1 Introduction

There are many applications with well-known Hardy’s inequality in analytics, which refers to the following: let ${a}_{k}>0$ ($k=1,2,\dots$), $p>1$, then

${\left(\frac{p}{p-1}\right)}^{p}\sum _{n=1}^{\mathrm{\infty }}{a}_{n}^{p}>\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^{n}{a}_{k}\right)}^{p}.$
(1.1)

In recent decades, there have also been many results due to the extension and refinement of this inequality (cf. [15]), especially the monograph [6], which summarized part of the research done before 2005. In research on the coefficient of (1.1), the following conclusion in the case $p=2$ was drawn in [6]:

$4\sum _{n=1}^{\mathrm{\infty }}\left(1-\frac{1}{3\sqrt{n}+5}\right){a}_{n}^{2}>\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^{n}{a}_{k}\right)}^{2}.$
(1.2)

In [7], by using the method of weight-coefficient, the following inequality is proved with $p\in \left[\frac{7}{6},2\right]$:

${\left(\frac{p}{p-1}\right)}^{p}\sum _{n=1}^{\mathrm{\infty }}\left(1-\frac{15}{196}\frac{1}{{n}^{1-1/p}+3,436}\right){a}_{n}^{p}>\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^{n}{a}_{k}\right)}^{p}.$
(1.3)

In the following section, let $p>1$ and

${Z}_{p}=\left\{\begin{array}{cc}p-1-\frac{{\left(p-1\right)}^{2}}{p}{2}^{\frac{1}{p}},\hfill & 12.\hfill \end{array}$

We shall strengthen Hardy’s inequality to

${\left(\frac{p-1}{p}\right)}^{p}\sum _{n=1}^{\mathrm{\infty }}\left(1-\frac{{Z}_{p}}{2{\left(n-\frac{1}{2}\right)}^{1-\frac{1}{p}}}\right){a}_{n}^{p}>\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^{n}{a}_{k}\right)}^{p}.$

2 Relevant lemmas

Some characters of a convex function will be cited in this section.

Definition 2.1 $L\subseteq \mathbb{R}$ is an interval, and $f:I\to \mathbb{R}$ is continuous. If

$f\left(\frac{x+y}{2}\right)\le \left(\ge \right)\frac{f\left(x\right)+f\left(y\right)}{2}$

holds for all $x,y\in I$, then f is called a convex (concave) function.

The sufficient and necessary condition for a second-order differential function f to be convex (concave) function is that ${f}^{″}\left(x\right)\ge \left(\le \right)0$ always holds for any $x\in I$. The famous Hadamard inequality is as follows. Let f be a convex (concave) function on $\left[a,b\right]$, then the equality

$f\left(\frac{b+a}{2}\right)\le \left(\ge \right)\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\le \left(\ge \right)\frac{f\left(b\right)+f\left(a\right)}{2}$
(2.1)

holds if and only if f is a linear function.

Lemma 2.1 Let $p>1$, and ${Z}_{p}$ is defined in the first section, then $0<{Z}_{p}<\frac{1}{2}$.

Proof The proof includes two parts.

Part 1: When $1, if we can obtain $f\left(p\right):={2}^{-\frac{1}{p}}-1+\frac{1}{p}>0$, then we can get ${Z}_{p}>0$ obviously. Since

${f}^{\prime }\left(p\right)=\frac{1}{{p}^{2}}{2}^{-\frac{1}{p}}ln2-\frac{1}{{p}^{2}}=\frac{1}{{p}^{2}}{2}^{-\frac{1}{p}}\left(ln2-{2}^{\frac{1}{p}}\right)<0,$

$f\left(p\right)$ is monotone decreasing on $\left(1,2\right]$, then $f\left(p\right)\ge f\left(2\right)=\frac{\sqrt{2}}{2}-\frac{1}{2}>0$.

Meanwhile, ${Z}_{p}<\frac{1}{2}$ is equivalent to

$p-\frac{3}{2}<\frac{{\left(p-1\right)}^{2}}{p}\cdot {2}^{\frac{1}{p}},\phantom{\rule{2em}{0ex}}-\frac{3}{2}<\left(p-2+\frac{1}{p}\right)\cdot {2}^{\frac{1}{p}}-p.$

The above two inequalities obviously hold with $1. If $\frac{3}{2}, then $p-2+\frac{1}{p}$ is increasing about p, ${2}^{\frac{1}{p}}$ and −p are decreasing in relation to p. Then

Part 2: When $p>2$, then it should be proved that ${\left(\frac{p-1}{p}\right)}^{p-1}{2}^{\frac{p-1}{p}}<1$ and ${\left(\frac{p-1}{p}\right)}^{p-1}{2}^{\frac{p-1}{p}}>\frac{1}{2}$ are respectively equivalent to $2<{\left(1+\frac{1}{p-1}\right)}^{p}$ and ${\left(1+\frac{1}{p-1}\right)}^{p}<{2}^{2+\frac{1}{p-1}}$. Since ${\left(1+\frac{1}{p-1}\right)}^{p}$ is strictly decreasing for $p\in \left(2,+\mathrm{\infty }\right)$, then it is proved that

${\left(1+\frac{1}{p-1}\right)}^{p}>\underset{p\to \mathrm{\infty }}{lim}{\left(1+\frac{1}{p-1}\right)}^{p}=e>2$

and

${\left(1+\frac{1}{p-1}\right)}^{p}<{\left(1+\frac{1}{2-1}\right)}^{2}=4<{2}^{2+\frac{1}{p-1}}.$

The proof of Lemma 2.1 is completed. □

Lemma 2.2

1. (i)

Let $x\ge 1$, $p>2$, then

${\left(1-{Z}_{p}{x}^{\frac{1}{p}-1}\right)}^{\frac{2-p}{p-1}}\left(1-\frac{p-2}{p-1}{Z}_{p}{x}^{\frac{1}{p}-1}\right)>1.$
(2.2)
2. (ii)

Let $p>1$, then $f:x\in \left[1,+\mathrm{\infty }\right)\to {x}^{\frac{1}{p}-2}-{Z}_{p}{x}^{\frac{2}{p}-3}$ is a convex function.

Proof (i) The proposition is equivalent to

$1-\frac{p-2}{p-1}{Z}_{p}{x}^{\frac{1}{p}-1}>{\left(1-{Z}_{p}{x}^{\frac{1}{p}-1}\right)}^{\frac{p-2}{p-1}}.$

Bernoulli’s inequality refers to the following: ${\left(1+t\right)}^{\alpha }\le 1+\alpha t$ ($t>-1$, $0<\alpha <1$) holds if $t=0$. If $t=-{Z}_{p}{x}^{-1+\frac{1}{p}}$, $\alpha =\frac{p-2}{p-1}$, then formula (2.2) holds.

1. (ii)
$\begin{array}{rcl}{f}^{″}\left(x\right)& =& \left(\frac{1}{p}-2\right)\left(\frac{1}{p}-3\right){x}^{\frac{1}{p}-4}-{Z}_{p}\left(\frac{2}{p}-3\right)\left(\frac{2}{p}-4\right){x}^{\frac{2}{p}-5}\\ =& \frac{\left(2p-1\right){x}^{\frac{2}{p}-5}}{{p}^{2}}\left[\left(3p-1\right){x}^{1-\frac{1}{p}}-2{Z}_{p}\left(3p-2\right)\right]\\ \ge & \frac{\left(2p-1\right){x}^{\frac{2}{p}-5}}{{p}^{2}}\left[\left(3p-1\right)-2{Z}_{p}\left(3p-2\right)\right].\end{array}$

According to Lemma 2.1,

${f}^{″}\left(x\right)\ge \frac{\left(2p-1\right){x}^{\frac{2}{p}-5}}{{p}^{2}}\left[\left(3p-1\right)-\left(3p-2\right)\right]>0,$

so, f is a convex function. The proof of Lemma 2.2 is completed. □

Lemma 2.3 Let n be a positive natural number, $p>1$.

1. (i)

If $1, then

$\sum _{k=1}^{n}\frac{1}{{\left(k-\frac{1}{2}\right)}^{\frac{1}{p}}}\le \frac{p}{p-1}\left({n}^{1-\frac{1}{p}}-\frac{{Z}_{p}}{p-1}\right).$
(2.3)
2. (ii)

If $p>2$, then

$\sum _{k=1}^{n}\frac{1}{{\left(k-\frac{1}{2}\right)}^{\frac{1}{p}}}\le \frac{p}{p-1}{n}^{1-\frac{2}{p}}{\left({n}^{1-\frac{1}{p}}-{Z}_{p}\right)}^{\frac{1}{p-1}}.$
(2.4)

Proof (i) If $n=1$, inequality (2.3) is proved easily. Assume that when $n=m\ge 1$, the following equality holds for $n=m+1$:

$\begin{array}{rcl}\sum _{k=1}^{m+1}\frac{1}{{\left(k-\frac{1}{2}\right)}^{\frac{1}{p}}}& \le & \frac{p}{p-1}\left({m}^{1-\frac{1}{p}}-\frac{{Z}_{p}}{p-1}\right)+\frac{1}{{\left(m+\frac{1}{2}\right)}^{\frac{1}{p}}}\\ =& \frac{p}{p-1}\left[{\left(m+1\right)}^{1-\frac{1}{p}}-\frac{{Z}_{p}}{p-1}\right]-{\int }_{m}^{m+1}{x}^{-\frac{1}{p}}\phantom{\rule{0.2em}{0ex}}dx+\frac{1}{{\left(m+\frac{1}{2}\right)}^{\frac{1}{p}}}.\end{array}$

Because ${x}^{-\frac{1}{p}}$ is a convex function on $\left[m,m+1\right]$, then according to Hadamard’s inequality of a convex function, formula (2.3) also holds if $n=m+1$.

1. (ii)

If $n=1$, inequality (2.4) is proved easily. Assume that when $n=m\ge 1$, the inequality holds. For $n=m+1$,

$\begin{array}{rcl}\sum _{k=1}^{m+1}\frac{1}{{\left(k-\frac{1}{2}\right)}^{\frac{1}{p}}}& \le & \frac{p}{p-1}{m}^{1-\frac{2}{p}}{\left({m}^{1-\frac{1}{p}}-{Z}_{p}\right)}^{\frac{1}{p-1}}+\frac{1}{{\left(m+\frac{1}{2}\right)}^{\frac{1}{p}}}\\ =& \frac{p}{p-1}{\left(m+1\right)}^{1-\frac{2}{p}}{\left({\left(m+1\right)}^{1-\frac{1}{p}}-{Z}_{p}\right)}^{\frac{1}{p-1}}\\ -{\int }_{m}^{m+1}{\left[\frac{p}{p-1}{x}^{1-\frac{2}{p}}{\left({x}^{1-\frac{1}{p}}-{Z}_{p}\right)}^{\frac{1}{p-1}}\right]}^{\prime }\phantom{\rule{0.2em}{0ex}}dx+\frac{1}{{\left(m+\frac{1}{2}\right)}^{\frac{1}{p}}}\\ =& \frac{p}{p-1}{\left(m+1\right)}^{1-\frac{2}{p}}{\left({\left(m+1\right)}^{1-\frac{1}{p}}-{Z}_{p}\right)}^{\frac{1}{p-1}}+\frac{1}{{\left(m+\frac{1}{2}\right)}^{\frac{1}{p}}}\\ -\frac{p}{p-1}{\int }_{m}^{m+1}\left[\left(1-\frac{2}{p}\right){x}^{-\frac{2}{p}}{\left({x}^{1-\frac{1}{p}}-{Z}_{p}\right)}^{\frac{1}{p-1}}\\ +\frac{1}{p-1}\left(1-\frac{1}{p}\right){x}^{1-\frac{2}{p}}{\left({x}^{1-\frac{1}{p}}-{Z}_{p}\right)}^{\frac{1}{p-1}-1}\cdot {x}^{-\frac{1}{p}}\right]\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{p}{p-1}{\left(m+1\right)}^{1-\frac{2}{p}}{\left({\left(m+1\right)}^{1-\frac{1}{p}}-{Z}_{p}\right)}^{\frac{1}{p-1}}+\frac{1}{{\left(m+\frac{1}{2}\right)}^{\frac{1}{p}}}\\ -\frac{p}{p-1}{\int }_{m}^{m+1}{x}^{-\frac{2}{p}}{\left({x}^{1-\frac{1}{p}}-{Z}_{p}\right)}^{\frac{2-p}{p-1}}\\ ×\left[\left(1-\frac{1}{p}\right){x}^{1-\frac{1}{p}}-\left(1-\frac{2}{p}\right){Z}_{p}\right]\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{p}{p-1}{\left(m+1\right)}^{1-\frac{2}{p}}{\left({\left(m+1\right)}^{1-\frac{1}{p}}-{Z}_{p}\right)}^{\frac{1}{p-1}}+\frac{1}{{\left(m+\frac{1}{2}\right)}^{\frac{1}{p}}}\\ -{\int }_{m}^{m+1}{x}^{-\frac{1}{p}}{\left(1-{Z}_{p}{x}^{-1+\frac{1}{p}}\right)}^{\frac{2-p}{p-1}}\left[1-\frac{p-2}{p-1}{Z}_{p}{x}^{-1+\frac{1}{p}}\right]\phantom{\rule{0.2em}{0ex}}dx.\end{array}$

By inequality (2.2) and the fact that ${x}^{-\frac{1}{p}}$ is a convex function on $\left[m,m+1\right]$, we have

$\begin{array}{rcl}\sum _{k=1}^{m+1}\frac{1}{{\left(k-\frac{1}{2}\right)}^{\frac{1}{p}}}& <& \frac{p}{p-1}{\left(m+1\right)}^{1-\frac{2}{p}}{\left({\left(m+1\right)}^{1-\frac{1}{p}}-{Z}_{p}\right)}^{\frac{1}{p-1}}+\frac{1}{{\left(m+\frac{1}{2}\right)}^{\frac{1}{p}}}-{\int }_{m}^{m+1}{x}^{-\frac{1}{p}}\phantom{\rule{0.2em}{0ex}}dx\\ \le & \frac{p}{p-1}{\left(m+1\right)}^{1-\frac{2}{p}}{\left({\left(m+1\right)}^{1-\frac{1}{p}}-{Z}_{p}\right)}^{\frac{1}{p-1}}+\frac{1}{{\left(m+\frac{1}{2}\right)}^{\frac{1}{p}}}-{\left(m+\frac{1}{2}\right)}^{-\frac{1}{p}}\\ =& \frac{p}{p-1}{\left(m+1\right)}^{1-\frac{2}{p}}{\left({\left(m+1\right)}^{1-\frac{1}{p}}-{Z}_{p}\right)}^{\frac{1}{p-1}}.\end{array}$

Thus, the inequality (2.4) also holds if $n=m+1$. □

Lemma 2.4 Let i be any positive natural number and $p>1$, then

$\sum _{n=i}^{\mathrm{\infty }}\left({n}^{-2+\frac{1}{p}}-{Z}_{p}{n}^{-3+\frac{2}{p}}\right)<\frac{p}{p-1}\left[{\left(i-\frac{1}{2}\right)}^{\frac{1}{p}-1}-\frac{{Z}_{p}}{2}{\left(i-\frac{1}{2}\right)}^{\frac{2}{p}-2}\right].$

Proof Let $i=1,2,\dots$ and

$f\left(i\right)=\frac{p}{p-1}\left[{\left(i-\frac{1}{2}\right)}^{\frac{1}{p}-1}-\frac{{Z}_{p}}{2}{\left(i-\frac{1}{2}\right)}^{\frac{2}{p}-2}\right]-\sum _{n=i}^{\mathrm{\infty }}\left({n}^{\frac{1}{p}-2}-{Z}_{p}{n}^{\frac{2}{p}-3}\right).$

Then

$\begin{array}{rcl}f\left(i\right)-f\left(i+1\right)& =& \frac{p}{p-1}\left[{\left(i-\frac{1}{2}\right)}^{\frac{1}{p}-1}-\frac{{Z}_{p}}{2}{\left(i-\frac{1}{2}\right)}^{\frac{2}{p}-2}\right]-\left({i}^{\frac{1}{p}-2}-{Z}_{p}{i}^{\frac{2}{p}-3}\right)\\ -\frac{p}{p-1}\left[{\left(i+\frac{1}{2}\right)}^{\frac{1}{p}-1}-\frac{{Z}_{p}}{2}{\left(i+\frac{1}{2}\right)}^{\frac{2}{p}-2}\right]\\ =& -\frac{p}{p-1}{\int }_{i-\frac{1}{2}}^{i+\frac{1}{2}}{\left({x}^{\frac{1}{p}-1}-\frac{{Z}_{p}}{2}{x}^{\frac{2}{p}-2}\right)}^{\prime }\phantom{\rule{0.2em}{0ex}}dx-\left({i}^{\frac{1}{p}-2}-{Z}_{p}{i}^{\frac{2}{p}-3}\right)\\ =& {\int }_{i-\frac{1}{2}}^{i+\frac{1}{2}}\left({x}^{\frac{1}{p}-2}-{Z}_{p}{x}^{\frac{2}{p}-3}\right)\phantom{\rule{0.2em}{0ex}}dx-\left({i}^{\frac{1}{p}-2}-{Z}_{p}{i}^{\frac{2}{p}-3}\right).\end{array}$

According to the conclusion of Lemma 2.2 and Hadamard’s inequality of a convex function, the sequence ${\left\{f\left(i\right)\right\}}_{i=1}^{+\mathrm{\infty }}$ is a strictly decreasing sequence. It is also known that ${lim}_{i\to +\mathrm{\infty }}f\left(i\right)=0$, then $f\left(i\right)>0$ always holds. The proof of Lemma 2.4 is completed. □

3 A new strengthened version of Hardy’s inequality

Theorem 3.1 Let ${a}_{k}>0$, $n\ge 1$, $n\in N$, $p>1$, and

${Z}_{p}=\left\{\begin{array}{cc}p-1-\frac{{\left(p-1\right)}^{2}}{p}{2}^{\frac{1}{p}},\hfill & 12,\hfill \end{array}$

then

$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^{n}{a}_{k}\right)}^{p}<{\left(\frac{p}{p-1}\right)}^{p}\sum _{n=1}^{\mathrm{\infty }}\left[1-\frac{{Z}_{p}}{2{\left(n-\frac{1}{2}\right)}^{1-\frac{1}{p}}}\right]{a}_{n}^{p}.$
(3.1)

Proof Let $r=\frac{p}{p-1}$, then $\frac{1}{r}+\frac{1}{p}=1$. According to Holder’s inequality,

$\begin{array}{rcl}{\left(\frac{1}{n}\sum _{k=1}^{n}{a}_{k}\right)}^{p}& =& \frac{1}{{n}^{p}}{\left(\sum _{k=1}^{n}{\left(k-\frac{1}{2}\right)}^{-\frac{p-1}{{p}^{2}}}{\left(k-\frac{1}{2}\right)}^{\frac{p-1}{{p}^{2}}}{a}_{k}\right)}^{p}\\ \le & \frac{1}{{n}^{p}}{\left(\sum _{k=1}^{n}{\left(k-\frac{1}{2}\right)}^{-\frac{p-1}{{p}^{2}}r}\right)}^{\frac{p}{r}}{\left(\sum _{k=1}^{n}{\left(k-\frac{1}{2}\right)}^{\frac{p-1}{{p}^{2}}p}{a}_{k}^{p}\right)}^{\frac{p}{p}}\\ =& \frac{1}{{n}^{p}}{\left(\sum _{k=1}^{n}{\left(k-\frac{1}{2}\right)}^{-\frac{1}{p}}\right)}^{p-1}\sum _{k=1}^{n}{\left(k-\frac{1}{2}\right)}^{\frac{p-1}{p}}{a}_{k}^{p}.\end{array}$

If ${\lambda }_{n}=\frac{1}{{n}^{p}}{\left({\sum }_{k=1}^{n}{\left(k-\frac{1}{2}\right)}^{-\frac{1}{p}}\right)}^{p-1}$ and ${\mu }_{k}={\left(k-\frac{1}{2}\right)}^{\frac{p-1}{p}}$, then

$\begin{array}{rcl}\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^{n}{a}_{k}\right)}^{p}& \le & \sum _{n=1}^{\mathrm{\infty }}{\lambda }_{n}\cdot \sum _{k=1}^{n}{\mu }_{k}{a}_{k}^{p}\\ =& \sum _{n=1}^{\mathrm{\infty }}\left(\sum _{k=n}^{\mathrm{\infty }}{\lambda }_{k}\right){\mu }_{n}{a}_{n}^{p}.\end{array}$
(3.2)

Thus,

$\sum _{k=n}^{\mathrm{\infty }}{\lambda }_{k}=\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{k}^{p}}{\left(\sum _{j=1}^{k}{\left(j-\frac{1}{2}\right)}^{-\frac{1}{p}}\right)}^{p-1}.$

If $1, according to formula (2.3) and Bernoulli’s inequality,

$\begin{array}{rcl}\sum _{k=n}^{\mathrm{\infty }}{\lambda }_{k}& =& {\left(\frac{p}{p-1}\right)}^{p-1}\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{k}^{p}}{\left({k}^{1-\frac{1}{p}}-\frac{{Z}_{p}}{p-1}\right)}^{p-1}\\ =& {\left(\frac{p}{p-1}\right)}^{p-1}\sum _{k=n}^{\mathrm{\infty }}{k}^{-2+\frac{1}{p}}{\left(1-\frac{{Z}_{p}}{p-1}{k}^{-1+\frac{1}{p}}\right)}^{p-1}\\ <& {\left(\frac{p}{p-1}\right)}^{p-1}\sum _{k=n}^{\mathrm{\infty }}{k}^{-2+\frac{1}{p}}\left(1-{Z}_{p}{k}^{-1+\frac{1}{p}}\right)\\ =& {\left(\frac{p}{p-1}\right)}^{p-1}\sum _{k=n}^{\mathrm{\infty }}\left({k}^{-2+\frac{1}{p}}-{Z}_{p}{k}^{-3+\frac{2}{p}}\right).\end{array}$

If $p>2$, by using formula (2.4), we obtain

$\begin{array}{rcl}\sum _{k=n}^{\mathrm{\infty }}{\lambda }_{k}& \le & {\left(\frac{p}{p-1}\right)}^{p-1}\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{k}^{p}}{\left({k}^{1-\frac{2}{p}}{\left({k}^{1-\frac{1}{p}}-{Z}_{p}\right)}^{\frac{1}{p-1}}\right)}^{p-1}\\ =& {\left(\frac{p}{p-1}\right)}^{p-1}\sum _{k=n}^{\mathrm{\infty }}\left({k}^{-2+\frac{1}{p}}-{Z}_{p}{k}^{-3+\frac{2}{p}}\right).\end{array}$

Therefore, for any $p>1$, the following inequality holds:

$\sum _{k=n}^{\mathrm{\infty }}{\lambda }_{k}\le {\left(\frac{p}{p-1}\right)}^{p-1}\sum _{k=n}^{\mathrm{\infty }}\left({k}^{-2+\frac{1}{p}}-{Z}_{p}{k}^{-3+\frac{2}{p}}\right).$

By Lemma 2.4, we get

$\sum _{n=k}^{\mathrm{\infty }}{\lambda }_{k}\le {\left(\frac{p}{p-1}\right)}^{p}\left({\left(n-\frac{1}{2}\right)}^{\frac{1}{p}-1}-\frac{{Z}_{p}}{2}{\left(n-\frac{1}{2}\right)}^{\frac{2}{p}-2}\right).$
(3.3)

So, from inequalities (3.2) and (3.3), the following result can be obtained:

$\begin{array}{rcl}\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^{n}{a}_{k}\right)}^{p}& \le & {\left(\frac{p}{p-1}\right)}^{p}\sum _{n=1}^{\mathrm{\infty }}\left({\left(n-\frac{1}{2}\right)}^{\frac{1}{p}-1}-\frac{{Z}_{p}}{2}{\left(n-\frac{1}{2}\right)}^{\frac{2}{p}-2}\right){\left(n-\frac{1}{2}\right)}^{\frac{p-1}{p}}{a}_{n}^{p}\\ =& {\left(\frac{p}{p-1}\right)}^{p}\sum _{n=1}^{\mathrm{\infty }}\left(1-\frac{{Z}_{p}}{2{\left(n-\frac{1}{2}\right)}^{1-\frac{1}{p}}}\right){a}_{n}^{p}.\end{array}$

The proof of Theorem 3.1 is completed. □