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GAPS AND THE EXPONENT OF CONVERGENCE OF AN INTEGER SEQUENCE 2 1 GEORGESGREKOS,MARTINSLEZIAK,ANDJANATOMANOVA´ 0 2 Abstract. Professor Tibor Sˇalát, at one of his seminars at Comenius Uni- n versity,Bratislava,askedtostudytheinfluenceofgapsofanintegersequence a A = {a1 < a2 < ··· < an < ...} on its exponent of convergence. The J exponent of convergence of A coincides with its upper exponential density. 0 In this paper we consider an extension of Professor Sˇalát’s question and we 1 study theinfluence ofthesequence ofratios(aam )∞m=1 andofthe sequence ] (am+a1m−am)∞m=1 ontheupperandonthelowerme+x1ponential densitiesofA. T N . h 1. Introduction t a m The concept of exponent of convergence is introduced in [10]. The authors of this treatise proved that for any real sequence r = (r ) , 0 < r r ... [ n ∞n=1 1 ≤ 2 ≤ ≤ r ...,with lim r =+ ,thereexistsτ [0,+ ],suchthattheseries ∞ r σ v1 isnc≤onvergentwnh→en∞evner σ >∞τ and divergent∈whenev∞er σ <τ ([10, PartI,ExnP=er1cisn−es 8 113,114]). Thenumberτ iscalledtheexponentofconvergence ofthesequencerand 5 1 denotedbyτ(r). Theexponentofconvergenceofrealnon-decreasingsequenceswas 2 also studied in [7, 8, 11]. It was proved by Po´lya and Szego¨ [10, Part I, Exercises . 113,114]that τ(r) can be calculated by the formula 1 0 logn 2 (1) τ(r)=limsup . logr 1 n→∞ n : Inparticular,ifr isaninteger sequence A= a <a <...<a <... (thatis, v A is an infinite subset of N = 1,2,... ), then{A1has a2n exponennt of con}vergence i X τ(A) [0,1]. { } ∈ r This simple observation indicates that when dealing with sequences of positive a integers,thentheexponentofconvergencecouldberelatedtothenumber-theoretic densities. We recall the notion of exponential density [3, 9]. Definition 1.1. The upper and lower exponential densities ofaninfinite subsetA of N are defined by logA(k) ε(A)=limsup , logk k →∞ logA(k) ε(A)=liminf , k logk →∞ where A(x) denotes A [1,x]. | ∩ | If ε(A) = ε(A), then we say that A has the exponential density ε(A) = ε(A) = ε(A). 1 2 GEORGESGREKOS,MARTINSLEZIAK,ANDJANATOMANOVA´ One caneasily see that, for aninfinite subsetA= a <a <...<a <... of 1 2 n { } N, we have ε(A)=τ(A)=limsup logn , and ε(A)=liminf logn . n→∞ logan n→∞ logan The purpose of this paper is the investigation of the influence of gaps g =a a n n+1 n − in the set A = a < a < < a < ... N on its exponent of convergence. 1 2 n This study was{suggested to·u··s by the late}P⊆rofessor Tibor Sˇala´t. We will be also concerned with a slightly more general question about the influence of gaps in A on both exponential densities. 2. Ratios of consecutive terms and exponent of convergence We are interested in the influence of gaps g =a a n n+1 n − in the set A on the exponent of convergence. Since a gap of given length has less influenceifitissituatedfarfromtheorigin,wemightexpectthat(atleasttosome extent) we can describe the behavior of the exponent of convergence in terms of the asymptotic behavior of the fractions g a n or n . a a n+1 n+1 Note that an + gn =1. an+1 an+1 Definition 2.1. We define the upper and the lower limit ratios of A by a ̺(A)=limsup n , a n n+1 →∞ a ̺(A)=liminf n . n→∞ an+1 We remark that several related concepts have been studied in various contexts. The gap density λ(A) = limsupan+1 was introduced in [5] to study properties of an the density set D(A) = (d(B),d(B));B A of A, and further studied in [4]. { ⊆ } Clearly ̺(A)= 1 (using the convention 1 =0). λ(A) The sets A with ̺(A)=0 (called thin set∞s) play a role in the study of measures whichcanberegardedascertainextensionsofasymptoticdensity[1]. Thesetswith ̺(A)< 1 (called almost thin sets) are studied in connection with some ultrafilters on N [2]. First we showthat the exponentofconvergenceof any setA N with ̺(A)<1 ⊆ is equal to zero. We will need the following well-knownresult (see e.g. [6, Problem 2.3.11], [12]): Theorem 2.2(StolzTheorem). Let(x ) , (y ) besequences ofrealnumbers n ∞n=1 n ∞n=1 such that (y ) is strictly increasing, unbounded and n ∞n=1 x x lim n+1− n =g, n→∞ yn+1−yn g R. Then ∈ x lim n =g. n→∞ yn Proposition 2.3. If ̺(A)<1, then τ(A)=0. GAPS AND THE EXPONENT OF CONVERGENCE OF AN INTEGER SEQUENCE 3 Proof. We will apply Theorem 2.2 to the sequences x = logn and y = loga . n n n Clearly, y is strictly increasing and unbounded. Note that lim (x x ) = n n+1 n n − lim logn+1 =0. Therefore →∞ n n →∞ x x logn+1 (2) lim n+1− n = lim n =0, n→∞ yn+1−yn n→∞logaann+1 wheneverlogan+1 isboundedfromzero. Thustheassumption̺(A)<1issufficient an to infer this. From (2) we get logn lim =0 n→∞logan by Stolz theorem. Thus τ(A)=0. (cid:3) It remainsonly to analysethe case̺(A)=1. The followingexamples showthat in this case nothing can be said about τ(A) in general. Example 2.4. Let a ]0,1], and let A = na1 ;n N . Then ̺(A) = 1 and ∈ {⌊ ⌋ ∈ } ε(A)=a. Example 2.5. Let A = 2n;n N 22N +1;N N . Then ̺(A) = 1 and { ∈ }∪{ ∈ } ε(A)=0. n Example 2.6. Let A= a = u ;n 1,u = 1+ 1 . Then ̺(A)=1 and { n ⌊ n⌋ ≥ n √n } ε(A)=0. (cid:16) (cid:17) Proof. First observe that u tends to infinity as n , since n →∞ 1 lim logu = lim nlog 1+ =+ . n n n √n ∞ →∞ →∞ (cid:18) (cid:19) This yields that a u and loga logu . n n n n ∼ ∼ We have logn logn logn lim = lim = lim =0 n→∞logan n→∞logun n→∞nlog 1+ 1 √n (cid:16) (cid:17) and so τ(A)=ε(A)=0. We will show that lim un =1, which implies ̺(A)=1. Note that n un+1 →∞ u t n = n , u 1+ 1 n+1 √n+1 where 1+ 1 n t = √n . n 1+ 1 ! √n+1 So it is sufficient to show that lim t =1 or equivalently lim lnt =0. n n n n →∞ →∞ 4 GEORGESGREKOS,MARTINSLEZIAK,ANDJANATOMANOVA´ Indeed 1 1 lnt =n ln 1+ ln 1+ = n √n − √n+1 (cid:18) (cid:18) (cid:19) (cid:18) (cid:19)(cid:19) 1 1 1 1 =n (1+o(1)) + (1+o(1)) = √n − 2n − √n+1 2(n+1) (cid:18) (cid:19) n 1 n = + 1 +o(1) √n√n+1(√n+√n+1) 2 n+1 − (cid:18) (cid:19) tends to 0 as n . →∞ Thusitremainstoprovethatu u >1forallsufficientlylargen,andhence n+1 n − the elements(integers)a ofthe setAarepairwisedistinctforn largeenough. We n shall prove that lim (u u )=+ using the function n+1 n n − ∞ →∞ x f(x)= 1+ 1 =exln(cid:16)1+√1x(cid:17). √x (cid:18) (cid:19) An easy computation gives 1 1 f(x) (3) f (x)=f(x) ln 1+ . ′ √x − 2(√x+1) ≥ 2(√x+1) (cid:18) (cid:18) (cid:19) (cid:19) From the inequality 1+ 1 t 2, which is valid for t 1, we get f(x) 2√x and t ≥ ≥ ≥ (cid:0) (cid:1) 2√x (4) f (x) ′ ≥ 2(√x+1) for x 1. This implies that lim f (x)=+ and ′ ≥ x ∞ →∞ lim (u u )= lim (f(n+1) f(n))=+ n+1 n n − n − ∞ →∞ →∞ by the mean value theorem. (cid:3) The above examples suggest the following question: Problem 2.7. Given a,b,c [0,1], a b, does there exist a subset A N such ∈ ≤ ⊆ that ε(A)=a, ε(A)=b, ̺(A)=c and ̺(A)=1? 3. Rate of proximity of an to 1 and the exponential densities an+1 The following three examples provide a motivation for the questions studied in this section. Example 3.1. Let a = αbn(1+o(1)), for some α > 0 and b > 1 (geometric-like n sequence). Then τ(A)=0 and g a 1 lim n = lim 1 n =1 ]0,1[. n→∞an+1 n→∞ − an+1 − b ∈ Example 3.2. If a =k+ln is an arithmetic sequence then τ(A)=1 and n g 1 n = (1+o(1)). a n n+1 GAPS AND THE EXPONENT OF CONVERGENCE OF AN INTEGER SEQUENCE 5 Example 3.3. If a =k+ln+tn2, then τ(A)= 1 and n 2 g 2 n = (1+o(1)). a n n+1 More generally, for any real d > 1 and any integer sequence (a ) satisfying n ∞n=1 a =tnd(1+o(1)) we get τ(A)= 1 and n d g d n = (1+o(1)). a n n+1 Thus for a real number d>0 and (a ) as above, we have τ(A)=max 1,1 . n ∞n=1 { d} The following corollary is a straightforwardconsequence of Proposition 2.3. Corollary 3.4. If liminf gn >0 then τ(A)=ε(A)=0. n an+1 →∞ This can be improved as follows. Proposition 3.5. If liminf gn >0 then τ(A)=ε(A)=0. n an →∞ Proof. Thehypothesisliminf gn >0guaranteesthatthereexistsaδ >0suchthat n an gn δ for each n. This i→m∞plies an ≥ a g n+1 =1+ n 1+δ. a a ≥ n n Hence loga logc+nlog(1+δ), for some constant c. n ≥ From this we get logn logn 0 lim lim =0. ≤n→∞logan ≤n→∞logc+nlog(1+δ) (cid:3) Examples 3.1, 3.2 and 3.3 show that there is a relationbetween the exponent of convergence and the limit behavior of gn/an+1. This is generalized in Proposition 1/n 3.7. We will need a slightly more general form of Stolz Theorem. For the sake of completeness we include the proof of this result. Lemma 3.6. Let (x ) and (y ) be sequences of positive real numbers such n ∞n=1 n ∞n=1 that (y ) is strictly increasing and unbounded. Then n ∞n=1 x x x x x x liminf n+1− n liminf n limsup n limsup n+1− n. n→∞ yn+1−yn ≤ n→∞ yn ≤ n→∞ yn ≤ n→∞ yn+1−yn lProoεf.foPrualtlln=>linnm→.i∞nUfsxyinnn++g11−−thyxinns.faGcitvweneεge>t,0fo,rthaellrene>xisnts,n0 such that xynn++11−−xynn ≥ 0 0 − x x (y y )(l ε). n− n0 ≥ n− n0 − Thus x liminf n l ε. n→∞ yn ≥ − Since this is true for any ε>0, we get x liminf n l. n→∞ yn ≥ The proof of the second part of this lemma is analogous. (cid:3) 6 GEORGESGREKOS,MARTINSLEZIAK,ANDJANATOMANOVA´ Usingthe well-knownequation 1 =lnn+γ+o 1 wegetthe alternative k n k n formulae for exponential densities ≤ P (cid:0) (cid:1) 1 (5) ε(A)=limsup k≤n k , 1 n→∞ Pk≤an k 1 (6) ε(A)=liminf Pk≤n k . n 1 →∞ Pk≤an k Proposition 3.7. Let A= a <a <...<Pa <... and 1 2 n { } g α(A)=liminfn n , n→∞ an+1 g β(A)=limsupn n. a n n →∞ Then 1 1 (7) ε(A) ε(A) . β(A) ≤ ≤ ≤ α(A) (We use the convention 1 =0 and 1 = .) 0 ∞ In particular, if α(A)∞=β(A), then ε(A)= 1 = 1 . α(A) β(A) Proof. Applying Lemma 3.6 to (5) we get 1 1 1 1 1 limsup Pkk≤≤annkk1 ≤limsup an<nk+≤1an+1 k1 ≤limsup ana+nn1++−11an =limsupn+1·angn+1. By a simPilar argument we gPet 1 1 1 1 1 linm→i∞nf Pkk≤≤annkk1 ≥linm→i∞nf an<nk+≤1an+1 k1 ≥linm→i∞nf an+na1+n−1an =linm→i∞nf n+1 agnn. (cid:3) P P The above lemma yields Theorem 3.8. Let α, β be real numbers with 1 α β. ≤ ≤ (i) If there exists a real sequences (e ) tending to zero such that, for each n ∞n=1 n 1, ≥ α g (1+e ) n , n n ≤ a n+1 then 1 ε(A) . ≤ α (ii) If there exists a real sequences (f ) tending to zero such that, for each n ∞n=1 n 1, ≥ g β n (1+f ), a ≤ n n n then 1 ε(A). β ≤ Example 3.9. Proposition3.7canbeusedtocomputethe exponentialdensityfor some sets from the above examples. By a straightforwardcomputation we get the following values. GAPS AND THE EXPONENT OF CONVERGENCE OF AN INTEGER SEQUENCE 7 A= a ;n N α(A) β(A) n { ∈ } A= na1 ;n N 1 1 {⌊ ⌋ ∈ } a a A= 2n;n N 22N +1;N N 0 + A={n2;n∈N,}n∪i{s not square∈ } 2 4∞ { ∈ } a =αbn(1+o(1)) + + n ∞ ∞ a =k+ln 1 1 n a =k+ln+tn2 2 2 n a =tnd(1+o(1)) d d n We have shown in Example 2.5 that ε(A) = 0 for the set in the second row from thistable. Itcanbeshowneasilythatε(A)= 1 forthesetinthefourthrow. These 2 two examples show that the inequalities in (7) can be strict. In all other rows we have α(A)=β(A), and the value of ε(A) can be computed using Proposition 3.7. ThecomputationforthesetinExample2.6isslightlymorecomplicated. Weuse x the function f(x)= 1+ 1 again. Notice that (3) implies that this function is √x increasing. (cid:16) (cid:17) Using the mean value theorem and (3) we get f(n+1) f(n) min f (c);c [n,n+1] n − n { ′ ∈ } f(n) ≥ f(n) ≥ min f(c) ;c [n,n+1] n f(n) n n {2(√c+1) ∈ } ≥ f(n) ≥ f(n) · 2(√n+1+1)2(√n+1+1) and the last expression tends to + as n . ∞ →∞ We found out that α(A)=β(A)=+ . Thus ε(A)=0 by Proposition 3.7. ∞ Remark 3.10. If the set A satisfies the hypotheses of Theorem 3.8 with α = β, then ε(A) = τ(A) = 1. This type of sets A generalizes the sets considered in α Example 3.2 (sequences increasing arithmetically; α = β = 1) and in Example 3.3 (sequences increasing polynomially; α=β >1). Now we state two refinements of Theorem 3.8, Theorems 3.12 and 3.13. Lemma 3.11. For any x [0,1] the inequality ∈ 2 ln(1 x) x+x2 − − ≤ holds. Proof. By studying f(x)=x+x2 [ ln(1 x)]. (cid:3) − − − Theorem 3.12. Let β be a real number, β 1. If there exists a real sequence ≥ (f ) tending to zero such that, for each n 1, n ∞n=1 ≥ g β n (1+f ), a ≤ n n n+1 then 1 ε(A) . ≥ β 8 GEORGESGREKOS,MARTINSLEZIAK,ANDJANATOMANOVA´ Proof. By the definition of ε(A) it is sufficient to show that for each 0 < ε < 1 β there exists n =n (ε) such that for each n n 0 0 0 ≥ lnn 1 ε. lna ≥ β − n We will verify that ln(n+1) lna n+1 ≤ 1 ε β − for every n large enough. The hypothesis gives a a β n+1− n (1+f ), a ≤ n n n+1 a 1 n+1 . an ≤ 1− nβ(1+fn) Let us choose n such that f < 1, for each n n . Then we get 1 | n| 2 ≥ 1 n n a β lna =lna + ln i+1 lna ln(1 (1+f )). n+1 n1 a ≤ n1 − − n n i iX=n1 iX=n1 Let n n be such that β(1+f ) 1 whenever n n . Then we get 2 ≥ 1 |n n |≤ 2 ≥ 2 n β n β2 lna c + (1+f )+ (1+f )2. n+1 ≤ 1 i i i2 i iX=n2 iX=n2 Put δ = 1 βε >0 and choose n n such that f δ, whenever i n . 21 βε 3 ≥ 2 | i|≤ ≥ 3 Then, for−all n n , 3 ≥ n β(1+δ) n β2(1+δ)2 lna c + + n+1 ≤ 2 i i2 ≤ iX=n3 iX=n3 n 1 c +β(1+δ) c +β(1+δ)+β(1+δ)ln(n+1) ≤ 3 i ≤ 3 i=1 X holds. Obviously 1 β(1+δ)< 1 ε β − and therefore, the right hand side of the above inequality is at most ln(n+1) for 1 ε every sufficiently large n. β− (cid:3) Theorem 3.13. Let α be a real number, α 1. Suppose that for all n ≥ g α n (1+e ), a ≥ n n n where lim e =0. Then n n →∞ 1 ε(A) . ≤ α Note that Proposition 3.5 can be deduced from the above theorem. GAPS AND THE EXPONENT OF CONVERGENCE OF AN INTEGER SEQUENCE 9 Proof. If α=1 we are done. Suppose that α>1. By the definition of ε(A) it suffices to show that for each ε > 0 there is an n =n (ε) such that 0 0 lnn 1 +ε, lna ≤ α n whenever n n . That is, lnn 1 +ε lna or equivalently ≥ 0 ≤ α n (cid:0) (cid:1) lnn lna . n ≥ 1 +ε α Fix ε>0. We shall prove that for all n sufficiently large ln(n+1) (8) lna . n+1 ≥ 1 +ε α Let δ = 1 αε . Obviously δ <1. Choose n suchthat e δ for all n n . The 21+αε 1 | n|≤ ≥ 1 hypothesis of the theorem implies that for all n a α n+1 1+ (1+e ). a ≥ n n n Then for n n , we have 1 ≥ n n a α lna =lna + ln i+1 lna + ln 1+ (1+e ) , n+1 n1 a ≥ n1 i i i iX=n1 iX=n1 (cid:16) (cid:17) and by the inequality x2 ln(1+x) x ≥ − 2 (valid for x 0), we have ≥ n α 1 n α2 lna lna + (1+e ) (1+e )2 n+1 ≥ n1 i i − 2 i2 i ≥ iX=n1 iX=n1 n 1 1 ∞ 1 lna +α(1 δ) (1+δ)2α2 . ≥ n1 − i − 2 i2 iX=n1 Xi=1 Finally n 1 lna c +α(1 δ) , n+1 ≥ 4 − i i=1 X where c does not depend on n. The last inequality implies that 4 lna c +α(1 δ)ln(n+1), n+1 4 ≥ − since n 1 >ln(n+1). i=1 i Now to deduce that (8) is valid for all n large enough, it suffices to verify that α(1 Pδ)> 1 , which is straightforward. (cid:3) − α1+ε NotethatbyProposition3.5thestudyofexponentialdensitiesisnon-trivialonly for sets A such that liminf gn =0. In view of this, the results stated in Theorems n an 3.12 and 3.13 are far fr→o∞m being complete since only comparison of gn or gn to an an+1 sequencesofthe type “constanttimes 1”wasconsidered. Neverthelessthese cases, n motivated by Examples 3.2 and 3.3, are the most important ones as the following observations show. 10 GEORGESGREKOS,MARTINSLEZIAK,ANDJANATOMANOVA´ Observation 1. If gn is approximatively 1 with α > 1, then a is ap- an nα n+1 n proximatively a (1+nα) and a is approximatively c 1+ 1 . The product n n iα i=1 ∞ 1+ 1 beingconvergent,wegetthattheset c n Q1(cid:0)+ 1 ;(cid:1)n N isfinite. iα {⌊ iα ⌋ ∈ } i=1 i=1 QO(cid:0)bserv(cid:1)ation 2. If gn is approximatively 1 withQ0(cid:0)< α <(cid:1)1, again a would an nα n n be close to u := c 1+ 1 . Then lnu n1−α, so that lim lnun = + . In n i=1 iα n ∼ 1−α n→∞ lnn ∞ other words, A has zQero(cid:0) expon(cid:1)ential density. Acknowledgement: A large part of this researchwas carried out during stays of the first named author in Bratislava, on invitation by the Slovak University of Technology in Bratislava (Slovenská Technická Univerzita v Bratislave) and the Slovak Academy of Sciences (Slovenská Akadémia Vied). The second author was supported by grants VEGA/1/0588/09 and UK/374/2009. The third author was supported by grant VEGA/2/7138/27. References [1] BLASS, A.–FRANKIEWICZ, R.– PLEBANEK,G. –RYLL-NARDZEWSKI, C.: A note on extensionsof asymptotic density,Proc.Amer.Math.Soc.,129(11):3313–3320, 2001. [2] FLASˇKOVA´, J.: Thin ultrafilters, Acta Univ. Math. Carolinae - Mathematica et Physica, 46:13–19,2005. [3] GREKOS G.: On various definitions of density (survey), Tatra Mt. Math. Publ., 31:17–27, 2005. [4] GREKOS, G. – SˇALA´T, T. – TOMANOVA´, J.: Gaps and densities, Bull. Math. Soc. Sci. 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UniversitéJeanMonnet,FacultédesSciencesetTechniques,DépartementdeMathématiques, 23Rue du Docteur PaulMichelon,42023Saint-EtienneCedex 2,France E-mail address: [email protected] DepartmentofAlgebra,GeometryandMathematicalEducation,FacultyofMathe- matics,PhysicsandInformatics,ComeniusUniversity,Mlynska´dolina,84248Bratislava, Slovakia E-mail address: [email protected], [email protected]