Gaps and the exponent of convergence of an integer sequence

0.15 MB·English

by  Georges Grekos

GAPS AND THE EXPONENT OF CONVERGENCE OF AN
INTEGER SEQUENCE
2
1 GEORGESGREKOS,MARTINSLEZIAK,ANDJANATOMANOVA´
0
2
Abstract. Professor Tibor Sˇal´at, 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 Sˇal´at’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 (Slovensk´a Technick´a Univerzita v Bratislave) and the
Slovak Academy of Sciences (Slovensk´a Akad´emia 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.
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Universit´eJeanMonnet,Facult´edesSciencesetTechniques,D´epartementdeMath´ematiques,
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]