Convergent subseries of divergent series

Let X\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathscr {X}$$\end{document} be the set of positive real sequences x=(xn)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$x=(x_n)$$\end{document} such that the series ∑nxn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\sum _n x_n$$\end{document} is divergent. For each x∈X\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$x \in \mathscr {X}$$\end{document}, let Ix\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {I}_x$$\end{document} be the collection of all A⊆N\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$A\subseteq \mathbf {N}$$\end{document} such that the subseries ∑n∈Axn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\sum _{n \in A}x_n$$\end{document} is convergent. Moreover, let A\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathscr {A}$$\end{document} be the set of sequences x∈X\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$x \in \mathscr {X}$$\end{document} such that limnxn=0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\lim _n x_n=0$$\end{document} and Ix≠Iy\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {I}_x\ne \mathcal {I}_y$$\end{document} for all sequences y=(yn)∈X\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$y=(y_n) \in \mathscr {X}$$\end{document} with lim infnyn+1/yn>0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\liminf _n y_{n+1}/y_n>0$$\end{document}. We show that A\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathscr {A}$$\end{document} is comeager and that contains uncountably many sequences x which generate pairwise nonisomorphic ideals Ix\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {I}_x$$\end{document}. 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 ⊆ , and it is different from the power set P( ) . 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 positive integers contains an infinite subset in I x if and only if lim n x n = 0 . Accordingly, define 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 1 n ; see, e.g., [11,14,15,17]. Roughly, the question that we are going to answer is the following: Is it true that for each x ∈ Z there exists y ∈ X such that I x = I y and (y n ) "does not oscillates too much"?
Hoping for a characterization of the class of summable ideals I x with x ∈ Z , M. Filipczak and G. Horbaczewska asked recently in [5] the following: Question 1.1 Is it true that for each x ∈ Z there exists y ∈ X such that I x = I y and We show in Theorem 1.3 below that the answer is negative in a strong sense. To this aim, define and note that A − 1 ∶= {a − 1 ∶ a ∈ A} belongs to I y whenever y ∈ Y and A ∈ I y , cf. (2) below. Let also ∼ be the equivalence relation on X so that two sequences are identified if they generate the same ideal, hence 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 is a G -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: 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 nonequivalent 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 -comeager subset) and Y is an uncountable F -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 ∶ → such that f [A] ∈ I if and only if A ∈ J for all A ⊆ ).

Theorem 1.5 There are 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 ∈ .

Proof of Proposition 1.2
Equivalently, we have to show that the set E ∶= (x, y) ∈ X 2 ∶ x ≁ y is analytic in X 2 . For, note that E is the projection on X 2 of E 1 ∪ E 2 , where and, similarly, Now, for each n ∈ , define the functions n , n ∶ P( ) × X 2 → by n (A, x, y) = ∑ x t and n (A, x, y) = ∑ y t , where each sum is extended over all t ∈ A such that t ≤ n . Since they are continuous, the set ( n ≤ k) ∶= {(A, x, y) ∈ P( ) × X 2 ∶ n (A, x, y) ≤ k} is closed and ( n > k) is open for all n, k ∈ . Therefore is the intersection of an F -set and a G -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 ∶= lim inf n y n+1 ∕y n > 0 and let us show that I x ≠ I y . Let 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 ∈ I x . In addition, − 1 ∉ I x because Lastly, suppose for the sake of contradiction that I x = I y . Then we should have that ∈ I y and, at the same time, − 1 ∉ I y . The latter means that However, this implies that contradicting that ∈ 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 Note that the open neighborhood of a sequence x ∈ Z with radius > 0 satisfies Since x ∈ Z ⊆ c 0 , there exists k 0 = k 0 (x, ) ∈ such that x n < 2 for all n ≥ k 0 . Hence ∑ n≥1 y p n −1 = ∞.
(2) ∑ n≥1 y p n ≫ ∑ n≥1 y p n −1 = ∞, For each m ∈ , suppose that the nonempty open set U m has been fixed by Player I. Hence, U m contains an open ball B m (x (m) ) , for some x (m) ∈ Z and m > 0 . In particular, thanks to (3), there exists a sufficiently large integer k 0 = k 0 (x (m) , m ) ∈ such that y n < m 2 for all y ∈ W m (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 ∈ , let t m ≥ k 0 (x (m) , m 2 ) be an integer such that max{x ⋆ p m , x ⋆ p m −1 } < t m ⋅ m 4 (we recall that p m stands for the mth prime number), and define the positive real Note that lim m m = 0 . Without loss of generality, we can assume also that k 0 (x (m+1) , m+1 ) > k 0 (x (m) , m ) + 2t m for all m ∈ . Now, define the set I m = ∩ [k 0 (x (m) , m ), k 0 (x (m) , m ) + 2t m ) (hence the sets I m are pairwise disjoint) and let z (m) be the sequence such that Lastly, set V m ∶= B m (z (m) ) and note that, for each sequence y ∈ V m , we have by construction: (In the last point, we used that if n > max I m ≥ t m ≥ k 0 (x (m) , m 2 ) then x (m) n ≤ 1 2 ⋅ m 2 .) This implies that y ∈ W m (x (m) ) . To sum up, we obtain 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 . Indeed, since the sequence Hence (z (m) ) 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 � ⧵ 2 . Proceeding as in the proof of Theorem 1.3, we see that hence A ∈ I z . Similarly, A − 1 ∉ I z since (3) B (x) ⊇ W (x) ∶= y ∈ Z ∶ ∀n ≥ k 0 (x, ), y n < 2 and ∀n < k 0 (x, ), |x n − y n | < 2 .
x ⋆ p m −1 ∕t m if n ∈ I m and n even, if n ∈ I m and n odd,