |I_{j}| ≥ j + 1. Let us define

H = N ∪ ⋃_{j=1}^{∞} I_{2j}.

We filled up long empty intervals to achieve N < H, so H ∈ H. We also retained infinitely many empty intervals where the sum is above ε, consequently H ∉ H_{b}.

□

Incompatibility of series

When examining the slicing in the previous chapter, we have found a latticelike structure on it. Now we move on to conditionally convergent series, where we will experience an opposite property.

It is meaningless to state that if a ∈ S_{0}, b ∈ S \ S_{0}, then K(a) ⊋ K(b) holds for the convergence sets. The interesting point is that this inclusion can only happen in this situation.

Before proving this, let us have a closer look on P_{0}. In [1] we have a characterization on permutations which preserves not only convergence, but also the value of the sum:

Theorem 2 (Agnew) Choose π ∈ P. Then π preserves the sum of all conditionally convergent series if and only if there is an integer M such that for all n ∈ ℕ, π([1, n]) is a union of M or fewer intervals of ℕ.

We will show that exactly these are the universally convergence preserving permutations.

Theorem 3 Choose π ∈ P. Then π ∈ P_{0} if and only if there is an integer M such that for all n, π([1, n]) is a union of M or fewer intervals of ℕ.

Proof: The condition is sufficient, this is obvious from the previous theorem. For the other implication, given a π ∈ P which does not satisfy the condition, we will build an a ∈ S such that πa will not be convergent, π ∉ K(a).

By the properties of π, we can choose N_{i} as follows. Let N_{1} be 1, this is the first number for which π([1, N_{1}]) consists of at least 1 intervals. If N_{i−1} is defined, let’s choose N_{i} such that π([1, N_{i}]) ⊃ [1, N_{i−1}], and π([1, N_{i}]) is the union of at least i intervals.

Fix an i ≥ 2. π([1, N_{i}]) = I ∪ J_{1} ∪ … ∪ J_{i−1}, where I is the interval containing [1, N_{i−1}], J_{k} are the following intervals except the last one which is the union of the remaining (one or more) intervals. Let the first elements of J_{k} be j_{k}. Now we define some of the elements of a:

a_{j_{k}−1} = \frac{(−1)^{i−1}}{\sqrt{i − 1}} ,

a_{j_{k}} = \frac{(−1)^{i}}{\sqrt{i − 1}}.

We set all remaining elements of [N_{i} + 1, N_{i+1}] to 0, then repeat this for all i ≥ 2 to define all elements of a.

In the end the series consists of a lot of 0-s, and when a nonzero element

q appears, it is immediately followed by −q. This implies |∑_{k=1}^{n} a_{k}| ≤ q.

Moreover, if n > N_{i}, then this holds for q = \frac{1}{\sqrt{i−1}}. So we got ∑_{k=1}^{∞} a_{k} = 0.

Let’s look at the permutated sum. Using π([1, N_{i}]) = I ∪ ⋃_{k=1}^{i−1} J_{k} as above,

and ∑_{k∈I} a_{k} = 0 and ∑_{k∈J′_{k}} a_{k} = \frac{(−1)^{i}}{\sqrt{i − 1}}. Adding these up we get

∑_{k=1}^{N_{i}} a_{π(k)} = (−1)^{i}\sqrt{i − 1}.

So we got

lim sup_{n→∞} (∑_{k=1}^{n} a_{π(k)}) = ∞.

In the end, the permutation ruined convergence for this specific a, π ∉ K(a), π ∉ P_{0}.

□

To quantify the relation between two series a and b, let us define the following measures on \overline{ℝ}:

μ_{+} = ∑_{j:a_{j}>0} a_{j}δ_{\frac{b_{j}}{a_{j}}} + ∑_{j:a_{j}=0, b_{j}≠0} a_{j}δ_{sgn(b_{j})∞},

μ_{−} = − ∑_{j:a_{j}<0} a_{j}δ_{\frac{b_{j}}{a_{j}}} .

These measures give a sight on which ratios are more likely to appear. We will investigate the places where the measure is ∞, more precisely

Definition 3 For a ν measure on \overline{ℝ}, c ∈ \overline{ℝ} is a limit point if for all c ∈ U neighborhoods, ν(U) = ∞.

Lemma 1 μ_{+} has at least one limit point (and similarly μ_{−}).

Proof: We use only μ_{+}(\overline{ℝ}) = ∞, which is a consequence of conditional convergence. If ∞ and −∞ are not limit points, then μ_{+}(\overline{ℝ} \ [−K, K]) < ∞ for an appropriate K.

Suppose there are no limit points in [−K, K]. This means for all x ∈ [−K, K] there is x ∈ U_{x}, μ_{+}(U_{x}) < ∞. Using compactness, finite of them

will also cover, [−K, K] ⊂ ⋃_{i=1}^{n} U_{x_{i}}.

This gives a finite upper bound ∑_{i=1}^{n} μ_{+}(U_{x_{i}}) on μ_{+}([−K, K]). Comparing

with μ_{+}(\overline{ℝ}) = ∞ ends up at a contradiction, so there must be at least one limit point.

□

We will have some interesting properties if special limit points appear.

Lemma 2 Suppose 0 is a limit point of μ_{+} (μ_{−}). Then for any ε > 0, δ > 0, T ≥ 0 (T ≤ 0), and J ⊂ ℕ finite set one can find a finite I ⊂ ℕ \ J such

that |T − ∑_{i∈I} a_{i}| < ε, |∑_{i∈I} b_{i}| < δ, and ∀i ∈ I a_{i} ≥ 0 (a_{i} ≤ 0).

Proof: The T = 0 case is trivial, otherwise without loss of generality, we may suppose T > 0. Discarding finite number of elements from the series does not change the limit point structure, so we may also suppose J = ∅. μ_{+} has 0 as a limit point, so

μ_{+} ((−\frac{δ}{T + ε}, \frac{δ}{T + ε})) = ∑_{i:| \frac{b_{i}}{a_{i}} |< \frac{δ}{T +ε} ,0≤a_{i}} a_{i} = ∞.

Using a_{i} → 0, ∑_{i:a_{i}≥ε} a_{i} < ∞, so it is also true that

∑_{i:| \frac{b_{i}}{a_{i}} |< \frac{δ}{T +ε} ,0≤a_{i}<ε} a_{i} = ∞.

Now let be

I′ = \{i ∈ ℕ|i : |\frac{b_{i}}{a_{i}} | < \frac{δ}{T + ε}, 0 ≤ a_{i} < ε\} ,

and choose minimal n sufficing ∑_{i∈I′∩[1,n]} a_{i} > T. For this I = I′ ∩ [1, n], auto-

matically ∑_{i∈I} a_{i} < T + ε. Moreover,

|∑_{i∈I} b_{i}| ≤ ∑_{i∈I} a_{i} |\frac{b_{i}}{a_{i}} | < \frac{δ}{T + ε} ∑_{i∈I} a_{i} < δ.

□

Lemma 3 Suppose ∞ is a limit point of μ_{+} (μ_{−}). Then for any ε > 0, Δ > 0, T ≥ 0 (T ≤ 0), and J ⊂ ℕ finite set one can find a finite I ⊂ ℕ \ J such

that |T − ∑_{i∈I} a_{i}| < ε, ∑_{i∈I} b_{i} > Δ, and ∀i ∈ I a_{i} ≥ 0 (a_{i} ≤ 0).

The proof is similar, the only difference is that instead of using an upper bound on |\frac{b_{i}}{a_{i}} | in the definition of I′, we need a lower bound on \frac{b_{i}}{a_{i}}. This is exactly what we can modify when changing 0 to ∞ as the limit point.

Now let’s get back to the theorem mentioned at the beginning of the chapter.

Theorem 4 Let a, b ∈ S \ S_{0} be nonequivalent conditionally convergent series. In this case K(a) \ K(b) ≠ ∅.

Proof: Let us define M = max_{i≥1} |b_{i}|.

We know μ_{+} has a limit point c, and μ_{−} has a limit point d. It may happen that c ≠ d. We can suppose c < d, and choose c < T < U < d. This means there are I, J ⊂ ℕ such that

i ∈ I ⇒ a_{i} ≥ 0, \frac{b_{i}}{a_{i}} < T,

j ∈ J ⇒ a_{j} < 0, \frac{b_{j}}{a_{j}} > U.

and ∑_{i∈I}a_{i} = ∞, ∑_{j∈J}a_{j} = −∞.

Now we build the permutation needed for the theorem, we will define the values of the permutation one by one. There will be three types of steps.

In the first, we choose the first few unused elements of J such that the partial sum goes below 0.

In the second, we continue with some unused elements of I to increase the sum above 0.

In the third step, we insert the first unused element of ℕ \ (I ∪ J).

We will repeat the first two steps a lot, this will hold the partial sums of a near 0 while unboundedly decreasing that of b. We will will perform the third step relatively rarely not to ruin the effect of the first two steps, but to provide every element would occur.

To show the recursive method, suppose we already determined a few elements of the permutation, and the partial sum of a until now is K. Let us repeat the first two steps multiple times, starting with the one corresponding to the sign of K and ending with the first. Denote the indices chosen from I and J by I^{∗} and J^{∗}, respectively. If we do this long enough, we can achieve

∑_{i∈I^{∗}} a_{i} > V + K, ∑_{j∈J^{∗}} −a_{j} > ∑_{i∈I^{∗}} a_{i} − K > V for any V, we will choose it

later. These new elements have the following addition to the partial sum of the rearranged b:

∑_{k∈I^{∗}∪J^{∗}} b_{k} = ∑_{i∈I^{∗}} b_{i} + ∑_{j∈J^{∗}} b_{j} < T ∑_{i∈I^{∗}} a_{i} + U ∑_{j∈J^{∗}} a_{j} =

= T (∑_{i∈I^{∗}} a_{i} + ∑_{j∈J^{∗}} a_{j}) + (U − T) ∑_{j∈J^{∗}} a_{j} < T K − (U − T)V.

And the right hand side will be less than −2M if we choose V large enough. If it is possible, we perform the third step once, in the end the partial sum of b decreased by at least M.

Now we start over. We can do this infinitely because I, J and the sums of a over I and J are infinite. The result must be a permutation as we use any element at most once and the possibility of third type steps provides we use them exactly once. The partial sum of πb decreases by M every

time, finally it diverges to −∞. It is clear that lim inf_{n→∞} ∑_{i=1}^{n} a_{π(i)} ≤ 0 and

lim sup_{n→∞} ∑_{i=1}^{n} a_{π(i)} ≥ 0. If lim inf_{n→∞} ∑_{i=1}^{n} a_{π(i)} < −3ε, then we cross the −2ε barrier

infinite times. Then there are infinitely many i such that a_{i} < −ε since the barrier will be passed after a first and/or a third type step. This is impossible, so the lim inf is nonnegative, similarly the lim sup is nonpositive, which means. s(a, π) = 0.

To sum up, this gives a legal permutation, one of K(a) \ K(b) in the case when μ_{+} and μ_{−} have different limit points.

The other possibility is that there is only one common limit point, this might be 0, a nonzero finite number, or ±∞.

Suppose this limit point is 0. We will describe an algorithm somewhat similar to the previous one to build an adequate permutation π.

We will describe step N, starting at N = 1. Choose I ⊂ ℕ a finite subset belonging to negative elements of b satisfying

∑_{i∈I} b_{i} < −4M.

This will help the sum of πb go to −∞, but we need to compensate the effect of the sum of corresponding elements of a. For each i ∈ I we choose (a finite) J_{i} ⊂ ℕ disjoint from each other and I such that

|∑_{j∈J_{i}} a_{j} + a_{i}| < \frac{M}{|I|N} ,

|∑_{j∈J_{i}} b_{j}| < \frac{M}{|I|}.

This is possible by using Lemma 2 multiple times. If all a_{i} are 0, let all J_{i} be empty.

|∑_{i∈I} (a_{i} + ∑_{j∈J_{i}} a_{j})| ≤ ∑_{i∈I} |a_{i} + ∑_{j∈J_{i}} a_{j}| ≤ |I|\frac{M}{N|I|} = \frac{M}{N} ,

∑_{i∈I} (b_{i} + ∑_{j∈J_{i}} b_{j}) ≤ ∑_{i∈I} b_{i} + ∑_{i∈I} |∑_{j∈J_{i}} b_{j}| < −4M + |I|\frac{M}{|I|} = −3M.

Using these relations we can tell the first part of the permutation. Insert the first element of I, then all of the corresponding J_{i}, then do this for all elements of I. After that, insert the first unused element l ∈ ℕ, to provide every index would appear in the permutation. Let A be the sum of elements of a used until now. In the end, we choose a J′ ⊂ ℕ finite subset of unused

element such that |A − ∑_{j∈J′} a_{j}| < \frac{M}{N}, but |∑_{j∈J′} b_{j}| < M. Put these elements

to the end of the permutation.

Suppose we determined the [K, L] elements of π during this step.

∑_{i=K}^{L} b_{π(i)} = ∑_{i∈I} b_{i} + ∑_{i∈I} ∑_{j∈J_{i}} b_{j} + b_{l} + ∑_{j∈J′} b_{j} ≤ −3M + M + M = −M,

|∑_{i=1}^{L} a_{π(i)}| < \frac{M}{N} .

For any K ≤ L′ ≤ L, π([K, L′]) consists of a few “blocks” - an i ∈ I and the corresponding J_{i} or l and J′, and of at most one partial block. The sum of a elements in a block is less then \frac{M}{|I|N}, or \frac{M}{N} in the last block. In a partial block, there is an initial element, and the first few compensating terms. So the sum in a partial block is less than max(max_{i∈I} |a_{i}|, |a_{l}|) ≤ max_{i≥N} |a_{i}|. This means

|∑_{i=K}^{L′} a_{π(i)}| ≤ |I|\frac{M}{|I|N} + \frac{M}{N} + max_{i≥N} |a_{i}| = \frac{2M}{N} + max_{i≥N} |a_{i}|.

After that we repeat this whole procedure on the remaining elements, using N = 2, 3, . . .. We will end up at a real permutation, because it uses every element exactly once.

In each step, the sum of the permutated b series decreases by at least M, so it will diverge to −∞. The sum of permutated a series is at most \frac{M}{N} at the end of step N, which converges to 0, and even when looking at the middle of a step, we make at most \frac{2M}{N} + max_{i≥N} |a_{i}| additional error, so in the end s(a, π) = 0.

This means π ∈ K(a) \ (b).

Now suppose the common limit point is c. Define b′ = b − ca. \frac{b′_{i}}{a_{i}} = \frac{b_{i}−ca_{i}}{a_{i}} = \frac{b_{i}}{a_{i}} − c, so switching b to b′ causes the limit point move to 0. From the previous case, there exists π ∈ K(a) \ K(b′). We only need π ∉ K(b), and this is true because πb = πb′+cπa is the sum of a series with convergent sum and one with divergent sum, so it must also have divergent sum.

The remaining case is when we have an infinite limit point, we may suppose this is ∞.

The construction will also be similar, only a bit simpler.

In step N, where N starts from 1, we choose the first unused element l ∈ ℕ. Let J be the indices used until now, and A be the sum of corresponding elements of a. Then using Lemma 3 we can get I ⊂ ℕ \ J such that

|A + ∑_{i∈I} a_{i}| < \frac{1}{N} ,

∑_{i∈I} b_{i} > M.

Now put l and then I to the end of the permutation.

It is again clear that all elements will be used exactly once. No matter how far we go, we can find an [N′, N′′] interval such that π([K, L]) is the I of a step. But

|∑_{i=K}^{L} b_{π(i)}| = |∑_{i∈I} b_{i}| > M.

Literally the oscillation will never go under M, so the permutated b series will diverge. The proof of convergence of the permutated a series is the same as in the previous case.

Finally, we could build permutations for all cases, so the proof is complete. □

This also gives an addition to the subject of [13]. There it is supposed

that we have a_{1}, a_{2}, …, a_{n} ∈ S such that ∑_{i=1}^{n} λ_{i}a_{i} ∈ S_{0} can only happen with

λ_{1} = λ_{2} = … = λ_{n} = 0. As stated there, if a_{n+1} ∈ S and there exist

λ_{1}, λ_{2}, …, λ_{n} such that a_{n+1} − ∑_{i=1}^{n} λ_{i}a_{i} ∈ S_{0}, then ⋂_{i=1}^{n} K(a_{i}) ⊆ K(a_{n+1}).

Now it turns out that ⋂_{i=1}^{n} K(a_{i}) = K(a_{n+1}) can not happen, this would

mean one nontrivial convergence set contains another, which is impossible.

We can prove an opposite statement telling that convergence sets cannot be completely different.

Theorem 5 If a^{i} ∈ S \ S_{0}, i ∈ ℕ, then

⋂_{i=1}^{∞} K(a^{i}) ⊋ P_{0}.

Proof:

⋂_{i=1}^{∞} K(a^{i}) ⊇ P_{0}.

is true because of the definition of P_{0}. We only have to find a π ∈ ⋂_{i=1}^{∞} K(a^{i}) \ P_{0}.

To satisfy π ∉ P_{0}, we have to provide for every n that there is an N such that π([1, N]) consists of at least n intervals. To achieve this, we will define π incrementally, fixing some elements for each n. We suppose after we are done with 2, 3, …, n − 1, we defined π on [1, N_{n−1}] such that π([1, N_{n−1}]) = [1, N_{n−1}].

We need π ∈ ⋂_{i=1}^{∞} K(a^{i}), so in the next step, we will ensure the first n series

to suffer only small changes regarding partial sums. We pick K > N_{n−1} such that

|a_{k}^{i}| < \frac{1}{n^{2}} , ∀1 ≤ i ≤ n, ∀k ≥ K.

Now let’s define

π(N_{n−1} + j) = K + 2j, 1 ≤ j ≤ n − 1,

and N_{n} to be K + 2n − 2. Define π on [N_{n−1} + n, N_{n}] to increasingly fill the remaining elements of [N_{n−1}, N_{n}]. After that we can start over the iteration.

We constructed π to avoid P_{0} as π([1, N_{n−1} + n − 1]) consists of exactly n intervals. We still have to check π ∈ K(a^{i}).

Let A^{i} = ∑_{j=1}^{∞} a_{j}^{i}, fix ε > 0. It is enough to prove the partial sum of the

rearranged series will approache A^{i} with at most ε error after a bound. Take

K such that whenever N > K, |A_{i} − ∑_{j=1}^{N} a_{j}^{i}| < \frac{ε}{2}. Choose n ≥ i to N_{n−1} > K

and n > \frac{2}{ε} to hold. This N_{n−1} will be the good bound for the rearranged series. To show this, suppose M > N_{n−1}. Then M ∈ [N_{n′−1}, N_{n′}] for some n′ ≥ n. From the construction π([1, M]) = [1, M′] ∪ H for an appropriate M′ ≥ N_{n′} and H ⊂ {K + 2, K + 4, …, K + 2n − 2}. This allows us to estimate the sum of the rearranged series.

|∑_{j=1}^{M} a_{π(j)}^{i} − A^{i}| ≤ |∑_{j=1}^{M} a_{π(j)}^{i} − ∑_{j=1}^{M′} a_{j}^{i}| + |∑_{j=1}^{M′} a_{j}^{i} − A^{i}| ≤

≤ | ∑_{j∈π([1,M])\[1,M′]} a_{j}^{i}| + \frac{ε}{2} < (n′ − 1) · \frac{1}{n^{′2}} + \frac{ε}{2} < ε.

This works for all i ∈ ℕ and ε > 0, so π ∈ K(a^{i}), which means π suffices our needs.

□

A natural continuation is to extend the previous theorems on S̃. However, only the first will remain true.

Theorem 6 Let a, b ∈ S̃ \ S_{0} nonequivalent series. In this case, K(a) \ K(b) ≠ ∅.

Proof: a can be rearranged to form a conditionally convergent series, choose such an adequate σ. This means σ ∈ K(a). If σ ∉ K(b), we are ready. If not, then σa, σb ∈ S \ S_{0} and are nonequivalent, so Theorem 4 can be used and provides a π ∈ K(σa) \ K(σb). Hence σπ ∈ K(a) \ K(b), we are ready.

□

Proposition 8 There is an H ⊂ S̃\S_{0}, |H| = c such that K(a)∩K(b) = ∅ for any a, b ∈ H, a ≠ b.

Proof: First we construct two series with disjoint convergence sets. An example is a_{n} = (−1)^{n} \frac{1}{n}, b_{2n} = \frac{2}{2n}, b_{2n+1} = −\frac{1}{2n+1}. In other words a_{n} = b_{n} if n is odd, and 2a_{n} = b_{n} is n is even. a, b ∈ S̃ \ S_{0} is obvious.

The idea is that even if positive and negative terms are balanced in the rearranged a series, the rearranged b has the positive terms doubled, and because positive and negative parts both cumulate to ∞, the sum of the rearranged b will go out to ∞.

Formally, pick any π ∈ P. Choose an arbitrary T. If n is large enough, π([1, n]) ⊃ [1, 2T].

∑_{i=1}^{n} b_{π(i)} − ∑_{i=1}^{n} a_{π(i)} = ∑_{{i∈[1,n] \atop π(i) is even}} \frac{1}{π(i)} ≥ ∑_{i=1}^{T} \frac{1}{2i} > \frac{log T − 1}{2}.

But for two series with convergent sums, this difference would remain bounded. Consequently π ∈ K(a) and π ∈ K(b) cannot happen simultaneously, by other words, K(a) ∩ K(b) = ∅.

We did use nothing special about the number 2 in the construction, only that it is different from 1. Therefore everything works for two series of the form a_{2n}^{r} = \frac{r}{2n}, a_{2n+1}^{r} = −\frac{1}{2n+1} with two different positive r. Consequently H = {a^{r} | r ∈ ℝ^{+}} is the set we searched for.

□

The construction in the proof can be generalized, we don’t need to start from the series (−1)^{n} \frac{1}{n}, any a^{1} ∈ S̃ \ S_{0} will suffice, we scale the positive elements the same way. This means we can even achieve a^{1} ∈ H for some prescribed a^{1}, in the end this shows K(a^{1}) is “small” in some sense.

An important consequence of the main theorem in this chapter is that nonequivalent series are really different when looking at possible convergencepreserving rearrangements. This will be the starting point of the next chapter.

Reconstruction of series

We turn to a natural problem about conditionally convergent series and permutations. The question is whether we can reconstruct the conditionally convergent series a given the information K(a).

Clearly we need to refine the problem. If a ≡ b, then K(a) = K(b), so it will be possible to reconstruct a only up to equivalency. On the other hand, Theorem 4 provides that nonequivalent series have different convergence sets, so this modified problem can be solved in theory.

Initially we make our work easier supposing we have some additional information.

We will need to combine multiple series. To describe a specific combination, we define indexing series. An acceptable indexing series to combine n series is an element of

K_{n} = {k = (k_{i})_{i=1}^{∞} ∈ [1, n]^{ℕ} | ∀i ∈ [1, n] ∥{j ∈ ℕ | k_{j} = i}∥ = ∞} .

Definition 4 Let us have series b^{1}, b^{2} …, b^{n}, and an indexing series k ∈ K_{n}. We define the slicing of these series b = k(b^{1}, b^{2}, …, b^{n}) recursively, let b_{i} be the first unused element of b^{k_{i}}.

Note that s(b) = ∑_{i=1}^{n} s(b^{i}) if all s(b^{i}) and their sum exists. If not, s(b)

may still exist. To be more precise,

∑_{j=1}^{N} b_{j} = ∑_{i=1}^{n} ( ∑_{j=1}^{|{k≤N:n_{k}=i}|} b_{j}^{i}) .

This means we add up the partial sums of the subseries, but the length of these partial sums might grow with different speed. These speed can be controlled via k. By fixing a slicing we mean fixing k, allowing b^{i} to be changed or rearranged.

First we try to determine the signs of the elements of a. For the same reason mentioned at the beginning of the chapter this is not possible, so instead we want to split a into two series a^{1} and a^{2} such that

∑_{j=1}^{∞} |a_{j}^{1}|^{+} = ∞, ∑_{j=1}^{∞} |a_{j}^{2}|^{+} < ∞,

∑_{j=1}^{∞} |a_{j}^{1}|^{−} < ∞, ∑_{j=1}^{∞} |a_{j}^{2}|^{−} = ∞,

or reversed. By other words, a^{1}, −a^{2} ∈ S̃^{+} \ S_{0}.

To do this, let us deduce some statements about splitting of a conditionally convergent series. In the following theorems, a^{1}, a^{2} will play the role of a splitting of a.

Proposition 9 a^{1}, −a^{2} ∈ S̃^{+}\S_{0} or reversed if and only if for any π_{1}, π_{2} ∈ P there exist k ∈ K_{2} slicing such that |s(k(π_{1}a^{1}, π_{2}a^{2}))| < ∞, but we can find ρ_{1}, ρ_{2} ∈ P and k′ ∈ K_{2} slicing that s(k′(ρ_{1}a^{1}, ρ_{2}a^{2})) does not exist or is infinite.

Proof:

Observe if a^{1}, a^{2} is a splitting of a, then this is a condition using only summability of some rearranged versions a.

If we have any type of summability, a_{j}^{1}, a_{j}^{2} will converge to 0, this is provided by both conditions, so we don’t have to care about it.

The easier part is if we have a^{1}, −a^{2} ∈ S̃^{+} \ S_{0}. This means a^{1} = b^{1} − c^{1}, a^{2} = c^{2} − b^{2}, where b^{1}, b^{2} ∈ S^{+}, c^{1}, c^{2} ∈ S_{0}^{+}. We described in the first chapter how to slice a positive and a negative series to have convergent sum. We have to do this for b^{1} = a^{1} + c^{1} and −b^{2} = a^{2} − c^{2}. We will get a slicing which is also good for a^{1} and a^{2} because the difference is only an absolutely convergent series. Of course the same thing works for π_{1}a^{1}, −π_{2}a^{2}.

If −a^{1}, a^{2} ∈ S̃^{+} \ S_{0}, we prove similarly.

For the reverse direction, let us suppose a^{1} ∈ S̃ \ S_{0}. We will show that the summability conditions will not hold. Choose an arbitrary π_{2} ∈ P.

If lim inf_{i→∞} ∑_{j=1}^{i} a_{π_{2}(j)}^{2} > −∞, then find a π_{1} ∈ P such that ∑_{j=1}^{i} a_{π_{1}(j)}^{1} = ∞.

Obviously any slicing of these rearranged series will sum up to ∞. We do the same but negated if the lim sup is less than ∞. The remaining case is when the lim inf and the lim sup are −∞ and ∞, respectively. We may then choose π_{1} ∈ K(a^{1}), clearly slicing this with π_{2}a^{2} will not change these values.

Now let us prove the reverse direction, suppose our summability conditions hold. A basic consequence is that the combined series, e.g. a = k(a^{1}, a^{2}) for some k ∈ K_{2} both has rearrangements which have convergent and which have divergent sum. So a ∈ S̃ \ S_{0}. Adding the previous observation we have

∑_{j=1}^{∞} |a_{j}^{1}|^{+} < ∞ or ∑_{j=1}^{∞} |a_{j}^{1}|^{−} < ∞,

∑_{j=1}^{∞} |a_{j}^{2}|^{+} < ∞ or ∑_{j=1}^{∞} |a_{j}^{2}|^{−} < ∞,

∑_{j=1}^{∞} |a_{j}|^{+} = ∞ and ∑_{j=1}^{∞} |a_{j}|^{−} = ∞.

This is true only in the cases which the proposition permits.

□

Proposition 10 a^{1}, a^{2} ∈ S̃ \ S_{0} if and only of there exists π_{1}, π_{2}, ρ_{1}, ρ_{2} ∈ P and k′, k′′ ∈ K_{2} slicing such that |s(k(π_{1}a^{1}, π_{2}a^{2}))| < ∞ for any k ∈ K_{2} slicing, but s(k′(ρ_{1}a^{1}, π_{2}a^{2})) and s(k′′(π_{1}a^{1}, ρ_{2}a^{2})) are infinite or does not exist.

Proof:

Suppose a^{1}, a^{2} ∈ S̃\S_{0}. Take π_{1} ∈ K(a^{1}), π_{2} ∈ K(a^{2}), ρ_{1} ∉ K(a^{1}), ρ_{2} ∉ K(a^{2}). In this case s(k(π_{1}a^{1}, π_{2}a^{2})) = s(a^{1}, π_{1}) + s(a^{2}, π_{2}), so it is finite. On the other hand, the combination of a convergent and a divergent series is divergent, and ρ_{1}a^{1} and ρ_{2}a^{2} are divergent, so we are done using arbitrary k′, k′′.

Now suppose the conditions hold. We will show |s(a^{1}, π_{1})|, |s(a^{2}, π_{2})| < ∞. If this is not true, we will be able to construct a slicing so that the sum will not be finite. For π_{1}a^{1} there are four possibilities, the sum might be ∞, −∞, the lim inf and lim sup might differ, or the sum exists and is finite. We will choose an n_{1}^{1} ≤ n_{2}^{1} ≤ …, lim_{i→∞} n_{i}^{1} = ∞, similarly (n_{i}^{2})_{i=1}^{∞} showing how long k should be 1, or 2.

In the first case, choose it to suffice ∑_{j=1}^{n_{i}^{1}}(π_{1}a^{1})_{j} > 3i.

In the second case to satisfy −i > ∑_{j=1}^{n_{i}^{1}}(π_{1}a^{1})_{j} > −2i.

In the third case, such that lim sup_{i→∞} ∑_{j=1}^{n_{2i}^{1}}(π_{1}a^{1})_{j} < lim inf_{i→∞} ∑_{j=1}^{n_{2i−1}^{1}}(π_{1}a^{1})_{j}.

In the fourth case, n_{i}^{1} can be arbitrary. Clearly all these n_{i}^{1} exist if i is large enough.

We generate n_{i}^{2} for a^{2} the same way. Now define k starting by n_{1}^{1} elements 1, then n_{1}^{2} elements 2, then n_{2}^{1} − n_{1}^{1} times 1 and so on. Now for the merged series a = k(π_{1}a^{1}, π_{2}a^{2}),

∑_{j=1}^{n_{i}^{1}+n_{i}^{2}} a_{j} = ∑_{j=1}^{n_{i}^{1}}(π_{1}a^{1})_{j} + ∑_{j=1}^{n_{i}^{2}}(π_{2}a^{2})_{j}.

It is easy to verify that this diverges to ∞ if the first case occur for one of the series. If not, it will diverge to −∞ if the second case happen at least once. Otherwise if the third case is true for a series we will still end up at a series with divergent sum. In order to avoid contradiction with the condition, the fourth case has to happen for both series, so both s(a^{1}, π_{1}) and s(a^{2}, π_{2}) must be finite.

On the other hand, if |s(a^{1}, ρ_{1})| < ∞, then s(k′(ρ_{1}a^{1}, π_{2}a^{2})) must be finite for any k′ ∈ K_{2} slicing. This is not the case, consequently s(a^{1}, ρ_{1}) does not exist or is not finite. We got two rearrangements of a_{1}, one with the sum converging, one without, this can only happen to a series in S̃ \ S_{0}. We can say the same about a_{2}.

□

Proposition 11 Supposing k′(a^{1}, a^{2}) ∈ S̃\S_{0}, a^{1} ∈ S_{0} if and only of there is π_{2} ∈ P and k ∈ K_{2} slicing such that |s(k(π_{1}a^{1}, π_{2}a^{2}))| < ∞ for any π_{1} ∈ P.

Proof: Suppose a^{1} ∈ S_{0}, this implies a^{2} ∈ S̃ \ S_{0}, and we can choose π_{2} ∈ K(a^{2}). Now for any π_{1} ∈ P = K(a^{1}) and k ∈ K_{2} |s(k(π_{1}a^{1}, π_{2}a^{2}))| < ∞ because we merged to series with convergent sums.

For the reverse direction, choose π_{1}, π′_{1} ∈ P.

s(k(π_{1}a^{1}, π_{2}a^{2})) − s(k(π′_{1}a^{1}, π_{2}a^{2})) = s(k(π_{1}a^{1} − π′_{1}a^{1}, π_{2}a^{2} − π_{2}a^{2})) =

= s(π_{1}a^{1} − π′_{1}a^{1}),

and the condition tells this sum is finite. If a^{1} ∈ S̃ \ S_{0}, then the choice π_{1} ∈ K(a^{1}), π′_{1} ∈ P \K(a^{1}) would lead to contradiction. If a^{1} ∈ S̃^{+}\S_{0}, then

for n_{1} large enough, ∑_{i=1}^{n_{1}} a_{i}^{1} > 10. If m_{1} is large enough, ∑_{i=m_{1}}^{m_{1}+n_{1}−1} a_{i}^{1} < 1. Let us define the first elements of π_{1} to be 1, 2, …, n_{1}, m_{1}, m_{1}+1, …, n_{1}+m_{1}−1, and the start of π′_{1} to be m_{1}, m_{1} + 1, …, n_{1} + m_{1} − 1, 1, 2, …, n_{1}. Therefore

∑_{i=1}^{n_{1}}(a_{π_{1}(i)}^{1} − a_{π′_{1}(i)}^{1}) > 9,

∑_{i=1}^{2n_{1}}(a_{π_{1}(i)}^{1} − a_{π′_{1}(i)}^{1}) = 0.

By words, the sum oscillates at least 9 between 0 and 2n_{1}. Now throw away used elements, and do this again to extend π_{1} and π′_{1}. The oscillation will reach 9 from time to time, so the sum will not converge, and this is contradiction. To sum up a^{1} ∉ S̃ \ S_{0}, a^{1} ∉ S̃^{+} \ S_{0}, similarly −a^{1} ∉ S̃^{+} \ S_{0}, only a^{1} ∈ S_{0} remained as a possibility.

□

The last three propositions made it possible to detect what type of series did we get at an arbitrary splitting. Take a decomposition of a to a^{1} and a^{2} for which a^{1}, −a^{2} ∈ S̃^{+} \ S_{0}. Adding an absolutely convergent series does not change anything, so we may suppose a^{1}, −a^{2} ∈ S^{+} \ S_{0}. By previous observations, for N ⊂ ℕ, N ∈ H_{a^{1}} exactly if there is a π ∈ K(a) where a^{1} is segmented as described by N. Consequently we can determine H_{a^{1}} and H_{a^{2}}, and this allows to get some bounds by comparing these with H_{b} for known b ∈ S^{+} \ S_{0}.

Although the problem of complete reconstruction remained unsolved, we could filter out some well formed numerical information by only knowing the convergence sets.

Random permutations

We can go one step further and ask whether a random permutation preserves the convergence of a conditionally convergent series. Suppose we work on a probability space (Ω, M, P).

The first problem we have to face is how to generate infinite random permutations. There might be several possibilities, however we will choose one which is natural in some sense.

We will start with a nontrivial distributions on ℕ, an element of:

Q = {Q = {q_{i}}_{i=1}^{∞} ∈ [0, 1]^{ℕ} | ∀i ∈ ℕ q_{i} > 0, ∑_{i=1}^{∞} q_{i} = 1} .

Definition 5 Suppose there is a Q ∈ Q distribution given on ℕ. Take X_{1}, X_{2}, … IID random variables with distribution Q. For any ω ∈ Ω, define Π_{Q} permutation as follows:

Π_{Q}(n) = min{X_{k} : |{X_{1}, X_{2}, …, X_{l}}| = n}.

By words, we write X_{1} to the first position of Π_{Q}, then we examine X_{2}, X_{3}, …, and write it to the next position of Π_{Q} if we find a new value. We need this is almost surely a permutation.

Proposition 12 If Q ∈ Q, then P (Π_{Q} ∈ P) = 1.

Proof: When Π_{Q}(n) = m, then in the definition above, X_{k} = m, but m ∉ {X_{1}, X_{2}, …, X_{k−1}} otherwise k would not be minimal. This also implies m ∉ Π_{Q}([1, n − 1]), so Π_{Q} is injective always. We miss an m value in the permutation only if none of X_{k} are m. Using Q(m) > 0, this happens with 0 probability. Thus the event that Π_{Q} is not surjective is the union of countably many zero-probability events, so we avoid it almost surely.

□

Our goal is to tell something about Π_{Q} ∈ K(a) for a specific a and about Π_{Q} ∈ P_{0}. We must start with ensuring measurability. Let’s fix the Q distribution for now.

Proposition 13 Fix a ∈ S̃. Then {Π_{Q} ∈ K(a)} is measurable with respect to F = σ(X_{1}, X_{2}, …).

Proof: Let’s define S_{n,m} = ∑_{i=n}^{m} a_{Π_{Q}(i)}, for n ≤ m. Π_{Q}(i) is measurable, so is

a_{Π_{Q}(i)}, and also a deterministic finite sum of these, namely S_{n,m}. Let’s focus on Cauchy-convergence. The event “N is a good bound for ε” can be written

as A_{N,ε} = ⋂_{n=N}^{∞} ⋂_{m=n}^{∞}{|S_{n,m}| < ε}, this does not take out from F. In the end,

the sum exist exactly on the set ⋂_{k=1}^{∞} ⋃_{N=1}^{∞} A_{N, \frac{1}{k}} ∈ F.

□

Proposition 14 {Π_{Q} ∈ P_{0}} is measurable with respect to F.

Proof: We will check the property found at Theorem 3. {Π_{Q}([1, n]) = A} for some A ⊂ ℕ is measurable, taking the union of countable many of these gives the measurable set A_{n,K} = {Π_{Q}([1, n]) consists of at most K intervals}.

Using the theorem {Π_{Q} ∈ P_{0}} = ⋃_{K=1}^{∞} ⋂_{n=1}^{∞} A_{n,K} ∈ F.

□

Proposition 15 For any a ∈ S̃, P (Π_{Q} ∈ K(a)), P (Π_{Q} ∈ P_{0}) are all 0 or 1.

Proof: This is an application of the Kolmogorov 0-1 rule. To use it, we have to show that {Π_{Q} ∈ K(a)}, {Π_{Q} ∈ P_{0}} ∈ G_{n}, where G_{n} = σ(X_{n+1}, X_{n+2}, …). Generate the permutation Π′_{Q} the same way as Π_{Q} from X_{n+1}, X_{n+2}, … setting X_{1} = X_{2} = … = X_{n} = X_{n+1}. {Π′_{Q} ∈ K(a)}, {Π′_{Q} ∈ P_{0}} are clearly G_{n} measurable. The point is that

Π′_{Q} ∈ K(a) ⇔ Π_{Q} ∈ K(a),
Π′_{Q} ∈ P_{0} ⇔ Π_{Q} ∈ P_{0}.

We have to check what does happen to Π′_{Q} when we put the real randomized values to X_{1}, X_{2}, …, X_{n}. It can be seen that nothing will change above max(Π_{Q}^{′−1}({X_{1}, X_{2}, …, X_{n+1}})). In the end we modified only finite elements of the permutation which does not effect convergence preserving in any sense. So {Π_{Q} ∈ K(a)}, {Π_{Q} ∈ P_{0}} are identical to G_{n} measurable sets, and this is enough to conclude.

□

In the next few theorems, we will show that this construction of random permutation is meaningful, both 0 and 1 will appear in the previous proposition. Notice that when looking at one a and one Q = {q_{1}, q_{2}, …}, we end up at exactly the same situation if we apply the same rearrangement on {a_{1}, a_{2}, …} and {q_{1}, q_{2}, …}. In fact we don’t need the order of the elements of a, only that which probability and element belong together.

Theorem 7 For any Q ∈ Q there exists an a ∈ S̃ such that P (Π_{Q} ∈ K(a)) = 0.

Moreover, we may prescribe the sign of elements of a, and even fix the values of the positive elements.

Proof: Denote the probabilities of positive elements by q′_{1}, q′_{2}, …, that of negative elements by q′′_{1}, q′′_{2}, …, the values of positive elements by a′_{1}, a′_{2}, …, their indices by j′_{1}, j′_{2}, . . .. We may suppose the series a′_{i} is nonincreasing.

The idea is to choose negative elements to be very small. So small that if big positive elements appear often enough, they cannot compensate and the sum goes to ∞.

When want to estimate the probability of a′_{1} to be in the first n_{1} elements of the rearranged series. This surely happens if j′_{1} appears within X_{1}, X_{2}, … X_{n_{1}}. The probability of this is 1 − (1 − q′_{1})^{n_{1}}. This can be done for multiple terms: a′_{2} will be among the first n_{1} + n_{2} elements, if j′_{2} appears within X_{n_{1}+1}, X_{n_{1}+2}, …, X_{n_{1}+n_{2}}. Even this is independent from what happens on X_{1}, X_{2}, …, X_{n_{1}}.

Therefore for any n_{1}, n_{2}, … the probability of j′_{i} ∈ Π_{Q}([1, n_{1}+n_{2}+…+n_{i}]) for all i is at least

∏_{i=1}^{∞}(1 − (1 − q′_{i})^{n_{i}}).

Let’s call this event A.

Let’s choose n_{1} ≤ n_{2} ≤ … such that this infinite product is positive, this can be easily achieved. This is what we call positive elements appear “often enough”.

To assign the values of negative elements, let’s choose a series b_{1}, b_{2}, … such that

∀i ∈ ℕ a′_{i} > b_{i} ≥ b_{i+1} > 0,

∑_{i=1}^{∞} b_{i} = ∑_{i=1}^{∞}(a′_{i} − b_{i}) = ∞.

This implies ∀i ∈ ℕ \frac{b_{i}}{n_{i}−1} ≥ \frac{b_{i+1}}{n_{i+1}−1}.

Now we can define the negative elements of a, for every i let’s have n_{i} − 1 pieces of −\frac{b_{i}}{n_{i}−1}.

When A holds, we want a lower bound on ∑_{i=1}^{n_{1}+...+n_{k}} a_{Π_{Q}(i)}. A means we can

find a′_{1}, a′_{2}, …, a′_{k} among the summands. In the worst case, the remaining terms are the possible smallest negative elements. These are n_{i} − 1 times −\frac{b_{i}}{n_{i}−1} for 1 ≤ i ≤ k. The resulting lower bound is

∑_{i=1}^{k} a′_{i} − ∑_{i=1}^{k}(n_{i} − 1)\frac{b_{i}}{n_{i} − 1} = ∑_{i=1}^{k}(a′_{i} − b_{i}) → ∞

if k → ∞, so on A, ∑_{i=1}^{∞} a_{Π_{Q}(i)} will diverge.

Using P (A) > 0, we can say P (Π_{Q} ∈ K(a)) < 1, but using Proposition 15 P (Π_{Q} ∈ K(a)) can only be 0.

□

Corollary 1 There exists Q ∈ Q so that P (Π_{Q} ∈ P_{0}) = 0. Moreover, any Q′ can be transformed to such a Q by permutating the appearing probabilities.

Proof: Fix any Q′ ∈ Q, and get an a ∈ S̃ from the theorem. Choose a π ∈ K(a), and define Q = πQ′. The meaning of πQ′ is obvious, we can look at Q′ as an element of S_{0}.

P (Π_{Q′} ∈ K(a)) = P (Π_{Q} ∈ K(πa)) = 0, but P_{0} ⊂ K(πa), so our claim holds.

□

To deal with the opposite question, we will have to work with the information when we first take an element of a fixed set.

Definition 6 Given an A ⊂ ℕ, define the following stopping times:

τ_{A}^{∗} = min{n : X_{n} ∈ A},
τ_{A} = min{n : Π_{Q}(n) ∈ A}.

The fact that P (Π_{Q} ∈ P) = 1 implies that τ_{A}^{∗}, τ_{A} will be finite for any nonempty A with probability one.

Lemma 4 Take a Q ∈ Q and nonempty A_{1}, A_{2} ⊂ ℕ, A_{1} ∩ A_{2} = ∅. Then

P (τ_{A_{1}} < τ_{A_{2}}) = \frac{Q(A_{1})}{Q(A_{1}) + Q(A_{2})}.

Moreover, a similar version is true for multiple terms. Take disjoint nonempty A_{1}, A_{2}, … A_{n} ⊂ ℕ. In this case,

P (τ_{A_{1}} < τ_{A_{2}} < … < τ_{A_{n}}) = ∏_{i=1}^{n−1} \frac{Q(A_{i})}{∑_{j=i}^{n} Q(A_{j})}.

Proof: The second statement contains the first, so it is enough to prove the second one. It is equivalent to compare τ_{A_{i}}^{∗} instead of τ_{A_{i}}, and it will be easier

to handle. Let us define B_{i} = ℕ \ ⋃_{j=i}^{n} A_{j} The event {τ_{A_{1}}^{∗} < τ_{A_{2}}^{∗} < … < τ_{A_{n}^{∗}}}

means that there are k_{1}, k_{2}, …, k_{n−1} positive integers, l_{1} = 0, l_{i} = ∑_{j=1}^{i−1} k_{j} for

1 ≤ i ≤ n − 1, which may depend on ω ∈ Ω of course, such that

X_{l_{1}+1}, X_{l_{1}+2}, …, X_{l_{1}+k_{1}−1} ∈ B_{1}, X_{l_{1}+k_{1}} ∈ A_{1}
X_{l_{2}+1}, X_{l_{2}+2}, …, X_{l_{2}+k_{2}−1} ∈ B_{2}, X_{l_{2}+k_{2}} ∈ A_{2}
X_{l_{n−1}+1}, X_{l_{n−1}+2}, …, X_{l_{n−1}+k_{n−1}−1} ∈ B_{n−1}, X_{l_{n−1}+k_{n−1}} ∈ A_{n−1}

Clearly these are disjoint events for different (k_{1}, k_{2}, …, k_{n−1}), and they cover {τ_{A_{1}}^{∗} < τ_{A_{2}}^{∗} < … < τ_{A_{n}^{∗}}}. The probability of this event for fixed (k_{1}, k_{2}, …, k_{n−1})

∏_{i=1}^{n−1} Q(B_{i})^{k_{i}−1}Q(A_{i}).

To get P (τ_{A_{1}}^{∗} < τ_{A_{2}}^{∗} < … < τ_{A_{n}}^{∗}) we have to sum up the previous expression for every (k_{1}, k_{2}, …, k_{n−1}) ∈ ℕ^{n−1}. This will be

∏_{i=1}^{n−1} (∑_{k_{i}=1}^{∞} Q(B_{i})^{k_{i}−1}Q(A_{i})) = ∏_{i=1}^{n−1} (Q(A_{i})\frac{1}{1 − Q(B_{i})}) .

By the definition of B_{i}, 1 − Q(B_{i}) = Q (⋃_{j=i}^{n} A_{j}) = ∑_{j=i}^{n} Q(A_{j}). Writing this

back to the previous expression results in the claim of the lemma. □ Theorem 8 For any Q ∈ Q there exists an a ∈ S \ S_{0} such that P (Π_{Q} ∈ K(a)) = 1.

Moreover, we can find a in the form a = πb, where b ∈ S̃ \ S_{0} is fixed, and we require π ∈ K(b).

Proof: Let’s choose an arbitrary c ∈ S \ S_{0}. First we will create a c′ series by inserting a lot of zeros into c. The idea is to do this in such a way that Π_{Q} retains the order of original nonzero elements with positive probability, therefore convergence remains.

Define n_{1} = 1, and choose n_{2} to meet \frac{q_{1}}{q_{1} + ∑_{i=n_{2}}^{∞} q_{i}} > \frac{1}{2}. By Lemma

4, 1 will appear before any n_{2}, n_{2} + 1, … in Π_{Q} with probability at least \frac{1}{2}.

To continue, choose n_{3} to satisfy \frac{q_{1}}{q_{1} + ∑_{i=n_{2}}^{∞} q_{i}} · \frac{q_{n_{2}}}{q_{n_{2}} + ∑_{i=n_{3}}^{∞} q_{i}} > \frac{1}{2}. Using

Lemma 4,

P (τ_{\{n_{1}\}} < τ_{\{n_{2}\}} < τ_{[n_{3},∞)}) = \frac{q_{1}}{q_{1} + q_{n_{2}} + ∑_{i=n_{3}}^{∞} q_{i}} · \frac{q_{n_{2}}}{q_{n_{2}} + ∑_{i=n_{3}}^{∞} q_{i}} >

> \frac{q_{1}}{q_{1} + ∑_{i=n_{2}}^{∞} q_{i}} · \frac{q_{n_{2}}}{q_{n_{2}} + ∑_{i=n_{3}}^{∞} q_{i}} > \frac{1}{2}.

We can repeat this infinitely which means we end up at n_{1} < n_{2} < … such that

P (τ_{\{n_{1}\}} < τ_{\{n_{2}\}} < … < τ_{\{n_{k−1}\}} < τ_{[n_{k},∞)}) > \frac{1}{2}

for any k ≥ 2. These are monotonically decreasing events in k, so for their intersection

P (τ_{\{n_{1}\}} < τ_{\{n_{2}\}} < …) ≥ \frac{1}{2}.

Let c′_{n_{i}} = c_{i}, and set all other elements of c′ to 0. With probability at least \frac{1}{2}, Π_{Q}c′ will contain the elements c_{i} in the original order, only the zeros will be mixed. This implies the sum will converge, therefore P (Π_{Q} ∈ K(c′)) ≥ \frac{1}{2}, in fact it is 1 by Proposition 15.

This is enough to prove the first part of the theorem, however we have to slightly modify it to get the second part. Take a b ∈ S̃ \ S_{0}. lim_{n→∞} b_{n} = 0 means we can take a b′′ ∈ S_{0} subseries of b. Rearrange the remaining elements to form a conditionally convergent series b′. Perform the previous construction with b′ playing the role of c, and insert the elements of b′′ instead of zeros. The construction still works because mixing an absolutely convergent series is irrelevant to the sum as much as mixing zeros in the original case. Moreover, now this is the rearrangement of b, a = πb. a has convergent sum, so π ∈ K(b) holds.

□ Corollary 2 For any a ∈ S̃\S_{0} there is a Q ∈ Q satisfying P (Π_{Q} ∈ K(a)) = 1.

Moreover, we can find Q as a rearrangement Q = πQ^{∗}, where Q^{∗} ∈ Q is fixed.

Proof: Use the stronger statement of the previous theorem on Q^{∗} and a. We obtain π such that P (Π_{Q^{∗}} ∈ K(πa)) = 1. But from a previous remark, Π_{Q^{∗}} ∈ K(πa) ⇔ Π_{π^{−1}Q^{∗}} ∈ K(π^{−1}πa), so for Q = π^{−1}Q^{∗} the claim will hold.

□

The same idea can give us even more:

Theorem 9 There exists Q ∈ Q with the property P (Π_{Q} ∈ P_{0}) = 1.

Proof: We will provide the identic permutation occurs with positive probability, that is

P (τ_{{1}} < τ_{{2}} < …) > 0.

Using the same train of thought, it is enough to suffice

\frac{n}{2n − 1} · \frac{q_{n}}{∑_{i=n}^{∞} q_{i}} ≥ \frac{n + 1}{2n + 1}

for all positive n. Equivalently

q_{n} ≥ \frac{2n^{2} + n + 1}{2n^{2} + n} ∑_{i=n}^{∞} q_{i} = (1 − \frac{1}{2n^{2} + n}) ∑_{i=n}^{∞} q_{i},

∑_{i=n+1}^{∞} q_{i} ≤ \frac{1}{2n^{2} + n} ∑_{i=n}^{∞} q_{i}.

This clearly holds if we achieve ∑_{i=n}^{∞} q_{i} = \frac{1}{3^{n−1}(n − 1)!^{2}} , so let us define

q_{n} = \frac{1}{3^{n−1}(n − 1)!^{2}} − \frac{1}{3^{n}n!^{2}} = \frac{3n^{2} − 1}{3^{n}n!^{2}}.

□

By a simple transformation, we can get more convergence-preserving random permutations for a specific series.

Proposition 16 For any choice of Q ∈ Q and π ∈ P, πΠ_{Q} and Π_{πQ} have the same distribution.

Proof: Suppose we generate Π_{Q} using the independent random variables X_{1}, X_{2}, …, which all have distribution Q. In this case, πX_{i} have distribution πQ, and are still independent, so they generate Π′_{πQ} which has the distribution of Π_{πQ}.

On the other hand, if we look at the definition of Π_{Q}, and put πX_{i} instead of X_{i}, the resulting permutation will be πΠ_{Q}. This finishes the proof.

□

If P (Π_{Q} ∈ P_{0}) = 1 for some Q ∈ Q, and π ∈ K(a) for some a ∈ S̃ \ S_{0}, then obviously Π_{Q}(πa) = (πΠ_{Q})a has convergent sum with probability 1. According to the previous Proposition, we may rephrase this as P (Π_{πQ} ∈ K(a)) = 1.

Although we generated a family of distributions which give convergencepreserving permutation for an a ∈ S̃\S_{0}, it is still not clear how to determine P (Π_{Q} ∈ K(a)) for general Q and a. Even if this is complex to solve, it would be interesting to know the answer for special distributions, eg. geometric distribution (which will not be in P_{0} almost surely) or Poisson distribution.

Another direction for going forward is to introduce other methods of generating random permutations.

An interesting method generates the same distribution class, therefore it gives a new sight on it. Let X_{n} be a q_{n}-exponential random variable, choose them to be independent. The permutation will be the ordering of X_{n}.

A direct generalization of the one we investigated until now is when we have a Q_{H} : 2^{H} → [0, 1] distribution on every infinite subset of ℕ, and we choose one element after another, always independently from previous choices by the distribution of the remaining elements. The situation we worked with until now fits this scheme with Q_{H}(i) = \frac{q_{i}}{∑_{j∈H} q_{j}} if i ∈ H.

A different possibility is to fix a series σ_{1}^{2}, σ_{2}^{2}, …, and take independent normal variables X_{n} ∼ N(n, σ_{n}^{2}). The permutation will be defined by the ordering of X_{n}.

Final words

We observed the diversity of problems originating from Riemann’s observation. Let us conclude with two more questions not covered here.

We can use randomness not only when choosing the permutation, but when taking a series. These two might be independent or not, one example on the latter one is the extension of the exponentially generated permutation defined in the previous chapter. The point is that we may choose the value a_{n} independently by a distribution F_{x}, this depends only on the value x taken by X_{n}.

To find new problems we can exchange summability to some other asymptotic property, which depends on the order of elements. An example for that is to require

lim_{n→∞} \frac{∑_{j=1}^{n} a_{j}}{n}

to exist. From this point we can do the similar analysis on permutations which does or does not preserve this property.

Bibliography

[1] R. P. Agnew, Permutations preserving convergence of series, Proceedings of the American Mathematical Society, 6 (1955), pp. 563–564.

[2] U. Elias, Rearrangement of a conditionally convergent series, American Mathematical Monthly, 110 (2003), p. 57.

[3] G. Giorgobiani, Convergent rearrangements of series of vector-valued functions, Georgian Mathematical Journal, 7 (2000), pp. 43–51.

[4] U. Guha, On Levi’s theorem on rearrangements of convergent series, Indian Journal of Mathematics, 9 (1967), pp. 91–93.

[5] F. Levi, Rearrangement of convergent series, Duke Mathematical Journal, 13 (1946), pp. 579–585.

[6] E. M. Nikishin, On convergent rearrangements of function series, Mat. Zametki, 1 (1967), pp. 129–136.

[7] P. Pleasants, Rearrangemenys that preserve convergence, Journal of the London Mathematical Society, 15 (1977), pp. 134–142.

[8] P. Schaefer, Sum-preserving rearrangements of infinite series, American Mathematical Monthly, 88 (1981), pp. 33–40.

[9] J. H. Smith, Rearrangements of conditionally convergent real series with preassigned cycle type, Proceedings of the American Mathematical Society, 47 (1975), pp. 167–170.

[10] J. R. Stefánsson, Forward shifts and backward shifts in a rearrangement of a conditionally convergent series, American Mathematical Monthly, 111 (2004), pp. 913–914.

[11] G. S. Stoller, The convergence-preserving rearrangements of real infinite series, Pacific Journal of Mathematics, 73 (1977), pp. 227–231.

[12] Q. F. Stout, On Levi’s duality between permutations and convergent series, Journal of the London Mathematical Society, 34 (1986), pp. 67– 80.

[13] G. Tusnády, On rearrangements of infinite series, Annales Universitatis Scientiarum Budapestinensis de Rolando Eotvos Nominatae, 9 (1966), pp. 121–126.