A sharpened version of Hardy’s inequality for parameter p
          =
          5
          /
          4

Yongping Deng; Shanhe Wu; Deng He

doi:10.1186/1029-242X-2013-63

Research
Open Access

A sharpened version of Hardy’s inequality for parameter $p = 5 / 4$

Yongping Deng¹,
Shanhe Wu¹Email author and
Deng He²

Journal of Inequalities and Applications20132013:63

https://doi.org/10.1186/1029-242X-2013-63

Received: 10 December 2012
Accepted: 1 February 2013
Published: 20 February 2013

Abstract

The purpose of this paper is to investigate a sharpened version of Hardy’s inequality for parameter $p = 5 / 4$ . By evaluating the weight coefficient $W (k, 5 / 4)$ , sharpened Hardy’s inequality that contains the best coefficient $η_{5 / 4} = 0.46 \dots$ is established.

MSC:26D15, 26D20, 26D07.

Keywords

Hardy’s inequality
weight coefficient
sharpened inequality
exponential parameter
best coefficient

1 Introduction

Let

p > 1

,

1 / p + 1 / q = 1

,

a_{n} \geq 0

(

n = 1, 2, \dots

),

0 < \sum_{n = 1}^{\infty} a_{n}^{p} < \infty

. Then

\sum_{n = 1}^{\infty} {(\frac{1}{n} \sum_{k = 1}^{n} a_{k})}^{p} < q^{p} \sum_{n = 1}^{\infty} a_{n}^{p},

(1)

where $q^{p} = {(\frac{p}{p - 1})}^{p}$ is the best coefficient. Inequality (1) is called Hardy’s inequality which is of great use in the field of modern mathematics (see [1, 2]).

A special case of (1) yields the following inequalities:

\sum_{n = 1}^{\infty} {(\frac{1}{n} \sum_{k = 1}^{n} a_{k})}^{2} < 4 \sum_{n = 1}^{\infty} a_{n}^{2},

(2)

\sum_{n = 1}^{\infty} {(\frac{1}{n} \sum_{k = 1}^{n} a_{k})}^{3} < \frac{27}{8} \sum_{n = 1}^{\infty} a_{n}^{3} .

(3)

In 1998, Yang and Zhu [3] evaluated the weight coefficient

W (k, p)

,

W (k, p) = k^{1 - 1 / p} \sum_{n = k}^{\infty} \frac{1}{n^{p}} {(\sum_{j = 1}^{n} \frac{1}{j^{1 / p}})}^{p - 1}, k = 1, 2, \dots,

(4)

and established an improved version of inequality (2) as follows:

\sum_{n = 1}^{\infty} {(\frac{1}{n} \sum_{k = 1}^{n} a_{k})}^{2} < 4 \sum_{n = 1}^{\infty} (1 - \frac{1}{3 \sqrt{n} + 5}) a_{n}^{2} .

(5)

With the same approach, that is, evaluating the weight coefficient

W (k, p)

, Huang [4–7] gave some improvements on Hardy’s inequality for

p = 3

and

p = 3 / 2

, i.e.,

\sum_{n = 1}^{\infty} {(\frac{1}{n} \sum_{k = 1}^{n} a_{k})}^{3} < \frac{27}{8} \sum_{n = 1}^{\infty} (1 - \frac{3}{19 n^{2 / 3}}) a_{n}^{3},

(6)

\sum_{n = 1}^{\infty} {(\frac{1}{n} \sum_{k = 1}^{n} a_{k})}^{3 / 2} \leq 3 \sqrt{3} \sum_{n = 1}^{\infty} (1 - \frac{1}{5} \cdot \frac{1}{\sqrt[3]{n} + 3}) a_{n}^{3 / 2} .

(7)

Some further extensions of Hardy’s inequality related to the range of parameter p were given in Huang [7, 8].

In 2005, Yang [9] proved an inequality for the weight coefficient

W (k, 2)

W (k, 2) = \sqrt{k} \sum_{n = k}^{\infty} \frac{1}{n^{2}} (\sum_{j = 1}^{n} \frac{1}{\sqrt{j}}) \leq 4 [1 - \frac{1}{\sqrt{k}} (1 - \frac{1}{4} W (1, 2))]

and established the following inequality:

\sum_{n = 1}^{\infty} {(\frac{1}{n} \sum_{k = 1}^{n} a_{k})}^{2} < 4 \sum_{n = 1}^{\infty} (1 - \frac{θ_{2}}{\sqrt{n}}) a_{n}^{2},

(8)

where $θ_{2} = 1 - \frac{1}{4} W (1, 2) = 0.13788928 \dots$ is the best coefficient under the weight coefficient $W (k, 2)$ .

In 2009, Zhang and Xu made use of the monotonicity theorem [10–13] and obtained an improvement of inequality (1):

\sum_{n = 1}^{\infty} {(\frac{1}{n} \sum_{k = 1}^{n} a_{k})}^{p} \leq {(\frac{p}{p - 1})}^{p} \sum_{n = 1}^{\infty} (1 - \frac{c_{p}}{2 {(n - 1 / 2)}^{1 - 1 / p}}) a_{n}^{p},

(9)

where

c_{p} = {\begin{cases} (p - 1) [1 - 2^{1 / p} (1 - 1 / p)], & 1 < p \leq 2, \\ 1 - 2^{1 - 1 / p} {(1 - 1 / p)}^{p - 1}, & p > 2 . \end{cases}

By evaluating the weight coefficient

W (k, p)

, and with the help of an inequality-proving package called BOTTEMA [14, 15], He [16] investigated a sharpened version of Hardy’s inequality for

p \in N

and obtained the following improved version of inequality (3):

\sum_{n = 1}^{\infty} {(\frac{1}{n} \sum_{k = 1}^{n} a_{k})}^{3} \leq \frac{27}{8} \sum_{n = 1}^{\infty} (1 - \frac{θ_{3}}{n^{2 / 3}}) a_{n}^{3},

(10)

where $θ_{3} = 1 - \frac{8}{27} W (1, 3) = 0.1673 \dots$ is the best coefficient under the weight coefficient $W (k, 3)$ .

In addition, in [16] the author wrote the computer program HDISCOVER to accomplish the automated verification of the following inequality for

p \in N

(N is the set of natural numbers):

\sum_{n = 1}^{\infty} {(\frac{1}{n} \sum_{k = 1}^{n} a_{k})}^{p} \leq {(\frac{p}{p - 1})}^{p} \sum_{n = 1}^{\infty} (1 - \frac{θ_{p} (1)}{n^{1 - 1 / p}}) a_{n}^{p},

(11)

where $θ_{p} (1) = 1 - {(\frac{p - 1}{p})}^{p} W (1, p)$ is the best coefficient of (11) under the weight coefficient $W (k, p)$ .

Recently, based on the program HDISCOVER 2012 written by Deng, He and Wu [17], an automated verification of inequality (11) is achieved for $p \in Q$ (Q is the set of rational numbers).

For more detailed information of Hardy’s inequality, we refer the interested readers to relevant research papers [10, 12, 18–23].

In this paper, by evaluating the weight coefficient

W (k, 5 / 4)

, we establish an improvement of Hardy’s inequality for parameter

p = 5 / 4

as follows:

\sum_{n = 1}^{\infty} {(\frac{1}{n} \sum_{k = 1}^{n} a_{k})}^{5 / 4} \leq 5^{5 / 4} \sum_{n = 1}^{\infty} (1 - \frac{1}{10} \cdot \frac{1}{n^{1 / 5} + η_{5 / 4}}) a_{n}^{5 / 4},

(12)

where $η_{5 / 4} = \frac{5^{5 / 4}}{10 [5^{5 / 4} - W (1, 5 / 4)]} - 1 = 0.46 \dots$ is the best coefficient under the weight coefficient $W (k, 5 / 4)$ .

2 Lemmas

To prove the main results in Section 3, we will use the following lemmas.

Lemma 1 (see[22])

If

p > 1

, then for all integers

n \geq 1

, it holds that

\begin{aligned} \frac{p}{p - 1} n^{1 - 1 / p} - \frac{p}{p - 1} + \frac{1}{2} + \frac{1}{2 n^{1 / p}} \\ \leq \sum_{j = 1}^{n} \frac{1}{j^{1 / p}} \leq \frac{p}{p - 1} n^{1 - 1 / p} - \frac{p}{p - 1} + \frac{1}{2} + \frac{1}{2 n^{1 / p}} + \frac{1}{12 p} - \frac{1}{12 p n^{1 + 1 / p}} . \end{aligned}

(13)

Lemma 2 (see[3])

If

p > 1

, then for all integers

n \geq k \geq 1

, it holds that

\frac{1}{(p - 1) k^{p - 1}} + \frac{1}{2 k^{p}} < \sum_{n = k}^{\infty} \frac{1}{n^{p}} < \frac{1}{(p - 1) k^{p - 1}} + \frac{1}{2 k^{p}} + \frac{p}{12 k^{p + 1}} .

Lemma 3 Let

p > 1

,

1 / p + 1 / q = 1

, and let

g_{r}

,

g_{l}

be the functions defined by

g_{r} (x) = - \frac{6 p + 12 q - 1}{12 p q x^{1 / q}} + \frac{1}{2 q x} - \frac{1}{12 p q x^{2}}, g_{l} (x) = - \frac{p + 2 q}{2 p q x^{1 / q}} + \frac{1}{2 q x}, x \in [1, + \infty) .

Then $- 1 < g_{r} (x) < 0$ , $- 1 < g_{l} (x) < 0$ .

Proof Since $p > 1$ , $1 / p + 1 / q = 1$ , hence $1 / x^{1 + 1 / p} \geq 1 / x^{2}$ for $x \in [1, + \infty)$ .

Further, we have

\begin{aligned} g_{r}^{'} (x) & = \frac{6 p + 12 q - 1}{12 p q^{2} x^{1 + 1 / q}} - \frac{1}{2 q x^{2}} + \frac{1}{6 p q x^{3}} \\ \geq \frac{6 p + 12 q - 1}{12 p q^{2} x^{2}} - \frac{1}{2 q x^{2}} + \frac{1}{6 p q x^{3}} \\ = \frac{(5 p x + x + 2 p) (p - 1)}{12 p^{3} x^{3}} > 0, \end{aligned}

and consequently, $g_{r}$ is strictly increasing on $[1, + \infty)$ .

Now, from $g_{r} (1) = - 1 / p > - 1$ and ${lim}_{x \to + \infty} g_{r} (x) = 0$ , it follows that $g_{r} (x) \geq g_{r} (1) = - 1 / p > - 1$ and $g_{r} (x) < 0$ .

Similarly, from

\begin{matrix} g_{l}^{'} (x) = \frac{p + 2 q}{2 p q^{2} x^{1 + 1 / q}} - \frac{1}{2 q x^{2}} \geq \frac{p + 2 q}{2 p q^{2} x^{2}} - \frac{1}{2 q x^{2}} = \frac{p - 1}{2 p^{2} x^{2}} > 0, \\ g_{l} (1) = - 1 / p > - 1 and lim_{x \to + \infty} g_{l} (x) = 0, \end{matrix}

we deduce that $- 1 < g_{l} (x) < 0$ .

Lemma 3 is proved. □

Lemma 4 Let

- 1 < g (x) < 0

. If

α \in (0, 1]

, then

\begin{aligned} (1 + g (x)) (1 + (α - 1) g (x) + \frac{(α - 1) (α - 2)}{2} g^{2} (x)) \\ \leq {(1 + g (x))}^{α} \leq 1 + α g (x) + \frac{α (α - 1)}{2} g^{2} (x) . \end{aligned}

If

α \in [1, 2]

, then

{(1 + g (x))}^{α} \geq 1 + α g (x) + \frac{α (α - 1)}{2} g^{2} (x) .

Proof When

α \in (0, 1]

. By using the Maclaurin formula

\begin{aligned} {(1 + g (x))}^{α} = & 1 + α g (x) + \frac{α (α - 1)}{2} g^{2} (x) \\ + \frac{α (α - 1) (α - 2) {(1 + θ g (x))}^{α - 3}}{6} g^{3} (x), θ \in (0, 1), \end{aligned}

and noticing

- 1 < g (x) < 0

, we find

\begin{matrix} 1 + θ g (x) > 1 + g (x) > 0, \\ \frac{α (α - 1) (α - 2) {(1 + θ g (x))}^{α - 3}}{6} g^{3} (x) \leq 0, \\ \frac{(α - 1) (α - 2) (α - 3) {(1 + θ g (x))}^{α - 4}}{6} g^{3} (x) \geq 0 . \end{matrix}

Thus

\begin{matrix} {(1 + g (x))}^{α} \leq 1 + α g (x) + \frac{α (α - 1)}{2} g^{2} (x), \\ \begin{aligned} {(1 + g (x))}^{α} & = (1 + g (x)) {(1 + g (x))}^{α - 1} \\ \geq (1 + g (x)) (1 + (α - 1) g (x) + \frac{(α - 1) (α - 2)}{2} g^{2} (x)) . \end{aligned} \end{matrix}

When

α \in [1, 2]

. We have

\frac{α (α - 1) (α - 2) {(1 + θ g (x))}^{α - 3}}{6} g^{3} (x) \geq 0 .

Thus

{(1 + g (x))}^{α} \geq 1 + α g (x) + \frac{α (α - 1)}{2} g^{2} (x) .

The proof of Lemma 4 is complete. □

Lemma 5 Let

p > 1

,

1 / p + 1 / q = 1

,

n \geq k \geq 1

, and let

[x]

denote the greatest integer less than or equal to the real number x. Then we have

\begin{aligned} W (k, p) \leq & q^{p - 1} k^{1 / q} \sum_{n = k}^{\infty} [\frac{1}{n^{1 + 1 / q}} {(1 + g_{r} (n))}^{[p] - 1} \\ \times (1 + (p - [p]) g_{r} (n) + \frac{(p - [p]) (p - [p] - 1)}{2} g_{r}^{2} (n))] . \end{aligned}

Proof By Lemma 1 and the identity

p q = p + q

,

q (p - 1) = p

, it follows that

\begin{aligned} W (k, p) & = k^{1 - 1 / p} \sum_{n = k}^{\infty} \frac{1}{n^{p}} {(\sum_{j = 1}^{n} \frac{1}{j^{1 / p}})}^{p - 1} \\ \leq k^{1 - 1 / p} \sum_{n = k}^{\infty} \frac{1}{n^{p}} {(\frac{p}{p - 1} n^{1 - 1 / p} - \frac{p}{p - 1} + \frac{1}{2} + \frac{1}{2 n^{1 / p}} + \frac{1}{12 p} - \frac{1}{12 p n^{1 + 1 / p}})}^{p - 1} \\ = k^{1 / q} \sum_{n = k}^{\infty} \frac{1}{n^{p}} {(q n^{1 / q} - \frac{6 p + 12 q - 1}{12 p} + \frac{1}{2 n^{1 / p}} - \frac{1}{12 p n^{1 + 1 / p}})}^{p - 1} \\ = k^{1 / q} \sum_{n = k}^{\infty} \frac{1}{n^{p}} q^{p - 1} n^{(p - 1) / q} {(1 - \frac{6 p + 12 q - 1}{12 p q n^{1 / q}} + \frac{1}{2 q n} - \frac{1}{12 p q n^{2}})}^{p - 1} \\ = q^{p - 1} k^{1 / q} \sum_{n = k}^{\infty} \frac{1}{n^{1 + 1 / q}} {(1 + g_{r} (n))}^{p - 1} . \end{aligned}

Combining Lemmas 3 and 4, we obtain

\begin{matrix} W (k, p) \\ \leq q^{p - 1} k^{1 / q} \sum_{n = k}^{\infty} [\frac{1}{n^{1 + 1 / q}} {(1 + g_{r} (n))}^{[p] - 1} {(1 + g_{r} (n))}^{p - [p]}] \\ \leq q^{p - 1} k^{1 / q} \sum_{n = k}^{\infty} [\frac{1}{n^{1 + 1 / q}} {(1 + g_{r} (n))}^{[p] - 1} \\ \times (1 + (p - [p]) g_{r} (n) + \frac{(p - [p]) (p - [p] - 1)}{2} g_{r}^{2} (n))] . \end{matrix}

This completes the proof of Lemma 5. □

Lemma 6 Let

1 / p + 1 / q = 1

,

n \geq k \geq 1

. If

p \in (1, 2)

, then

\begin{aligned} W (k, p) \geq & q^{p - 1} k^{1 / q} \sum_{n = k}^{\infty} [\frac{1}{n^{1 + 1 / q}} {(1 + g_{l} (n))}^{[p]} \\ \times (1 + (p - [p] - 1) g_{l} (n) + \frac{(p - [p] - 1) (p - [p] - 2)}{2} g_{l}^{2} (n))] . \end{aligned}

If

p \in [2, + \infty)

, then

\begin{aligned} W (k, p) \geq & q^{p - 1} k^{1 / q} \sum_{n = k}^{\infty} [\frac{1}{n^{1 + 1 / q}} {(1 + g_{l} (n))}^{[p] - 2} \\ \times (1 + (p - [p] + 1) g_{l} (n) + \frac{(p - [p] + 1) (p - [p])}{2} g_{l}^{2} (n))] . \end{aligned}

Proof Since

p q = p + q

,

q (p - 1) = p

, using Lemma 1 gives

\begin{aligned} W (k, p) & = k^{1 - 1 / p} \sum_{n = k}^{\infty} \frac{1}{n^{p}} {(\sum_{j = 1}^{n} \frac{1}{j^{1 / p}})}^{p - 1} \\ \geq k^{1 - 1 / p} \sum_{n = k}^{\infty} \frac{1}{n^{p}} {(\frac{p}{p - 1} n^{1 - 1 / p} - \frac{p}{p - 1} + \frac{1}{2} + \frac{1}{2 n^{1 / p}})}^{p - 1} \\ = k^{1 / q} \sum_{n = k}^{\infty} \frac{1}{n^{p}} {(q n^{1 / q} - \frac{p + 2 q}{2 p} + \frac{1}{2 n^{1 / p}})}^{p - 1} \\ = k^{1 / q} \sum_{n = k}^{\infty} \frac{1}{n^{p}} q^{p - 1} n^{(p - 1) / q} {(1 - \frac{p + 2 q}{2 p q n^{1 / q}} + \frac{1}{2 q n})}^{p - 1} \\ = q^{p - 1} k^{1 / q} \sum_{n = k}^{\infty} \frac{1}{n^{1 + 1 / q}} {(1 + g_{l} (n))}^{p - 1} . \end{aligned}

When

p \in (1, 2)

. From Lemmas 3 and 4, we have

\begin{aligned} W (k, p) \geq & q^{p - 1} k^{1 / q} \sum_{n = k}^{\infty} \frac{1}{n^{1 + 1 / q}} {(1 + g_{l} (n))}^{[p] - 1} {(1 + g_{l} (n))}^{p - [p]} \\ \geq & q^{p - 1} k^{1 / q} \sum_{n = k}^{\infty} [\frac{1}{n^{1 + 1 / q}} {(1 + g_{l} (n))}^{[p]} \\ \times (1 + (p - [p] - 1) g_{l} (n) + \frac{(p - [p] - 1) (p - [p] - 2)}{2} g_{l}^{2} (n))] . \end{aligned}

When

p \in [2, + \infty)

. Using Lemmas 3 and 4, we obtain

\begin{aligned} W (k, p) \geq & q^{p - 1} k^{1 / q} \sum_{n = k}^{\infty} \frac{1}{n^{1 + 1 / q}} {(1 + g_{l} (n))}^{[p] - 2} {(1 + g_{l} (n))}^{p - [p] + 1} \\ \geq & q^{p - 1} k^{1 / q} \sum_{n = k}^{\infty} [\frac{1}{n^{1 + 1 / q}} {(1 + g_{l} (n))}^{[p] - 2} \\ \times (1 + (p - [p] + 1) g_{l} (n) + \frac{(p - [p] + 1) (p - [p])}{2} g_{l}^{2} (n))] . \end{aligned}

Lemma 6 is proved. □

Lemma 7 (see[3])

Let

p > 1

,

a_{n} \geq 0

(

n = 1, 2, \dots

),

0 < \sum_{n = 1}^{\infty} a_{n}^{p} < \infty

. Then

\sum_{n = 1}^{\infty} {(\frac{1}{n} \sum_{k = 1}^{n} a_{k})}^{p} \leq \sum_{k = 1}^{\infty} [k^{1 - 1 / p} \sum_{n = k}^{\infty} \frac{1}{n^{p}} {(\sum_{j = 1}^{n} \frac{1}{j^{1 / p}})}^{p - 1} a_{k}^{p}] = \sum_{k = 1}^{\infty} W (k, p) a_{k}^{p} .

3 Main results

Theorem 1 For an arbitrary natural number k, the following inequality holds true:

W (k, 5 / 4) < R_{5 / 4} (k),

where

\begin{aligned} R_{5 / 4} (k) = & 5^{1 / 4} (5 - \frac{133}{240 k^{1 / 5}} - \frac{17, 689}{144, 000 k^{2 / 5}} + \frac{25}{48 k} - \frac{19}{192 k^{6 / 5}} - \frac{17, 689}{480, 000 k^{7 / 5}} + \frac{467}{4, 224 k^{2}} \\ + \frac{133}{18, 000 k^{11 / 5}} + \frac{97}{38, 400 k^{3}} + \frac{133}{60, 000 k^{16 / 5}} + \frac{61}{504, 000 k^{4}} + \frac{19}{240, 000 k^{5}}) . \end{aligned}

Proof

Using Lemma 5 gives

W (k, 5 / 4) \leq 5^{1 / 4} k^{1 / 5} \sum_{n = k}^{\infty} [\frac{1}{n^{6 / 5}} (1 + \frac{1}{4} g_{r} (n) - \frac{3}{32} g_{r}^{2} (n))] = 5^{1 / 4} k^{1 / 5} \sum_{n = k}^{\infty} r_{5 / 4} (n),

where

\begin{aligned} r_{5 / 4} (n) = & \frac{1}{n^{6 / 5}} - \frac{133}{600 n^{7 / 5}} - \frac{17, 689}{240, 000 n^{8 / 5}} + \frac{1}{40 n^{11 / 5}} + \frac{133}{8, 000 n^{12 / 5}} - \frac{41}{9, 600 n^{16 / 5}} \\ - \frac{133}{60, 000 n^{17 / 5}} + \frac{1}{4, 000 n^{21 / 5}} - \frac{1}{60, 000 n^{26 / 5}} . \end{aligned}

Hence

\begin{aligned} W (k, 5 / 4) \leq & 5^{1 / 4} k^{1 / 5} \sum_{n = k}^{\infty} (\frac{1}{n^{6 / 5}} - \frac{133}{600 n^{7 / 5}} - \frac{17, 689}{240, 000 n^{8 / 5}} + \frac{1}{40 n^{11 / 5}} + \frac{133}{8, 000 n^{12 / 5}} \\ - \frac{41}{9, 600 n^{16 / 5}} - \frac{133}{60, 000 n^{17 / 5}} + \frac{1}{4, 000 n^{21 / 5}} - \frac{1}{60, 000 n^{26 / 5}}) . \end{aligned}

Using Lemma 2 and taking

p = 6 / 5, 7 / 5, 8 / 5, 11 / 5, 12 / 5, 16 / 5, 17 / 5, 21 / 5, 26 / 5

in the right-hand side of inequality (13), respectively, we get

\begin{matrix} \sum_{n = k}^{\infty} \frac{1}{n^{6 / 5}} < \frac{5}{k^{1 / 5}} + \frac{1}{2 k^{6 / 5}} + \frac{1}{10 k^{11 / 5}}, \\ - \sum_{n = k}^{\infty} \frac{133}{600 n^{7 / 5}} < - \frac{133}{240 k^{2 / 5}} - \frac{133}{1, 200 k^{7 / 5}}, \\ \dots, \\ \sum_{n = k}^{\infty} \frac{1}{4, 000 n^{21 / 5}} < \frac{1}{12, 800 k^{16 / 5}} + \frac{1}{8, 000 k^{21 / 5}} + \frac{7}{80, 000 k^{26 / 5}}, \\ - \sum_{n = k}^{\infty} \frac{1}{60, 000 n^{26 / 5}} < - \frac{1}{252, 000 k^{21 / 5}} - \frac{1}{120, 000 k^{26 / 5}} . \end{matrix}

Adding up the above inequalities, we obtain

W (k, 5 / 4) < R_{5 / 4} (k) .

Theorem 1 is proved. □

Theorem 2 For an arbitrary natural number k, the following inequality holds true:

W (k, 5 / 4) > L_{5 / 4} (k),

where

\begin{aligned} L_{5 / 4} (k) = & 5^{1 / 4} (5 - \frac{9}{16 k^{1 / 5}} - \frac{81}{640 k^{2 / 5}} - \frac{15, 309}{25, 600 k^{3 / 5}} + \frac{25}{48 k} - \frac{45}{448 k^{6 / 5}} + \frac{3, 159}{51, 200 k^{7 / 5}} \\ - \frac{15, 309}{64, 000 k^{8 / 5}} + \frac{17}{1, 408 k^{2}} - \frac{129}{5, 120 k^{11 / 5}} + \frac{891}{12, 800 k^{12 / 5}} - \frac{45, 927}{640, 000 k^{13 / 5}} \\ - \frac{27}{102, 400 k^{3}} - \frac{567}{64, 000 k^{16 / 5}} + \frac{1}{12, 800 k^{4}} - \frac{3, 213}{640, 000 k^{21 / 5}}) . \end{aligned}

Proof

Utilizing Lemma 6 gives

\begin{aligned} W (k, 5 / 4) & \geq 5^{1 / 4} k^{1 / 5} \sum_{n = k}^{\infty} [\frac{1}{n^{6 / 5}} (1 + g_{l} (n)) (1 - \frac{3}{4} g_{l} (n) + \frac{21}{32} g_{l}^{2} (n))] \\ = 5^{1 / 4} k^{1 / 5} \sum_{n = k}^{\infty} l_{5 / 4} (n), \end{aligned}

where

\begin{aligned} l_{5 / 4} (n) = & \frac{1}{n^{6 / 5}} - \frac{9}{40 n^{7 / 5}} - \frac{243}{3, 200 n^{8 / 5}} - \frac{15, 309}{32, 000 n^{9 / 5}} + \frac{1}{40 n^{11 / 5}} + \frac{27}{1, 600 n^{12 / 5}} \\ + \frac{5, 103}{32, 000 n^{13 / 5}} - \frac{3}{3, 200 n^{16 / 5}} - \frac{567}{32, 000 n^{17 / 5}} + \frac{21}{32, 000 n^{21 / 5}} . \end{aligned}

Hence

\begin{aligned} W (k, 5 / 4) \geq & 5^{1 / 4} k^{1 / 5} \sum_{n = k}^{\infty} (\frac{1}{n^{6 / 5}} - \frac{9}{40 n^{7 / 5}} - \frac{243}{3, 200 n^{8 / 5}} - \frac{15, 309}{32, 000 n^{9 / 5}} + \frac{1}{40 n^{11 / 5}} \\ + \frac{27}{1, 600 n^{12 / 5}} + \frac{5, 103}{32, 000 n^{13 / 5}} - \frac{3}{3, 200 n^{16 / 5}} \\ - \frac{567}{32, 000 n^{17 / 5}} + \frac{21}{32, 000 n^{21 / 5}}) . \end{aligned}

Using Lemma 2 and taking

p = 6 / 5, 7 / 5, 8 / 5, 9 / 5, 11 / 5, 12 / 5, 13 / 5, 16 / 5, 17 / 5, 21 / 5

in the left-hand side of inequality (13), respectively, we get

\begin{matrix} \sum_{n = k}^{\infty} \frac{1}{n^{6 / 5}} > \frac{5}{k^{1 / 5}} + \frac{1}{2 k^{6 / 5}}, \\ - \sum_{n = k}^{\infty} \frac{9}{40 n^{7 / 5}} > - \frac{9}{16 k^{2 / 5}} - \frac{9}{80 k^{7 / 5}} - \frac{21}{800 k^{12 / 5}}, \\ \dots, \\ - \sum_{n = k}^{\infty} \frac{567}{32, 000 n^{17 / 5}} > - \frac{189}{25, 600 k^{12 / 5}} - \frac{567}{64, 000 k^{17 / 5}} - \frac{3, 213}{640, 000 k^{22 / 5}}, \\ \sum_{n = k}^{\infty} \frac{21}{32, 000 n^{21 / 5}} > \frac{21}{102, 400 k^{16 / 5}} + \frac{21}{64, 000 k^{21 / 5}} . \end{matrix}

Adding up the above inequalities, we obtain

W (k, 5 / 4) > L_{5 / 4} (k) .

Theorem 2 is proved. □

Theorem 3 Let

a_{n} \geq 0

(

n = 1, 2, \dots

),

0 < \sum_{n = 1}^{\infty} a_{n}^{5 / 4} < \infty

. Then

\sum_{n = 1}^{\infty} {(\frac{1}{n} \sum_{k = 1}^{n} a_{k})}^{5 / 4} \leq 5^{5 / 4} \sum_{n = 1}^{\infty} (1 - \frac{1}{10} \cdot \frac{1}{n^{1 / 5} + η_{5 / 4}}) a_{n}^{5 / 4},

(14)

where $η_{5 / 4} = \frac{5^{5 / 4}}{10 [5^{5 / 4} - W (1, 5 / 4)]} - 1 = 0.46 \dots$ is the best possible under the weight coefficient $W (k, 5 / 4)$ .

Proof

By Lemma 7, we have

\sum_{n = 1}^{\infty} {(\frac{1}{n} \sum_{k = 1}^{n} a_{k})}^{5 / 4} \leq \sum_{k = 1}^{\infty} W (k, 5 / 4) a_{k}^{5 / 4} .

Therefore, to prove inequality (14), it suffices to show that

W (k, 5 / 4) \leq 5^{5 / 4} (1 - \frac{1}{10} \cdot \frac{1}{k^{1 / 5} + η_{5 / 4}}) .

(15)

Obviously, inequality (15) becomes an equality for $k = 1$ . In what follows, we will assume that $k \geq 2$ .

By Theorem 1

W (k, 5 / 4) < R_{5 / 4} (k)

, we need only to prove that

R_{5 / 4} (k) \leq 5^{5 / 4} (1 - \frac{1}{10} \cdot \frac{1}{k^{1 / 5} + η_{5 / 4}}) .

Note that

η_{5 / 4} = \frac{5^{5 / 4}}{10 [5^{5 / 4} - W (1, 5 / 4)]} - 1 = 0.44 \dots > \frac{11}{25},

it suffices to show

R_{5 / 4} (k) \leq 5^{5 / 4} (1 - \frac{1}{10} \cdot \frac{1}{k^{1 / 5} + 11 / 25}) .

(16)

Substituting

k = x^{5}

in (16), inequality (16) becomes

R_{5 / 4} (x^{5}) \leq 5^{5 / 4} (1 - \frac{1}{10} \cdot \frac{1}{x + 11 / 25}), where x \geq \sqrt[5]{2},

which is equivalent to the following inequality:

\begin{array}{r} \begin{aligned} 5^{1 / 4} (5 - \frac{133}{240 x} - \frac{17, 689}{144, 000 x^{2}} + \frac{25}{48 x^{5}} - \frac{19}{192 x^{6}} - \frac{17, 689}{480, 000 x^{7}} + \frac{467}{4, 224 x^{10}} + \frac{133}{18, 000 x^{11}} \\ + \frac{97}{38, 400 x^{15}} + \frac{133}{60, 000 x^{16}} + \frac{61}{504, 000 x^{20}} + \frac{19}{240, 000 x^{25}}) \\ \leq 5^{5 / 4} (1 - \frac{1}{10} \cdot \frac{1}{x + 11 / 25}) \end{aligned} \\ \begin{aligned} \Leftrightarrow 5 - \frac{133}{240 x} - \frac{17, 689}{144, 000 x^{2}} + \frac{25}{48 x^{5}} - \frac{19}{192 x^{6}} - \frac{17, 689}{480, 000 x^{7}} + \frac{467}{4, 224 x^{10}} + \frac{133}{18, 000 x^{11}} \\ + \frac{97}{38, 400 x^{15}} + \frac{133}{60, 000 x^{16}} + \frac{61}{504, 000 x^{20}} + \frac{19}{240, 000 x^{25}} \leq 5 - \frac{1}{2} \cdot \frac{1}{x + 11 / 25} \\ \Leftrightarrow - \frac{133}{240 x} - \frac{17, 689}{144, 000 x^{2}} + \frac{25}{48 x^{5}} - \frac{19}{192 x^{6}} - \frac{17, 689}{480, 000 x^{7}} + \frac{467}{4, 224 x^{10}} + \frac{133}{18, 000 x^{11}} \\ + \frac{97}{38, 400 x^{15}} + \frac{133}{60, 000 x^{16}} + \frac{61}{504, 000 x^{20}} + \frac{19}{240, 000 x^{25}} + \frac{1}{2} \cdot \frac{1}{x + 11 / 25} \leq 0 \\ \Leftrightarrow - \frac{f (x)}{221, 760, 000 x^{25} (25 x + 11)} \leq 0, \end{aligned} \end{array}

(17)

where

\begin{aligned} f (x) = & 300, 300, 000 x^{25} + 2, 032, 838, 500 x^{24} + 299, 651, 660 x^{23} - 2, 887, 500, 000 x^{21} \\ - 721, 875, 000 x^{20} + 445, 702, 950 x^{19} + 89, 895, 498 x^{18} - 612, 937, 500 x^{16} \\ - 310, 656, 500 x^{15} - 18, 024, 160 x^{14} - 14, 004, 375 x^{11} - 18, 451, 125 x^{10} \\ - 5, 407, 248 x^{9} - 671, 000 x^{6} - 295, 240 x^{5} - 438, 900 x - 193, 116 . \end{aligned}

From the hypothesis

x \geq \sqrt[5]{2} > 1.14

, we have

\begin{matrix} 300, 300, 000 x^{25} + 2, 032, 838, 500 x^{24} + 299, 651, 660 x^{23} - 2, 887, 500, 000 x^{21} \\ - 721, 875, 000 x^{20} + 445, 702, 950 x^{19} + 89, 895, 498 x^{18} \\ > (300, 300, 000 \times {1.14}^{4} + 2, 032, 838, 500 \times {1.14}^{3} + 299, 651, 660 \times {1.14}^{2} \\ - 2, 887, 500, 000) x^{21} - 721, 875, 000 x^{20} + 445, 702, 950 x^{19} + 89, 895, 498 x^{18} \\ = 1, 020, 861, 716 x^{21} - 721, 875, 000 x^{20} + 445, 702, 950 x^{19} + 89, 895, 498 x^{18} \\ = [(1, 020, 861, 716 x - 721, 875, 000) x^{2} + 445, 702, 950 x + 89, 895, 498] x^{18} \\ > [(1, 020, 861, 716 \times 1.14 - 721, 875, 000) \times {1.14}^{2} \\ + 445, 702, 950 \times 1.14 + 89, 895, 498] x^{18} \\ = 1, 172, 299, 661 x^{18} . \end{matrix}

Further, we have

\begin{aligned} f (x) > & 1, 172, 299, 661 x^{18} - 612, 937, 500 x^{16} - 310, 656, 500 x^{15} \\ - 18, 024, 160 x^{14} - 14, 004, 375 x^{11} - 18, 451, 125 x^{10} \\ - 5, 407, 248 x^{9} - 671, 000 x^{6} - 295, 240 x^{5} - 438, 900 x - 193, 116 \\ > & 1, 172, 299, 661 x^{18} - 612, 937, 500 x^{18} - 310, 656, 500 x^{18} \\ - 18, 024, 160 x^{18} - 14, 004, 375 x^{18} \\ - 18, 451, 125 x^{18} - 5, 407, 248 x^{18} - 671, 000 x^{18} - 295, 240 x^{18} \\ - 438, 900 x^{18} - 193, 116 x^{18} \\ = & 191, 220, 497 x^{18} > 0 . \end{aligned}

Consequently, inequality (17) holds true, and inequality (14) is proved.

Let us now show that $η_{5 / 4} = \frac{5^{5 / 4}}{10 [5^{5 / 4} - W (1, 5 / 4)]} - 1 = 0.46 \dots$ is the best possible under the weight coefficient $W (k, 5 / 4)$ .

Consider inequality (14) in a general form as

W (k, 5 / 4) \leq 5^{5 / 4} (1 - \frac{1}{10} \cdot \frac{1}{k^{1 / 5} + η_{5 / 4}}) .

(18)

Putting

k = 1

in (18) yields

η_{5 / 4} \geq \frac{5^{5 / 4}}{10 [5^{5 / 4} - W (1, 5 / 4)]} .

Thus the best possible value for $η_{5 / 4}$ in (18) should be $η_{min} = \frac{5^{5 / 4}}{10 [5^{5 / 4} - W (1, 5 / 4)]}$ .

This completes the proof of Theorem 3. □

Remark 1 From the definition of

W (k, p)

and in the same way as in [17], we can establish the following accurate estimates of

W (1, 5 / 4)

:

6.965042829 < W (1, 5 / 4) < 6.967740323 .

(19)

Further, the approximation of

η_{5 / 4}

can be derived as follows:

η_{5 / 4} = \frac{5^{5 / 4}}{10 [5^{5 / 4} - W (1, 5 / 4)]} - 1 = 0.46 \dots .

Remark 2 For

p = 5 / 4

, inequality (11) can be written as

\sum_{n = 1}^{\infty} {(\frac{1}{n} \sum_{k = 1}^{n} a_{k})}^{5 / 4} \leq 5^{5 / 4} \sum_{n = 1}^{\infty} (1 - \frac{1 - {(\frac{1}{5})}^{5 / 4} W (1, 5 / 4)}{n^{1 / 5}}) a_{n}^{5 / 4} .

(20)

It is easy to observe that

\frac{1}{10} \cdot \frac{1}{n^{1 / 5} + η_{5 / 4}} > \frac{1}{10} \cdot \frac{1}{n^{1 / 5} + 4, 711 / 10, 000} > \frac{7}{100 n^{1 / 5}}

and

1 - {(\frac{1}{5})}^{5 / 4} W (1, 5 / 4) < 1 - {(\frac{1}{5})}^{5 / 4} \times 6.967740323 = 0.06808 \dots < \frac{7}{100},

hence

\frac{1 - {(\frac{1}{5})}^{5 / 4} W (1, 5 / 4)}{n^{1 / 5}} < \frac{1}{10} \cdot \frac{1}{n^{1 / 5} + η_{5 / 4}} .

This implies that inequality (14) is stronger than inequality (11).

Declarations

Acknowledgements

The present investigation was supported, in part, by the Natural Science Foundation of Fujian province of China under Grant 2012J01014 and, in part, by the Foundation of Scientific Research Project of Fujian Province Education Department of China under Grant JK2012049.

Authors’ Affiliations

(1)

Department of Mathematics and Computer Science, Longyan University, Longyan, Fujian, 364012, P.R. China

(2)

Gangtou Middle School, Fuqing, Fujian, 350317, P.R. China

References

Hardy GH, Littlewood JE, Pólya G: Inequalities. 2nd edition. Cambridge University Press, Cambridge; 1952.Google Scholar
Mitrinović DS, Pečarić JE, Fink AM: Inequalities Involving Functions and Thier Integrals and Derivatives. Kluwer Academic, Dordrecht; 1991.View ArticleGoogle Scholar
Yang BC, Zhu YH: An improvement on Hardy’s inequality. Acta Sci. Natur. Univ. Sunyatseni 1998, 37(1):41–44.MathSciNetGoogle Scholar
Huang QL:An improvement of series Hardy’s inequality for $p = 3$ . J. Hubei Inst. Natl. 1999, 17(3):54–57.Google Scholar
Huang QL:A sharpness and improvement of Hardy’s inequality for $p = 3 / 2$ . J. Guangxi Norm. Univ. (Nat. Sci. Ed.) 2000, 18(1):38–41.Google Scholar
Huang QL: An improvement of Hardy’s inequality in an interval. Acta Sci. Natur. Univ. Sunyatseni 2000, 39(3):20–24.MathSciNetGoogle Scholar
Huang QL:An improvement of Hardy’s series inequality in the interval $[2, 5]$ . J. South China Univ. Technol. (Natl. Sci. Ed.) 2000, 28(2):64–68.Google Scholar
Huang QL: An extension of a strengthened inequality. J. Guangdong Educ. Inst. 2001, 21(2):17–20.Google Scholar
Yang BC: On a strengthened version of Hardys inequality. J. Guangdong Educ. Inst. 2005, 25(5):5–8.Google Scholar
Kuang JC: Applied Inequalities. 4th edition. Shangdong Science and Technology Press, Jinan; 2010:576.Google Scholar
Zhang XM: A new proof method of analytic inequality. RGMIA Research Report Collection 2009, 12(1):18–29.Google Scholar
Zhang XM, Chu YM: New Discussion to Analytic Inequalities. Harbin Institute of Technology press, Harbin; 2009:260–275.Google Scholar
Zhang, XM, Chu, YM: A new method to study analytic inequalities. J. Inequal. Appl. (2010). http://www.hindawi.com/journals/jia/2010/698012Google Scholar
Yang L, Xia BC: Automated Discovering and Proving for Mathematical Inequalities. Science Press, Beijing; 2008:33–42. 117–142Google Scholar
Yang L, Zhang JZ, Hou XR: Nonlinear Algebraic Equation Systems and Automated Theorem Proving. Shanghai Science’s and Technology’s Education Publishing House, Shanghai; 1996:137–166.Google Scholar
He D:The automatic verification of Hardy inequality’s improvements and its improving type of $P \in N$ . J. Guangdong Educ. Inst. 2012, 32(3):28–35.Google Scholar
Deng, YP, He, D, Wu, SH: A sharpened version of Hardy’s inequality and the automated verification program. J. Math. Pract. Theory (2012, submitted)Google Scholar
Huang YZ, He D:An improvement of Hardy’s inequality for $p = 3 / 2$ . J. Shantou Univ. (Natl. Sci. Ed.) 2012, 27(3):15–22.Google Scholar
Yang BC: On a strengthened Hardy’s inequality. J. Xinyang Teach. Coll. (Natl. Sci. Ed.) 2002, 15(1):37–39.Google Scholar
Huang QL:A strengthened Hardy’s inequality in the interval $[1, 2]$ . J. Guangdong Educ. Inst. 2002, 22(2):20–23.Google Scholar
Huang QL, Yang BC: A strengthened dual form of Hardy’s inequality. J. Math. (PRC) 2005, 25(3):307–311.Google Scholar
Zhu YH, Yang BC: Improvement on Euler’s summation formula and some inequalities on sums of powers. Acta Sci. Natur. Univ. Sunyatseni 1997, 36(4):21–26.MathSciNetGoogle Scholar
Qian X, Meixiu Z, Zhang X: On a strengthened version of Hardy’s inequality. J. Inequal. Appl. 2012, 2012: 300. 10.1186/1029-242X-2012-300View ArticleGoogle Scholar

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A sharpened version of Hardy’s inequality for parameter p = 5 / 4

Acknowledgements

A sharpened version of Hardy’s inequality for parameter $p = 5 / 4$