# On Faster Convergent Infinite Series

Authors: Dušan Holý; Ladislav Matejíčka; Ľudovít Pinda
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

Suffcient conditions, necessary conditions for faster convergent infinite series, fasterτ-convergent infinite series are studied. The faster convergence of infinite series of Kummer's type is proved.


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## Body

## 1. Introduction

There have been papers devoted to the study of faster
convergence of sequences. Certain methods concerning acceleration of
convergence of sequences of partial sums of fixed series using linear or
nonlinear transformations of partial sums of series are studied in [1]. The acceleration field of
subsequence matrix transformations with respect to the convergence rate of the
sequence being accelerated are studied in [2]. In [3], it is discussed a class of methods for summing
sequences which are generalizations of a method due to Salzer [4], which accelerate some
convergent sequences especially monotone sequences. In [5], is characterized the
summability field of a matrix A by showing A is convergence preserving over the set of all
sequences which converge faster than some fixed sequence x, A is convergence preserving over the set of all
sequences, or A only preserves the limit of a set of constant
sequences. Statistical acceleration convergence of sequences was discussed in [6]. The notion of
faster convergent series with positive terms is defined in [7] and the notion of τ-convergent series is defined in [8–10]. The statistical convergence
of infinite series as a special case of τ-convergence of infinite series is discussed
in [11–14].In this paper, some questions related to sufficient
conditions and necessary conditions for faster convergent infinite series are
studied, fasterτ-convergent series are defined and studied, and faster
convergence of series of Kummer's type is proved. In [7, page 146], it is mentioned the
Kummer's result: if ∑n=1∞an is convergent series with positive terms and
with an unknown sum a, ∑n=1∞cn is a convergent series with positive terms and
with a known sum c,
and limn→∞an/cn=p>0,
then the Kummer series ∑n=1∞bn has the same sum as ∑n=1∞an and is a faster convergent series than ∑n=1∞an,
if all terms of ∑n=1∞bn are positive. Hence we can calculate the
unknown sum a faster by summation terms of the Kummer
series ∑n=1∞bn,
which is constructed by ∑n=1∞an, ∑n=1∞cn,
and p.
In Lemma 4.1, we proved a faster convergence of the Kummer series ∑n=1∞bn to the unknown sum a of ∑n=1∞an without conditions of positivity of p and terms of ∑n=1∞an,∑n=1∞bn,∑n=1∞cn.We denote byℕ the set of all positive integers and by ℝ the set of all real numbers.Definition 1.1 (See [2]).
Let∑n=1∞an, ∑n=1∞bn be convergent real series with the same sum,
with nonzero terms and such that bn+bn+1+…≠0,n∈ℕ.
The series ∑n=1∞an is called faster convergent than ∑n=1∞bn if limn→∞(an+an+1+…)/(bn+bn+1+…)=0.Lemma 1.2 (See [7]).
Let∑n=1∞an, ∑n=1∞bn be convergent real series with positive terms
and with the same sum. If limn→∞(an/bn)=0,
then limn→∞(an+an+1+…)/(bn+bn+1+…)=0.In what follows, we do not assume equality of sum of
convergent series∑n=1∞an,∑n=1∞bn because we can construct series ∑n=1∞an*,∑n=1∞bn*,
where an*=an and bn*=bn for n≥2 with the same sum.


## 2. Faster Convergent Series

Lemma 2.1.
Let∑n=1∞an, ∑n=1∞bn be convergent real series with positive terms.
If(2.1)limn→∞an+an+1+…bn+bn+1+…=0,thenliminfn→∞anbn=0.Proof.
By the way of contradiction, we
suppose thatliminfn→∞(an/bn)=q>0,
then there exists n0∈ℕ such that for every n>n0 we have 0<q⋆<an/bn, where 0<q⋆<q. From this follows q⋆(bn+bn+1+…)<(an+an+1+…) for n>n0, which is a contradiction with limn→∞(an+an+1+…)/(bn+bn+1+…)=0, similarly for q=∞.Lemma 2.2.
Let∑n=1∞an, ∑n=1∞bn be convergent real series with positive terms.
Let limn→∞(an/bn) exist. Then(2.2)limn→∞anbn=0ifflimn→∞an+an+1+…bn+bn+1+…=0.The next example shows that under the conditions of
Lemma2.1, for all 0<r≤∞,  there exist a series ∑n=1∞an,∑n=1∞bn such that limsupn→∞(an/bn)=r.Example 2.3.
Let{αn;n∈ℕ} be any sequence of real positive numbers which
satisfy αn≤1/n4 for n≥1.
We define the series ∑n=1∞an=rα1+α1/12+rα2+α2/22+…+rαk+αk/k2+… and ∑n=1∞bn=α1+12α1+α2+22α2+…+αk+k2αk+….
Then we have(2.3)limn→∞a2n−1+a2n+…b2n−1+b2n+…=limn→∞r(αn+αn+1+…)+(1/n2)(αn+αn+1+…)(αn+αn+1+…)+n2(αn+αn+1+…)=0,limn→∞a2n+a2n+1+…b2n+b2n+1+…=limn→∞r(αn+1+αn+2+…)+(1/n2)(αn+αn+1+…)(αn+1+αn+2+…)+n2(αn+αn+1+…)=0,limn→∞a2n−1b2n−1=r,limn→∞a2nb2n=0. If ∑n=1∞an=α1+α1/12+2α2+α2/22+…+kαk+αk/k2+…,
then limn→∞(a2n−1/b2n−1)=∞.Remark 2.4.
It is evident that in general fromlimn→∞(an+an+1+…)/(bn+bn+1+…)=0 does not follow liminfn→∞(an/bn)=0. For example, if we put a2n=b2n=1/(n(n+1)),n∈ℕ, and a2n−1=−b2n−1=−1/(n(n+1)),n∈ℕ, we obtain limn→∞(an+an+1+…)/(bn+bn+1+…)=0 and {−1,1} is the set of all cluster points of the
sequence {an/bn;n∈ℕ}.In general, the conditionlimn→∞(an/bn)=0 does not imply the condition limn→∞(an+an+1+…)/(bn+bn+1+…)=0 as it follows from the next example.Example 2.5.
Let{αn;n∈ℕ} be a sequence such that αn=qn,n∈ℕ,
0<q<1/2.
We define convergent series ∑n=1∞an=α1+α1+α2/22+α2/22+…+αk/k2+αk/k2+… and ∑n=1∞bn=α1−α1/2+α2/3−α2/4+…+αk/(2k−1)−αk/2k+…. From the definition, we obtain (2.4)b2n+b2n+1+…=−αn2n+αn+1(2n+1)(2n+2)+αn+2(2n+3)(2n+4)+…<12n(−α+αn+1+αn+2+…)=12n(−qn+qn+1+qn+2+…)=qn(2q−1)2n(1−q)<0.   It is obvious that b2n+1+b2n+2+…>0 and limn→∞(an/bn)=0. From (2(k+m)−1)(2(k+m))>(k+m)2,k,m∈ℕ we have(2.5)a2k−1+a2k+…b2k−1+b2k+…=2(αk/k2+αk+1/(k+1)2+…)αk/(2k(2k−1))+αk+1/((2k+1)(2k+2))+…>2fork≥1.Thus limn→∞(an+an+1+…)/(bn+bn+1+…)≠0.Lemma 2.6.
Let∑n=1∞an, ∑n=1∞bn be convergent real series with nonzero terms.
Let bn+bn+1+…≠0 for all n∈ℕ.
Let li(a)=liminfn→∞|1+an+1/an+an+2/an+…|,ls(a)=limsupn→∞|1+an+1/an+an+2/an+…|,li(b)=liminfn→∞|1+bn+1/bn+bn+2/bn+…|,ls(b)=limsupn→∞|1+bn+1/bn+bn+2/bn+…|, then(1)
ifls(a)<∞,li(b)>0,limn→∞(an/bn)=0,thenlimn→∞(an+an+1+…)/(bn+bn+1+…)=0,(2)
ifan+an+1+…≠0for alln∈ℕ,li(a)>0,ls(b)<∞,limn→∞(an+an+1+…)/(bn+bn+1+…)=0,thenlimn→∞(an/bn)=0,(3)
ifan+an+1+…≠0for alln∈ℕ,0<li(a),ls(a)<∞,0<li(b),ls(b)<∞,thenlimn→∞(an/bn)=0 if  and only if
limn→∞(an+an+1+…)/(bn+bn+1+…)=0.Proof.
For everyn∈ℕ, we have(2.6)|an+an+1+…bn+bn+1+…|=|1+an+1/an+an+2/an+…||bn/an+bn+1/an+bn+2/an+…|=|1+an+1/an+an+2/an+…||bn/an||1+bn+1/bn+bn+2/bn+…|=|an/bn||1+an+1/an+an+2/an+an+3/an+…||1+bn+1/bn+bn+2/bn+bn+3/bn+…|.From this follows our
assertion.Remark 2.7.
The conditionliminfn→∞|1+bn+1/bn+bn+2/bn+…|>0 can be satisfied, for example, if limsupn→∞|bn+1/bn|<1/2. In fact, if limsupn→∞|bn+1/bn|<r<1/2. then there exists n0∈ℕ such that for every n>n0, we have(2.7)|bn+1bn+bn+2bn+bn+3bn+…|≤|bn+1bn|+|bn+2bn+1||bn+1bn|+|bn+3bn+2||bn+2bn+1||bn+1bn|+…≤r+r2+r3+…=r1−r,and thus |1+bn+1/bn+bn+2/bn+…|≥(1−2r)/(1−r)>0. The condition limsupn→∞|1+an+1/an+an+2/an+…|<∞ can be satisfied, for example, if limsupn→∞|an+1/an|<1. Indeed, if limsupn→∞|an+1/an|=α<1, then there exists n0∈ℕ such that for every n>n0, we have(2.8)|1+an+1an+an+2an+…|=|1+an+1an+an+2an+1an+1an+an+3an+2an+2an+1an+1an+…|≤1+β+β2+β3+…=11−β<∞,whereα<β<1.Conversely, from the condition liminfn→∞|1+bn+1/bn+bn+2/bn+…|>0 need not follow the condition limsupn→∞|(bn+1/bn)|<1/2.  For example, if we put a1≠0, a2n=1/2n, a2n+1=−1/2n, n=1,2,…, then ∑n=1∞an is convergent series and limsupn→∞|an+1/an|=1, limsupn→∞|1+an+1/an+an+2/an+…|=1.


## 3.τ-Convergent Series

Definition 3.1 (See [10]).
We say that a sequence{an;n∈ℕ} has τ-limit a real number L and we write τ-limn→∞an=L,
if for each ε>0 the set A(ε)={n;|an−L|≥ε} belongs to the ideal τ,
where τ is an admissible ideal of subsets of ℕ which is additive (if A,B∈τ,
then A∪B∈τ), hereditary (if B⊂A∈τ,
then B∈τ), containing all singletons and not
containing ℕ.We denote byτf the ideal of all finite subsets of ℕ.Definition 3.2 (See [10]).
We say that∑n=1∞anτ-converges to a real number L and we write  τ-∑n=1∞an=L if for each ε>0 the set A(ε)={n;|∑k=1nak−L|≥ε} belongs to the ideal τ,
where τ is an admissible ideal of subsets of ℕ.Definition 3.3.
Let∑n=1∞an, ∑n=1∞bn be τ-convergent real series with nonzero terms
such that bn+bn+1+…≠0,n∈ℕ. A series ∑n=1∞an is called τ-faster convergent than ∑n=1∞bn if τ-limn→∞(an+1+an+2+…)/(bn+1+bn+2+…)=0.Definition 3.4 (See [9]).
Letτ be an admissible ideal of subsets of ℕ.
A number x∈R is said to be a τ-cluster point of ∑n=1∞xn if for each ε>0 the set {n∈ℕ;|∑k=1nxk−x|<ε} is not from τ.Remark 3.5.
Of course, ifτ is an admissible ideal of subsets of ℕ,
by τ-cluster point of a real sequence {xn;n∈ℕ}, we mean a number x∈ℝ, where for each ε>0 the set {n∈ℕ;|xn−x|<ε} is not from τ. Moreover, we say that ∞ (−∞) is the τ-cluster point of a real sequence {xn;n∈ℕ} if for each c>0 (c<0) the set {n∈ℕ;xn>c} ({n∈ℕ;xn<c}) is not from τ.Remark 3.6.
If{xn;n∈ℕ} is a real sequence, τ is an admissible ideal, and X={x;xis aτ-cluster point of{xn;n∈ℕ}}, then X≠∅., Indeed, if {xn;n∈ℕ} is bounded, then by [8] there exists a τ-cluster point of {xn;n∈ℕ}. If {xn;n∈ℕ} is not bounded, then either +∞ or −∞ is τ-cluster point of {xn;n∈ℕ} according to Remark 3.5 or there exists l∈ℝ such that {n∈ℕ;|xn|>l}∈τ. If for some l∈ℝ{n∈ℕ;|xn|>l}∈τ,
then Kl={n∈ℕ;|xn|≤l}∉τ. Consider the ideal υ={Kl∩A;A∈τ}. Because {xn;n∈Kl} is a bounded set, there exists x⋆∈ℝ,|x⋆|≤l such that x⋆ is a υ-cluster point of {xn;n∈Kl}. Let ε>0.
Since x⋆ is a υ-cluster point of {xn;n∈Kl},
then the set {n∈Kl;|xn−x⋆|<ε}∉υ and then also is not from τ.
Since τ has a hereditary property, then {n∈ℕ;|xn−x⋆|<ε}∉τ.
So x⋆ is a cluster point of {xn;n∈ℕ}.Definition 3.7.
Letτ be an admissible ideal of subsets of ℕ.
Let ∑n=1∞xn be an infinite series of real numbers and let X={x∈ℝ;xis aτ-cluster point of∑n=1∞xn}. If X is bounded, then s=supX (s=infX) is said to be a τ-limsupn→∞∑n=1∞xn (τ-liminfn→∞∑n=1∞xn). If supX=∞ or ∞ is τ-cluster point (infX=−∞ or −∞ is τ-cluster point), then τ-limsupn→∞∑n=1∞xn=∞ (τ-liminfn→∞∑n=1∞xn=−∞).Definition 3.8.
We say that a sequence{xn;n∈ℝ} is τ-bounded above (τ-bounded bellow), if there exist m∈ℝ such that {n∈ℕ;xn>m}∈τ ({n∈ℕ;xn<m}∈τ) and is τ-bounded if it is τ-bounded above and below simultaneously.It is obvious that if∑n=1∞an is faster convergent than ∑n=1∞bn, then ∑n=1∞an is τ-faster convergent, the ∑n=1∞bn.
Generally, from the fact that ∑n=1∞an is τ-faster convergent, ∑n=1∞bn does not hold that ∑n=1∞an is faster convergent than ∑n=1∞bn.It is obvious that forτ-convergent series where τ has the property  p1:(3.1)ifM∈τ,thenM+1={n+1;n∈M∩N}∈τ,M−1={n−1>0;n∈M∩N}∈τ(τf or τst={A⊂N; asymptotic density (A)=0} have p1), we obtain similar lemma as Lemma 2.6.Lemma 3.9.
Letτ be an admissible ideal of subsets of ℕ with property p1. Let ∑n=1∞an, ∑n=1∞bn be τ-convergent real series with nonzero terms.
Let bn+bn+1+…≠0,  for all n∈ℕ.
Let
(3.2)ti(a)=τ-liminfn→∞|1+an+1an+an+2an+…|,ts(a)=τ-limsupn→∞|1+an+1an+an+2an+…|,ti(b)=τ-liminfn→∞|1+bn+1bn+bn+2bn+…|,ts(b)=τ-limsupn→∞|1+bn+1bn+bn+2bn+…|. Then (1)
ifts(a)<∞,ti(b)>0,τ-limn→∞(an/bn)=0, thenτ-limn→∞(an+an+1+…)/(bn+bn+1+…)=0,(2)
ifan+an+1+…≠0for alln∈ℕ,ti(a)>0,ts(b)<∞,τ-limn→∞(an+an+1+…)/(bn+bn+1+…)=0,thenτ-limn→∞(an/bn)=0,(3)
ifan+an+1+…≠0for alln∈ℕ,0<ti(a),ts(a)<∞,0<ti(b),ts(b)<∞,thenτ-limn→∞(an/bn)=0,ifandonlyifτ-limn→∞(an+an+1+…)/(bn+bn+1+…)=0.Proof.
First, we note that ifτ-limsupn→∞xn<∞,
then sequence {xn;n∈ℕ} is τ-bounded above (similarly for τ-liminf). Next, we show that if {an;n∈ℕ} is a τ-convergent sequence to 0 and {bn;n∈ℕ} is a τ-bounded sequence, then {anbn:n∈ℕ} is a τ-convergent sequence to 0. There exist k∈ℝ such that the set Nb={n;|bn|≥k} is from τ.
Let ε>0. The set Na(ε)={n;|an|≥ε/k} is from τ.
If n∈ℕ∖(Na(ε)∪Nb),
then |an|<ε/K and |bn|<k and so |anbn|<ε.
From this and from properties of ideal τ,  follows that {anbn:n∈ℕ}τ-converges to 0. Then the proof follows from
Lemma 2.6.The following examples show that we cannot replace lim
byτ-lim in Lemmas 1.2, 2.1, and 2.2.Example 3.10.
Let∑n=1∞bn be a convergent series with positive terms
such that ∑n=1∞(bn+bn+1+bn+2+…) is convergent (e.g., ∑n=1∞bn=∑n=1∞αn−1,  0<α<1). Let M={ij;1<ij<ij+1,j∈ℕ}, M∈τ be an infinite subset of ℕ,
where τ is an arbitrary admissible ideal with property
p1 different from τf.
We define a real series ∑n=1∞an in the following way:(3.3)an={1nbn,n>1,n≠ij,ij∈M,j=2,3,…bij−1+bij−1+1+…n∈M∖{i1},c,n=1,where c be a real number such that ∑n=1∞an=∑n=1∞bn.
We obtain τ-limn→∞(an/bn)=0, (an+an+1+…)/(bn+bn+1+…)>1 for all n≥i2.Example 3.11.
LetM={nj;n1=1,nj<nj+1,nj+1−nj≥2,j∈ℕ}, M∈τ be an infinite subset of ℕ,
where τ is an arbitrary admissible ideal with property
p1 different from τf.
Let {Bnj}j=1∞ be a sequence of positive real numbers such
that Bnj+1<Bnj<mBnj+1,limj→∞Bnj=0, where m>1.
Put Ak=Bk/2k,k∈M.
Let Ak,Bk for k∉M be defined as follows: if nj<k<nj+1,
where j∈ℕ we put εnj=(Anj−Anj+1)/(nj+1−nj),Ak=Anj−(k−nj)εnj,Bk=Bnj−(k−nj)εnj. For k∉M, we have Ak/Bk<Anj/Bnj+1=Bnj/2njBnj+1<m/2nj,
where nj<k<nj+1. From this and from the definition Ak,k∈M, it follows limk→∞(Ak/Bk)=0, hence τ-limk→∞(Ak/Bk)=0. It is obvious that (Ak−Ak+1)/(Bk−Bk+1)=1 for k≠nj−1,j>1. Let A0>A1 and B0>B1 be  real numbers. Put an=An−1−An,bn=Bn−1−Bn,n∈ℕ. The series ∑n=1∞an,∑n=1∞bn are convergent with positive terms and τ-limn→∞(an+an+1+…)/(bn+bn+1+…)=0,τ-limn→∞(an/bn)=1.Remark 3.12.
Ifτ have not the property p1, we get similar lemma as Lemma 3.9, but with
shift indices of given series. (In general, it does not hold that if τ-limn→∞an=a, then τ-limn→∞an+1=a,
and so on.)


## 4. Kummer Series

In [7], is showed: if ∑n=1∞an is a convergent real series with positive
terms and with an unknown sum a and ∑n=1∞cn is a convergent real series with positive
terms and with a known sum c,
then if limn→∞(an/cn)=p>0 the Kummer series ∑n=1∞bn,
where b1=a1+p(c−c1) and bn=an−pcn for n≥2,
has the same sum as ∑n=1∞an and if bn>0 is a faster convergent series then ∑n=1∞an.In the following lemma, we prove by using Lemma2.6 the
faster convergence of the Kummer series ∑n=1∞bn to the unknown sum of ∑n=1∞an without conditions of positivity of p and terms of ∑n=1∞an,∑n=1∞bn,∑n=1∞cn.Lemma 4.1.
Let∑n=1∞an be a convergent real series with nonzero
terms and with the sum a.
Let liminfn→∞|1+an+1/an+an+2/an+…|>0. Let an+an+1+…≠0 for all n∈ℕ.
Let ∑n=1∞cn be a convergent series with nonzero terms and
with the sum c.
Let limn→∞an/cn=p≠0.
If for the series ∑n=1∞bn,
where b1=a1+p(c−c1) and bn=an−pcn, for n≥2,
is bn≠0, for n≥2, and limsupn→∞|1+bn+1/bn+bn+2/bn+…|<∞,
then∑n=1∞bn=a and ∑n=1∞bn is a faster convergent series than ∑n=1∞an.Proof.
Sincelimn→∞(bn/an)=0,
the proof follows from Lemma 2.6.In the next example, we construct, by using of Remark2.7 and Lemma 4.1, the Kummer series ∑n=1∞bn faster convergent than the series ∑n=1∞an such that series ∑n=1∞an,∑n=1∞cn,∑n=1∞bn does not have positive terms.Example 4.2.
Let∑n=1∞an=∑n=1∞(−1)n/(4n(3n2+√n)), (the sum is unknown). It is evident that limsupn→∞|an+1/an|=1/4<1/2. Let ∑n=1∞cn=∑n=1∞(−1)n/(4nn2), then limn→∞an/cn=1/3  and from ∑n=1∞(xn/n2)=−∫0x(ln(1−t)/t)dt, x∈(−1,1), it follows ∑n=1∞(−1)n/(4nn2)=∫−1/40(ln(1−t)/t)dt   (the sum is known). The Kummer series is ∑n=1∞bn=b1+∑n=2∞(an−(1/3)cn), where ∑n=2∞bn=∑n=2∞(−1)n+1/(4n3n√n(3n2+√n)), b1=1/48+1/3∫−1/40(ln(1−t)/t)dt.
It is obvious that limsupn→∞|bn+1/bn|=1/4<1. Because ∑n=1∞an is series with alternating signs such that for
all n∈ℕ we have that |an+1|<|an|,
we get for n∈ℕ a2n+a2n+1+…>0,a2n+1+a2n+2+…<0. Hence ∑n=1∞bn is a faster convergent series than ∑n=1∞an and it has the same sum.



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