# A new discrete Hilbert-type inequality involving partial sums

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## Abstract

In this paper, we derive a new discrete Hilbert-type inequality involving partial sums. Moreover, we show that the constant on the right-hand side of this inequality is the best possible. As an application, we consider some particular settings.

## Introduction

The Hilbert inequality  asserts that

$$\sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{a_{m}b_{n}}{m+n}< \frac{ \pi }{\sin (\pi /p)} \Biggl( \sum_{m=1}^{\infty }a_{m}^{p} \Biggr)^{ \frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }b_{n}^{q} \Biggr)^{\frac{1}{q}},$$
(1)

holds for non-negative sequences $$a_{m}$$ and $$b_{n}$$, provided that $$(\sum_{m=1}^{\infty }a_{m}^{p} )^{\frac{1}{p}}>0$$ and $$(\sum_{n=1}^{\infty }b_{n}^{q} )^{\frac{1}{q}}>0$$. The parameters p and q appearing in (1) are mutually conjugate, i.e. $${\frac{1}{p}}+{\frac{1}{q}}=1$$, where $$p>1$$. In addition, the constant $$\frac{\pi }{\sin (\pi /p)}$$ is the best possible in the sense that it can not be replaced with a smaller constant so that (1) still holds.

The Hilbert inequality is one of the most interesting inequalities in mathematical analysis. For a detailed review of the starting development of the Hilbert inequality the reader is referred to monograph . The most important recent results regarding Hilbert-type inequalities are collected in monographs  and .

In 2006, Krnić and Pečarić , obtained the following generalization of classical Hilbert inequality.

### Theorem 1

Let $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, and let $$2< s\leq 14$$. Suppose that $$\alpha _{1}\in [-\frac{1}{q},\frac{1}{q} )$$, $$\alpha _{2}\in [-\frac{1}{p},\frac{1}{p} )$$ and $$p\alpha _{2}+q \alpha _{1}=2-s$$. If $$\sum_{m=1}^{\infty }m^{pq\alpha _{1}-1}a_{m}^{p}< \infty$$ and $$\sum_{n=1}^{\infty }n^{pq\alpha _{2}-1}b_{n}^{q}<\infty$$, then

$$\sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{(m+n)^{s}}< B(1-p \alpha _{2}, p\alpha _{2}+s-1) \Biggl(\sum _{m=1}^{\infty }m^{pq\alpha _{1}-1}a_{m}^{p} \Biggr)^{\frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }n ^{pq\alpha _{2}-1}b_{n}^{q} \Biggr)^{\frac{1}{q}},$$
(2)

where the constant $$B(1-p\alpha _{2}, p\alpha _{2}+s-1)$$ is the best possible.

In the last few years, considerable attention is given to a class of Hilbert-type inequalities where the functions and sequences are replaced by certain integral or discrete operators. For example: in 2013, Azar  introduced a new Hilbert-type integral inequality including functions $$F(x)=\int _{0}^{x} f(t)\,dt$$ and $$g(y)=\int _{0}^{y} g(t)\,dt$$. For some related Hilbert-type inequalities where the functions and sequences are replaced by certain integral or discrete operators, the reader is referred to  and .

The main objective of this paper is to derive a discrete Hilbert-type inequality involving partial sums, similar to a result of Azar . Such inequality is derived by virtue of inequality (2) and some well-known classical inequalities. As an application, we consider some particular settings.

## Preliminaries and lemma

Recall that the Gamma function $$\varGamma (\theta )$$ and the Beta function $$B ( \mu ,\nu )$$ are defined, respectively, by

\begin{aligned}& \varGamma (\theta )= \int _{0}^{\infty }t^{\theta -1}e^{-t}\,dt, \quad \theta >0, \\& B ( \mu ,\nu ) = \int _{0}^{\infty }\frac{t^{\mu -1}}{ ( t+1 ) ^{\mu +\nu }}\,dt, \quad \mu ,\nu >0, \end{aligned}

and they satisfy the following relation

$$B ( \mu ,\nu ) =\frac{\varGamma (\mu )\varGamma (\nu )}{\varGamma (\mu +\nu )}.$$

By the definition of the Gamma function, the following equality holds:

$$\frac{1}{ ( m+n ) ^{\lambda }}=\frac{1}{\varGamma ( \lambda ) } \int _{0}^{\infty }t^{\lambda -1}e^{- ( m+n ) t}\,dt.$$
(3)

To prove our main results we need the following lemma.

### Lemma 2

Let $$a_{m}>0$$, $$a_{m}\in \ell ^{1}$$, $$A_{m}=\sum_{k=1}^{m} a_{k}$$, then for $$t>0$$, we have

$$\sum_{m=1}^{\infty }e^{-tm}a_{m} \leq t\sum_{m=1}^{\infty }e^{-tm}A _{m}.$$
(4)

### Proof

Using Abel’s summation by parts formula and the inequality $$1-\frac{1}{e ^{t}}\leq t$$, we have

\begin{aligned} \sum_{m=1}^{\infty }e^{-tm}a_{m} =&\lim_{m\to \infty }A_{m}e^{-t(m+1)}+ \sum _{m=1}^{\infty }A_{m}\bigl(e^{-tm}-e^{-t(m+1)} \bigr) \\ =& \biggl(1-\frac{1}{e^{t}} \biggr)\sum_{m=1}^{\infty }e^{-tm}A_{m} \\ \leq &t\sum_{m=1}^{\infty }e^{-tm}A_{m}. \end{aligned}

The lemma is proved. □

## Main results

### Theorem 3

Let $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$\lambda >0$$, $$a_{m}, b_{n}>0$$, $$a _{m}, b_{n}\in \ell ^{1}$$, define $$A_{m}=\sum_{k=1}^{m}a_{k}$$, $$B_{n}= \sum_{k=1}^{n}b_{k}$$. If $$\sum_{m=1}^{\infty }m^{pq\alpha _{1}-1}A_{m} ^{p}<\infty$$ and $$\sum_{n=1}^{\infty }n^{pq\alpha _{2}-1}B_{n}^{q}< \infty$$, then

$$\sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{(m+n)^{ \lambda }}< C \Biggl(\sum _{m=1}^{\infty }m^{pq\alpha _{1}-1}A_{m}^{p} \Biggr) ^{\frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }n^{pq\alpha _{2}-1}B_{n}^{q} \Biggr) ^{\frac{1}{q}},$$
(5)

where $$\alpha _{1}\in [-\frac{1}{q},0 )$$, $$\alpha _{2}\in [-\frac{1}{p},0 )$$ and $$p\alpha _{2}+q\alpha _{1}=-\lambda$$. In addition, the constant $$C=pq\alpha _{1}\alpha _{2}B(-p\alpha _{2}, -q\alpha _{1})$$ is the best possible in (5).

### Proof

Using (3), the left-hand side of inequality (5) can be expressed in the following form:

\begin{aligned} \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{(m+n)^{ \lambda }} =&\frac{1}{\varGamma (\lambda )}\sum _{m=1}^{\infty }\sum_{n=1} ^{\infty }a_{m}b_{n} \biggl( \int _{0}^{\infty }t^{\lambda -1}e^{-(m+n)t}\,dt \biggr) \\ =&\frac{1}{\varGamma (\lambda )} \int _{0}^{\infty }t^{\lambda -1} \Biggl(\sum _{m=1}^{\infty }e^{-tm}a_{m} \Biggr) \Biggl(\sum_{n=1}^{\infty }e^{-tn}b _{n} \Biggr)\,dt. \end{aligned}
(6)

Now, by applying inequality (4) and equality (3) to the previous equality, we have

\begin{aligned} \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{(m+n)^{ \lambda }} \leq &\frac{1}{\varGamma (\lambda )} \int _{0}^{\infty }t^{ \lambda +1} \Biggl(\sum _{m=1}^{\infty }e^{-tm}A_{m} \Biggr) \Biggl(\sum_{n=1} ^{\infty }e^{-tn}B_{n} \Biggr)\,dt \\ =&\frac{1}{\varGamma (\lambda )}\sum_{m=1}^{\infty }\sum _{n=1}^{\infty }A_{m}B_{n} \biggl( \int _{0}^{\infty }t^{\lambda +1}e^{-(m+n)t}\,dt \biggr) \\ =&\frac{\varGamma (\lambda +2)}{\varGamma (\lambda )}\sum_{m=1}^{\infty } \sum _{n=1}^{\infty }\frac{A_{m}B_{n}}{(m+n)^{\lambda +2}}. \end{aligned}
(7)

Moreover, the last double series represents the left-hand side of the Hilbert-type inequality (2) for $$s=2+\lambda$$, that is, we have the inequality

$$\sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{A_{m}B_{n}}{(m+n)^{ \lambda +2}}< B(1-p\alpha _{2}, p\alpha _{2}+\lambda +1) \Biggl(\sum _{m=1} ^{\infty }m^{pq\alpha _{1}-1}A_{m}^{p} \Biggr)^{\frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }n^{pq\alpha _{2}-1}B_{n}^{q} \Biggr)^{\frac{1}{q}},$$

so by (7) we get

$$\sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{(m+n)^{ \lambda }}< C \Biggl(\sum _{m=1}^{\infty }m^{pq\alpha _{1}-1}A_{m}^{p} \Biggr) ^{\frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }n^{pq\alpha _{2}-1}B_{n}^{q} \Biggr) ^{\frac{1}{q}}.$$

Now we shall prove that the constant factor is the best possible. Assuming that the constant C is not the best possible, then there exists a positive constant K such that $$K< C$$ and (5) still remains valid if C is replaced by K. Further, consider the $$\tilde{a}_{m}=m^{-q\alpha _{1}-1-\frac{\varepsilon }{p}}$$ and $$\tilde{b}_{n}=n^{-p\alpha _{2}-1-\frac{\varepsilon }{q}}$$, where $$\varepsilon >0$$ is sufficiently small number. Then, we have

$$\tilde{A}_{m}=\sum_{k=1}^{m} \tilde{a}_{k}=\sum_{k=1}^{m}m^{-q\alpha _{1}-1-\frac{\varepsilon }{p}} \leq \int _{0}^{m}x^{-q\alpha _{1}-1-\frac{ \varepsilon }{p}}\,dx= \frac{m^{-q\alpha _{1}-\frac{\varepsilon }{p}}}{-q \alpha _{1}-\frac{\varepsilon }{p}},$$

and similarly

$$\tilde{B}_{n}=\sum_{k=1}^{n} \tilde{b}_{k}\leq \frac{n^{-p\alpha _{2}-\frac{ \varepsilon }{q}}}{-p\alpha _{2}-\frac{\varepsilon }{q}}.$$

Inserting the above sequences in (5), the right-hand side of (5) becomes

\begin{aligned} &K \Biggl(\sum_{m=1}^{\infty }m^{pq\alpha _{1}-1} \tilde{A}_{m}^{p} \Biggr) ^{\frac{1}{p}} \Biggl(\sum _{n=1}^{\infty }n^{pq\alpha _{2}-1}\tilde{B} _{n}^{q} \Biggr)^{\frac{1}{q}} \\ & \quad \leq K \Biggl(\sum_{m=1}^{\infty }m^{pq\alpha _{1}-1} \frac{m^{-pq\alpha _{1}-\varepsilon }}{(-q\alpha _{1}-\frac{\varepsilon }{p})^{p}} \Biggr) ^{\frac{1}{p}} \Biggl(\sum _{n=1}^{\infty }n^{pq\alpha _{2}-1}\frac{n ^{-pq\alpha _{2}-\varepsilon }}{(-p\alpha _{2}-\frac{\varepsilon }{q})^{q}} \Biggr) ^{\frac{1}{q}} \\ & \quad =\frac{K}{(-q\alpha _{1}-\frac{\varepsilon }{p})(-p\alpha _{2}-\frac{ \varepsilon }{q})} \Biggl(\sum_{m=1}^{\infty }m^{-1-\varepsilon } \Biggr) ^{\frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }n^{-1-\varepsilon } \Biggr) ^{\frac{1}{q}} \\ & \quad =\frac{K}{(-q\alpha _{1}-\frac{\varepsilon }{p})(-p\alpha _{2}-\frac{ \varepsilon }{q})} \Biggl(1+\sum_{m=2}^{\infty }m^{-1-\varepsilon } \Biggr) \\ & \quad \leq \frac{K}{(-q\alpha _{1}-\frac{\varepsilon }{p})(-p\alpha _{2}-\frac{ \varepsilon }{q})} \biggl(1+ \int _{1}^{\infty }x^{-1-\varepsilon }\,dx \biggr) \\ & \quad =\frac{K(1+\varepsilon )}{\varepsilon (-q\alpha _{1}-\frac{\varepsilon }{p})(-p\alpha _{2}-\frac{\varepsilon }{q})}. \end{aligned}
(8)

Now, let us estimate the left-hand side of inequality (5). Namely, by inserting the above defined sequences $$\tilde{a}_{m}$$ and $$\tilde{b}_{n}$$ in the left-hand side of inequality (5), we get the inequality

\begin{aligned} \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{\tilde{a}_{m}\tilde{b} _{n}}{(m+n)^{\lambda }} &=\sum _{m=1}^{\infty }\sum_{n=1}^{\infty } \frac{m ^{-q\alpha _{1}-1-\frac{\varepsilon }{p}}n^{-p\alpha _{2}-1-\frac{ \varepsilon }{q}}}{(m+n)^{\lambda }} \\ &\geq \int _{1}^{\infty } \int _{1}^{\infty }\frac{x^{-q\alpha _{1}-1-\frac{ \varepsilon }{p}}y^{-p\alpha _{2}-1-\frac{\varepsilon }{q}}}{(x+y)^{ \lambda }}\,dx \,dy \\ &= \int _{1}^{\infty }x^{-1-\varepsilon } \int _{1/x}^{\infty }\frac{u ^{-p\alpha _{2}-1-\frac{\varepsilon }{q}}}{(1+u)^{\lambda }}\,du \,dx \\ &= \int _{1}^{\infty }x^{-1-\varepsilon } \biggl( \int _{0}^{\infty }\frac{u ^{-p\alpha _{2}-1-\frac{\varepsilon }{q}}}{(1+u)^{\lambda }}\,du- \int _{0}^{1/x}\frac{u^{-p\alpha _{2}-1-\frac{\varepsilon }{q}}}{(1+u)^{ \lambda }}\,du \biggr)\,dx \\ &\geq \int _{1}^{\infty }x^{-1-\varepsilon } \biggl( \int _{0}^{\infty }\frac{u ^{-p\alpha _{2}-1-\frac{\varepsilon }{q}}}{(1+u)^{\lambda }}\,du- \int _{0}^{1/x}u^{-p\alpha _{2}-1-\frac{\varepsilon }{q}}\,du \biggr)\,dx \\ &=\frac{1}{\varepsilon }\cdot B \biggl(-p\alpha _{2}- \frac{\varepsilon }{q},-q\alpha _{1}+\frac{\varepsilon }{q} \biggr)- \frac{1}{ (p \alpha _{2}+\frac{\varepsilon }{q} ) (p\alpha _{2}-\frac{ \varepsilon }{p} )}. \end{aligned}
(9)

It follows from inequalities (8) and (9) that

$$B \biggl(-p\alpha _{2}-\frac{\varepsilon }{q},-q\alpha _{1}+\frac{\varepsilon }{q} \biggr)-\frac{\varepsilon }{ (p\alpha _{2}+\frac{\varepsilon }{q} ) (p\alpha _{2}-\frac{\varepsilon }{p} )}\leq \frac{K(1+ \varepsilon )}{(-q\alpha _{1}-\frac{\varepsilon }{p})(-p\alpha _{2}-\frac{ \varepsilon }{q})}.$$
(10)

Now, letting $$\varepsilon \to 0+$$, relation (10) yields a contradiction with the assumption $$K< C$$. So the constant C, in inequality (5) is the best possible. □

Considering Theorem 3, equipped with parameters $$\lambda =1$$, $$\alpha _{1}=-\frac{1}{q^{2}}$$, $$\alpha _{2}=-\frac{1}{p^{2}}$$, we obtain the following result.

### Corollary 4

Let $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, and let $$a_{m}, b_{n}>0$$ with $$a_{m}, b_{n}\in \ell ^{1}$$. If $$\sum_{m=1}^{\infty }m^{-p}A_{m}^{p}< \infty$$ and $$\sum_{n=1}^{\infty }n^{-q}B_{n}^{q}<\infty$$, then

$$\sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{m+n}< \frac{ \pi }{pq\sin (\pi /p)} \Biggl( \sum_{m=1}^{\infty } \biggl(\frac{A_{m}}{m} \biggr) ^{p} \Biggr)^{\frac{1}{p}} \Biggl(\sum _{n=1}^{\infty } \biggl(\frac{B _{n}}{n} \biggr)^{q} \Biggr)^{\frac{1}{q}}.$$
(11)

where the constant $$\frac{\pi }{pq\sin (\pi /p)}$$ is the best possible.

### Remark 5

It should be noticed here that if sequences $$a_{m}, b_{n}\in \ell ^{1}$$ such that $$\sum_{m=1}^{\infty }a_{m}^{p}<\infty$$ and $$\sum_{n=1}^{\infty }b_{n}^{q}<\infty$$, inequality (11) provides refinement of the Hilbert inequality. Indeed, by Hardy’s inequality, the series $$\sum_{m=1}^{\infty } (\frac{A_{m}}{m} ) ^{p}$$ and $$\sum_{n=1}^{\infty } (\frac{B_{n}}{n} )^{q}$$ are converge. So, inequality (11) holds. The Hilbert inequality becomes after applying Hardy’s inequality on the right-hand side of inequality (11).

Letting $$\alpha _{1}=\alpha _{2}=\frac{-\lambda }{pq}$$ in Theorem 3, we can obtain the following Hilbert-type inequality.

### Corollary 6

Let $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, and let $$0<\lambda \leq \min \{p,q \}$$, $$a_{m}, b_{n}>0$$ with $$a_{m}, b_{n}\in \ell ^{1}$$. If $$\sum_{m=1} ^{\infty }m^{-\lambda -1}A_{m}^{p}<\infty$$ and $$\sum_{n=1}^{\infty }n ^{-\lambda -1}B_{n}^{q}<\infty$$, then

$$\sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{(m+n)^{ \lambda }}< \frac{\lambda ^{2}}{pq} B \biggl(\frac{\lambda }{q}, \frac{ \lambda }{p} \biggr) \Biggl(\sum _{m=1}^{\infty }m^{-\lambda -1}A_{m} ^{p} \Biggr)^{\frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }n^{-\lambda -1}B _{n}^{q} \Biggr)^{\frac{1}{q}},$$
(12)

where the constant $$\frac{\lambda ^{2}}{pq} B (\frac{\lambda }{q}, \frac{\lambda }{p} )$$ is the best possible.

## Conclusion

In the present study, we have established a discrete Hilbert-type inequality involving partial sums. Moreover, we have proved that the constant on the right-hand side of this inequality is the best possible. As an application, we considered some particular settings.

## References

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## Author information

All authors read and approved the final version of the manuscript.

Correspondence to Tserendorj Batbold.

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### Competing interests

The authors declare that they have no competing interests. 