# A Multiparameter Hardy–Hilbert-Type Inequality Containing Partial Sums as the Terms of Series

Authors: Jianquan Liao; Shanhe Wu; Bicheng Yang
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this study, a multiparameter Hardy–Hilbert-type inequality for double series is established, which contains partial sums as the terms of one of the series. Based on the obtained inequality, we discuss the equivalent statements of the best possible constant factor related to several parameters. Moreover, we illustrate how the inequality obtained can generate some new Hardy–Hilbert-type inequalities.


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## Body

## 1. Introduction

In [1], Krnić and Pečarić proposed an interesting result on the extension of Hardy–Hilbert inequality with two parameters, as follows:(1)∑m=1∞∑n=1∞ambnm+nλ<Bλ1,λ2∑m=1∞mp1−λ1−1amp1/p∑n=1∞nq1−λ2−1bnq1/q,where p>1,1/p+1/q=1,am,bn≥0,0<∑m=1∞mp1−λ1−1amp<∞, 0<∑n=1∞nq1−λ2−1bnq<∞, λi∈0,2i=1,2,λ1+λ2=λ∈0,4. The constant factor Bλ1,λ2 in (1) is the best possible, which is represented by the beta function.(2)Bu,v=∫0∞tu−11+tu+vdt=ΓuΓvΓu+v,u,v>0.Obviously, forλ1=1−1/p,λ2=1−1/q, inequality (1) reduces to the classical Hardy–Hilbert inequality ([2], Theorem 315):(3)∑m=1∞∑n=1∞ambnm+n<πsinπ/p∑m=1∞amp1/p∑n=1∞bnq1/q.In a special case, whenλ1=λ2=λ/2,p=q=2, inequality (1) leads to a generalization of Hilbert inequality, i.e.,(4)∑m=1∞∑n=1∞ambnm+nλ<Bλ2,λ2∑m=1∞m1−λam21/2∑n=1∞n1−λbn21/2.Recently, by introducing more parameters, Yang et al. [3] established a further extension of inequality (1), as follows:(5)∑n=1∞∑m=1∞ambnmα+nβλ<1βkλλ21/p1αkλλ11/q×∑m=1∞mp1−αλ−λ2/p+λ1/q−1amp1/p∑n=1∞nq1−βλ−λ1/q+λ2/p−1bnq1/q.Inspired by the inequality (5) above, in this study, we construct and prove a new Hardy–Hilbert-type inequality by replacing the term of the series bn with Bn=∑k=1nbkn=1,2,… in the right-hand side of inequality (5). Our method is mainly based on real analysis techniques and the applications of the Euler–Maclaurin summation formula and Abel’s partial summation formula. For details of various clever uses of these techniques, we refer the readers to [4–10].The rest of the study is organized as follows. We first give some lemmas on the construction of weight function and several identities and inequalities related to the weight function. The results are then applied to derive a multiparameter Hardy–Hilbert-type inequality containing partial sums as the terms of one of the series. Finally, we illustrate the applications of the obtained inequality in discovering new Hardy–Hilbert-type inequalities.


## 2. Some Lemmas

Let us first state the following specified conditions (C1) that we will use in what follows. We suppose that(i)
(C1)p>1,1/p+1/q=1α,β∈0,1,λ∈0,5,λ1∈0,2/α∩0,λ+1,λ2∈0,2/β−1∩0,λ+1,λ^1≔λ−λ2/p+λ1/q, λ^2≔λ−λ1/q+λ2/p, kλ+1λi≔Bλi,λ+1−λii=1,2. Furthermore, for am,bn≥0, the partial sum Bn is defined by Bn≔∑k=1nbkn∈1,2,…, such that Bn=oetnβt>0;n⟶∞ with(6)0<∑m=1∞mp1−αλ^1−1amp<∞,0<∑n=1∞nq1−β1+λ^2−1Bnq<∞.Lemma 1 (See [11]).
(i)
Let−1idi/dtigt>0, t∈m,∞m∈N with gi∞=0gi∞=0i=0,1,2,3, and let Pit and Bii∈N denote, respectively, the Bernoulli functions and the Bernoulli numbers of i-order. Then,(7)∫m∞P2q−1tgtdt=−εqB2q2qgm,0<εq<1;q=1,2,….(ii)
In particular, whenq=1, in view of B2=1/6, we have(8)−112gm<∫m∞P1tgtdt<0.(iii)
Whenq=2, in view of B4=−1/30, we have(9)0<∫m∞P3tgtdt<1120gm.(iv)
(See [11]). If ft>0∈C3m,∞,fi∞=0i=0,1,2,3, then we have the following Euler–Maclaurin summation formulas:(10)∑k=m∞fk=∫m∞ftdt+12fm+∫m∞P1tf′tdt,(11)∫m∞P1tf′tdtt=−112f′m+16∫m∞P3tf‴tdt.Lemma 2.
Fors∈0,6,s2∈0,2/β∩0,s,kss2=Bs2,s−s2, we define the weight coefficient as follows:(12)ϖαs2,m≔mαs−s2∑n=1∞βnβs2−1mα+nβs,m∈N.
Then, the following inequalities hold:(13)0<kss21−O1mαs2<ϖαs2,m<kss2,m∈N,where O1/mαs2≔1/kss2∫01/mαus2−1/1+usdu>0.Proof.
For fixedm∈N, we define the function gm,t by(14)gm,t≔βtβs2−1mα+tβs,t>0.
By virtue of (10), we have(15)∑n=1∞gm,n=∫1∞gm,tdt+12gm,1+∫1∞P1tg′m,tdt=∫0∞gm,tdt−hm,hm≔∫01gm,tdt−12gm,1−∫1∞P1tg′m,tdt.
Note that−1/2gm,1=−β/2mα+1s. Integration by parts, we find(16)∫01gm,tdt=β∫01tβs2−1mα+tβsdt=u=tβ∫01us2−1mα+usdu=1s2∫01dus2mα+us=1s2us2mα+us|01+ss2∫01us2mα+us+1du=1s21mα+1s+ss2s2+1∫01dus2+1mα+us+1>1s21mα+1s+ss2s2+1us2+1mα+us+101+ss+1s2s2+1mα+1s+2∫01us2+1du=1s21mα+1s+λs2s2+11mα+1s+1+ss+1s2s2+1s2+2s+21mα+1s+1,−g′m,t=−ββs2−1tβs2−2mα+tβs+β2stβ+βs2−2mα+tβs+1=−ββs2−1tβs2−2mα+tβs+β2smα+tβ−mαtβs2−2mα+tβs+1=ββs−βs2+1tβs2−2mα+tβs−β2smαtβs2−2mα+tβs+1,then from 0<s2≤2/β,0<β≤1,s2<s≤6, it follows that(17)−1ididtitβs2−2mα+tβs>0,−1ididtitβs2−2mα+tβs+1>0,i=0,1,2,3.
By (8), (9), (10), and (11), we obtain(18)ββs−βs2+1∫1∞P1ttβs2−2mα+tβsdt>−ββs−βs2+112mα+1s,−β2mαs∫1∞P1ttβs2−2mα+tβs+1dt=β2mαs12mα+1s+1−β2mαs6∫1∞P3ttβs2−2mα+tβs+1″dt>β2mαs12mα+1s+1−β2mαs720tβs2−2mα+tβs+1t=1″>β2mα+1−1s12mα+1s+1−β2mα+1s720s+1s+2β2mα+1s+3+βs+15−β−2βs2mα+1s+2+2−βs23−βs2mα+1s+1=β2s12mα+1s−β2s12mα+1s+1−β2s720s+1s+2β2mα+1s+2+βs+15−β−2βs2mα+1s+1+2−βs23−βs2mα+1s.
Thus, we have(19)hm>1mα+1sh1+λmα+1s+1h2+ss+1mα+1s+2h3,where(20)h1≔1s2−β2−β−β2s212−β2s2−βs23−βs2720,h2≔1s2s2+1−β212−β3s+15−β−2βs2720,h3≔1s2s2+1s2+2−β4s+2720.
It is easy to observe thath1≥1/s2−β/2−β−β2s2/12−sβ22−βs23−βs2/720=gs2/720s2, where the function gσσ∈0,2/β is defined by(21)gσ≔720−420β+6sβ2σ+60β2+5sβ3σ2−sβ4σ3.
Therefore, we deduce that, forβ∈0,1,s∈0,6,(22)g′σ=−420β+6sβ2+260β2+5sβ3σ−3β4σ2≤−420β−6sβ2+260β2+5sβ32β=14sβ−180β<0,thus, it follows that h1≥gs2/720s2≥g2/β/720s2=1/6s2>0.
We find that fors2∈0,2/β, h2>β2/6−β2/12−5s+1β2/720=1/12−s+1/140β2>0 and(23)h3≥124−s+2720β3>0,0<s≤6.
Hence, we havehm>0, and then settingt=mα/βu1/β, it follows that(24)ϖαs2,m=mαs−s2∑n=1∞gm,n<mαs−s2∫0∞gm,tdt=βmαs−s2∫0∞tβs2−1mα+tβsdt=∫0∞us2−11+usdu=kss2.
On the other hand, by (10), we have(25)∑n=1∞gm,n=∫1∞gm,tdt+12gm,1+∫1∞P1tg′m,tdt=∫1∞gm,tdt+Hm,Hm≔12gm,1+∫1∞P1tg′m,tdt.
We have obtained that1/2gm,1=β/2mα+1s and g′m,t=−ββs−βs2+1tβs2−2/mα+tβs+β2smαtβs2−2/mα+tβs+1.
Fors2∈0,2/β∩0,s,0<s≤6, by (8), we find −ββs−βs2+1∫1∞P1ttβs2−2/mα+tβsdt>0 and(26)β2mαs∫1∞P1ttβs2−2mα+tβs+1dt>−β2mαs12mα+1s+1>−β2s12mα+1s.
Hence, we have(27)Hm>β2mα+1s−β2s12mα+1s≥β2mα+1s−6β12mα+1s=0,then, we obtain(28)ϖαλ2,m=mαs−s2∑n=1∞gm,n>mαs−s2∫1∞gm,tdt=mαs−s2∫0∞gm,tdt−mαs−s2∫01gm,tdt=kss21−1kss2∫01/mαus2−11+usdu>0,where we set O1/mαs2=1/kss2∫01/mαus2−1/1+usdu satisfying(29)0<∫01/mαus2−11+usdu<∫01/mαus2−1du=1s2mαs2.
Therefore, we obtain the required inequalities (13). This completes the proof of Lemma 2.Lemma 3.
Fors∈0,6,s1∈0,2/α∩0,s,s2∈0,2/β∩0,s, we have the following Hardy–Hilbert-type inequality with the internal variables:(30)I=∑n=1∞∑m=1∞ambnmα+nβs≤1βkss21/p1αkss11/q×∑m=1∞mp1−αs−s2/p+s1/q−1amp1/p∑n=1∞nq1−βs−s1/q+s2/p−1bnq1/q.Proof.
In the same way of proving inequality (13), we can prove the following inequalities for the weight coefficient ωβs1,n:(31)kss11−O1nβs1<ωβs1,n≔nβs−s1∑m=1∞αmαs1−1mα+nβs<kss1,n∈N.
By Hӧlder’s inequality ([12]), we obtain(32)I=∑n=1∞∑m=1∞1mα+nβsmα1−s1/qβnβ−11/pnβ1−s2/pαmα−11/qamnβ1−s2/pαmα−11/qmα1−s1/qβnβ−11/pbn≤∑m=1∞∑n=1∞βmα+nβsmα1−s1p−1nβ−1ampnβ1−s2αmα−1p−11/p∑n=1∞∑m=1∞αmα+nβsnβ1−s2−1mα−1bnqmα1−s1βnβ−1q−11/q=1α1/qβ1/p∑m=1∞ϖαs2,mmp1−αs−s2/p+s1/q−1amp1/p×∑n=1∞ωβs,1nnq1−βs−s1/q+s2/p−1bnq1/q.
Furthermore, by using (13) and (31), we get inequality (30). Lemma 3 is proved.Remark 1.
In particular, for(33)s=λ+1∈1,6,λ∈0,5,s1=λ1∈0,2α∩0,λ+1,s2=λ2+1∈1,2β,λ2∈0,2β−1∩0,λ+1,in (30), replacing bn by Bn, in view of (6), one has(34)∑n=1∞∑m=1∞amBnmα+nβλ+1<1βkλ+1λ2+11/p1αkλ+1λ11/q×∑m=1∞mp1−αλ^1−1amp1/p∑n=1∞nq1−β1+λ^2−1Bnq1/q.Lemma 4.
Fort>0, the following inequality holds:(35)∑n=1∞e−tnβbn≤t∑n=1∞e−tnβBn.Proof.
In view ofBne−tnβ=o1t>0;n⟶∞ and applying Abel’s summation by parts formula, we find(36)∑n=1∞e−tnβbn=limn⟶∞Bne−tnβ+∑n=1∞Bne−tnβ−e−tn+1β=∑n=1∞Bne−tnβ−e−tn+1β.
Since1−e−t<tt>0 and the fact that for β∈0,1,(37)e−tn+1β≥e−tnβ+1⇔etn+1β−nβ−1≤1⇔n+1β−nβ−1=βn+θnβ−1−1≤0,θn∈0,1,by (36), we have(38)∑n=1∞e−tnβbn≤∑n=1∞Bne−tnβ−e−tnβ+1=1−e−t∑n=1∞Bne−tnβ≤t∑n=1∞Bne−tnβ.
Hence, we derive the inequality (35). The proof of Lemma 4 is complete.


## 3. Main Results

Theorem 1.
Under the assumptions described in (C1), we have the following inequality containing partial sums as the terms of series:(39)I≔∑m=1∞∑n=1∞ambnmα+nβλ<λ1βkλ+1λ2+11/p1αkλ+1λ11/q×∑m=1∞mp1−αλ^1−1amp1/p∑n=1∞nq1−β1+λ^2−1Bnq1/q.
In particular, forλ1+λ2=λ with(40)λ1∈0,2α∩0,λ,λ2∈0,2β−1∩0,λ,λ∈0,50<∑m=1∞mp1−αλ1−1amp<∞,0<∑n=1∞nq1−β1+λ2−1Bnq<∞,we have the following inequality:(41)I=∑m=1∞∑n=1∞ambnmα+nβλ<λ2β1/pα1/qBλ1,λ2×∑m=1∞mp1−αλ1−1amp1/p∑n=1∞nq1−β1+λ2−1Bnq1/q.Proof.
By virtue of the fact that(42)1mα+nβλ=1Γλ∫0∞tλ−1e−mα+nβtdt,from (35), we obtain(43)I=1Γλ∑m=1∞∑n=1∞ambn∫0∞tλ−1e−mα+nβtdt=1Γλ∫0∞tλ−1∑m=1∞e−mαtam∑n=1∞e−nβtbndt≤1Γλ∫0∞tλ−1∑m=1∞e−mαtamt∑n=1∞e−nβtBndt=1Γλ∑m=1∞∑n=1∞amBn∫0∞tλ+1−1e−mα+nβtdt=Γλ+1Γλ∑m=1∞∑n=1∞amBnmα+nβλ+1=λ∑m=1∞∑n=1∞amBnmα+nβλ+1.
Then, by (34), we derive inequality (39). Theorem 1 is proved.Remark 2.
Puttingβ=1 in (39), we have(44)∑m=1∞∑n=1∞ambnmα+nλ<λα1/qkλ+1λ2+11/pkλ+1λ11/q×∑m=1∞mp1−αλ^1−1amp1/p∑n=1∞n−qλ^2−1Bnq1/q.Theorem 2.
Ifλ1+λ2=λ∈0,5, then the constant factor(45)λα1/qkλ+1λ2+11/pkλ+1λ11/q,in (44) is the best possible. Moreover, if(46)λ1,λ−λ2∈0,2α∩0,λ,λ2,λ−λ1∈0,1∩0,λ,and the constant factor λ/α1/qkλ+1λ2+11/pkλ+1λ11/q in (44) is the best possible, then we have λ1+λ2=λ..Proof.
Ifλ1+λ2=λ∈0,5, then we find λ1∈0,2/α∩0,λ,λ2∈0,1∩0,λ,(47)kλ+1λ2+1=kλ+1λ1=Bλ1,λ2+1=λ2λBλ1,λ2,and then inequality (44) reduces to(48)∑m=1∞∑n=1∞ambnmα+nλ<λα1/qBλ1,λ2∑m=1∞mp1−αλ1−1amp1/p∑n=1∞n−qλ2−1Bnq1/q.
For any0<ε<qλ2, we set a˜m≔mαλ1−ε/p−1,b˜n≔nλ2−ε/q−1m,n∈N. Then, we have(49)B˜n≔∑k=1nb˜k=∑k=1nkλ2−ε/p−1<∫0ntλ2−ε/q−1dt=1λ2−ε/qnλ2−ε/q,n∈N,B˜n=oetn,t>0;n⟶∞.
If there exists a positive constantM≤λ2/α1/qBλ1,λ2, (48) is valid when we replace λ2/α1/qBλ1,λ2 by M. Then, we have(50)I˜≔∑n=1∞∑m=1∞ambnmα+nλa˜mb˜n<M∑m=1∞mp1−αλ1−1a˜mp1/p∑n=1∞n−qλ2−1B˜nq1/q.
By (50) and the decreasing property of series, we obtain(51)I˜<M∑m=1∞mp1−αλ1−1mpαλ1−αε−p1/p1λ2−ε/q∑n=1∞n−qλ2−1nqλ2−ε1/q=Mλ2−ε/q1+∑m=1∞m−αε−11/p1+∑n=2∞n−ε−11/q<Mλ2−ε/q1+∫1∞x−αε−1dx1/q1+∫1∞y−ε−1dy1/q=1εα1/pMλ2−ε/qαε+11/pε+11/q.
By employing (31) (for β=1,s=λ,s1=λ1) and setting(52)λ˜1≔λ1−εp∈0,2α∩0,λ,0<λ˜2=λ2+εp=λ−λ˜1<λ,we deduce that(53)I˜=1α∑n=1∞nλ˜2∑m=1∞αm+nλmαλ˜1−1n−ε−1=1α∑n=1∞ω1λ˜1,nn−ε−1>1αkλλ˜1∑n=1∞1−O1nλ˜1n−ε−1=1αBλ˜1,λ˜2∑n=1∞n−ε−1−∑n=1∞O1nλ1−ε/p+1>1αBλ˜1,λ˜2∫1∞y−ε−1dy−O1=1εαBλ1−εp,λ2+εp1−εO1.
Thus, we have(54)1αBλ1−εp,λ2+εp1−εO1<εI˜<1α1/pMλ2−ε/qαε+11/pε+11/q.
Forε⟶0+, in view of the continuity of the beta function, we find λ2/α1/pBλ1,λ2≤M. Hence, M=λ2/α1/pBλ1,λ2 is the best possible constant factor of (48).
On the other hand, sinceλ1,λ−λ2∈0,2/α∩0,λ,λ2,λ−λ1∈0,1∩0,λ, we find(55)λ^1+λ^2=λ−λ2p+λ1q+λ−λ1q+λ1p=λ,0<λ^1,λ^2<λp+λq=λ,λ^1≤2/αp+2/α1=2α,λ^2≤1,and λ^2kλλ^1=λ^2Bλ^1,λ^2∈R+=0,∞.
By using Hӧlder’s inequality ([12]), we obtain(56)kλ+1λ^1=kλ+1λ−λ2p+λ1q=∫0∞11+uλ+1uλ−λ2/p+λ1/q−1du=∫0∞11+uλ+1uλ−λ2−1/puλ1−1/qdu≤∫0∞11+uλ+1uλ−λ2−1du1/p∫0∞11+uλ+1uλ1−1du1/q=∫0∞11+vλ+1vλ2+1−1dv1/p∫0∞11+uλ+1uλ1−1du1/q=kλ+1λ2+11/pkλ+1λ11/q.
If the constant factorλ/α1/qkλ+1λ2+11/pkλ+1λ11/q in (44) is the best possible, then by (48) (for λi=λ^ii=1,2), we have the following inequality:(57)λα1/qkλ+1λ2+11/pkλ+1λ11/q≤λ^2α1/qBλ^1,λ^2=λα1/qkλ+1λ^1,namely, kλ+1λ^1≥kλ+1λ2+11/pkλ+1λ11/q. Hence, (56) keeps the form of equality.
We observe that (56) keeps the form of equality if and only if there exist constants A and B, such that they are not all zero satisfying ([12]) Auλ−λ2−1=Buλ1−1 a.e. in R+.
Assuming thatA≠0, we have uλ−λ2−λ1=B/A a. e. in R+, and then, λ−λ2−λ1=0, that is, λ1+λ2=λ. The proof of Theorem 2 is complete.


## 4. Some Applications

Theorem 3.
Under the assumptions described in (C1), we have the following inequality which is equivalent to inequality (39):(58)J≔∑m=1∞mqαλ^1−1∑n=1∞bnmα+nβλq1/q<λ1βkλ+1λ2+11/p1αkλ+1λ11/q∑n=1∞nq1−β1+λ^2−1Bnq1/q.
In particular, forλ1+λ2=λ with(59)λ1∈0,2α∩0,λ,λ2∈0,2β−1∩0,λ,λ∈0,5,we have the following inequality which is equivalent to (41):(60)∑m=1∞mqαλ^1−1∑n=1∞bnmα+nβλq1/q<λ2β1/pα1/qBλ1,λ2∑n=1∞nq1−β1+λ2−1Bnq1/q.Proof.
Assuming that (58) is valid, by using Hӧlder’s inequality ([12]), we obtain(61)I=∑m=1∞m1/q−αλ^1amm−1/q+αλ^1∑n=1∞bnmα+nβλ≤∑m=1∞mp1−αλ^1−1amp1/pJ.
Then, by (58), we have (39). On the other hand, assuming that (39) is valid, we set(62)am≔mqαλ^1−1∑n=1∞bnmα+nβλq−1,m∈N.
IfJ=0, then (58) is naturally valid; if J=∞, then it is impossible that makes (58) valid, namely, J<∞. Suppose that 0<J<∞. By (39), we have(63)∑m=1∞mp1−αλ^1−1amp=Jq=I<λ1βkλ+1λ2+11/p1αkλ+1λ11/qJq−1∑n=1∞nq1−β1+λ^2−1Bnq1/q,J=∑m=1∞mp1−αλ^1−1amp1/q<λ1βkλ+1λ2+11/p1αkλ+1λ11/q∑n=1∞nq1−β1+λ^2−1Bnq1/q.
Hence, (58) is valid. Therefore, inequality (58) is equivalent to inequality (39). Theorem 3 is proved.Remark 3.
Puttingβ=1 in (58), we get the following inequality which is equivalent to (44).(64)∑m=1∞mqαλ^1−1∑n=1∞bnmα+nλq1/q<λα1/qkλ+1λ2+11/pkλ+1λ11/q∑n=1∞n−qλ^2−1Bnq1/q.Theorem 4.
Ifλ1+λ2=λ∈0,5, then the constant factor(65)λα1/qkλ+1λ2+11/pkλ+1λ11/q,in (64) is the best possible. Moreover, if(66)λ1,λ−λ2∈0,2α∩0,λ,λ2,λ−λ1∈0,1∩0,λ,and the constant factor λ/α1/qkλ+1λ2+11/pkλ+1λ11/q in (64) is the best possible, then we have λ1+λ2=λ.Proof.
If the constant factorλ/α1/qkλ+1λ2+11/pkλ+1λ11/q in (64) is not the best possible, then by (60) (for β=1), we would reach a contradiction that the same constant factor in (44) is not the best possible. On the other hand, if(67)λ1,λ−λ2∈0,2α∩0,λ,λ2,λ−λ1∈0,1∩0,λ,and the constant factor λ/α1/qkλ+1λ2+11/pkλ+1λ11/q in (64) is the best possible, then, in view of the equivalency of (44) and (64) and I=Jq, the same constant factor in (20) is the best possible. Furthermore, by Theorem 2, we have λ1+λ2=λ. This completes the proof of Theorem 4.Remark 4.
If we takeβ=1 in (41) and (59), respectively, then for(68)λ1∈0,2α∩0,λ,λ2∈0,1∩0,λ,λ∈0,5,we have the following inequalities with the best possible constant factor λ2/α1/qBλ1,λ2:(69)∑m=1∞∑n=1∞ambnmα+nλ<λ2α1/qBλ1,λ2∑m=1∞mp1−αλ1−1amp1/p∑n=1∞n−qλ2−1Bnq1/q,(70)∑m=1∞mqαλ1−1∑n=1∞bnmα+nλq1/q<λ2α1/qBλ1,λ2∑n=1∞n−qλ2−1Bnq1/q.(i)
Takingλ=α=1,λ1=1/q,λ2=1/p in (69) and (70), respectively, we obtain the following inequalities with the best possible constant factor π/psinπ/p:(71)∑m=1∞∑n=1∞ambnm+n<πpsinπ/p∑m=1∞amp1/p∑n=1∞Bnnq1/q,(72)∑m=1∞∑n=1∞bnm+nq1/q<πpsinπ/p∑n=1∞Bnnq1/q.(ii)
Takingλ=α=1,λ1=1/p,λ2=1/q in (69) and (70), respectively, we get the following inequalities with the best possible constant factor π/qsinπ/p:(73)∑m=1∞∑n=1∞ambnm+n<πqsinπ/p∑m=1∞mp−2amp1/p∑n=1∞Bnqn21/q,(74)∑m=1∞∑n=1∞bnm+nq1/q<πqsinπ/p∑n=1∞Bnqn21/q.
In particular, if we takep=q=2, then both (71) and (73) reduce to(75)∑m=1∞∑n=1∞ambnm+n<π2∑m=1∞am2∑n=1∞Bnn21/2,and both (72) and (74) reduce to(76)∑m=1∞∑n=1∞bnm+n21/2<π2∑n=1∞Bnn21/2,which is the equivalent form of inequality (75).


## 5. Conclusions

In this study, by using the weight coefficients, idea of introducing parameters, Euler–Maclaurin summation formula, and Abel’s partial summation formula, a multiparameter Hardy–Hilbert-type inequality is established, which contains partial sums as terms in one of the series. As applications, several equivalent statements of the best possible constant factor related to several parameters are provided. Moreover, we illustrate that our main results can generate more Hardy–Hilbert-type inequalities such as the inequalities (69)–(76) asserted by Remark 4.



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