Convergent subseries of divergent series

Let $\mathscr{X}$ be the set of positive real sequences $x=(x_n)$ such that the series $\sum_n x_n$ is divergent. For each $x \in \mathscr{X}$, let $\mathcal{I}_x$ be the collection of all $A\subseteq \mathbf{N}$ such that the subseries $\sum_{n \in A}x_n$ is convergent. Moreover, let $\mathscr{A}$ be the set of sequences $x \in \mathscr{X}$ such that $\lim_n x_n=0$ and $\mathcal{I}_x\neq \mathcal{I}_y$ for all sequences $y=(y_n) \in \mathscr{X}$ with $\liminf_n y_{n+1}/y_n>0$. We show that $\mathscr{A}$ is comeager and that contains uncountably many sequences $x$ which generate pairwise nonisomorphic ideals $\mathcal{I}_x$. This answers, in particular, an open question recently posed by M. Filipczak and G. Horbaczewska.


Introduction
Let X be the set of positive real sequences x = (x n ) with divergent series n x n . For each x ∈ X , let I x the collection of sets of positive integers A such that the (possibly finite) subseries indexed by A is convergent, that is, Note that each I x is closed under finite unions and subsets, i.e., it is an ideal. Moreover, it contains the collection Fin of finite sets A ⊆ N, and and it is different from the power set P(N). Following [4], a collection of sets of the type (1) is called summable ideal. It is not difficult to see that every infinite set of N contains an infinite subset in I x if and only if lim n x n = 0. Accordingly, define Z := X ∩ c 0 = {x ∈ X : lim n→∞ x n = 0}.
It is known that the families I x defined in (1) are "small", both in the measure-theoretic sense and the categorical sense, meaning that "almost all" subseries diverge, see [3,6,13,16]. Related results in the context of filter convergence have been given in [1,2,10]. The set of limits of convergent subseries of a given series n x n , which is usually called "achievement set", has been studied in [7,9,12]. Of special interest have been specific subseries of the harmonic series n Question 1.1. Is it true that for each x ∈ Z there exists y ∈ X such that I x = I y and ∀n ∈ N, y n+1 y n ≥ n n + 2 ?
We show in Theorem 1.3 below that the answer is negative in a strong sense. To this aim, define Y := y ∈ X : lim inf n→∞ y n+1 y n > 0 , and let ∼ be the equivalence relation on X so that two sequences are identified if they generate the same ideal, so that First, we show that the set of pairs (x, y) ∈ X 2 such that x is ∼-equivalent to y is topologically well behaved: Then, we answer Question 1.1 by showing that: In light of the explicit example which will be given in the proof of Theorem 1.3, one may ask about the topological largeness of the set of such sequences. To be precise, is it true that is a set of second Baire category, i.e., not topologically small? Note that the question is really meaningful since Z is completely metrizable (hence by Baire's category theorem Z is not meager in itself): this follows by Alexandrov's theorem [8,Theorem 3.11] and the fact that {x ∈ c 0 : x 1 + · · · + x k > m} is a G δ -subset of the Polish space c 0 . With the premises, we show that A is comeager, that is, Z \ A is a set of first Baire category: Theorem 1.4. A is comeager in Z . In particular, A is uncountable.
We remark that Theorem 1.4 gives an additional information on relation ∼. Since it is a coanalytic equivalence relation by Proposition 1.2, we can appeal to the deep theorem by Silver [8,Theorem 35.20] which states that every coanalytic equivalence relation on a Polish space either has countably many equivalence classes or there is a perfect set consisting of non-equivalent pairs. Thanks to Theorem 1.4, the latter holds for the relation ∼ in a strong form. Indeed, every pair in A × Y does not belong to ∼, where A is comeager (hence it contains a G δ -comeager subset) and Y is an uncountable F σ -set. Therefore A × Y contains a product of two perfect sets by [8,Theorem 13.6].
Lastly, on a similar direction, we strenghten the fact that A is uncountable by proving that exist uncountably many sequences in A which generate pairwise nonisomorphic ideals (here, recall that two ideals I, J are isomorphic if there exists a bijection f : N → N such that f [A] ∈ I if and only if A ∈ J for all A ⊆ N). Theorem 1.5. There are c sequences in A which generate pairwise nonisomorphic ideals.
Hereafter, we use the convention that n≥1 a n ≪ n≥1 b n , with each a n , b n > 0, is a shorthand for the existence of C > 0 such that n≤k a n ≤ C n≤k b n for all k ∈ N.

Proof of Proposition 1.2
Equiavalently, we have to show that the set E : and, similarly, is the intersection of an F σ -set and a G δ -set, hence it is Borel. Analogously, E 2 is Borel. This proves that E is analytic subset of X 2 .

Proof of Theorem 1.3
Define the sequence x = (x n ) so that x n = 1 n if n is even and x n = 1 n log(n+1) if n is odd. Note that lim n x n = 0 and that n x n = ∞, hence x ∈ Z . At this point, fix y ∈ Y such that κ := lim inf n y n+1 /y n > 0 and let us show that I x = I y .
Let P be the set of prime numbers, with increasing enumeration (p n ). By the prime number theorem we have p n is asymptotically equal to n log(n) as n → ∞, hence with the consequence that P ∈ I x . In addition, P − 1 / ∈ I x because n∈P−1 Lastly, suppose for the sake of contradiction that I x = I y . Then we should have that P ∈ I y and, at the same time, P − 1 / ∈ I y . The latter means that n≥1 y pn−1 = ∞.
However, this implies that contradicting that P ∈ I y .

Proof of Theorem 1.4
Consider the Banach-Mazur game defined as follows: Players I and II choose alternatively nonempty open subsets of Z as a nonincreasing chain U 1 ⊇ V 1 ⊇ U 2 ⊇ V 2 ⊇ · · · , where Player I chooses the sets U m . Player II has a winning strategy if m V m ⊆ A . Thanks to [8,Theorem 8.33], Player II has a winning strategy if and only if A is comeager. Hence, the rest of the proof consists in showing that Player II has a winning strategy.
Note that the open neighborhood of a sequence x ∈ Z with radius ε > 0 satisfies For each m ∈ N, suppose that the nonempty open set U m has been fixed by Player I. Hence, U m contains an open ball B εm (x (m) ), for some x (m) ∈ Z and ε m > 0. In particular, thanks to (3), there exists a sufficiently large integer k 0 = k 0 (x (m) , ε m ) ∈ N such that y n < ε /2 for all y ∈ W εm (x (m) ) and n ≥ k 0 . Without loss of generality, let us suppose that k 0 is even. At this point, let x ⋆ be the sequence in A defined in the proof of Theorem 1.3. Then, for each m ∈ N, let t m be an integer such that max{x ⋆ pm , x ⋆ pm−1 } < t m · εm 2 (we recall that p m stands for the mth prime number), and define the positive real Lastly, set V m := B δm (z (m) ) and note that by construction hence V m is a nonempty open set contained in U m . In addition, the sequence of centers (z (m) ) is a Cauchy sequence in the complete metric space Z . Hence it is convergent to some z ∈ Z and it is straighforward to see that {z} = m V m .
To complete the proof, we need to show that z ∈ A . Set A := ( m I m ) \ 2N. Proceeding as in the proof of Theorem 1.3, we see that Now, fix y ∈ Y such that κ := lim inf n y n+1 /y n > 0 and suppose that I z = I y . Then we would have that A ∈ I y and A − 1 / ∈ I y , which is impossible reasoning as in (2).

Proof of Theorem 1.5
Let x ⋆ be the sequence defined in the proof of Theorem 1.3. For each r ∈ (0, 1], let x (r) be the sequence defined by x (r) n = (x ⋆ n ) r for all n ∈ N. Replacing the set of primes P with {⌊p 1/r n ⌋ : n ∈ N} and reasoning as in the proof of Theorem 1.3, we obtain that x (r) ∈ A . To complete the proof, fix reals r, s such that 0 < r < s ≤ 1. Then, it is sufficient to show that the ideals generated by x (r) and x (s) are not isomorphic. To this aim, let f : N → N be a bijection and assume for the sake contradiction that Fix t ∈ (1, s /r) and define S := {n ∈ N : f (n) > n t } and T := N \ S. We have n∈S Considering that f is a bijection and the harmonic series is divergent, we obtain that n∈T 1 f (n) = ∞ (in particular, T is infinite). In addition, since r < 1 and we get that n∈T x (r) f (n) = ∞ and, thanks to (4), also that n∈T x (s) n = ∞. Note that, if n ∈ T , then f (n) ≤ n t , which implies that ∀n ∈ T, x (s) n x (r) f (n) ≤ 1/n s 1/(f (n) log(f (n) + 1)) r ≪ (log n) r n s−tr , which has limit 0 if n → ∞ (and belongs to T ). In particular, for each k ∈ N, there exists n k ∈ N such that x (s) n /x (r) f (n) ≤ 1 /k 2 for all n ≥ n k . Let (A k ) be a sequence of finite subsets of T defined recursively as follows: for each k ∈ N, let A k be a finite subset of T such that min A k ≥ n k + max A k−1 and n∈A k x (r) f (n) ∈ ( 1 2 , 1) where, by convention, we assume max A 0 := 0 (note that it is really possible to define such sequence). Finally, define A := k A k so that we obtain This contradicts (4), concluding the proof.

Concluding Remarks
We remark that the ideal I x defined in the proof above is just (an isomorphic copy of) the Fubini sum I s ⊕ I t , where s, t ∈ Z are sequences defined by s n = 1 n and t n = 1 n log(n+1) for all n ∈ N. Here, we recall that the Fubini sum of two ideals I and J on N is the ideal I ⊕ J on {0, 1} × N of all sets A such that {n ∈ N : (0, n) ∈ A} ∈ I and {n ∈ N : (1, n) ∈ A} ∈ J , cf. e.g. [4, p. 8] Also, some comments are in order about the simplifications. The assumption that the sequence x has positive elements (instead of nonnegative elements) is rather innocuous. Indeed, in the opposite, if x n = 0 for infinitely many n, then the summable ideal I x would be (isomorphic to) the Fubini sum P(N) ⊕ I y , for some y ∈ X . Lastly, also the hypothesis x ∈ Z in Question 1.1 (instead of x ∈ X ) has a similar justification. Indeed, if x ∈ X \ Z , then I x would be the Fubini sum Fin ⊕ I y , for some y ∈ Z .
We conclude with two open questions: Question 6.1. Is it true that for each x ∈ Z there are (possibly infinite) sequences y 1 , y 2 , . . . ∈ Y such that I x = ⊕ i I y i ? Question 6.2. Is it true that A is an analytic subset of Z ?