Convergent Subseries of Divergent Series

Rendiconti del Circolo Matematico di Palermo Series 2 https://doi.org/10.1007/s12215-021-00629-3

Convergent subseries of divergent series

Marek Balcerzak1 · Paolo Leonetti2

Received: 20 November 2020 / Accepted: 1 June 2021 © The Author(s) 2021

Abstract X Let be the set of positive real sequences x =(xn) such that the series ∑n xn is diver‑ X I ⊆ gent. For each x ∈ , let x be the collection of all A such that the subseries ∑n∈A xn A X lim 0 is convergent. Moreover, let be the set of sequences x ∈ such that n xn = and I I X lim inf 0 A x ≠ y for all sequences y =(yn)∈ with n yn+1∕yn > . We show that is comeager and that contains uncountably many sequences x which generate pairwise noni‑ I somorphic ideals x . This answers, in particular, an open question recently posed by M. Filipczak and G. Horbaczewska.

Keywords Summable ideal · Fubini sum · Convergent series

Mathematics Subject Classifcation Primary: 40A05 · Secondary: 54A20

1 Introduction

X Let be the set of positive real sequences x =(xn) with divergent series ∑n xn . For each X I x ∈ , let x the collection of sets of positive integers A such that the (possibly fnite) sub‑ series indexed by A is convergent, that is, I ⊆ < x ∶= A ∶ xn ∞ . (1) n∈A I Note that each x is closed under fnite unions and subsets, i.e., it is an ideal. Moreover, it contains the collection Fin of fnite sets A ⊆ , and it is diferent from the power set P( ) . Following [4], a collection of sets of the type (1) is called summable ideal. It is not difcult

* Paolo Leonetti [email protected] https://sites.google.com/site/leonettipaolo/ Marek Balcerzak [email protected]

1 Institute of Mathematics, Lodz University of Technology, ul. Wólczańska 215, 93‑005 Lodz, Poland 2 Department of Decision Sciences, Università Bocconi, via Roentgen 1, 20136 Milan, Italy

Vol.:(0123456789)1 3 M. Balcerzak, P. Leonetti

I to see that every infnite set of positive integers contains an infnite subset in x if and only if limn xn = 0 . Accordingly, defne Z X X ∶= ∩ c0 ={x ∈ ∶ lim xn = 0}. n→∞ I It is known that the families x defned 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 flter convergence have been given in [1, 2, 10]. The set of limits of convergent subseries of a given series ∑n xn , which is usually called “achievement set”, has been studied in [7, 9, 12]. Of special interest have been specifc sub‑ ∑ 1 series of the harmonic series n 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 Z X I I each x ∈ there exists y ∈ such that x = y and (yn) “does not oscillates too much”? I Z Hoping for a characterization of the class of summable ideals x with x ∈ , M. Filipc‑ zak and G. Horbaczewska asked recently in [5] the following:

Z X I I Question 1.1 Is it true that for each x ∈ there exists y ∈ such that x = y and y n ∀n ∈ , n+1 ≥ ? yn n + 2

We show in Theorem 1.3 below that the answer is negative in a strong sense. To this aim, defne y Y ∶= y ∈ X ∶ lim inf n+1 > 0 , n→ ∞ yn I Y I and note that A − 1 ∶= {a − 1 ∶ a ∈ A} belongs to y whenever y ∈ and A ∈ y , cf. (2) below. Let also ∼ be the equivalence relation on X so that two sequences are identifed if they generate the same ideal, hence X ⟺ ⟺ ∀x, y ∈ , x ∼ y ∀A ⊆ , xn < ∞ yn < ∞ . n∈A n∈A

2 First, we show that the set of pairs (x, y)∈X such that x is ∼-equivalent to y is topologi‑ cally well behaved:

Proposition 1.2 ∼ is a coanalytic relation on X .

Then, we answer Question 1.1 by showing that:

Theorem 1.3 There exists x ∈ Z such that x ≁ y for all y ∈ Y .

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 A ∶= {x ∈ Z ∶∀y ∈ Y, x ≁ y}

1 3 Convergent subseries of divergent series 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 Z ⋯ = x ∈ c0 ∶ xn > 0 ∩ x ∈ c0 ∶ x1 + + xk > m n≥1 m≥1 k≥1 A is a G -subset of the Polish space c0 . With the premises, we show that is comeager, that is, Z ⧵ A is a set of frst 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 Sil‑ ver [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 Y A Y G -comeager subset) and is an uncountable F -set. Therefore × 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 an ∑n≥1 bn , with each an, bn 0 , is a short‑ > hand for the existence of C 0 such that ∑n≤k an ≤ C ∑n≤k bn for all k ∈ .

2 Proof of Proposition 1.2

2 2 Equivalently, we have to show that the set E ∶= (x, y)∈X ∶ x ≁ y is analytic in X . For, X2 note that E is the projection on of E1 ∪ E2 , where P X2 I I E1 ∶= (A, x, y)∈ ( )× ∶ A ∈ x ⧵ y and, similarly, P X2 I I E2 ∶= (A, x, y)∈ ( )× ∶ A ∈ y ⧵ x. P X2 → Now, for each n ∈ , defne the functions n, n ∶ ( )× by n(A, x, y)=∑ xt and n(A, x, y)=∑ yt , where each sum is extended over all t ∈ A such that t ≤ n . Since they P X2 are continuous, the set ( n ≤ k) ∶= {(A, x, y)∈ ( )× ∶ n(A, x, y) ≤ k} is closed and (�n > k) is open for all n, k ∈ . Therefore

1 3 M. Balcerzak, P. Leonetti

E1 = (�n ≤ k) ∩ (�n > k) k≥1 n≥1 k≥1 n≥1 is the intersection of an F -set and a G -set, hence it is Borel. Analogously, E2 is Borel. 2 This proves that E is analytic subset of X .

3 Proof of Theorem 1.3

x =(x ) x = 1 x = 1 Defne the sequence n so that n n if n is even and n n log(n+1) if n is odd. Note that Z Y limn xn = 0 and that ∑n xn =∞ , hence x ∈ . At this point, fx y ∈ such that I I � ∶= lim infn yn+1∕yn > 0 and let us show that x ≠ y. Let be the set of prime numbers, with increasing enumeration (pn) . By the prime number → theorem we have pn is asymptotically equal to n log(n) as n ∞ , hence 1 1 x = x ≪ ≪ < ∞, n pn p p 2 n∈ n≥1 n≥1 n log( n) n≥2 n log (n) I I with the consequence that ∈ x . In addition, − 1 ∉ x because 1 1 x = x ≫ ≫ =∞. n pn−1 p n n n∈ −1 n≥1 n≥1 n n≥2 log( ) I I I Lastly, suppose for the sake of contradiction that x = y . Then we should have that ∈ y I and, at the same time, − 1 ∉ y . The latter means that y =∞. pn−1 n≥1

However, this implies that y ≫ � y =∞, pn pn−1 n≥1 n≥1 (2) I contradicting that ∈ y.

4 Proof of Theorem 1.4

Consider the Banach–Mazur game defned as follows: Players I and II choose alternatively Z ⋯ nonempty open subsets of as a nonincreasing chain U1 ⊇ V1 ⊇ U2 ⊇ V2 ⊇ , where ⊆ A Player I chooses the sets Um . Player II has a winning strategy if ⋂m Vm . 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 �>0 satisfes � B (x) ∶= {y ∈ Z ∶ x − y <�} ⊇ �y ∈ Z ∶∀n ∈ , x − y < �. � ‖ ‖ � n n� 2 x ∈ Z ⊆ c k = k (x, )∈ x < � n ≥ k Since 0 , there exists 0 0 such that n 2 for all 0 . Hence

1 3 Convergent subseries of divergent series

� � B (x) ⊇ W (x) ∶= y ∈ Z ∶∀n ≥ k (x, �), y < and ∀n < k (x, �), x − y < . � � 0 n 2 0 n n 2 (3) For each m ∈ , suppose that the nonempty open set Um has been fxed by Player I. Hence, (m) (m) Z Um B (x ) x ∈ �m > 0 contains an open ball m , for some and . In particular, thanks� to k = k (x(m), )∈ y < m (3), there exists a sufciently large integer 0 0 m such that n 2 for all y ∈ W (x(m)) n ≥ k k m and 0 . Without loss of generality, let us suppose that 0 is even. x⋆ A At this point, let be the sequence in defned in the proof of Theorem 1.3� . Then, for (m) m ⋆ ⋆ ⋅ m m ∈ tm ≥ k0(x , ) max{x , x } < tm each , let 2 be an integer such that pm pm−1 4 (we recall that pm stands for the mth prime number), and defne the positive real ⋆ ⋆ � max{x , x } 1 pm pm−1 � ∶= min , m − . m 2 m tm 4 tm

Note that limm m = 0 . Without loss of generality, we can assume also (m+1) (m) that k0(x , �m+1) > k0(x , �m)+2tm for all m ∈ . Now, defne the set (m) (m) Im = ∩[k0(x , m), k0(x , m)+2tm) (hence the sets Im are pairwise disjoint) and let (m) z be the sequence such that

⋆ x ∕tm if n ∈ Im and n even, ⎧ pm−1 ∀n ∈ z(m) = ⎪ x⋆ ∕t if n ∈ I and n odd, n pm m m ⎨ x(m) if n ∉ I . ⎪ n m ⎩ V ∶= B (z(m)) y ∈ V Lastly, set m m and note that, for each sequence m , we have by construction:

(m) (m) �m yn − x = yn − z <�m < n < min Im (i) n n 2 if � ; < (m) � ≤ ⋆ ⋆ � < m (ii) yn zn + m max{xp −1, xp }∕tm + m if n ∈ Im; m � m � � 2 y < z(m) + � = x(m) + � < m + m = m n > max I (iii) n n m n m 4 4 2 if m. � n > max I ≥ t ≥ k (x(m), m ) x(m) ≤ 1 ⋅ m (In the last point, we used that if m m 0 2 then n 2 2 .) This y ∈ W (x(m)) implies that m . To sum up, we obtain ∀m ∈ , V ⊆ W (x(m)) ⊆ B (x(m)) ⊆ U , m �m �m m

(m) hence Vm is a nonempty open set contained in Um . In addition, the sequence of centers (z ) Z is a Cauchy sequence in the complete metric space . Indeed, since the sequence (Vn) is (m) (m�) ≤ ≤ 1 ≥ ≥ (m) decreasing, we have z − z m 2 for all integers m m 1 . Hence (z ) is ‖ Z ‖ m convergent to some z ∈ and it is straighforward to see that {z}=⋂m Vm. A To complete the proof, we need to show that z ∈ . Set A ∶= �⋃m Im� ⧵ 2 . Proceeding as in the proof of Theorem 1.3, we see that x⋆ pm 1 z = z ≤ (z(m) + � ) ≤ I + < ∞, n n n m m 2 tm m tm n∈A m≥1 n∈Im⧵2 m≥1 n∈Im⧵2 m≥1 I I hence A ∈ z . Similarly, A − 1 ∉ z since

1 3 M. Balcerzak, P. Leonetti

x⋆ pm−1 1 z ≥ (z(m) − � ) ≫ I − =∞. n n m m 2 tm m tm n∈A−1 m≥1 n∈Im∩2 m≥1 Y I I Now, fx y ∈ such that � ∶= lim infn yn+1∕yn > 0 and suppose that z = y . Then we I I would have that A ∈ y and A − 1 ∉ y , which is impossible reasoning as in (2).

5 Proof of Theorem 1.5

⋆ (r) Let x be the sequence defned in the proof of Theorem 1.3. For each r ∈(0, 1] , let x be (r) ⋆ r the sequence defned by xn =(xn ) for all n ∈ . Replacing the set of primes with

1∕r ∶= 2 pn + 1 ∶ n ∈ , r 2 we obtain that r 1 1 x(r) ≪ ≪ < ∞, n 1∕r 1∕r 1+r n (log(n)) n∈ r n≥1 pn log(pn ) n≥1 (r) ≫ 1 and ∑n∈ −1 xn ∑n≥1 =∞ . Reasoning as in the proof of Theorem 1.3, we conclude r pn (r) that x ∈ A for each r ∈(0, 1]. To complete the proof, fx reals r, s such that 0 < r < s ≤ 1 . Then, it is sufcient to show (r) (s) that the ideals generated by x and x are not isomorphic. To this aim, let f ∶ → be a bijection and assume for the sake contradiction that ⊆ (r) < (s) < ∀A , xf (n) ∞ if and only if xn ∞. n∈A n∈A (4) t ∈(1, s ) S ∶= {n ∈ ∶ f (n) > nt} T ∶= ⧵ S Fix r and defne and . We have ∑ 1 ≤ ∑ 1 < ∞ n∈S f (n) n∈S nt . Considering that f is a bijection and the harmonic series is diver‑ ∑ 1 =∞ r < 1 gent, we obtain that n∈T f (n) (in particular, T is infnite). In addition, since and 1 1 1 ≪ ≤ x(r) ≤ , f (n) (f (n) log(f (n)+1))r f (n) f r(n)

(r) (s) we get that ∑n∈T xf (n) =∞ and, thanks to (4), also that ∑n∈T xn =∞ . Note that, if n ∈ T , t then f (n) ≤ n , which implies that x(s) 1∕ns (log n)r ∀n ∈ T, n ≤ ≪ , (r) 1∕(f (n) log(f (n)+1))r ns−tr xf (n) which has limit 0 if n → ∞ (and belongs to T). In particular, for each k ∈ , there exists n ∈ x(s)∕x(r) ≤ 1 n ≥ n (A ) k such that n f (n) k2 for all k . Let k be a sequence of fnite subsets of T defned recursively as follows: for each k ∈ , let Ak be a fnite subset of T such that min A ≥ n + max A ∑ x(r) ∈(1 ,1) k k k−1 and n∈Ak f (n) 2 where, by convention, we assume max A0 ∶= 0 (note that it is really possible to defne such sequence). Finally, defne A ∶= ⋃k Ak so that we obtain

1 3 Convergent subseries of divergent series

x(r) f (n) 1 x(r) = x(r) =∞ and x(s) ≤ ≤ < ∞. f (n) f (n) n k2 k2 n∈A k≥1 n∈Ak n∈A k≥1 n∈Ak k≥1

This contradicts (4), concluding the proof.

6 Concluding remarks

I We remark that the ideal x defned in the proof above is just (an isomorphic copy of) the I ⊕ I s, t ∈ Z s = 1 t = 1 Fubini sum s t , where are sequences defned by n n and n n log(n+1) for all n ∈ . Here, we recall that the Fubini sum of two ideals I and J on is the ideal I ⊕ J on {0, 1}× of all sets A such that {n ∈ ∶(0, n)∈A}∈I and {n ∈ ∶(1, n)∈A}∈J , cf. e.g. [4, p. 8] Also, some comments are in order about the simplifcations. The assumption that the sequence x has positive elements (instead of nonnegative elements) is rather innocu‑ I ous. Indeed, in the opposite, if xn = 0 for infnitely many n, then the summable ideal x P I X would be (isomorphic to) the Fubini sum ( ) ⊕ y , for some y ∈ . Lastly, also the hypothesis x ∈ Z in Question 1.1 (instead of x ∈ X ) has a similar justifcation. Indeed, if X Z I I Z x ∈ ⧵ , then x would be the Fubini sum Fin ⊕ y , for some y ∈ . We conclude with two open questions:

Question 6.1 Is it true that for each x ∈ Z there are (possibly infnite) sequences 1 2 Y I I y , y , …∈ such that x = ⊕i yi?

Question 6.2 Is it true that A is an analytic subset of Z ?

Acknowledgements The authors are grateful to two anonymous referees for several constructive comments which improved the exposition of the article.

Funding Open access funding provided by Università Commerciale Luigi Bocconi within the CRUI-CARE Agreement. P.L. is grateful to PRIN 2017 (grant 2017CY2NCA) for fnancial support.

Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Com‑ mons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.

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