A note on Hardy’s inequality

Abstract

In this paper, we prove that the inequality ${\sum }_{n=1}^{\mathrm{\infty }}{(\frac{1}{n}{\sum }_{k=1}^n{a}_k)}^p\le {(\frac{p}{p-1})}^p{\sum }_{n=1}^{\mathrm{\infty }}(1-\frac{d(p)}{{(n-1/2)}^{1-1/p}})a_n^p$ holds for $p\le -1$ and $d(p)=(1+(2^{-1/p}-1)p)/[8(1+(2^{-1/p}-1)p)+2]$ if $a_n>0$ ($n=1,2,\dots$), and ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n^p<+\mathrm{\infty }$.

MSC:26D15.

1 Introduction

Let $p>1$ and $a_n>0$ ($n=1,2,\dots$) with ${\sum }_{n=1}^{\mathrm{\infty }}{a_n}^p<+\mathrm{\infty }$, then Hardy’s well-known inequality [1] is given by

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

(1.1)

Recently, the refinement, improvement, generalization, extension, and application for Hardy’s inequality have attracted the attention of many researchers [2–10].

Yang and Zhu [11] presented an improvement of Hardy’s inequality (1.1) for $p=2$ as follows:

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

For $7/6\le p\le 2$, Huang [12] proved that

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

In [13], Wen and Zhang proved that the inequality

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

(1.2)

holds for $p>1$ if $a_n>0$ ($n=1,2,\dots$), with ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n^p<+\mathrm{\infty }$, where $C_p=1-{(1-1/p)}^{p-1}$ for $p\ge 2$ and $C_p=1-1/p$ for $1<p\le 2$.

Xu et al. [14] gave a further improvement of the inequality (1.2):

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

where $Z_p=p-1-\frac{{(p-1)}^2}{p}{2}^{1/p}$ for $1<p\le 2$ and $Z_p=1-{(\frac{p-1}{p})}^{p-1}{2}^{\frac{p-1}{p}}$ for $p>2$.

For the special parameter $p=5/4$, Deng et al. [15] established

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

where ${\eta }_{5/4}=5^{5/4}/[10(5^{5/4}-({\sum }_{n=1}^{\mathrm{\infty }}\frac{1}{n^{5/4}}{({\sum }_{m=1}^n{m}^{-4/5})}^{1/4}))-1]=0.46\cdots$ .

In [16], Long and Linh discussed Hardy’s inequality with the parameter $p<0$, and proved that

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

(1.3)

for $p\le -1$ and

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

for $-1<p<0$ if $a_n>0$ ($n=1,2,\dots$) with ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n^p<+\mathrm{\infty }$.

It is the aim of this paper to present an improvement of inequality (1.3) for the parameter $p\le -1$. Our main result is Theorem 1.1.

Theorem 1.1 Let $p\le -1$, $d(p)=(1+(2^{-1/p}-1)p)/[8(1+(2^{-1/p}-1)p)+2]$ and $a_n>0$ ($n=1,2,\dots$) with ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n^p<+\mathrm{\infty }$, then

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

2 Lemmas

In order to establish our main result we need several lemmas, which we present in this section.

Lemma 2.1 (see [[17], Corollary 1.3])

Suppose that $a,b\in \mathbb{R}$ with $a<b$, $f:{[a,b]}^n\to \mathbb{R}$ has continuous partial derivatives and

$$D_m=\{x=(x_1,x_2,\dots ,x_n)|a\le \underset{1\le k\le n}{min}\{x_k\}<x_m=\underset{1\le k\le n}{max}\{x_k\}\le b\},m=1,2,\dots ,n.$$

If $\frac{\partial f(x)}{\partial x_m}>0$ holds for all $x=(x_1,x_2,\dots ,x_n)\in D_m$ and $m=1,2,\dots ,n$, then

$$f(x_1,x_2,\dots ,x_n)\ge f(x_{min},x_{min},\dots ,x_{min})$$

for all $x_m\in [a,b]$ ($m=1,2,\dots ,n$), where $x_{min}={min}_{1\le k\le n}\{x_k\}$.

Lemma 2.2 Let $n\in \mathbb{R}$ be a positive natural number and $r\in \mathbb{R}$ with $r\ge 1$. Then

$$\sum _{k=1}^n{(k-\frac{1}{2})}^{1/r}\ge \frac{r}{r+1}(n^{1+1/r}-1)+2^{-1/r}.$$

(2.1)

Proof We use mathematical induction to prove inequality (2.1). We clearly see that inequality (2.1) becomes equality for $n=1$. We assume that inequality (2.1) holds for $n=i$ ($i\in \mathbb{N}$, $i\ge 1$), namely

$$\sum _{k=1}^i{(k-\frac{1}{2})}^{1/r}\ge \frac{r}{r+1}(i^{1+1/r}-1)+2^{-1/r}.$$

Then for $n=i+1$ we have

$$\begin{array}{rl}\sum _{k=1}^{i+1}{(k-\frac{1}{2})}^{1/r}& =\sum _{k=1}^i{(k-\frac{1}{2})}^{1/r}+{(i+\frac{1}{2})}^{1/r}\\ \ge \frac{r}{r+1}(i^{1+1/r}-1)+2^{-1/r}+{(i+\frac{1}{2})}^{1/r}\\ =\frac{r}{r+1}[{(i+1)}^{1+1/r}-1]+2^{-1/r}+{(i+\frac{1}{2})}^{1/r}-{\int }_i^{i+1}{x}^{1/r}dx.\end{array}$$

(2.2)

Note that $x^{1/r}$ ($r\ge 1$) is concave on $(0,+\mathrm{\infty })$, therefore Hermite-Hadamard’s inequality implies that

$${(i+\frac{1}{2})}^{1/r}\ge {\int }_i^{i+1}{x}^{1/r}dx.$$

(2.3)

From (2.2) and (2.3) we know that inequality (2.1) holds for $n=i+1$. □

Remark 2.1 The inequality

$$2^{-1/r}\ge \frac{r}{r+1}$$

(2.4)

holds for all $r\ge 1$ with equality if and only if $r=1$.

Proof We clearly see that inequality (2.4) becomes equality for $r=1$.

If $r>1$, then it is well known that the function ${(1+1/r)}^r$ is strictly increasing on $(1,+\mathrm{\infty })$, so we get

$${(1+\frac{1}{r})}^r>2.$$

(2.5)

Therefore, inequality (2.4) follows from (2.5). □

Lemma 2.3 The inequality

$$\frac{(3r+1)2^{-1-1/r}}{r+1}>(2^{-1/r}-\frac{r}{r+1})\frac{3r^2+1}{r}$$

(2.6)

holds for all $r\ge 1$.

Proof Let $r\ge 1$, then we clearly see that

$$(6log2-3)r^2+r+2log2-2\ge 4(2log2-1)>0.$$

(2.7)

Inequality (2.7) leads to

$$e^{(3r^2-r+2)/(6r^2+2)}<2.$$

(2.8)

It follows from the well-known inequality ${(1+x)}^{1/x}<e$ ($x>0$) that

$$e>{(1+\frac{3r^2-r+2}{6r^3+2r})}^{(6r^3+2r)/(3r^2-r+2)}.$$

(2.9)

From (2.8) and (2.9) we have

$$2^{-1/r}<\frac{6r^3+2r}{6r^3+3r^2+r+2}.$$

(2.10)

Therefore, inequality (2.6) follows easily from (2.10). □

Lemma 2.4 Let $r\ge 1$ and

$$f(x)=\frac{x^r}{{(\frac{r}{r+1}{x}^{1+1/r}+2^{-1/r}-\frac{r}{r+1})}^{r+1}}.$$

(2.11)

Then f is convex on $[1/2,+\mathrm{\infty })$.

Proof From (2.11) we have

$$\begin{array}{r}{f}^{\prime }(x)=\frac{-\frac{2r+1}{r+1}{x}^{r+1/r}+(2^{-1/r}-\frac{r}{r+1})r{x}^{r-1}}{{(\frac{r}{r+1}{x}^{1+1/r}+2^{-1/r}-\frac{r}{r+1})}^{r+2}},\\ f^{″}(x)=\frac{\frac{(3r+1)(2r+1)}{{(r+1)}^2}{x}^{2+2/r}-(2^{-1/r}-\frac{r}{r+1})\frac{(3r^2+1)(2r+1)}{r(r+1)}{x}^{1+1/r}+r(r-1){(2^{-1/r}-\frac{r}{r+1})}^2}{{(\frac{r}{r+1}{x}^{1+1/r}+2^{-1/r}-\frac{r}{r+1})}^{r+3}}{x}^{r-2}.\end{array}$$

(2.12)

It follows from Lemma 2.3 and (2.12) that

$$f^{″}(x)\ge \frac{\frac{(3r+1)(2r+1)}{{(r+1)}^2}{2}^{-1-1/r}-(2^{-1/r}-\frac{r}{r+1})\frac{(3r^2+1)(2r+1)}{r(r+1)}}{{(\frac{r}{r+1}{x}^{1+1/r}+2^{-1/r}-\frac{r}{r+1})}^{r+3}}{x}^{r+1/r-1}>0$$

(2.13)

for all $x\in [1/2,+\mathrm{\infty })$.

Therefore, Lemma 2.4 follows from inequality (2.13). □

Lemma 2.5 Let $r\ge 1$, $0\le t\le 4$ and $c=(r+1-2^{1/r}r)/[8(r+1-2^{1/r}r)+2]$, then

$$(r+1)(2^{-1/r}-\frac{r}{r+1})(1-ct)t\ge 1-{[1+(\frac{1+r}{2^{1/r}r}-1)t]}^{-r}.$$

(2.14)

Proof If $r=1$, then we clearly see that inequality (2.14) becomes equality. Next, we assume that $r>1$. Let

$$f(t)=(r+1)(2^{-1/r}-\frac{r}{r+1})(1-ct)t-1+{[1+(\frac{1+r}{2^{1/r}r}-1)t]}^{-r}.$$

(2.15)

Then simple computations lead to

$$f(0)=0,$$

(2.16)

$$f^{\prime }(t)=\frac{(r+1)(2^{-1/r}-\frac{r}{r+1})}{{[1+(\frac{r+1}{2^{1/r}r}-1)t]}^{r+1}}\{(1-2ct){[1+(\frac{r+1}{2^{1/r}r}-1)t]}^{r+1}-1\}.$$

(2.17)

Note that

$$1-2ct\ge 1-8c=\frac{1}{4(r+1-2^{1/r}r)+1}>0,$$

(2.18)

$${[1+(\frac{r+1}{2^{1/r}r}-1)t]}^{r+1}\ge 1+(r+1)(\frac{r+1}{2^{1/r}r}-1)t.$$

(2.19)

It follows from Remark 2.1 and (2.17)-(2.19) that

$$f^{\prime }(t)\ge \frac{(r+1)(2^{-1/r}-\frac{r}{r+1})}{{[1+(\frac{r+1}{2^{1/r}r}-1)t]}^{r+1}}\{(1-2ct)[1+(r+1)(\frac{r+1}{2^{1/r}r}-1)t]-1\}.$$

(2.20)

Let

$$g(t)=(1-2ct)[1+(r+1)(\frac{r+1}{2^{1/r}r}-1)t]-1.$$

(2.21)

Then from $g(0)=0$ and $g(4)=4{(r+1-2^{1/r}r)}^2/[2^{1/r}r(4(r+1-2^{1/r}r)+1)]\ge 0$ together with the fact that $g(t)$ is a concave parabola we know that

$$g(t)\ge 0$$

(2.22)

for $t\in [0,4]$.

Therefore, Lemma 2.5 follows easily from (2.15) and (2.16) together with (2.20)-(2.22). □

Lemma 2.6 Let $r\ge 1$, $c=(r+1-2^{1/r}r)/[8(r+1-2^{1/r}r)+2]$, N is a positive natural number, $a_k>0$ ($k=1,2,\dots ,N$) and $B_N={min}_{1\le k\le N}\{{(k-1/2)}^{1/r}{a}_k\}$, then

$$\begin{array}{r}{\left(\frac{r+1}{r}\right)}^r\sum _{n=1}^N(1-\frac{c}{{(n-1/2)}^{1+1/r}})a_n^r-\sum _{n=1}^N{\left(\frac{n}{{\sum }_{k=1}^{n}1/a_k}\right)}^r\\ \ge B_N^r[{\left(\frac{r+1}{r}\right)}^r\sum _{n=1}^N(1-\frac{c}{{(n-1/2)}^{1+1/r}})\frac{1}{n-1/2}\\ -\sum _{n=1}^N{\left(\frac{n}{{\sum }_{k=1}^n{(k-1/2)}^{1/r}}\right)}^r].\end{array}$$

(2.23)

Proof Let $a_k=b_k/{(k-1/2)}^{1/r}$ ($k=1,2,\dots ,N$), then $B_N={min}_{1\le k\le N}\{b_k\}$ and inequality (2.23) becomes

$$\begin{array}{r}{\left(\frac{r+1}{r}\right)}^r\sum _{n=1}^N(1-\frac{c}{{(n-1/2)}^{1+1/r}})\frac{b_n^r}{n-1/2}-\sum _{n=1}^N{\left(\frac{n}{{\sum }_{k=1}^n\frac{{(k-1/2)}^{1/r}}{b_k}}\right)}^r\\ \ge B_N^r[{\left(\frac{r+1}{r}\right)}^r\sum _{n=1}^N(1-\frac{c}{{(n-1/2)}^{1+1/r}})\frac{1}{n-1/2}\\ -\sum _{n=1}^N{\left(\frac{n}{{\sum }_{k=1}^n{(k-1/2)}^{1/r}}\right)}^r].\end{array}$$

(2.24)

Let $D_m=\{\mathbf{b}=(b_1,b_2,\dots ,b_N)|b_m={max}_{1\le k\le N}\{b_k\}>{min}_{1\le k\le N}\{b_k\}\}$ ($m=1,2,\dots ,N$), and

$$\begin{array}{rl}f(b_1,b_2,\dots ,b_N)=& {\left(\frac{r+1}{r}\right)}^r\sum _{n=1}^N(1-\frac{c}{{(n-1/2)}^{1+1/r}})\frac{b_n^r}{n-1/2}\\ -\sum _{n=1}^N{\left(\frac{n}{{\sum }_{k=1}^n\frac{{(k-1/2)}^{1/r}}{b_k}}\right)}^r.\end{array}$$

(2.25)

Then for any $\mathbf{b}\in D_m$ ($m=1,2,\dots ,N$) we have

$$\begin{array}{rl}\frac{\partial f(\mathbf{b})}{\partial b_m}=& (1-\frac{c}{{(m-1/2)}^{1+1/r}})\frac{{(r+1)}^r{b}_m^{r-1}}{(m-1/2)r^{r-1}}\\ -\frac{r{(m-1/2)}^{1/r}}{b_m^2}\sum _{n=m}^N\frac{n^r}{{({\sum }_{k=1}^n\frac{{(k-1/2)}^{1/r}}{b_k})}^{r+1}}\\ >& (1-\frac{c}{{(m-1/2)}^{1+1/r}})\frac{{(r+1)}^r{b}_m^{r-1}}{(m-1/2)r^{r-1}}\\ -r{(m-1/2)}^{1/r}{b}_m^{r-1}\sum _{n=m}^{+\mathrm{\infty }}\frac{n^r}{{({\sum }_{k=1}^n{(k-1/2)}^{1/r})}^{r+1}}.\end{array}$$

(2.26)

From Lemma 2.2 and (2.26) one has

$$\begin{array}{rl}\frac{1}{r{(m-1/2)}^{1/r}{b}_m^{r-1}}\frac{\partial f(\mathbf{b})}{\partial b_m}>& {\left(\frac{r+1}{r}\right)}^r(1-\frac{c}{{(m-1/2)}^{1+1/r}})\frac{1}{{(m-1/2)}^{1+1/r}}\\ -\sum _{n=m}^{+\mathrm{\infty }}\frac{n^r}{{(\frac{r}{r+1}{n}^{1+1/r}+2^{-1/r}-\frac{r}{r+1})}^{r+1}}.\end{array}$$

(2.27)

It clearly follows from Lemma 2.4 and the Hermite-Hadamard inequality that

$${\int }_{m-1/2}^{m+1/2}\frac{x^r}{{(\frac{r}{r+1}{x}^{1+1/r}+2^{-1/r}-\frac{r}{r+1})}^{r+1}}\ge \frac{m^r}{{(\frac{r}{r+1}{m}^{1+1/r}+2^{-1/r}-\frac{r}{r+1})}^{r+1}}$$

and

$${\int }_{m-1/2}^{+\mathrm{\infty }}\frac{x^r}{{(\frac{r}{r+1}{x}^{1+1/r}+2^{-1/r}-\frac{r}{r+1})}^{r+1}}\ge \sum _{n=m}^{+\mathrm{\infty }}\frac{n^r}{{(\frac{r}{r+1}{n}^{1+1/r}+2^{-1/r}-\frac{r}{r+1})}^{r+1}}.$$

(2.28)

Note that

$$\begin{array}{r}{\int }_{m-1/2}^{+\mathrm{\infty }}\frac{x^r}{{(\frac{r}{r+1}{x}^{1+1/r}+2^{-1/r}-\frac{r}{r+1})}^{r+1}}\\ ={\left(\frac{1+r}{r}\right)}^r\frac{2^{1/r}}{r+1-2^{1/r}r}\{1-{[1+(\frac{r+1}{2^{1/r}r}-1){(m-1/2)}^{-1-1/r}]}^{-r}\},\end{array}$$

(2.29)

$$0<{(m-1/2)}^{-1-1/r}\le 2^{1+1/r}\le 4.$$

(2.30)

From Lemma 2.5 and (2.30) one has

$$\begin{array}{r}{\left(\frac{r+1}{r}\right)}^r(1-\frac{c}{{(m-1/2)}^{1+1/r}})\frac{1}{{(m-1/2)}^{1+1/r}}\\ \ge {\left(\frac{1+r}{r}\right)}^r\frac{2^{1/r}}{r+1-2^{1/r}r}\{1-{[1+(\frac{r+1}{2^{1/r}r}-1){(m-1/2)}^{-1-1/r}]}^{-r}\}.\end{array}$$

(2.31)

Inequalities (2.27), (2.28), and (2.31) together with (2.29) lead to the conclusion that

$$\frac{\partial f(\mathbf{b})}{\partial b_m}>0$$

(2.32)

for any $\mathbf{b}=(b_1,b_2,\dots ,b_N)\in D_m$ and $m=1,2,\dots ,N$.

It follows from Lemma 2.1 and (2.32) that

$$f(b_1,b_2,\dots ,b_N)\ge f(B_N,B_N,\dots ,B_N).$$

(2.33)

Therefore, inequality (2.24) follows from (2.25) and (2.33). □

Lemma 2.7 Let $r\ge 1$, $c=(r+1-2^{1/r}r)/[8(r+1-2^{1/r}r)+2]$, then

$$\left(\frac{r+1}{r}\right)(1-2^{1+1/r}c)>1.$$

(2.34)

Proof We clearly see that inequality (2.34) holds for $r=1$. Next, we assume that $r>1$, let $t=2^{1+1/r}$, then $0<t<4$ and Lemma 2.5 leads to

$$\begin{array}{rl}{\left(\frac{r+1}{r}\right)}^r(1-2^{1+1/r}c)-1& \ge \frac{{(\frac{r+1}{r})}^r}{2(r+1-2^{1/r}r)}[1-{(1+\frac{2(r+1-2^{1/r}r)}{r})}^{-r}]-1\\ \ge \frac{{(\frac{r+1}{r})}^r}{2(r+1-2^{1/r}r)}\frac{2(r+1-2^{1/r}r)}{1+2(r+1-2^{1/r}r)}-1\\ =\frac{{(\frac{r+1}{r})}^r}{1+2(r+1-2^{1/r}r)}-1.\end{array}$$

(2.35)

Note that

$$r+1-2^{1/r}r<1-log2$$

(2.36)

for all $r\ge 1$. In fact, let $x\ge 1$ and

$$f(x)=x-2^{1/x}x+1.$$

(2.37)

Then

$$f^{\prime }(x)=1+(\frac{log2}{x}-1)2^{1/x},$$

(2.38)

$$f^{″}(x)=-\frac{{(log2)}^2}{x^3}{2}^{1/x}<0.$$

(2.39)

It follows from (2.38) and (2.39) that

$$f^{\prime }(x)>\underset{x\to +\mathrm{\infty }}{lim}[1+(\frac{log2}{x}-1)2^{1/x}]=0.$$

(2.40)

Equation (2.37) and inequality (2.40) lead to the conclusion that

$$f(x)<\underset{x\to +\mathrm{\infty }}{lim}(x-2^{1/x}x+1)=1-log2.$$

(2.41)

From (2.35) and (2.36) together with the fact that ${[(r+1)/r]}^r\ge 2$ we have

$${\left(\frac{r+1}{r}\right)}^r(1-2^{1+1/r}c)-1>\frac{2}{1+2(1-log2)}-1=\frac{2log2-1}{3-2log2}>0.$$

(2.42)

Therefore, inequality (2.34) follows from (2.42). □

Lemma 2.8 Let $r\ge 1$, $c=(r+1-2^{1/r}r)/[8(r+1-2^{1/r}r)+2]$, N is a positive natural number, $a_k>0$ ($k=1,2,\dots ,N$) and $B_N={min}_{1\le k\le N}\{{(k-1/2)}^{1/r}{a}_k\}$, then

$$\begin{array}{r}{\left(\frac{r+1}{r}\right)}^r\sum _{n=1}^N(1-\frac{c}{{(n-1/2)}^{1+1/r}})a_n^r-\sum _{n=1}^N{\left(\frac{n}{{\sum }_{k=1}^{n}1/a_k}\right)}^r\\ \ge 2B_N^r[{\left(\frac{r+1}{r}\right)}^r(1-2^{1+1/r}c)-1].\end{array}$$

(2.43)

Proof Let $m\in \{1,2,\dots ,N\}$, $f(0)=0$ and

$$f(m)={\left(\frac{r+1}{r}\right)}^r\sum _{n=1}^m(1-\frac{c}{{(n-1/2)}^{1+1/r}})\frac{1}{n-1/2}-\sum _{n=1}^m{\left(\frac{n}{{\sum }_{k=1}^n{(k-1/2)}^{1/r}}\right)}^r.$$

(2.44)

Then

$$f(1)=2[{\left(\frac{1+r}{r}\right)}^r(1-2^{1+1/r}c)-1],$$

(2.45)

$$f(m)-f(m-1)=\frac{{(\frac{1+r}{r})}^r}{m-1/2}(1-\frac{c}{{(m-1/2)}^{1+1/r}})-{\left(\frac{m}{{\sum }_{k=1}^m{(k-1/2)}^{1/r}}\right)}^r.$$

(2.46)

It follows from Lemma 2.2 and (2.46) together with Remark 2.1 that

$$\begin{array}{r}f(m)-f(m-1)\\ \ge \frac{{(\frac{1+r}{r})}^r}{m-1/2}(1-\frac{c}{{(m-1/2)}^{1+1/r}})-{\left(\frac{m}{\frac{r}{r+1}(m^{1+1/r}-1)+2^{-1/r}}\right)}^r\\ \ge \frac{{(\frac{1+r}{r})}^r}{m-1/2}(1-\frac{c}{{(m-1/2)}^{1+1/r}})-{\left(\frac{m}{\frac{r}{r+1}{m}^{1+1/r}}\right)}^r\\ =\frac{{(\frac{1+r}{r})}^r[(4(r+1-2^{1/r}r)+1){(m-1/2)}^{1+1/r}-m(r+1-2^{1/r}r)]}{m{(m-1/2)}^{2+1/r}[8(r+1-2^{1/r}r)+2]}.\end{array}$$

(2.47)

Let

$$g(t)=[4(r+1-2^{1/r}r)+1]{(t-1/2)}^{1+1/r}-(r+1-2^{1/r}r)t.$$

(2.48)

Then

$$\begin{array}{rl}g(1)& =[4(r+1-2^{1/r}r)+1]2^{-1-1/r}-(r+1-2^{1/r}r)\\ >(2^{1-1/r}-1)(r+1-2^{1/r}r)\ge 0,\end{array}$$

(2.49)

$$\begin{array}{rl}{g}^{\prime }(t)& =(1+\frac{1}{r})[4(r+1-2^{1/r}r)+1]{(t-1/2)}^{1/r}-(r+1-2^{1/r}r)\\ >(2^{2-1/r}-1)(r+1-2^{1/r}r)\ge 0\end{array}$$

(2.50)

for $t\ge 1$.

From (2.47)-(2.50) we get

$$f(1)<f(2)<\cdots <f(N-1)<f(N).$$

(2.51)

Therefore, Lemma 2.8 follows easily from Lemma 2.6, (2.44), (2.45), and (2.51). □

3 Proof of Theorem 1.1

Let $r=-p$, $c=c(r)=d(-r)$ and $b_n=1/a_n$ ($n=1,2,\dots$), then $r\ge 1$, $c=(r+1-2^{1/r}r)/[8(r+1-2^{1/r}r)+2]$, $b_n>0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{b}_n^r<+\mathrm{\infty }$.

It follows from Lemmas 2.7 and 2.8 that one has

$$\sum _{n=1}^N{\left(\frac{n}{{\sum }_{k=1}^{n}1/b_k}\right)}^r\le {\left(\frac{r+1}{r}\right)}^r\sum _{n=1}^N(1-\frac{c}{{(n-1/2)}^{1+1/r}})b_n^r.$$

(3.1)

Letting $n\to +\mathrm{\infty }$, (3.1) leads to

$$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{n}{{\sum }_{k=1}^{n}1/b_k}\right)}^r\le {\left(\frac{r+1}{r}\right)}^r\sum _{n=1}^{\mathrm{\infty }}(1-\frac{c}{{(n-1/2)}^{1+1/r}})b_n^r.$$

(3.2)

Therefore, Theorem 1.1 follows immediately from (3.2) and $r=-p$ together with $b_n=1/a_n$.