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

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 ${\eta }_{5/4}=0.46\dots$ is established.

MSC:26D15, 26D20, 26D07.

1 Introduction

Let $p>1$, $1/p+1/q=1$, $a_n\ge 0$ ($n=1,2,\dots$), $0<{\sum }_{n=1}^{\mathrm{\infty }}{a}_n^p<\mathrm{\infty }$. Then

$$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^p<q^p\sum _{n=1}^{\mathrm{\infty }}{a}_n^p,$$

(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:

$$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^2<4\sum _{n=1}^{\mathrm{\infty }}{a}_n^2,$$

(2)

$$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^3<\frac{27}{8}\sum _{n=1}^{\mathrm{\infty }}{a}_n^3.$$

(3)

In 1998, Yang and Zhu [3] evaluated the weight coefficient $W(k,p)$,

$$W(k,p)=k^{1-1/p}\sum _{n=k}^{\mathrm{\infty }}\frac{1}{n^p}{\left(\sum _{j=1}^n\frac{1}{j^{1/p}}\right)}^{p-1},k=1,2,\dots ,$$

(4)

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

$$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^2<4\sum _{n=1}^{\mathrm{\infty }}(1-\frac{1}{3\sqrt{n}+5})a_n^2.$$

(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.,

$$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^3<\frac{27}{8}\sum _{n=1}^{\mathrm{\infty }}(1-\frac{3}{19n^{2/3}})a_n^3,$$

(6)

$$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^{3/2}\le 3\sqrt{3}\sum _{n=1}^{\mathrm{\infty }}(1-\frac{1}{5}\cdot \frac{1}{\sqrt[3]{n}+3})a_n^{3/2}.$$

(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 $W(k,2)$

$$W(k,2)=\sqrt{k}\sum _{n=k}^{\mathrm{\infty }}\frac{1}{n^2}\left(\sum _{j=1}^n\frac{1}{\sqrt{j}}\right)\le 4[1-\frac{1}{\sqrt{k}}(1-\frac{1}{4}W(1,2))]$$

and established the following inequality:

$$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^2<4\sum _{n=1}^{\mathrm{\infty }}(1-\frac{{\theta }_2}{\sqrt{n}})a_n^2,$$

(8)

where ${\theta }_2=1-\frac{1}{4}W(1,2)=0.13788928\dots$ 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):

$$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^p\le {\left(\frac{p}{p-1}\right)}^p\sum _{n=1}^{\mathrm{\infty }}(1-\frac{c_p}{2{(n-1/2)}^{1-1/p}})a_n^p,$$

(9)

where

$$c_p=\{\begin{array}{ll}(p-1)[1-2^{1/p}(1-1/p)],& 1<p\le 2,\\ 1-2^{1-1/p}{(1-1/p)}^{p-1},& p>2.\end{array}$$

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):

$$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^3\le \frac{27}{8}\sum _{n=1}^{\mathrm{\infty }}(1-\frac{{\theta }_3}{n^{2/3}})a_n^3,$$

(10)

where ${\theta }_3=1-\frac{8}{27}W(1,3)=0.1673\dots$ 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):

$$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^p\le {\left(\frac{p}{p-1}\right)}^p\sum _{n=1}^{\mathrm{\infty }}(1-\frac{{\theta }_p(1)}{n^{1-1/p}})a_n^p,$$

(11)

where ${\theta }_p(1)=1-{(\frac{p-1}{p})}^{p}W(1,p)$ 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:

$$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^{5/4}\le 5^{5/4}\sum _{n=1}^{\mathrm{\infty }}(1-\frac{1}{10}\cdot \frac{1}{n^{1/5}+{\eta }_{5/4}})a_n^{5/4},$$

(12)

where ${\eta }_{5/4}=\frac{5^{5/4}}{10[5^{5/4}-W(1,5/4)]}-1=0.46\dots$ 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 $n\ge 1$, it holds that

$$\begin{array}{r}\frac{p}{p-1}{n}^{1-1/p}-\frac{p}{p-1}+\frac{1}{2}+\frac{1}{2n^{1/p}}\\ \le \sum _{j=1}^n\frac{1}{j^{1/p}}\le \frac{p}{p-1}{n}^{1-1/p}-\frac{p}{p-1}+\frac{1}{2}+\frac{1}{2n^{1/p}}+\frac{1}{12p}-\frac{1}{12p{n}^{1+1/p}}.\end{array}$$

(13)

Lemma 2 (see[3])

If $p>1$, then for all integers $n\ge k\ge 1$, it holds that

$$\frac{1}{(p-1)k^{p-1}}+\frac{1}{2k^p}<\sum _{n=k}^{\mathrm{\infty }}\frac{1}{n^p}<\frac{1}{(p-1)k^{p-1}}+\frac{1}{2k^p}+\frac{p}{12k^{p+1}}.$$

Lemma 3 Let $p>1$, $1/p+1/q=1$, and let $g_r$, $g_l$ be the functions defined by

$$g_r(x)=-\frac{6p+12q-1}{12pq{x}^{1/q}}+\frac{1}{2qx}-\frac{1}{12pq{x}^2},g_l(x)=-\frac{p+2q}{2pq{x}^{1/q}}+\frac{1}{2qx},x\in [1,+\mathrm{\infty }).$$

Then $-1<g_r(x)<0$, $-1<g_l(x)<0$.

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

Further, we have

$$\begin{array}{rl}{g}_r^{\prime }(x)& =\frac{6p+12q-1}{12p{q}^2{x}^{1+1/q}}-\frac{1}{2q{x}^2}+\frac{1}{6pq{x}^3}\\ \ge \frac{6p+12q-1}{12p{q}^2{x}^2}-\frac{1}{2q{x}^2}+\frac{1}{6pq{x}^3}\\ =\frac{(5px+x+2p)(p-1)}{12p^3{x}^3}>0,\end{array}$$

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

Now, from $g_r(1)=-1/p>-1$ and ${lim}_{x\to +\mathrm{\infty }}{g}_r(x)=0$, it follows that $g_r(x)\ge g_r(1)=-1/p>-1$ and $g_r(x)<0$.

Similarly, from

$$\begin{array}{c}{g}_l^{\prime }(x)=\frac{p+2q}{2p{q}^2{x}^{1+1/q}}-\frac{1}{2q{x}^2}\ge \frac{p+2q}{2p{q}^2{x}^2}-\frac{1}{2q{x}^2}=\frac{p-1}{2p^2{x}^2}>0,\hfill \\ g_l(1)=-1/p>-1\text{and}\underset{x\to +\mathrm{\infty }}{lim}{g}_l(x)=0,\hfill \end{array}$$

we deduce that $-1<g_l(x)<0$.

Lemma 3 is proved. □

Lemma 4 Let $-1<g(x)<0$. If $\alpha \in (0,1]$, then

$$\begin{array}{r}(1+g(x))(1+(\alpha -1)g(x)+\frac{(\alpha -1)(\alpha -2)}{2}{g}^2(x))\\ \le {(1+g(x))}^{\alpha }\le 1+\alpha g(x)+\frac{\alpha (\alpha -1)}{2}{g}^2(x).\end{array}$$

If $\alpha \in [1,2]$, then

$${(1+g(x))}^{\alpha }\ge 1+\alpha g(x)+\frac{\alpha (\alpha -1)}{2}{g}^2(x).$$

Proof When $\alpha \in (0,1]$. By using the Maclaurin formula

$$\begin{array}{rl}{(1+g(x))}^{\alpha }=& 1+\alpha g(x)+\frac{\alpha (\alpha -1)}{2}{g}^2(x)\\ +\frac{\alpha (\alpha -1)(\alpha -2){(1+\theta g(x))}^{\alpha -3}}{6}{g}^3(x),\theta \in (0,1),\end{array}$$

and noticing $-1<g(x)<0$, we find

$$\begin{array}{c}1+\theta g(x)>1+g(x)>0,\hfill \\ \frac{\alpha (\alpha -1)(\alpha -2){(1+\theta g(x))}^{\alpha -3}}{6}{g}^3(x)\le 0,\hfill \\ \frac{(\alpha -1)(\alpha -2)(\alpha -3){(1+\theta g(x))}^{\alpha -4}}{6}{g}^3(x)\ge 0.\hfill \end{array}$$

Thus

$$\begin{array}{c}{(1+g(x))}^{\alpha }\le 1+\alpha g(x)+\frac{\alpha (\alpha -1)}{2}{g}^2(x),\hfill \\ \begin{array}{rl}{(1+g(x))}^{\alpha }& =(1+g(x)){(1+g(x))}^{\alpha -1}\\ \ge (1+g(x))(1+(\alpha -1)g(x)+\frac{(\alpha -1)(\alpha -2)}{2}{g}^2(x)).\end{array}\hfill \end{array}$$

When $\alpha \in [1,2]$. We have

$$\frac{\alpha (\alpha -1)(\alpha -2){(1+\theta g(x))}^{\alpha -3}}{6}{g}^3(x)\ge 0.$$

Thus

$${(1+g(x))}^{\alpha }\ge 1+\alpha g(x)+\frac{\alpha (\alpha -1)}{2}{g}^2(x).$$

The proof of Lemma 4 is complete. □

Lemma 5 Let $p>1$, $1/p+1/q=1$, $n\ge k\ge 1$, and let $[x]$ denote the greatest integer less than or equal to the real number x. Then we have

$$\begin{array}{rl}W(k,p)\le & q^{p-1}{k}^{1/q}\sum _{n=k}^{\mathrm{\infty }}[\frac{1}{n^{1+1/q}}{(1+g_r(n))}^{[p]-1}\\ \times (1+(p-[p])g_r(n)+\frac{(p-[p])(p-[p]-1)}{2}{g}_r^2(n))].\end{array}$$

Proof By Lemma 1 and the identity $pq=p+q$, $q(p-1)=p$, it follows that

$$\begin{array}{rl}W(k,p)& =k^{1-1/p}\sum _{n=k}^{\mathrm{\infty }}\frac{1}{n^p}{\left(\sum _{j=1}^n\frac{1}{j^{1/p}}\right)}^{p-1}\\ \le k^{1-1/p}\sum _{n=k}^{\mathrm{\infty }}\frac{1}{n^p}{(\frac{p}{p-1}{n}^{1-1/p}-\frac{p}{p-1}+\frac{1}{2}+\frac{1}{2n^{1/p}}+\frac{1}{12p}-\frac{1}{12p{n}^{1+1/p}})}^{p-1}\\ =k^{1/q}\sum _{n=k}^{\mathrm{\infty }}\frac{1}{n^p}{(q{n}^{1/q}-\frac{6p+12q-1}{12p}+\frac{1}{2n^{1/p}}-\frac{1}{12p{n}^{1+1/p}})}^{p-1}\\ =k^{1/q}\sum _{n=k}^{\mathrm{\infty }}\frac{1}{n^p}{q}^{p-1}{n}^{(p-1)/q}{(1-\frac{6p+12q-1}{12pq{n}^{1/q}}+\frac{1}{2qn}-\frac{1}{12pq{n}^2})}^{p-1}\\ =q^{p-1}{k}^{1/q}\sum _{n=k}^{\mathrm{\infty }}\frac{1}{n^{1+1/q}}{(1+g_r(n))}^{p-1}.\end{array}$$

Combining Lemmas 3 and 4, we obtain

$$\begin{array}{c}W(k,p)\hfill \\ \le q^{p-1}{k}^{1/q}\sum _{n=k}^{\mathrm{\infty }}\left[\frac{1}{n^{1+1/q}}{(1+g_r(n))}^{[p]-1}{(1+g_r(n))}^{p-[p]}\right]\hfill \\ \le q^{p-1}{k}^{1/q}\sum _{n=k}^{\mathrm{\infty }}[\frac{1}{n^{1+1/q}}{(1+g_r(n))}^{[p]-1}\hfill \\ \times (1+(p-[p])g_r(n)+\frac{(p-[p])(p-[p]-1)}{2}{g}_r^2(n))].\hfill \end{array}$$

This completes the proof of Lemma 5. □

Lemma 6 Let $1/p+1/q=1$, $n\ge k\ge 1$. If $p\in (1,2)$, then

$$\begin{array}{rl}W(k,p)\ge & q^{p-1}{k}^{1/q}\sum _{n=k}^{\mathrm{\infty }}[\frac{1}{n^{1+1/q}}{(1+g_l(n))}^{[p]}\\ \times (1+(p-[p]-1)g_l(n)+\frac{(p-[p]-1)(p-[p]-2)}{2}{g}_l^2(n))].\end{array}$$

If $p\in [2,+\mathrm{\infty })$, then

$$\begin{array}{rl}W(k,p)\ge & q^{p-1}{k}^{1/q}\sum _{n=k}^{\mathrm{\infty }}[\frac{1}{n^{1+1/q}}{(1+g_l(n))}^{[p]-2}\\ \times (1+(p-[p]+1)g_l(n)+\frac{(p-[p]+1)(p-[p])}{2}{g}_l^2(n))].\end{array}$$

Proof Since $pq=p+q$, $q(p-1)=p$, using Lemma 1 gives

$$\begin{array}{rl}W(k,p)& =k^{1-1/p}\sum _{n=k}^{\mathrm{\infty }}\frac{1}{n^p}{\left(\sum _{j=1}^n\frac{1}{j^{1/p}}\right)}^{p-1}\\ \ge k^{1-1/p}\sum _{n=k}^{\mathrm{\infty }}\frac{1}{n^p}{(\frac{p}{p-1}{n}^{1-1/p}-\frac{p}{p-1}+\frac{1}{2}+\frac{1}{2n^{1/p}})}^{p-1}\\ =k^{1/q}\sum _{n=k}^{\mathrm{\infty }}\frac{1}{n^p}{(q{n}^{1/q}-\frac{p+2q}{2p}+\frac{1}{2n^{1/p}})}^{p-1}\\ =k^{1/q}\sum _{n=k}^{\mathrm{\infty }}\frac{1}{n^p}{q}^{p-1}{n}^{(p-1)/q}{(1-\frac{p+2q}{2pq{n}^{1/q}}+\frac{1}{2qn})}^{p-1}\\ =q^{p-1}{k}^{1/q}\sum _{n=k}^{\mathrm{\infty }}\frac{1}{n^{1+1/q}}{(1+g_l(n))}^{p-1}.\end{array}$$

When $p\in (1,2)$. From Lemmas 3 and 4, we have

$$\begin{array}{rl}W(k,p)\ge & q^{p-1}{k}^{1/q}\sum _{n=k}^{\mathrm{\infty }}\frac{1}{n^{1+1/q}}{(1+g_l(n))}^{[p]-1}{(1+g_l(n))}^{p-[p]}\\ \ge & q^{p-1}{k}^{1/q}\sum _{n=k}^{\mathrm{\infty }}[\frac{1}{n^{1+1/q}}{(1+g_l(n))}^{[p]}\\ \times (1+(p-[p]-1)g_l(n)+\frac{(p-[p]-1)(p-[p]-2)}{2}{g}_l^2(n))].\end{array}$$

When $p\in [2,+\mathrm{\infty })$. Using Lemmas 3 and 4, we obtain

$$\begin{array}{rl}W(k,p)\ge & q^{p-1}{k}^{1/q}\sum _{n=k}^{\mathrm{\infty }}\frac{1}{n^{1+1/q}}{(1+g_l(n))}^{[p]-2}{(1+g_l(n))}^{p-[p]+1}\\ \ge & q^{p-1}{k}^{1/q}\sum _{n=k}^{\mathrm{\infty }}[\frac{1}{n^{1+1/q}}{(1+g_l(n))}^{[p]-2}\\ \times (1+(p-[p]+1)g_l(n)+\frac{(p-[p]+1)(p-[p])}{2}{g}_l^2(n))].\end{array}$$

Lemma 6 is proved. □

Lemma 7 (see[3])

Let $p>1$, $a_n\ge 0$ ($n=1,2,\dots$), $0<{\sum }_{n=1}^{\mathrm{\infty }}{a}_n^p<\mathrm{\infty }$. Then

$$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^p\le \sum _{k=1}^{\mathrm{\infty }}\left[k^{1-1/p}\sum _{n=k}^{\mathrm{\infty }}\frac{1}{n^p}{\left(\sum _{j=1}^n\frac{1}{j^{1/p}}\right)}^{p-1}{a}_k^p\right]=\sum _{k=1}^{\mathrm{\infty }}W(k,p)a_k^p.$$

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

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

Proof

Using Lemma 5 gives

$$W(k,5/4)\le 5^{1/4}{k}^{1/5}\sum _{n=k}^{\mathrm{\infty }}\left[\frac{1}{n^{6/5}}(1+\frac{1}{4}{g}_r(n)-\frac{3}{32}{g}_r^2(n))\right]=5^{1/4}{k}^{1/5}\sum _{n=k}^{\mathrm{\infty }}{r}_{5/4}(n),$$

where

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

Hence

$$\begin{array}{rl}W(k,5/4)\le & 5^{1/4}{k}^{1/5}\sum _{n=k}^{\mathrm{\infty }}(\frac{1}{n^{6/5}}-\frac{133}{600n^{7/5}}-\frac{17\text{,}689}{240\text{,}000n^{8/5}}+\frac{1}{40n^{11/5}}+\frac{133}{8\text{,}000n^{12/5}}\\ -\frac{41}{9\text{,}600n^{16/5}}-\frac{133}{60\text{,}000n^{17/5}}+\frac{1}{4\text{,}000n^{21/5}}-\frac{1}{60\text{,}000n^{26/5}}).\end{array}$$

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

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

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

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

Proof

Utilizing Lemma 6 gives

$$\begin{array}{rl}W(k,5/4)& \ge 5^{1/4}{k}^{1/5}\sum _{n=k}^{\mathrm{\infty }}\left[\frac{1}{n^{6/5}}(1+g_l(n))(1-\frac{3}{4}{g}_l(n)+\frac{21}{32}{g}_l^2(n))\right]\\ =5^{1/4}{k}^{1/5}\sum _{n=k}^{\mathrm{\infty }}{l}_{5/4}(n),\end{array}$$

where

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

Hence

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

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

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

Adding up the above inequalities, we obtain

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

Theorem 2 is proved. □

Theorem 3 Let $a_n\ge 0$ ($n=1,2,\dots$), $0<{\sum }_{n=1}^{\mathrm{\infty }}{a}_n^{5/4}<\mathrm{\infty }$. Then

$$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^{5/4}\le 5^{5/4}\sum _{n=1}^{\mathrm{\infty }}(1-\frac{1}{10}\cdot \frac{1}{n^{1/5}+{\eta }_{5/4}})a_n^{5/4},$$

(14)

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

Proof

By Lemma 7, we have

$$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^{5/4}\le \sum _{k=1}^{\mathrm{\infty }}W(k,5/4)a_k^{5/4}.$$

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

$$W(k,5/4)\le 5^{5/4}(1-\frac{1}{10}\cdot \frac{1}{k^{1/5}+{\eta }_{5/4}}).$$

(15)

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

By Theorem 1 $W(k,5/4)<R_{5/4}(k)$, we need only to prove that

$$R_{5/4}(k)\le 5^{5/4}(1-\frac{1}{10}\cdot \frac{1}{k^{1/5}+{\eta }_{5/4}}).$$

Note that

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

it suffices to show

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

(16)

Substituting $k=x^5$ in (16), inequality (16) becomes

$$R_{5/4}\left(x^5\right)\le 5^{5/4}(1-\frac{1}{10}\cdot \frac{1}{x+11/25}),\text{where }x\ge \sqrt[5]{2},$$

which is equivalent to the following inequality:

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

(17)

where

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

From the hypothesis $x\ge \sqrt[5]{2}>1.14$, we have

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

Further, we have

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

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

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

Consider inequality (14) in a general form as

$$W(k,5/4)\le 5^{5/4}(1-\frac{1}{10}\cdot \frac{1}{k^{1/5}+{\eta }_{5/4}}).$$

(18)

Putting $k=1$ in (18) yields

$${\eta }_{5/4}\ge \frac{5^{5/4}}{10[5^{5/4}-W(1,5/4)]}.$$

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

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 ${\eta }_{5/4}$ can be derived as follows:

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

Remark 2 For $p=5/4$, inequality (11) can be written as

$$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^{5/4}\le 5^{5/4}\sum _{n=1}^{\mathrm{\infty }}(1-\frac{1-{(\frac{1}{5})}^{5/4}W(1,5/4)}{n^{1/5}})a_n^{5/4}.$$

(20)

It is easy to observe that

$$\frac{1}{10}\cdot \frac{1}{n^{1/5}+{\eta }_{5/4}}>\frac{1}{10}\cdot \frac{1}{n^{1/5}+4\text{,}711/10\text{,}000}>\frac{7}{100n^{1/5}}$$

and

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

hence

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

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