# On the Constants and Extremal Function and Sequence for Hardy Inequalities
in $L_{p}$ and $l_{p}$

Ivan Gadjev  Department of Mathematics and Informatics, University of Sofia, 5
James Bourchier Blvd., 1164 Sofia, Bulgaria<EMAIL_ADDRESS>
###### Abstract.

We study the behavior of the smallest possible constants $d(a,b)$ and $d_{n}$
in Hardy inequalities

$\int_{a}^{b}\left(\frac{1}{x}\int_{a}^{x}f(t)dt\right)^{p}\,dx\leq
d(a,b)\,\int_{a}^{b}[f(x)]^{p}dx$

and

$\sum_{k=1}^{n}\Big{(}\frac{1}{k}\sum_{j=1}^{k}a_{j}\Big{)}^{p}\leq
d_{n}\,\sum_{k=1}^{n}a_{k}^{p}.$

The exact rate of convergence of $d(a,b)$ and $d_{n}$ is established and the
“almost extremal” function and sequence are found.

###### Key words and phrases:

Hardy inequality, exact constant, extremal function, extremal sequence.

###### 2010 Mathematics Subject Classification:

Primary 26D10, 26D15; Secondary 33C45, 15A42

Research supported by the Bulgarian National Research Fund through Contract
KP-06-N62/4.

##  1\. Introduction and statement of the results

Between 1919 and 1925, in the series of papers [1, 2, 3] G. H. Hardy
established the following inequalities which are nowadays known as the
celebrated Hardy’s ones. Let $p>1$, then the integral inequality states that

(1.1)
$\int_{0}^{\infty}\left(\frac{1}{x}\int_{0}^{x}f(t)dt\right)^{p}\,dx\leq\left(\frac{p}{p-1}\right)^{p}\,\int_{0}^{\infty}f^{p}(x)dx$

holds for every $f$, such that $f(x)\geq 0$ for $x\in(0,\infty)$ and $f^{p}$
is integrable over $(0,\infty)$.

The corresponding discrete version claims that

(1.2)
$\sum_{k=1}^{\infty}\Big{(}\frac{1}{k}\sum_{j=1}^{k}a_{j}\Big{)}^{p}\leq\left(\frac{p}{p-1}\right)^{p}\,\sum_{k=1}^{\infty}{a_{k}^{p}}$

for every sequence $\\{a_{k}\\}$ of non-negative numbers, for which the series
on the right-hand side converges.

In his 1920 paper [2] Hardy claimed that he and Marcel Riesz derived (1.2)
independently, but in their results the larger constant $(p^{2}/(p-1))^{p}$
appears on the right-hand side. E. Landau, in the letter [4] dated 1921,
published later in [5], was the first to establish (1.2) with the exact
constant $(p/(p-1))^{p}$ in the sense that there is no smaller one for which
(1.2) holds for every sequence of non-negative numbers $a_{k}$. For the latter
statement he considered the sequence $a_{k}^{\ast}=k^{-1/p-\varepsilon}$,
suggested earlier by Hardy, and showed that

$\left(\frac{a_{1}^{\ast}+\cdots+a_{k}^{\ast}}{k}\right)^{p}>\left(\frac{p}{p-1}\right)^{p}\left((a_{k}^{\ast})^{p}-\frac{p}{k^{2-1/p}}\right).$

Since $\sum_{k=1}^{\infty}(a_{k}^{\ast})^{p}\to\infty$ as $\varepsilon\to 0$,
the summation of the latter inequalities implies the sharpness of
$(p/(p-1))^{p}$ for (1.2). In the same letter Landau pointed out that equality
in (1.2) occurs only for the trivial sequence, that is, when $a_{k}=0$ for
every $k\in\mathbb{N}$. Similarly, equality in (1.1) occurs if and only if
$f(x)\equiv 0$ almost everywhere.

The lack of nontrivial extremizers and the fact that the above argument of
Landau does not work for finite sequences motivates one to consider, for any
$a$ and $b$ with $-\infty\leq a<b\leq\infty$ and weight positive a.e.
functions $u(x),v(x)$, the so-called general Hardy’s integral inequality

(1.3) $\int_{a}^{b}\left(\int_{a}^{x}f(t)dt\right)^{p}u(x)\,dx\leq
d(a,b)\,\int_{a}^{b}f^{p}(x)v(x)dx,\quad f\in L^{p}[v;a,b]$

and its discrete counterpart

(1.4) $\sum_{k=1}^{n}\Big{(}\sum_{j=1}^{k}a_{j}\Big{)}^{p}u_{k}\leq
d_{n}\,\sum_{k=1}^{n}{a_{k}^{p}}v_{k},\qquad a_{k},u_{k},v_{k}\geq
0,\,\,\,k=1,2,...,n.$

Obviously the equations (1.3) and (1.4) are the“finite versions” of (1.1) and
(1.2). The natural questions are: what are the best constants $d(a,b)$ and
$d_{n}$ and the corresponding extremizers. This is exactly our endeavor in
this paper, mainly because of the importance of Hardy’s inequalities, their
far reaching generalizations, especially to the so-called Hardy-Sobolev
inequalities, and thus the necessity of understanding them more thoroughly.
Answering the above questions in satisfactory manner for arbitrary weight
functions and sequences such that the inequalities (1.3) and (1.4) hold, is
impossible. In this paper we consider the important ”unweighted” versions of
inequalities (1.3) and (1.4), i.e.

(1.5) $\int_{a}^{b}\left(\frac{1}{x}\int_{a}^{x}f(t)dt\right)^{p}\,dx\leq
d(a,b)\,\int_{a}^{b}[f(x)]^{p}dx,\qquad f(x)\geq 0,\ \ f\in L^{p}[a,b]$

and

(1.6) $\sum_{k=1}^{n}\Big{(}\frac{1}{k}\sum_{j=1}^{k}a_{j}\Big{)}^{p}\leq
d_{n}\,\sum_{k=1}^{n}{a_{k}^{p}},\qquad a_{k}\geq 0,\,\,\,k=1,2,...,n.$

The behavior of the constant $d(a,b)$ was studied in many papers - see, for
instance, [6], [7],[8]. The best results about $d(a,b)$ for $p>1$ could be
summarized in the following way (see, for instance [9] or [10]). Let

$B=\sup_{a<x<b}\left\\{(x-a)^{p-1}\left(x^{1-p}-b^{1-p}\right)\right\\}$

then for the constant $d(a,b)$ the next estimations are true

$\frac{1}{p-1}B\leq d(a,b)\leq\left(\frac{p}{p-1}\right)^{p}B.$

It is easy to see that only the right estimation gives asymptotically (when
$a\rightarrow 0$ or $b\rightarrow\infty$ or both) the exact constant. But not
the rate of convergence.

In [11] we studied the inequality (1.5) for $p=2$ and established the exact
constant $d(a,b)$ and the extremal function.

######  Theorem 1.1.

[11] Let $a$ and $b$ be any fixed numbers with $0<a<b<\infty$. Then the
inequality

(1.7)
$\int_{a}^{b}\left(\frac{1}{x}\int_{a}^{x}f(t)dt\right)^{2}dx\leq\frac{4}{1+4\alpha^{2}}\,\int_{a}^{b}f^{2}(x)\,dx,$

where $\alpha$ is the only solution of the equation

$\tan\left(\alpha\ln\frac{b}{a}\right)+2\alpha=0\ \ \mathrm{in\ the\
interval}\ \
\left(\frac{\pi}{2\ln\frac{b}{a}},\frac{\pi}{\ln\frac{b}{a}}\right),$

holds for every $f\in L^{2}[a,b]$. Moreover, equality in (1.7) is attained for

$f_{a,b}(x)=x^{-1/2}\left(2\alpha\cos(\alpha\ln x)+\sin(\alpha\ln x)\right).$

The behavior of the constant $d_{n}$ for $p=2$ as a function of $n$ was also
studied extensively - see, for instance, [12], [13],[14], [15], [16]. In [14]
Herbert S. Wilf established the exact rate of convergence of the constant
$d_{n}$ for $p=2$

$d_{n}=4-\frac{16\pi^{2}}{\ln^{2}n}+O\left(\frac{\ln\ln n}{\ln^{3}n}\right).$

In [17] F. Stampach gave slightly better estimation, i.e.

$d_{n}=4-\frac{16\pi^{2}}{\ln^{2}n}+\frac{32\pi^{2}(\gamma+6\log
2)}{\log^{3}n}+O\left(\frac{1}{\ln^{4}n}\right).$

In [18] we also studied the asymptotic behavior of the constant $d_{n}$ for
$p=2$. It was proved there that $d_{n}$ can be expressed in terms of the
smallest zero of a continuous dual Hahn polynomial of degree $n$ (see [19]),
for a specific choice of the parameters, in terms of which these polynomials
are defined. Despite that nice interpretation of $d_{n}$, it was only proved
in [18, Theorem 1.1] that the next inequalities are true for every natural
$n\geq 3$

(1.8) $4\Bigg{(}1-\frac{4}{\ln n+4}\Bigg{)}\leq d_{n}\leq
4\Bigg{(}1-\frac{8}{(\ln n+4)^{2}}\Bigg{)}.$

In all proofs of the above mentioned estimations for the constant $d_{n}$, the
authors substantially used the special properties of the space $l_{2}$. In
[11] and [20] we applied a different approach which allowed us to give a
simpler proof of some of the mentioned estimations and to find an almost
extremal sequence. We proved the next theorem.

######  Theorem 1.2.

[11] Let

(1.9) $a_{k}=\int_{k}^{k+1}h(x)dx,$

where

$h(x)=x^{-1/2}\left(2\alpha\cos(\alpha\ln x)+\sin(\alpha\ln x)\right),\quad
1\leq x\leq n+1,$

and $\alpha$ is the only solution of the equation

$\tan(\alpha\ln(n+1))+2\alpha=0\ \ \mathrm{in\ the\ interval}\ \
\left(\frac{\pi}{2\ln(n+1)},\frac{\pi}{\ln(n+1)}\right).$

Then

(1.10)
$\sum_{k=1}^{n}\Big{(}\frac{1}{k}\sum_{j=1}^{k}a_{j}\Big{)}^{2}\geq\frac{4}{1+4\alpha^{2}}\,\sum_{k=1}^{n}{a_{k}^{2}}.$

By combining the results (1.8) and (1.10) the exact rate of convergence of
$\\{d_{n}\\}$ is established and the next very sharp estimates for $d_{n}$,
i.e. the next inequalities

$4-\frac{16\pi^{2}}{\ln^{2}(n+1)}\leq d_{n}\leq 4-\frac{32}{(\ln n+4)^{2}}$

hold for every natural $n\geq 3$. Also the sequence $\\{a_{k}\\}_{1}^{n}$
defined in (1.9) is the almost extremal sequence.

In this paper we establish very sharp estimates for $d(a,b)$ and $d_{n}$ for
$2\leq p<\infty$, as well as obtain an “almost extremal” function for (1.5)
and “almost extremal” sequence for (1.6). Our main results are summarized in
the following two theorems.

######  Theorem 1.3.

Let $2\leq p<\infty$ and $0<a<b<\infty$. Then there exist positive constants
$c_{1}=c_{1}(p)$ and $c_{2}=c_{2}(p)$, depending only on $p$, such that the
next estimates for the constant $d(a,b)$ in (1.5) hold

(1.11)
$\left(\frac{p}{p-1}\right)^{p}\left(1-\frac{c_{1}}{\ln^{2}\frac{b}{a}}\right)\leq
d(a,b)\,\leq\left(\frac{p}{p-1}\right)^{p}\left(1+\frac{c_{2}}{\ln^{2}\frac{b}{a}}\right)^{-1}.$

Moreover, the function

$f^{\ast}(x)=\frac{1}{x^{1/p}}\left(\frac{\alpha p}{p-1}\cos(\alpha\ln
x)+\sin(\alpha\ln x)\right)$

where $\alpha$ is the only solution of the equation

$\tan\left(\alpha\ln\frac{b}{a}\right)+\frac{\alpha p}{p-1}=0$

in the interval
$\left(\frac{\pi}{2\ln\frac{b}{a}},\frac{\pi}{\ln\frac{b}{a}}\right)$, is an
”almost extremal” function in the sense that

$\int_{a}^{b}\left(\frac{1}{x}\int_{a}^{x}f^{\ast}(t)dt\right)^{p}\,dx\geq\left(\frac{p}{p-1}\right)^{p}\left(1-\frac{c_{1}}{\ln^{2}\frac{b}{a}}\right)\,\int_{a}^{b}[f^{\ast}(x)]^{p}dx.$

An immediate consequence of Theorem 1.3 is the next corollary.

######  Corollary 1.4.

When either of the limits relations $a\rightarrow 0$, $b\rightarrow\infty$, or
both hold, i.e. $\ln(b/a)\rightarrow\infty$, then

$d(a,b)\sim\left(\frac{p}{p-1}\right)^{p}-\frac{C}{\ln^{2}\frac{b}{a}}.$

More precisely there exist constants $C_{1}(p)>0$ and $C_{2}(p)>0$ depending
only on $p$ such that

$\left(\frac{p}{p-1}\right)^{p}-\frac{C_{1}}{\ln^{2}\frac{b}{a}}\leq
d(a,b)\leq\left(\frac{p}{p-1}\right)^{p}-\frac{C_{2}}{\ln^{2}\frac{b}{a}}.$

Also, for the constant $C_{1}$ the next estimation holds

$C_{1}\leq\left(\frac{p}{p-1}\right)^{p+1}p\pi^{2}$

which is the exact constant for $p=2$.

######  Remark 1.5.

Because of the close connection to the maximal function also it is natural to
consider the inequality

$\int_{a}^{b}\left(\frac{1}{x-a}\int_{a}^{x}f(t)dt\right)^{p}\,dx\leq
d(a,b)\,\int_{a}^{b}[f(x)]^{p}dx,\qquad f(x)\geq 0,\ \ f\in L^{p}[a,b]$

which is equivalent (by change of variables) to

$\int_{0}^{b}\left(\frac{1}{x}\int_{0}^{x}f(t)dt\right)^{p}\,dx\leq
d(b)\,\int_{0}^{b}[f(x)]^{p}dx,\qquad f(x)\geq 0,\ \ f\in L^{p}[0,b].$

Then from the above corollary we obtain that
$d(b)=\left(\frac{p}{p-1}\right)^{p}$ and the only function for which the
equality is attained is $f=0$ a.e.

######  Theorem 1.6.

Let $2\leq p<\infty$. Then there exist positive constants $c_{3}=c_{3}(p)$ and
$c_{4}=c_{4}(p)$, depending only on $p$, such that for every natural $n\geq 2$
the next estimates for the constant $d_{n}$ in (1.6) hold

(1.12) $\left(\frac{p}{p-1}\right)^{p}-\frac{c_{3}}{\ln^{2}n}\leq
d_{n}\,\leq\left(\frac{p}{p-1}\right)^{p}-\frac{c_{4}}{\ln^{2}n}.$

Moreover, the sequence

$a_{k}^{\ast}=\int_{k}^{k+1}f^{*}(x)\,dx,\quad k=1,2,...,n,$

where

(1.13) $f^{*}(x)=\frac{1}{x^{1/p}}\left(\frac{\alpha p}{p-1}\cos(\alpha\ln
x)+\sin(\alpha\ln x)\right),\quad 1\leq x\leq n+1$

and $\alpha$ is the only solution of the equation

$\tan(\alpha\ln(n+1))+\frac{\alpha p}{p-1}=0$

in the interval $\left(\frac{\pi}{2\ln(n+1)},\frac{\pi}{\ln(n+1)}\right)$, is
an ”almost extremal” sequence in the sense that

(1.14)
$\sum_{k=1}^{n}\left(\frac{1}{k}\sum_{j=1}^{k}a_{j}^{\ast}\right)^{p}\geq\left(\left(\frac{p}{p-1}\right)^{p}-\frac{c_{3}}{\ln^{2}n}\right)\,\sum_{k=1}^{n}{[a_{k}^{\ast}]^{p}}.$

Also, leting $n\rightarrow\infty$ for the constant $c_{3}$ the next estimation
holds

$c_{3}\leq\left(\frac{p}{p-1}\right)^{p+1}p\pi^{2}$

which is the exact constant for $p=2$.

######  Remark 1.7.

The constants $c_{2}(p)$ and $c_{4}(p)$ in Theorem 1.3 and Theorem 1.6 are by
no means the best ones. They could be improved in a lot of ways but that could
have made the proofs longer and more complicated. Our goal was to establish
the exact rate of convergence and to keep the proofs as simple as possible.

Henceforth, the constant $c=c(p)$ will always be an absolute constant, which
means it will depend only on $p$. Also, it may be different on each
occurrence. The relation $O(f)$ means that there exists a positive constant
$c(p)$, depending only on $p$ such that $\left|O(f)\right|\leq c(p)|f|$.

The paper is organized as follows. In Section 2, some technical results are
proved, in Section 3, the right inequality of Theorem 1.3 is proved, in
Section 4, the left inequality of Theorem 1.3 is proved, in Section 5, the
left inequality of Theorem 1.6 is proved and in Section 6, the right
inequality of Theorem 1.6 is proved.

##  2\. Auxiliary Results

We need some technical lemmas.

######  Lemma 2.1.

For $0\leq x\leq 1$ and $\alpha\geq 0$ the next inequality is true:

(2.1) $(1-x)^{\alpha}\leq 1-\alpha x+\frac{1}{2}(\alpha x)^{2}.$

###### Proof.

From $1-x\leq e^{-x}$ we have for $\alpha\geq 0$ and $\alpha x<1$

$(1-x)^{\alpha}\leq e^{-\alpha x}=1-\alpha x+\frac{(\alpha
x)^{2}}{2}+\sum_{k=3}^{\infty}(-1)^{k}\frac{(\alpha x)^{k}}{k!}\leq 1-\alpha
x+\frac{1}{2}(\alpha x)^{2}.$

For $\alpha x\geq 1$ the function $f(x)=1-\alpha x+\frac{1}{2}(\alpha
x)^{2}-(1-x)^{\alpha}$ is increasing and consequently $f(x)\geq f(0)=0$. ∎

######  Lemma 2.2.

For $x\geq-1$, $0\leq\alpha\leq 1$ the next inequality is true:

(2.2) $(1+x)^{\alpha}\leq 1+\alpha x.$

###### Proof.

Let $\alpha=1/\beta$. Then (2.2) follows from the Bernoulli inequality. ∎

######  Lemma 2.3.

For $x\geq 0$, $\alpha\geq 0$ and $\alpha x<1$ the next inequality is true:

(2.3) $(1+x)^{\alpha}\leq 1+\alpha x+(\alpha x)^{2}.$

###### Proof.

$(1+x)^{\alpha}<e^{\alpha x}=1+\alpha x+\sum_{k=2}^{\infty}\frac{(\alpha
x)^{k}}{k!}\leq 1+\alpha x+(\alpha x)^{2}\sum_{k=2}^{\infty}\frac{1}{k!}\leq
1+\alpha x+(\alpha x)^{2}.$

∎

######  Lemma 2.4.

Let $b>1$, $p\geq 2$, $\frac{1}{p}+\frac{1}{q}=1$ and

(2.4) $g(x)=\frac{1}{x^{1/(pq)}}\left(\cos(\alpha\ln x)\right)^{1/q}$

where $\alpha=\frac{1}{\ln b}\arctan\frac{1}{p}$ . Then there exists a
constant $c=c(p)>0$, depending only on $p$ such that for $1\leq x\leq b$ the
next inequality holds:

(2.5) $\left(\frac{\cos(\alpha\ln x)+\alpha q\sin(\alpha\ln
x)}{1+\alpha^{2}q^{2}}\right)^{p/q}\leq-q\left(1+c\alpha^{2}\right)^{-1}x^{1+1/q}\left[(g(x))^{p}\right]^{\prime}.$

###### Proof.

The above inequality (2.5) is equivalent to

$\displaystyle\left(\frac{\cos(\alpha\ln x)+\alpha q\sin(\alpha\ln
x)}{1+\alpha^{2}q^{2}}\right)^{p/q}$
$\displaystyle\leq\left(1+c\alpha^{2}\right)^{-1}\left(\cos(\alpha\ln
x)\right)^{p/q-1}(\cos(\alpha\ln x)+\alpha p\sin(\alpha\ln x))$

or

$\left(\frac{1+\alpha qy}{1+\alpha^{2}q^{2}}\right)^{p/q}\leq\frac{1+\alpha
py}{1+c\alpha^{2}}$

where $y=\tan(\alpha\ln x)$. Obviously $y=\tan(\alpha\ln x)\leq\tan(\alpha\ln
b)=1/p$. We consider two cases.

Case 1. $\alpha\geq 1.$
Since

$\frac{1+\alpha qy}{1+\alpha^{2}q^{2}}\leq\frac{1+\alpha
q/p}{1+\alpha^{2}}\leq\frac{1+\alpha}{1+\alpha^{2}}\leq 1$

we have

$\left(\frac{1+\alpha qy}{1+\alpha^{2}q^{2}}\right)^{p/q}=\frac{1+\alpha
qy}{1+\alpha^{2}q^{2}}\left(\frac{1+\alpha
qy}{1+\alpha^{2}q^{2}}\right)^{p/q-1}\leq\frac{1+\alpha
py}{1+\alpha^{2}q^{2}}.$

Case 2. $\alpha<1.$
From lemma 2.3

$(1+\alpha qy)^{p/q}\leq 1+\alpha py+(\alpha py)^{2}$

and from Bernoulli’s inequality

$\left(1+\alpha^{2}q^{2}\right)^{p/q}\geq 1+\alpha^{2}pq.$

So, it is enough to prove that there exists a constant $c=c(p)>0$ such that

$1+\alpha py+(\alpha py)^{2}\leq\frac{1+\alpha^{2}pq}{1+c\alpha^{2}}(1+\alpha
py)$

which after some simplifications is

$(py)^{2}\leq\frac{pq-c}{1+c\alpha^{2}}.$

Since $py\leq 1$ and $\alpha<1$ the above inequality is true if we take, for
instance, $c=(pq-1)/2$.

By taking $c=\min\\{q^{2},(pq-1)/2\\}$ we complete the proof of the lemma. ∎

######  Remark 2.5.

The inequality (2.5) could be written in the following way

(2.6) $\left(\frac{\cos(\alpha\ln x)+\alpha q\sin(\alpha\ln
x)}{1+\alpha^{2}q^{2}}\right)^{p/q}\leq-q\left(1+\frac{c}{\ln^{2}b}\right)^{-1}x^{1+1/q}\left(g^{p}(x)\right)^{\prime}$

where $c=\arctan(1/p)\min\\{q^{2},(pq-1)/2\\}$.

######  Lemma 2.6.

Let $p\geq 2$, $\frac{1}{p}+\frac{1}{q}=1$, $0<\epsilon<1$,

$b_{0}=e^{\pi/\sqrt{\min\\{q(p-q)^{-1}(pq+1)^{-2},4(pq)^{-2}\\}\epsilon}}$

and

(2.7) $f^{*}(x)=\frac{1}{x^{1/p}}(\alpha q\cos(\alpha\ln x)+\sin(\alpha\ln
x))$

where $\alpha$ is the only solution of the equation

$\tan\left(\alpha\ln b\right)+\alpha q=0$

in the interval $\left(\frac{\pi}{2\ln b},\frac{\pi}{\ln b}\right)$. Then for
every $b>b_{0}$ the next inequality holds

(2.8) $\displaystyle(\sin(\alpha\ln
x))^{p/q}\geq-\frac{qx^{1+1/q}\left[(f^{*}(x))^{p/q}\right]^{\prime}}{1+(pq+\epsilon)\alpha^{2}}$

######  Remark 2.7.

The function $f^{*}$ defined by (2.7) is well defined. Indeed, if $\alpha\ln
x\in\left(0,\frac{\pi}{2}\right]$ it is obvious. If $\alpha\ln
x\in\left(\frac{\pi}{2},\pi\right)$ since the function $h(x)=\alpha
q\cos(\alpha\ln x)+\sin(\alpha\ln x)$ is decreasing and $h(b)=0$ it follows
that $h(x)>h(b)=0$, i.e. $h(x)>0$ for $1\leq x\leq b$ and consequently
$f^{*}(x)>0$.

###### Proof.

It is easy to see that for every $b>b_{0}$

$\alpha\leq\sqrt{\min\left\\{\frac{q}{(p-q)(pq+1)^{2}},\frac{4}{(pq)^{2}}\right\\}\epsilon}.$

Since

$\displaystyle-\frac{qx^{1+1/q}}{1+\alpha^{2}pq}\left[(f^{*}(x))^{p/q}\right]^{\prime}$
(2.9) $\displaystyle=(\alpha q\cos(\alpha\ln x)+\sin(\alpha\ln
x))^{p/q-1}\left(\sin(\alpha\ln
x)-\frac{\alpha(p-q)}{1+\alpha^{2}pq}\cos(\alpha\ln x)\right)$

we need to prove that for every $b>b_{0}$ the next inequality is true

$\displaystyle\frac{1+(pq+\epsilon)\alpha^{2}}{1+pq\alpha^{2}}(\sin(\alpha\ln
x))^{p/q}$ (2.10) $\displaystyle\geq(\alpha q\cos(\alpha\ln x)+\sin(\alpha\ln
x))^{p/q-1}\left(\sin(\alpha\ln
x)-\frac{\alpha(p-q)}{1+\alpha^{2}pq}\cos(\alpha\ln x)\right).$

We consider two cases.

Case 1. $\alpha\ln x\in\left[\frac{\pi}{2},\pi\right)$.
Let $\alpha\ln x=\pi-\phi,\,0<\phi\leq\pi/2$. Then(2) is equivalent to

$\displaystyle\frac{1+(pq+\epsilon)\alpha^{2}}{1+pq\alpha^{2}}(\sin(\alpha\ln
x))^{p/q}\geq(-\alpha
q\cos\phi+\sin\phi)^{p/q-1}\left(\sin\phi+\frac{\alpha(p-q)}{1+\alpha^{2}pq}\cos\phi\right)$

or

(2.11)
$\displaystyle\left[1+(pq+\epsilon)\alpha^{2}\right]y^{p/q}\geq\left(y-\alpha
q\right)^{p/q-1}\left[\left(1+\alpha^{2}pq\right)y+\alpha(p-q)\right]$

where $y=\tan\phi$. Since $y>\alpha q$ in this case, we have from Bernoulli’s
inequality

$\left(\frac{y}{y-\alpha q}\right)^{p/q}\geq\frac{y+\alpha(p-q)}{y-\alpha q}.$

So, it is enough to prove that

$\left[1+(pq+\epsilon)\alpha^{2}\right][y+\alpha(p-q)]\geq\left(1+\alpha^{2}pq\right)y+\alpha(p-q)$

which is easy to verify.

Case 2. $\alpha\ln x\in\left[0,\frac{\pi}{2}\right)$.
In this case (2) is equivalent to

(2.12)
$\displaystyle\left(1+(pq+\epsilon)\alpha^{2}\right)y^{p/q}\geq(y+\alpha
q)^{p/q-1}\left[\left(1+\alpha^{2}pq\right)y-\alpha(p-q)\right]$

where $y=\tan(\alpha\ln x)$. If $y\leq\frac{\alpha(p-q)}{1+\alpha^{2}pq}$ then
(2.12) is obvious. Let $y>\frac{\alpha(p-q)}{1+\alpha^{2}pq}$.

Case 2.1. $2\leq p<3$
(2.12) is equivalent to

$\displaystyle 1+(pq+\epsilon)\alpha^{2}\geq\left(1+\frac{\alpha
q}{y}\right)^{p/q-1}\left(1+\alpha^{2}pq-\frac{\alpha(p-q)}{y}\right).$

Since

$\left(1+\frac{\alpha q}{y}\right)^{p/q-1}\leq 1+\frac{\alpha(p-q)}{y}$

it is enough to prove that

$\displaystyle
1+(pq+\epsilon)\alpha^{2}\geq\left(1+\frac{\alpha(p-q)}{y}\right)\left(1+\alpha^{2}pq-\frac{\alpha(p-q)}{y}\right).$

Simplifying it

$\displaystyle\epsilon\geq\frac{\alpha pq(p-q)}{y}-\frac{(p-q)^{2}}{y^{2}}$

which is true since $\alpha pq<2\sqrt{\epsilon}$.

Case 2.2. $p\geq 3$
(2.12) is equivalent to

$\displaystyle\left[1+(pq+\epsilon)\alpha^{2}\right]\left(\frac{y}{y+\alpha
q}\right)^{p/q-1}\geq 1+\alpha^{2}pq-\frac{\alpha(p-q)}{y}.$

From Bernoulli’s inequality

$\displaystyle\left(\frac{y}{y+\alpha q}\right)^{p/q-1}\geq
1-\frac{\alpha(p-q)}{y+\alpha q}.$

So, it is enough to prove that

$\displaystyle\left[1+(pq+\epsilon)\alpha^{2}\right]\left(1-\frac{\alpha(p-q)}{y+\alpha
q}\right)\geq 1-\frac{\alpha(p-q)}{(1+\alpha^{2}pq)y}.$

Simplifying it

$\epsilon+\frac{q(p-q)}{y(y+\alpha
q)}\geq\frac{(p-q)(pq+\epsilon)\alpha}{y+\alpha q}.$

If $y\leq q[(pq+\epsilon)\alpha]^{-1}$ it is obvious. For
$y>q[(pq+\epsilon)\alpha]^{-1}$

$\displaystyle\frac{(p-q)(pq+\epsilon)\alpha}{y+\alpha
q}<\frac{(p-q)(pq+\epsilon)\alpha}{q[(pq+\epsilon)\alpha]^{-1}+\alpha
q}<\epsilon$

since

$\alpha^{2}<\frac{q\epsilon}{(p-q)(pq+1)^{2}}.$

The lemma is proved. ∎

######  Lemma 2.8.

For $p\geq 2$, $\frac{1}{p}+\frac{1}{q}=1$ and every natural numbers $i$ and
$n$ such that $i\leq n$ the next inequality is true:

(2.13)
$\sum_{k=i}^{n}\frac{1}{k^{1+1/q}}\left(1-\frac{1}{pk^{1/q}}\right)^{p/q}\leq
q\left(\frac{1}{i^{1/q}}-\frac{1}{(n+1)^{1/q}}\right).$

###### Proof.

We will prove that there is a $k_{0}\in\mathbb{N}$ such that for every $k\geq
k_{0}$ the next inequality is true:

(2.14) $\frac{1}{k^{1+1/q}}\left(1-\frac{1}{pk^{1/q}}\right)^{p/q}\leq
q\left(\frac{1}{k^{1/q}}-\frac{1}{(k+1)^{1/q}}\right).$

From (2.1)

$\left(1-\frac{1}{pk^{1/q}}\right)^{p/q}\leq
1-\frac{1}{qk^{1/q}}+\frac{1}{2q^{2}k^{2/q}}$

so, it is enough to prove that there is a $k_{0}\in\mathbb{N}$ such that for
every $k\geq k_{0}$

$\frac{1}{k^{1+1/q}}\left(1-\frac{1}{qk^{1/q}}+\frac{1}{2q^{2}k^{2/q}}\right)\leq
q\left(\frac{1}{k^{1/q}}-\frac{1}{(k+1)^{1/q}}\right)$

i.e.

$1-\frac{1}{qk^{1/q}}+\frac{1}{2q^{2}k^{2/q}}\leq
qk\left[1-\left(\frac{k}{k+1}\right)^{1/q}\right].$

But from (2.2)

$\left(\frac{k}{k+1}\right)^{1/q}=\left(1-\frac{1}{k+1}\right)^{1/q}\leq
1-\frac{1}{q(k+1)}.$

and then

$qk\left[1-\left(\frac{k}{k+1}\right)^{1/q}\right]\geq\frac{k}{k+1}=1-\frac{1}{k+1}.$

So, it is enough to prove that there is a $k_{0}\in\mathbb{N}$ such that for
every $k\geq k_{0}$

$1-\frac{1}{qk^{1/q}}+\frac{1}{2q^{2}k^{2/q}}\leq 1-\frac{1}{k+1},$

i.e.

$\frac{1}{k+1}+\frac{1}{2q^{2}k^{2/q}}\leq\frac{1}{qk^{1/q}}$

which is obviously true for $k$ big enough. Actually, by considering the
function $f(x)=\frac{1}{xk^{1/x}}+\frac{1}{2x^{2}k^{2/x}}$ it is not difficult
to prove that it is true for every $k\geq 8$. From the above it follows that
there is $i_{0}$ such that for every $i\geq i_{0}$ (2.14) is true and
consequently (2.13) as well.

Now, let (2.13) is true for $i+1$. We will prove that it is true for $i$ as
well. We have

$\sum_{k=i}^{n}\frac{1}{k^{1+1/q}}\left(1-\frac{1}{pk^{1/q}}\right)^{p/q}\leq\frac{1}{i^{1+1/q}}\left(1-\frac{1}{pi^{1/q}}\right)^{p/q}+q\left(\frac{1}{(i+1)^{1/q}}-\frac{1}{(n+1)^{1/q}}\right)$

So, it is enough to prove that

$\frac{1}{i^{1+1/q}}\left(1-\frac{1}{pi^{1/q}}\right)^{p/q}+\frac{q}{(i+1)^{1/q}}\leq\frac{q}{i^{1/q}}$

i.e.

$\frac{1}{qi}\left(1-\frac{1}{pi^{1/q}}\right)^{p/q}+\left(\frac{i}{i+1}\right)^{1/q}\leq
1.$

and from (2.2) it follows that it is enough to prove that

$\left(1-\frac{1}{pi^{1/q}}\right)^{p/q}\leq\frac{i}{i+1}.$

From Bernoulli’s inequality for $\alpha\geq 1$ and $0\leq x<1$

$(1-x)^{\alpha}\leq 1-\frac{\alpha x}{1+(\alpha-1)x}$

we have

$\left(1-\frac{1}{pi^{1/q}}\right)^{p/q}\leq 1-\frac{p}{pqi^{1/q}+p-q}.$

So, it is enough to prove

$\frac{1}{i+1}\leq\frac{p}{pqi^{1/q}+p-q}$

i.e.

$i^{1/q}\leq\frac{i}{q}+\frac{1}{p}$

which is easy to verify.

The proof of the lemma is complete. ∎

######  Lemma 2.9.

For $p\geq 2$, $\frac{1}{p}+\frac{1}{q}=1$ and every natural numbers $i$ and
$n$ such that $i\leq n$ the next inequality is true:

$\displaystyle\sum_{k=i}^{n}$ $\displaystyle\frac{\ln^{2}k-2q\ln
k+2q^{2}}{k^{1+1/q}}$ (2.15)
$\displaystyle>\frac{q\left(\ln^{2}i+2q^{2}\right)}{i^{1/q}}+\frac{\ln^{2}i-2q\ln
i+2q^{2}}{2i^{1+1/q}}-\frac{q\left(\ln^{2}n+2q^{2}\right)}{n^{1/q}}.$

###### Proof.

For the function

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

we have

$f^{\prime}(x)=\frac{1}{x^{2+1/q}}\left[-\left(1+\frac{1}{q}\right)\ln^{2}x+2(q+2)\ln
x-2q(q+2)\right]$

and

$f^{\prime\prime}(x)=\frac{1}{x^{3+1/q}}\left[\left(2+\frac{3}{q}+\frac{1}{q^{2}}\right)\ln^{2}x-2(6+2q+\frac{3}{q})\ln
x+4\left(q^{2}+3q+2\right)\right].$

It is easy to see that $f^{\prime}(x)<0$, $f^{\prime\prime}(x)>0$ and
consequently $f^{\prime}(x)$ is increasing and $|f^{\prime}(x)|$ is
decreasing. From Euler’s summation formula (see, for instance [21, p.149])

$\sum_{k=i}^{n}f(k)=\int_{i}^{n}f(x)dx+\frac{f(i)+f(n)}{2}+\int_{i}^{n}\left(x-[x]-\frac{1}{2}\right)f^{\prime}(x)dx$

where $[x]$ is the floor function.

$\int_{i}^{n}\left(x-[x]-\frac{1}{2}\right)f^{\prime}(x)dx=\sum_{k=i}^{n-1}\left(\int_{k}^{k+1/2}+\int_{k+1/2}^{k+1}\right)>0$

since

$\int_{k}^{k+1/2}>0,\,\,\int_{k+1/2}^{k+1}<0\quad\mbox{and}\quad\int_{k}^{k+1/2}>\left|\int_{k+1/2}^{k+1}\right|.$

Then

$\displaystyle\sum_{k=i}^{n}f(k)$
$\displaystyle>\int_{i}^{n}f(x)dx+\frac{f(i)+f(n)}{2}>\int_{i}^{n}f(x)dx+\frac{\ln^{2}i-2q\ln
i+2q^{2}}{2i^{1+1/q}}$
$\displaystyle=\frac{q\left(\ln^{2}i+2q^{2}\right)}{i^{1/q}}+\frac{\ln^{2}i-2q\ln
i+2q^{2}}{2i^{1+1/q}}-\frac{q\left(\ln^{2}n+2q^{2}\right)}{n^{1/q}}.$

The lemma is proved. ∎

######  Lemma 2.10.

For $p\geq 2$, $\frac{1}{p}+\frac{1}{q}=1$ and every natural numbers $i$ and
$n$ such that $i\leq n$ the next inequality is true:

(2.16) $\displaystyle\sum_{k=i}^{n}$ $\displaystyle\frac{\ln^{2}k-2q\ln
k+2q^{2}}{k^{1+2/q}}<\frac{\ln^{2}i-2q\ln
i+2q^{2}}{i^{1+2/q}}+\frac{q\left(\ln^{2}i-q\ln i+3/2q^{2}\right)}{2i^{2/q}}.$

###### Proof.

For the function

$g(x)=\frac{\ln^{2}x-2q\ln x+2q^{2}}{x^{1+2/q}}$

we have

$g^{\prime}(x)=\frac{1}{x^{2+2/q}}\left[-\left(1+\frac{2}{q}\right)\ln^{2}x+2(q+3)\ln
x-2q(q+3)\right]<0$

and consequently $g(x)$ is decreasing and

$\displaystyle\sum_{k=i}^{n}g(k)$ $\displaystyle<\frac{\ln^{2}i-2q\ln
i+2q^{2}}{i^{1+2/q}}+\int_{i}^{n}g(x)dx$ $\displaystyle<\frac{\ln^{2}i-2q\ln
i+2q^{2}}{i^{1+2/q}}+\frac{q\left(\ln^{2}i-q\ln i+3/2q^{2}\right)}{2i^{2/q}}.$

The lemma is proved. ∎

######  Lemma 2.11.

For $p\geq 2$, $\frac{1}{p}+\frac{1}{q}=1$ and every natural numbers $i$ and
$n$ such that $i\leq n$ the next inequality is true:

$\displaystyle\sum_{k=i}^{n}\frac{1}{k^{p}}$
$\displaystyle\left[\left(k^{1/q}\right)^{p/q-1}-\left(k^{1/q}-\frac{1}{p}\right)^{p/q-1}\right]\left[k^{1/q}\left(\ln^{2}k-2q\ln
k+2q^{2}\right)-2q^{2}\right]$ (2.17) $\displaystyle<\frac{\ln^{2}i-2q\ln
i+2q^{2}}{qi^{1+2/q}}+\frac{\ln^{2}i-q\ln
i+3/2q^{2}}{2i^{2/q}}-\frac{2q^{2}}{3i^{3/q}}+\frac{2q^{2}}{3n^{3/q}}.$

###### Proof.

Since

$\left(1-\frac{1}{pk^{1/q}}\right)^{p/q-1}>\left(1-\frac{1}{pk^{1/q}}\right)^{p/q}>1-\frac{1}{qk^{1/q}}$

we have

$\displaystyle\left(k^{1/q}\right)^{p/q-1}-\left(k^{1/q}-\frac{1}{p}\right)^{p/q-1}=\left(k^{1/q}\right)^{p/q-1}\left[1-\left(1-\frac{1}{pk^{1/q}}\right)^{p/q-1}\right]<\frac{1}{q}\left(k^{1/q}\right)^{p/q-2}.$

Then

$\displaystyle\sum_{k=i}^{n}\frac{1}{k^{p}}$
$\displaystyle\left[\left(k^{1/q}\right)^{p/q-1}-\left(k^{1/q}-\frac{1}{p}\right)^{p/q-1}\right]\left[k^{1/q}\left(\ln^{2}k-2q\ln
k+2q^{2}\right)-2q^{2}\right]$
$\displaystyle<\frac{1}{q}\sum_{k=i}^{n}\frac{1}{k^{1+3/q}}\left[k^{1/q}\left(\ln^{2}k-2q\ln
k+2q^{2}\right)-2q^{2}\right]$
$\displaystyle=\frac{1}{q}\sum_{k=i}^{n}\frac{\ln^{2}k-2q\ln
k+2q^{2}}{k^{1+2/q}}-2q\sum_{k=i}^{n}\frac{1}{k^{1+3/q}}$
$\displaystyle<\frac{1}{q}\sum_{k=i}^{n}\frac{\ln^{2}k-2q\ln
k+2q^{2}}{k^{1+2/q}}-2q\int_{i}^{n}\frac{dx}{x^{1+3/q}}$
$\displaystyle=\frac{1}{q}\sum_{k=i}^{n}\frac{\ln^{2}k-2q\ln
k+2q^{2}}{k^{1+2/q}}-\frac{2q^{2}}{3i^{3/q}}+\frac{2q^{2}}{3n^{3/q}}$

and the lemma follows from (2.16). ∎

######  Lemma 2.12.

For $p\geq 2$, $\frac{1}{p}+\frac{1}{q}=1$ and every natural $i$ there is a
constant $c=c(p)>0$, depending only on $p$ such that for every natural $i\geq
2$ the next inequality is true:

$\displaystyle
2q^{2}-\frac{3}{4}+\frac{2q}{3i^{2/q}}-\frac{2q}{i^{1+1/q}}-\frac{q^{2}}{i^{1/q}}-\frac{\ln^{2}i-q\ln
i+3/2q^{2}}{2qi^{1/q}}>c(p).$

###### Proof.

It is obvious that there is a $i_{0}$ and a constant $c(p,i_{0})$ such that
for every $i>i_{0}$ the above inequality is true. For $i\leq i_{0}$ we have

$\displaystyle
2q^{2}-\frac{3}{4}+\frac{2q}{3i^{2/q}}-\frac{2q}{i^{1+1/q}}-\frac{q^{2}}{i^{1/q}}-\frac{\ln^{2}i-q\ln
i+3/2q^{2}}{2qi^{1/q}}$
$\displaystyle>2q^{2}-\frac{3}{4}-\frac{4q}{3i^{1+1/q}}-\frac{q^{2}}{i^{1/q}}-\frac{\left(\ln
i-q/2\right)^{2}+5/4q^{2}}{2qi^{1/q}}$
$\displaystyle=\frac{1}{i^{1/q}}\left[\left(2q^{2}-\frac{3}{4}\right)i^{1/q}-\frac{4q}{3i}-q^{2}-\frac{\left(\ln
i-q/2\right)^{2}+5/4q^{2}}{2q}\right]$

Now we will prove that for every $i\leq i_{0}$

$\displaystyle\left(2q^{2}-\frac{3}{4}\right)i^{1/q}>q^{2}+\frac{4q}{3i}+\frac{\left(\ln
i-q/2\right)^{2}+5/4q^{2}}{2q}.$

We consider the cases $i=2$ and $i\geq 3$ separately.

Case 1. $i=2$
We need to prove that

$\displaystyle\left(2q^{2}-\frac{3}{4}\right)2^{1/q}>q^{2}+\frac{2}{3}q+\frac{\left(\ln
2-q/2\right)^{2}+5/4q^{2}}{2q}.$

Since $\left(\ln 2-q/2\right)^{2}<0.1$ it is enough to prove

$\displaystyle\left(2q^{2}-\frac{3}{4}\right)2^{1/q}>q^{2}+\frac{2}{3}q+\frac{1}{20q}+\frac{5}{8}q=q^{2}+\frac{31}{24}q+\frac{1}{20q}.$

Considering for $1\leq x\leq 2$ the function

$f(x)=\left(2x^{2}-\frac{3}{4}\right)2^{1/x}-x^{2}-\frac{31}{24}x-\frac{1}{20x}$

we have

$\displaystyle f^{\prime}(x)$
$\displaystyle=4x2^{1/x}-\left(2-\frac{3}{4x^{2}}\right)2^{1/x}\ln
2-2x-\frac{31}{24}+\frac{1}{20x^{2}}$
$\displaystyle>4x2^{1/x}-\frac{7}{10}\left(2-\frac{3}{4x^{2}}\right)2^{1/x}-2x-\frac{31}{24}$
$\displaystyle=\left(4x-\frac{7}{5}+\frac{21}{40x^{2}}\right)2^{1/x}-2x-\frac{31}{24}$
$\displaystyle>4\sqrt{2}x-\frac{7\sqrt{2}}{5}-2x-\frac{31}{24}>4\sqrt{2}-\frac{7\sqrt{2}}{5}-2-\frac{31}{24}>0$

and consequently the function $f(x)$ is increasing and $f(x)>f(1)>0$.

Case 2. $i\geq 3$
We have

$i^{1/q}=e^{\ln i/q}>1+\frac{\ln
i}{q}+\frac{\ln^{2}i}{2q^{2}}\quad\mbox{and}\quad\left(\ln
i-q/2\right)^{2}<\ln^{2}i$

so it is enough to prove that for $3\leq i$ the next inequality is true

$\left(2q^{2}-\frac{3}{4}\right)\left(1+\frac{\ln
i}{q}+\frac{\ln^{2}i}{2q^{2}}\right)>q^{2}+\frac{4q}{3i}+\frac{\ln^{2}i}{2q}+\frac{5}{8}q$

i.e.

$q^{2}-\frac{3}{4}+\left(2q-\frac{3}{4q}\right)\ln
i>\frac{4q}{3i}+\frac{5}{8}q$

because

$\frac{2q^{2}-\frac{3}{4}}{2q^{2}}>\frac{1}{2q}.$

But $2q-\frac{3}{4q}>5/4$ and

$q^{2}-\frac{3}{4}+\left(2q-\frac{3}{4q}\right)\ln
i>q^{2}-\frac{3}{4}+\frac{5}{4}\ln 3>q^{2}+\frac{1}{2}.$

Also, $\frac{4q}{3i}+\frac{5}{8}q<\frac{4q}{9}+\frac{5}{8}q=\frac{77}{72}q$
and since $q^{2}+\frac{1}{2}>\frac{77}{72}q$ the lemma is proved. ∎

######  Lemma 2.13.

For $p\geq 2$, $\frac{1}{p}+\frac{1}{q}=1$ there is a constant $c=c(p)>0$ such
that for every natural $n$ the next inequality is true:

$\displaystyle\frac{\ln^{2}2+2q^{2}}{2^{1/q}}+\frac{2}{3}\frac{q}{2^{3/q}}-\ln^{2}2-2q\sum_{k=2}^{n}\frac{1}{k^{1+2/q}}-\frac{\ln^{2}2-q\ln
2+3/2q^{2}}{q2^{1+2/q}}>c(p).$

###### Proof.

We have

$\sum_{k=2}^{n}\frac{1}{k^{1+2/q}}=\frac{1}{2^{1+2/q}}+\sum_{k=3}^{n}\frac{1}{k^{1+2/q}}<\frac{1}{2^{1+2/q}}+\int_{2}^{\infty}\frac{dx}{x^{1+2/q}}=\frac{1+q}{2.2^{2/q}}$

and

$\frac{\ln^{2}2-q\ln 2+3/2q^{2}}{q2^{1+2/q}}=\frac{\left(\ln
2-q/2\right)^{2}+5/4q^{2}}{q2^{1+2/q}}<\frac{1}{20q2^{2/q}}+\frac{5q}{2^{3+2/q}}.$

Since

$\frac{2}{3}\frac{q}{2^{3/q}}>\frac{1}{20q2^{2/q}}$

and

$\left(1-\frac{1}{2^{1/q}}\right)\ln^{2}2<\frac{1}{4}$

it follows

$\displaystyle\frac{\ln^{2}2+2q^{2}}{2^{1/q}}+\frac{2}{3}\frac{q}{2^{3/q}}-\ln^{2}2-2q\sum_{k=2}^{n}\frac{1}{k^{1+2/q}}-\frac{\ln^{2}2-q\ln
2+3/2q^{2}}{q2^{1+2/q}}$
$\displaystyle>\frac{q}{2^{2/q}}\left[\left(2^{1+1/q}-1\right)q-\frac{13}{8}\right]-\frac{1}{4}>\frac{1}{4}\left[\left(2^{1+1/q}-1\right)q-\frac{21}{8}\right]>0$

since the function $f(x)=\left(2^{1+1/x}-1\right)x$ is increasing and
consequently
$\left(2^{1+1/q}-1\right)q>f(1)=3$. ∎

######  Lemma 2.14.

For $p\geq 2$, $\frac{1}{p}+\frac{1}{q}=1$ and every natural numbers $i$ and
$n$ such that $2\leq i\leq n$ the next inequality is true:

$\displaystyle\sum_{k=i}^{n}\frac{1}{k^{p}}$
$\displaystyle\left(k^{1/q}-\frac{1}{p}\right)^{p/q-1}\left[k^{1/q}\left(\ln^{2}k-2q\ln
k+2q^{2}\right)-2q^{2}\right]$
$\displaystyle>\frac{q\left(\ln^{2}i+2q^{2}\right)}{i^{1/q}}-\frac{q\left(\ln^{2}n+2q^{2}\right)}{n^{1/q}}-2q^{2}\sum_{k=i}^{n}\frac{1}{k^{1+2/q}}$
(2.18) $\displaystyle-\frac{\ln^{2}i-q\ln
i+3/2q^{2}}{2i^{2/q}}+\frac{2q^{2}}{3i^{3/q}}-\frac{2q^{2}}{3n^{3/q}}.$

###### Proof.

$\displaystyle\sum_{k=i}^{n}\frac{1}{k^{p}}$
$\displaystyle\left(k^{1/q}-\frac{1}{p}\right)^{p/q-1}\left[k^{1/q}\left(\ln^{2}k-2q\ln
k+2q^{2}\right)-2q^{2}\right]=J-I$

where

$\displaystyle J$
$\displaystyle=\sum_{k=i}^{n}\frac{1}{k^{p}}\left(k^{1/q}\right)^{p/q-1}\left[k^{1/q}\left(\ln^{2}k-2q\ln
k+2q^{2}\right)-2q^{2}\right]$
$\displaystyle=\sum_{k=i}^{n}\frac{\ln^{2}k-2q\ln
k+2q^{2}}{k^{1+1/q}}-2q^{2}\sum_{k=i}^{n}\frac{1}{k^{1+2/q}}$

and

$\displaystyle I$
$\displaystyle=\sum_{k=i}^{n}\frac{1}{k^{p}}\left[\left(k^{1/q}\right)^{p/q-1}-\left(k^{1/q}-\frac{1}{p}-\right)^{p/q-1}\right]\left[k^{1/q}\left(\ln^{2}k-2q\ln
k+2q^{2}\right)-2q^{2}\right].$

From (2.9)

$\displaystyle J$
$\displaystyle>\frac{q\left(\ln^{2}i+2q^{2}\right)}{i^{1/q}}+\frac{\ln^{2}i-2q\ln
i+2q^{2}}{2i^{1+1/q}}-\frac{q\left(\ln^{2}n+2q^{2}\right)}{n^{1/q}}-2q^{2}\sum_{k=i}^{n}\frac{1}{k^{1+2/q}}$

and from (2.11)

$\displaystyle I$ $\displaystyle<\frac{\ln^{2}i-2q\ln
i+2q^{2}}{qi^{1+2/q}}+\frac{\ln^{2}i-q\ln
i+3/2q^{2}}{2i^{2/q}}-\frac{2q^{2}}{3i^{3/q}}+\frac{2q^{2}}{3n^{3/q}}.$

For $i\geq 2$ we have $qi^{1/q}\geq 2$ and consequently

$\displaystyle\frac{\ln^{2}i-2q\ln
i+2q^{2}}{2i^{1+1/q}}\geq\frac{\ln^{2}i-2q\ln i+2q^{2}}{qi^{1+2/q}}.$

The lemma is proved. ∎

######  Lemma 2.15.

For $p\geq 2$, $\frac{1}{p}+\frac{1}{q}=1$ and every natural $n$ the next
inequality is true:

$\displaystyle\sum_{k=1}^{n}\frac{1}{k^{p}}$
$\displaystyle\left(k^{1/q}-\frac{1}{p}\right)^{p/q-1}\left[k^{1/q}\left(\ln^{2}k-2q\ln
k+2q^{2}\right)-2q^{2}\right]$
$\displaystyle>\frac{q\left(\ln^{2}2+2q^{2}\right)}{2^{1/q}}-\frac{q\left(\ln^{2}n+2q^{2}\right)}{n^{1/q}}-2q^{2}\sum_{k=2}^{n}\frac{1}{k^{1+2/q}}$
(2.19) $\displaystyle-\frac{\ln^{2}2-q\ln
2+3/2q^{2}}{2^{1+2/q}}+\frac{2}{3}\frac{q^{2}}{2^{3/q}}-\frac{2q^{2}}{3n^{3/q}}.$

###### Proof.

Since

$\displaystyle\sum_{k=1}^{n}\frac{1}{k^{p}}\left(k^{1/q}-\frac{1}{p}\right)^{p/q-1}\left[k^{1/q}\left(\ln^{2}k-2q\ln
k+2q^{2}\right)-2q^{2}\right]$
$\displaystyle=\sum_{k=2}^{n}\frac{1}{k^{p}}\left(k^{1/q}-\frac{1}{p}\right)^{p/q-1}\left[k^{1/q}\left(\ln^{2}k-2q\ln
k+2q^{2}\right)-2q^{2}\right]$

the lemma follows from the previous one. ∎

######  Lemma 2.16.

Let $p\geq 2$, $\frac{1}{p}+\frac{1}{q}=1$ and $A>1$. Then for every natural
numbers $i$ and $n$ such that $i\leq n$ the next equality is true:

$\displaystyle\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{1}{A^{j}\ln^{2j}(n+1)}\sum_{k=i}^{n}\frac{\ln^{2j}k-2qj\ln^{2j-1}k}{k^{1+1/q}}$
(2.20)
$\displaystyle=\frac{q}{i^{1/q}}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}i}{A^{j}\ln^{2j}(n+1)}+O\left(\frac{1}{A^{2}\ln^{2}(n+1)i^{1/q}}\right)+O\left(\frac{1}{A^{2}n^{1/q}}\right).$

###### Proof.

Let us denote by $g$ the function

$g(x)=\frac{\ln^{2j}x-2qj\ln^{2j-1}x}{x^{1+1/q}}.$

Again, from Euler’s summation formula

$\sum_{k=i}^{n}g(k)=\int_{i}^{n}g(x)dx+\frac{g(i)+f(n)}{2}+\int_{i}^{n}\left(x-[x]-\frac{1}{2}\right)g^{\prime}(x)dx$

where $[x]$ is the floor function. Now

$|g(i)|=\left|\frac{\ln^{2j}i-2qj\ln^{2j-1}i}{i^{1+1/q}}\right|<\frac{c(p)j\ln^{2j}i}{i^{1+1/q}}<\frac{c(p)j^{2}\ln^{2j-2}n}{i^{1/q}}$

and

$|g(n)|=\left|\frac{\ln^{2j}n-2qj\ln^{2j-1}n}{n^{1+1/q}}\right|<\frac{c(p)j\ln^{2j}n}{n^{1+1/q}}<\frac{c(p)j^{2}\ln^{2j-2}n}{i^{1/q}}.$

For $g(x)$ we have

$g^{\prime}(x)=\frac{\ln^{2j-2}x}{x^{2+1/q}}\left[-\left(1+\frac{1}{q}\right)\ln^{2}x+2j(q+2)\ln
x-2qj(2j-1)\right]$

and consequently

$|g^{\prime}(x)|<\frac{c(p)j^{2}\ln^{2j}x}{x^{2+1/q}}<\frac{c(p)j^{2}\ln^{2j-2}x}{x^{1+1/q}}<\frac{c(p)j^{2}\ln^{2j-2}n}{x^{1+1/q}}.$

Then

$\left|\int_{i}^{n}\left(x-[x]-\frac{1}{2}\right)g^{\prime}(x)dx\right|<c(p)j^{2}\ln^{2j-2}n\int_{i}^{n}\frac{1}{x^{1+1/q}}<\frac{c(p)j^{2}\ln^{2j-2}n}{i^{1/q}}.$

Also

$\int_{i}^{n}g(x)dx=\frac{q\ln^{2j}i}{i^{1/q}}-\frac{q\ln^{2j}n}{n^{1/q}}.$

Consequently

$\displaystyle\sum_{k=i}^{n}\frac{\ln^{2j}k-2qj\ln^{2j-1}k}{k^{1+1/q}}=\frac{q\ln^{2j}i}{i^{1/q}}-\frac{q\ln^{2j}n}{n^{1/q}}+O\left(\frac{j^{2}\ln^{2j-2}n}{i^{1/q}}\right).$

Then

$\displaystyle\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{1}{A^{j}\ln^{2j}(n+1)}\sum_{k=i}^{n}\frac{\ln^{2j}k-2qj\ln^{2j-1}k}{k^{1+1/q}}$
$\displaystyle=\frac{q}{i^{1/q}}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}i}{A^{j}\ln^{2j}(n+1)}+O\left(\frac{1}{A^{2}\ln^{2}(n+1)i^{1/q}}\right)+O\left(\frac{1}{A^{2}n^{1/q}}\right).$

∎

##  3\. Proof of the right inequality in theorem 1.3
$d(a,b)\,\leq\left(\frac{p}{p-1}\right)^{p}\left(1+\frac{c}{\ln^{2}\frac{b}{a}}\right)^{-1}\,,\qquad
c=c(p)>0.$

By simple change of variables and notations it is easy to see that it is
enough to prove (1.11) for the interval $(1,b)$.

From Holder’s inequality we have for every two functions $f(x)\geq 0$ and
$g(x)>0,\,x\in(1,b)$ and $f^{p}(x)$ and $g^{q}(x)$ are integrable over (1,b),

$\left(\int_{1}^{x}f(t)dt\right)^{p}\leq\left(\int_{1}^{x}g^{q}(t)dt\right)^{p/q}\left(\int_{1}^{x}\frac{f^{p}(t)}{g^{p}(t)}dt\right).$

After multiplying both sides by $x^{-p}$, integrating from 1 to $b$ and
changing the order of integration in the right side we get

$\displaystyle\int_{1}^{b}\left(\frac{1}{x}\int_{1}^{x}f(t)dt\right)^{p}dx\leq\int_{1}^{b}\left[\frac{1}{g^{p}(t)}\int_{t}^{b}\left(\int_{1}^{x}g^{q}(u)du\right)^{p/q}\frac{dx}{x^{p}}\right]f^{p}(t)dt.$

Let us denote for brevity $M(g,t)=g^{-p}(t)M^{*}(g,t)$ where

$M^{*}(g,t)=\int_{t}^{b}\left(\int_{1}^{x}g^{q}(u)du\right)^{p/q}\frac{dx}{x^{p}}.$

Then for every two functions $f(x)\geq 0$ and $g(x)>0$, $1<x<b$ such that
$f^{p}(x)$ and $g^{q}(x)$ are integrable the next upper estimation holds

$\displaystyle\int_{1}^{b}\left(\frac{1}{x}\int_{1}^{x}f(t)dt\right)^{p}dx\leq\max_{1<t<b}M(g,t)\int_{1}^{b}f^{p}(t)dt$

and consequently for every function $g(x)>0,\,1<x<b$

$d(1,b)\leq\max_{1<t<b}M(g,t).$

Now we want to minimize

$\max_{1<t<b}M(g,t)$

over all functions $g(x)>0$ on the interval $(1,b)$ or to find

$\min_{g(x)>0}\,\max_{1<t<b}\frac{1}{g^{p}(t)}\int_{t}^{b}\left(\int_{1}^{x}g^{q}(u)du\right)^{p/q}\frac{dx}{x^{p}}.$

######  Remark 3.1.

For $g(x)=x^{-1/(pq)}$ we obtain the original Hardy inequality. Indeed, we
have

$\int_{1}^{x}g^{q}(u)du=qx^{1/q}-q<qx^{1/q},$
$\int_{t}^{b}\left(\int_{1}^{x}g^{q}(u)du\right)^{p/q}\frac{dx}{x^{p}}<\int_{t}^{b}\left(qx^{1/q}\right)^{p/q}\frac{dx}{x^{p}}=q^{p}\left(t^{-1/q}-b^{-1/q}\right)<q^{p}t^{-1/q}$

for every $1<t<b$. Consequently $M(g,t)<q^{p}$ for every $1<t<b$, which means
that

$\max_{1<t<b}M(g,t)<q^{p}\quad\mbox{i.e.}\quad d(1,b)\leq q^{p}.$

Now, for the function $g(x)$, defined by (2.4) we have

$\displaystyle\int_{1}^{x}g^{q}(u)du$
$\displaystyle=\frac{q}{1+\alpha^{2}q^{2}}\left[x^{1/q}(\cos(\alpha\ln
x)+\alpha q\sin(\alpha\ln x))-1\right]$
$\displaystyle<\frac{qx^{1/q}}{1+\alpha^{2}q^{2}}\left[\cos(\alpha\ln
x)+\alpha q\sin(\alpha\ln x)\right]$

and for every $1<t<b$

$M^{*}(g,t)<q^{p/q}\int_{t}^{b}\left(\frac{\cos(\alpha\ln x)+\alpha
q\sin(\alpha\ln x)}{1+\alpha^{2}q^{2}}\right)^{p/q}\frac{dx}{x^{1+1/q}}.$

Then from (2.6) it follows that for every $1<t<b$

$M^{*}(g,t)<-q^{p}\left(1+\frac{c}{\ln^{2}b}\right)^{-1}\int_{t}^{b}\left(g^{p}(x)\right)^{\prime}dx\leq
q^{p}\left(1+\frac{c}{\ln^{2}b}\right)^{-1}g^{p}(t)$

and consequently

$M(g,t)\leq
q^{p}\left(1+\frac{c}{\ln^{2}b}\right)^{-1}=\left(\frac{p}{p-1}\right)^{p}\left(1+\frac{c}{\ln^{2}\frac{b}{a}}\right)^{-1}.$

The last means that

$\max_{1<t<b}M(g,t)\leq\left(\frac{p}{p-1}\right)^{p}\left(1+\frac{c}{\ln^{2}\frac{b}{a}}\right)^{-1}$

i.e.

$d(1,b)\leq\left(\frac{p}{p-1}\right)^{p}\left(1+\frac{c}{\ln^{2}\frac{b}{a}}\right)^{-1}.$

##  4\. Proof of the left inequality in theorem 1.3
$d(a,b)\,\geq\left(\frac{p}{p-1}\right)^{p}-\frac{c}{\ln^{2}\frac{b}{a}}\,,\qquad
c=c(p)>0.$

By changing the order of integration, we write the left side of (1.5) for
$a=1$ and $f(x)>0,\,\,1<x<b$ in the following way

$\displaystyle\int_{1}^{b}\left(\frac{1}{x}\int_{1}^{x}f(t)dt\right)^{p}\,dx=\int_{1}^{b}M(t)f^{p}(t)dt$

where

$\displaystyle
M(t)=\frac{1}{[f(t)]^{p/q}}\int_{t}^{b}\left(\int_{1}^{x}f(u)du\right)^{p/q}\frac{dx}{x^{p}}.$

Obviously

$d(1,b)\geq\min_{1<t<b}M(t).$

Then for the function $f^{*}(x)$ defined in (2.7) we have

$\int_{1}^{x}f^{*}(u)du=qx^{1/q}\sin(\alpha\ln x)$

and

(4.1)
$\displaystyle\int_{t}^{b}\left(\int_{1}^{x}f^{*}(u)du\right)^{p/q}\frac{dx}{x^{p}}=q^{p/q}\int_{t}^{b}(\sin(\alpha\ln
x))^{p/q}\frac{dx}{x^{1+1/q}}.$

Now let $0<\epsilon<1$. Then from (4.1) and (2.8) it follows that for
$b>b_{0}$ where

$b_{0}=e^{\pi/\sqrt{\min\\{q(p-q)^{-1}(pq+1)^{-2},4(pq)^{-2}\\}\epsilon}}$

the next inequality holds

$\int_{t}^{b}\left(\int_{1}^{x}f^{*}(u)du\right)^{p/q}\frac{dx}{x^{p}}\geq-\frac{q^{p}}{1+(pq+\epsilon)\alpha^{2}}\int_{t}^{b}\left[(f^{*}(x))^{p/q}\right]^{\prime}dx=\frac{q^{p}[f^{*}(t)]^{p/q}}{1+(pq+\epsilon)\alpha^{2}}.$

Consequently for every $b>b_{0}$

$\displaystyle M(t)\geq\frac{q^{p}}{1+(pq+\epsilon)\alpha^{2}}\geq
q^{p}\left(1-(pq+\epsilon)\alpha^{2}\right)\geq
q^{p}-\frac{q^{p}(pq+\epsilon)\pi^{2}}{\ln^{2}b}$

i.e.

$d(1,b)\geq q^{p}-\frac{q^{p}(pq+\epsilon)\pi^{2}}{\ln^{2}b}.$

Since for $b\leq b_{0}$ we have

$d(1,b)\geq q^{p}-\frac{q^{p}\ln^{2}b_{0}}{\ln^{2}b}$

by taking $c=\max\\{q^{p}(pq+\epsilon)\pi^{2},q^{p}\ln^{2}b_{0}\\}$ we
complete the proof.

##  5\. Proof of the left inequality in theorem 1.6
$d_{n}\geq\left(\frac{p}{p-1}\right)^{p}-\frac{c}{\ln^{2}n}\,,\qquad
c=c(p)>0.$

By changing the order of summation we write the left side of (1.6) in the
following way

$\sum_{k=1}^{n}\left(\frac{1}{k}\sum_{j=1}^{k}a_{j}\right)^{p}=\sum_{i=1}^{n}\left[\frac{1}{a_{i}^{p/q}}\sum_{k=i}^{n}\frac{1}{k^{p}}\left(\sum_{j=1}^{k}a_{j}\right)^{p/q}\right]a_{i}^{p}=\sum_{i=1}^{n}M_{i}a_{i}^{p}$

where

$M_{i}=\frac{1}{a_{i}^{p/q}}M_{i}^{*}\quad\mbox{and}\quad
M_{i}^{*}=\sum_{k=i}^{n}\frac{1}{k^{p}}\left(\sum_{j=1}^{k}a_{j}\right)^{p/q}.$

Then

$\sum_{k=1}^{n}\left(\frac{1}{k}\sum_{j=1}^{k}a_{j}\right)^{p}\geq\min_{1\leq
i\leq n}M_{i}\sum_{k=1}^{n}a_{i}^{p}$

and consequently

$d_{n}\geq\min_{1\leq i\leq n}M_{i}.$

Now, we will prove that the sequence $a_{k}^{*}$ defined in Theorem 1.6 is the
“almost extremal” sequence, i. e. the inequality (1.14) holds. We have

$\sum_{j=1}^{k}a_{j}^{*}=\int_{1}^{k+1}f^{*}(x)dx=q(k+1)^{1/q}\sin(\alpha\ln(k+1))$

where $f^{*}$ is the function defined by (1.13). It is easy to see that the
function $x^{p-1-1/q}\left(\sin(\alpha\ln
x)\right)^{p/q}=\left(x^{1/q}\sin(\alpha\ln x)\right)^{p/q}$ is increasing and
consequently

$\displaystyle M_{i}^{*}$
$\displaystyle=q^{p/q}\sum_{k=i}^{n}\frac{(k+1)^{p-1-1/q}}{k^{p}}\left(\sin(\alpha\ln(k+1))\right)^{p/q}$
(5.1) $\displaystyle\geq q^{p/q}\int_{i}^{n+1}(\sin(\alpha\ln
x))^{p/q}\frac{dx}{x^{1+1/q}}.$

Since the function $f^{*}(x)$ is continuous there exists a point
$\eta_{i}\in[i,i+1]$ such that $a_{i}=f^{*}(\eta_{i})$. Now let
$0<\epsilon<1$. Then from (5) and Lemma 2.6 it follows that for every integer

$n>n_{0}=e^{\pi/\sqrt{\min\\{q(p-q)^{-1}(pq+1)^{-2},4(pq)^{-2}\\}\epsilon}}$
$\displaystyle
M_{i}^{*}\geq-\frac{q^{p}}{1+(pq+\epsilon)\alpha^{2}}\int_{\eta_{i}}^{b}\left[(f^{*}(x))^{p/q}\right]^{\prime}dx=\frac{q^{p}\left[f^{*}(\eta_{i})\right]^{p/q}}{1+(pq+\epsilon)\alpha^{2}}$

and consequently

$\displaystyle M_{i}$
$\displaystyle\geq\frac{q^{p}}{1+(pq+\epsilon)\alpha^{2}}\geq
q^{p}\left(1-(pq+\epsilon)\alpha^{2}\right)\geq
q^{p}-\frac{q^{p}(pq+\epsilon)\pi^{2}}{\ln^{2}(n+1)}$

i.e.

$d_{n}\geq q^{p}-\frac{q^{p}(pq+\epsilon)\pi^{2}}{\ln^{2}(n+1)}.$

Since for $n\leq n_{0}$ we have

$d_{n}\geq q^{p}-\frac{q^{p}\ln^{2}n_{0}}{\ln^{2}n}$

by taking $c=\max\\{q^{p}(pq+\epsilon)\pi^{2},q^{p}\ln^{2}n_{0}\\}$ we
complete the proof.

##  6\. Proof of the right inequality in theorem 1.6
$d_{n}<\left(\frac{p}{p-1}\right)^{p}-\frac{c}{\ln^{2}n}\,,\qquad c=c(p)>0.$

From Holder’s inequality we have for every two sequences $\mu_{i}>0$ and
$\eta_{i}\geq 0$, $i=1,...n$

$\sum_{i=1}^{k}\mu_{i}\eta_{i}\leq\left(\sum_{i=1}^{k}\eta_{i}^{p}\right)^{1/p}\left(\sum_{i=1}^{k}\mu_{i}^{q}\right)^{1/q}$

or

$\left(\frac{1}{k}\sum_{i=1}^{k}\mu_{i}\eta_{i}\right)^{p}\leq\frac{1}{k^{p}}\left(\sum_{i=1}^{k}\eta_{i}^{p}\right)\left(\sum_{i=1}^{k}\mu_{i}^{q}\right)^{p/q}.$

Denoting $a_{i}=\mu_{i}\eta_{i}$ and after changing the order of summation we
get

$\sum_{k=1}^{n}\left(\frac{1}{k}\sum_{i=1}^{k}a_{i}\right)^{p}\leq\sum_{i=1}^{n}M_{i}a_{i}^{p}\leq\left(\max_{1\leq
i\leq n}M_{i}\right)\sum_{i=1}^{n}a_{i}^{p},$

where

$M_{i}=\frac{1}{\mu_{i}^{p}}M_{i}^{*},\quad
M_{i}^{*}=\sum_{k=i}^{n}\frac{1}{k^{p}}\left(\sum_{j=1}^{k}\mu_{j}^{q}\right)^{p/q}.$

Obviously

$d_{n}\leq\max_{1\leq i\leq n}M_{i},\quad\mbox{so we want to minimize
}\quad\max_{1\leq i\leq n}M_{i}$

over all sequences $\mu=\\{\mu_{i}>0\\},\,i=1,2,...,n$, i.e. to find

$\min_{\mu>0}\,\max_{1\leq i\leq n}M_{i}$

or, at least, to make it as small as possible.

######  Remark 6.1.

By choosing, for instance,

$\mu_{k}=k^{-1/(pq)},\quad k=1,2,\,...\,n$

we obtain the Hardy’s inequality with $d_{n}=\left(\frac{p}{p-1}\right)^{p}$.

Indeed,

(6.1)
$\sum_{j=1}^{k}\mu_{j}^{q}=1+\sum_{j=2}^{k}\frac{1}{j^{1/p}}<1+\int_{1}^{k}\frac{dx}{x^{1/p}}=qk^{1/q}-q+1=qk^{1/q}\left(1-\frac{1}{pk^{1/q}}\right)$

and from (2.13) of Lemma 2.8

$M_{i}^{*}\leq
q^{p/q}\sum_{k=i}^{n}\frac{1}{k^{1+1/q}}\left(1-\frac{1}{pk^{1/q}}\right)^{p/q}\leq\left(\frac{p}{p-1}\right)^{p}i^{-1/q}.$

Consequently

$M_{i}\leq\left(\frac{p}{p-1}\right)^{p}\quad\mbox{and}\quad
d_{n}\leq\left(\frac{p}{p-1}\right)^{p}.$

But in order to prove the right inequality of (1.12) we need to make a more
complicated choise of the sequence $\mu_{k}$. Let

$\mu_{k}=\left(\frac{A}{k^{1/p}}-\frac{1}{\ln^{2}(n+1)}\int_{k}^{k+1}\frac{\ln^{2}x}{x^{1/p}}dx\right)^{1/q}$

where $A=A(p)>2$ is a constant which depends only on $p$ and will be chosen
later. It is obvious that the sequence $\mu_{k},\,k=1,...,n$ is well defined.
Then for every $i$ such that $1\leq i\leq n$

(6.2) $\mu_{i}^{p}<\frac{A^{p/q}}{i^{1/q}}=\frac{c(p)}{i^{1/q}}$

and

$\displaystyle\mu_{i}^{p}$
$\displaystyle>\left(\frac{A}{i^{1/p}}-\frac{\ln^{2}(i+1)}{i^{1/p}\ln^{2}(n+1)}\right)^{p/q}=\frac{A^{p/q}}{i^{1/q}}\left(1-\frac{\ln^{2}(i+1)}{A\ln^{2}(n+1)}\right)^{p/q}$
(6.3)
$\displaystyle=\frac{A^{p/q}}{i^{1/q}}\sum_{j=0}^{\infty}(-1)^{j}\binom{p/q}{j}\left(\frac{\ln^{2}(i+1)}{A\ln^{2}(n+1)}\right)^{j}.$

Now

$\displaystyle\sum_{j=1}^{k}\mu_{j}^{q}$
$\displaystyle=A\sum_{j=1}^{k}\frac{1}{j^{1/p}}-\frac{1}{\ln^{2}(n+1)}\int_{1}^{k+1}\frac{\ln^{2}x}{x^{1/p}}dx$
$\displaystyle<A\sum_{j=1}^{k}\frac{1}{j^{1/p}}-\frac{1}{\ln^{2}(n+1)}\int_{1}^{k}\frac{\ln^{2}x}{x^{1/p}}dx.$

Since

$\int_{1}^{k}\frac{\ln^{2}x}{x^{1/p}}dx=qk^{1/q}\left(\ln^{2}k-2q\ln
k+2q^{2}\right)-2q^{3}$

and from (6.1) we have

$\displaystyle\sum_{j=1}^{k}\mu_{j}^{q}$
$\displaystyle<Aq\left(k^{1/q}-\frac{1}{p}\right)-\frac{q}{\ln^{2}(n+1)}\left[k^{1/q}\left(\ln^{2}k-2q\ln
k+2q^{2}\right)-2q^{2}\right]$
$\displaystyle=Aq\left(k^{1/q}-\frac{1}{p}\right)(1-S(k))$

where for brevity we denoted by

$S(k)=\frac{k^{1/q}\left(\ln^{2}k-2q\ln
k+2q^{2}\right)-2q^{2}}{A\left(k^{1/q}-\frac{1}{p}\right)\ln^{2}(n+1)}.$

We have

$\displaystyle(Aq)^{-p/q}M_{i}^{*}$
$\displaystyle<(Aq)^{-p/q}\sum_{k=i}^{n}\frac{1}{k^{p}}\left[Aq\left(k^{1/q}-\frac{1}{p}\right)\right]^{p/q}\left(1-S(k)\right)^{p/q}$
$\displaystyle=\sum_{k=i}^{n}\frac{1}{k^{p}}\left(k^{1/q}-\frac{1}{p}\right)^{p/q}\sum_{j=0}^{\infty}(-1)^{j}\binom{p/q}{j}S^{j}(k)$
$\displaystyle=\sum_{k=i}^{n}\frac{1}{k^{p}}\left(k^{1/q}-\frac{1}{p}\right)^{p/q}-\frac{p}{q}\sum_{k=i}^{n}\frac{1}{k^{p}}\left(k^{1/q}-\frac{1}{p}\right)^{p/q}S(k)$
(6.4)
$\displaystyle+\sum_{k=i}^{n}\frac{1}{k^{p}}\left(k^{1/q}-\frac{1}{p}\right)^{p/q}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}S^{j}(k)=L_{1}-L_{2}+L_{3}.$

From 2.13 of Lemma 2.8

(6.5)
$L_{1}=\sum_{k=i}^{n}\frac{1}{k^{p}}\left(k^{1/q}-\frac{1}{p}\right)^{p/q}\leq
q\left(\frac{1}{i^{1/q}}-\frac{1}{(n+1)^{1/q}}\right).$

From 2.14 of Lemma 2.14 we have for $2\leq i\leq n$

$\displaystyle L_{2}(i\geq 2)$
$\displaystyle=\frac{p}{q}\sum_{k=i}^{n}\frac{1}{k^{p}}\left(k^{1/q}-\frac{1}{p}\right)^{p/q}S(k)$
$\displaystyle>\frac{p}{qA\ln^{2}(n+1)}\Big{[}\frac{q\left(\ln^{2}i+2q^{2}\right)}{i^{1/q}}-\frac{q\left(\ln^{2}n+2q^{2}\right)}{n^{1/q}}-2q^{2}\sum_{k=i}^{n}\frac{1}{k^{1+2/q}}$
(6.6) $\displaystyle-\frac{\ln^{2}i-q\ln
i+3q^{2}/2}{2i^{2/q}}+\frac{2q^{2}}{3i^{3/q}}-\frac{2q^{2}}{3n^{3/q}}\Big{]}$

and since $S(1)=0$

$\displaystyle L_{2}(i=1)$
$\displaystyle=\frac{p}{q}\sum_{k=1}^{n}\frac{1}{k^{p}}\left(k^{1/q}-\frac{1}{p}\right)^{p/q}S(k)$
$\displaystyle>\frac{p}{qA\ln^{2}(n+1)}\Big{[}\frac{q\left(\ln^{2}2+2q^{2}\right)}{2^{1/q}}-\frac{q\left(\ln^{2}n+2q^{2}\right)}{n^{1/q}}-2q^{2}\sum_{k=2}^{n}\frac{1}{k^{1+2/q}}$
(6.7) $\displaystyle-\frac{\ln^{2}2-q\ln
2+3q^{2}/2}{2^{1+2/q}}+\frac{2}{3}\frac{q^{2}}{2^{3/q}}-\frac{2q^{2}}{3n^{3/q}}\Big{]}$

for $i=1$.

For $k\geq 2$ from (2.1) of Lemma 2.1 and since

$\left|\frac{2q}{\ln
k}-\frac{2q^{2}-2q^{2}k^{-1/q}}{\ln^{2}k}\right|<\frac{c(p)}{\ln k}$

it follows that for every natural $j\geq 2$

$\displaystyle S^{j}(k)$
$\displaystyle=\left[\frac{k^{1/q}\ln^{2}k}{A\left(k^{1/q}-\frac{1}{p}\right)\ln^{2}(n+1)}\right]^{j}\left[1-\frac{2qj}{\ln
k}+\frac{\left(2q^{2}-2q^{2}k^{-1/q}\right)j}{\ln^{2}k}+O\left(\frac{j^{2}}{\ln^{2}k}\right)\right]$
$\displaystyle=\left[\frac{k^{1/q}\ln^{2}k}{A\left(k^{1/q}-\frac{1}{p}\right)\ln^{2}(n+1)}\right]^{j}\left[1-\frac{2qj}{\ln
k}+O\left(\frac{j^{2}}{\ln^{2}k}\right)\right]$
$\displaystyle=\left[\frac{k^{1/q}}{A\left(k^{1/q}-\frac{1}{p}\right)\ln^{2}(n+1)}\right]^{j}\left[\ln^{2j}k-2qj\ln^{2j-1}k\right]$
$\displaystyle+\left[\frac{k^{1/q}\ln^{2}k}{A\left(k^{1/q}-\frac{1}{p}\right)\ln^{2}(n+1)}\right]^{j}O\left(\frac{j^{2}}{\ln^{2}k}\right)$
$\displaystyle=\left[\frac{k^{1/q}}{A\left(k^{1/q}-\frac{1}{p}\right)\ln^{2}(n+1)}\right]^{j}\left[\ln^{2j}k-2qj\ln^{2j-1}k\right]+\left(\frac{2}{A}\right)^{j}O\left(\frac{j^{2}}{\ln^{2}(n+1)}\right).$

Then for $i\geq 2$

$\displaystyle L_{3}$
$\displaystyle=\sum_{k=i}^{n}\frac{1}{k^{p}}\left(k^{1/q}-\frac{1}{p}\right)^{p/q}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}S^{j}(k)$
$\displaystyle=\sum_{k=i}^{n}\frac{1}{k^{p}}\left(k^{1/q}-\frac{1}{p}\right)^{p/q}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{k^{j/q}\left(\ln^{2j}k-2qj\ln^{2j-1}k\right)}{A^{j}\left(k^{1/q}-\frac{1}{p}\right)^{j}\ln^{2j}(n+1)}$
$\displaystyle+\sum_{k=i}^{n}\frac{1}{k^{p}}\left(k^{1/q}-\frac{1}{p}\right)^{p/q}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\left(\frac{2}{A}\right)^{j}O\left(\frac{j^{2}}{\ln^{2}(n+1)}\right)=L_{31}+L_{32}.$

Now

$\displaystyle\left|L_{32}\right|$
$\displaystyle=\left|\sum_{k=i}^{n}\frac{1}{k^{p}}\left(k^{1/q}-\frac{1}{p}\right)^{p/q}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\left(\frac{2}{A}\right)^{j}O\left(\frac{j^{2}}{\ln^{2}(n+1)}\right)\right|$
$\displaystyle<\frac{c(p)}{\ln^{2}(n+1)}\sum_{k=i}^{n}\frac{1}{k^{p}}\left(k^{1/q}\right)^{p/q}\sum_{j=2}^{\infty}\left|\binom{p/q}{j}\right|\left(\frac{2}{A}\right)^{j}j^{2}$
(6.8)
$\displaystyle=\frac{c(p)}{A^{2}\ln^{2}(n+1)}\sum_{k=i}^{n}\frac{1}{k^{1+1/q}}<\frac{c(p)}{i^{1/q}A^{2}\ln^{2}(n+1)}=\frac{1}{i^{1/q}}O\left(\frac{1}{A^{2}\ln^{2}(n+1)}\right).$

For $\alpha\geq 1$ and $0\leq x\leq 1/2$ the next inequality holds

$\frac{1}{(1-x)^{\alpha}}\leq 1+\alpha 2^{\alpha+1}x$

which is easy to verify. Then for $p\geq 2$ and $j\geq 1$ we have

$\frac{k^{j/q}}{\left(k^{1/q}-\frac{1}{p}\right)^{j}}=\frac{1}{\left(1-\frac{1}{pk^{1/q}}\right)^{j}}=1+O\left(\frac{2^{j}j}{k^{1/q}}\right)=1+O\left(\frac{2^{j}j}{\ln^{2}k}\right).$

Consequently

$\displaystyle\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{k^{j/q}\left(\ln^{2j}k-2qj\ln^{2j-1}k\right)}{A^{j}\left(k^{1/q}-\frac{1}{p}\right)^{j}\ln^{2j}(n+1)}$
$\displaystyle=\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}k-2qj\ln^{2j-1}k}{A^{j}\ln^{2j}(n+1)}+O\left(\frac{1}{A^{2}\ln^{2}(n+1)}\right).$

Then by (2.16) of Lemma 2.16

$\displaystyle L_{31}$
$\displaystyle=\sum_{k=i}^{n}\frac{1}{k^{p}}\left(k^{1/q}-\frac{1}{p}\right)^{p/q}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}k-2qj\ln^{2j-1}k}{A^{j}\ln^{2j}(n+1)}$
$\displaystyle+\sum_{k=i}^{n}\frac{1}{k^{p}}\left(k^{1/q}-\frac{1}{p}\right)^{p/q}O\left(\frac{1}{A^{2}\ln^{2}(n+1)}\right)$
$\displaystyle=\sum_{k=i}^{n}\frac{1}{k^{p}}\left(k^{1/q}-\frac{1}{p}\right)^{p/q}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}k-2qj\ln^{2j-1}k}{A^{j}\ln^{2j}(n+1)}$
$\displaystyle+\frac{1}{i^{1/q}}O\left(\frac{1}{A^{2}\ln^{2}(n+1)}\right)$
$\displaystyle=\sum_{k=i}^{n}\frac{1}{k^{1+1/q}}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}k-2qj\ln^{2j-1}k}{A^{j}\ln^{2j}(n+1)}$
$\displaystyle+\sum_{k=i}^{n}\frac{1}{k^{1+1/q}}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}k-2qj\ln^{2j-1}k}{A^{j}\ln^{2j}(n+1)}O\left(\frac{1}{k^{1/q}}\right)$
$\displaystyle+\frac{1}{i^{1/q}}O\left(\frac{1}{A^{2}\ln^{2}(n+1)}\right)$
$\displaystyle=\sum_{k=i}^{n}\frac{1}{k^{1+1/q}}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}k-2qj\ln^{2j-1}k}{A^{j}\ln^{2j}(n+1)}$
$\displaystyle+\frac{1}{i^{1/q}}O\left(\frac{1}{A^{2}\ln^{2}(n+1)}\right)$
$\displaystyle=\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{1}{A^{j}\ln^{2j}(n+1)}\sum_{k=i}^{n}\frac{\ln^{2j}k-2qj\ln^{2j-1}k}{k^{1+1/q}}$
$\displaystyle+\frac{1}{i^{1/q}}O\left(\frac{1}{A^{2}\ln^{2}(n+1)}\right)$
(6.9)
$\displaystyle=\frac{q}{i^{1/q}}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}i}{A^{j}\ln^{2j}(n+1)}+\frac{1}{i^{1/q}}O\left(\frac{1}{A^{2}\ln^{2}(n+1)}\right)+O\left(\frac{1}{A^{2}n^{1/q}}\right)$

because

$\displaystyle\frac{1}{k^{p}}\left(k^{1/q}-\frac{1}{p}\right)^{p/q}$
$\displaystyle=\frac{1}{k^{1+1/q}}\left(1-\frac{1}{pk^{1/q}}\right)^{p/q}=\frac{1}{k^{1+1/q}}\left(1+O\left(\frac{1}{k^{1/q}}\right)\right),$

$\displaystyle\left|\binom{p/q}{j}\frac{(-1)^{j}\left(\ln^{2j}k-2qj\ln^{2j-1}k\right)}{A^{j}\ln^{2j}(n+1)}O\left(\frac{1}{k^{1/q}}\right)\right|<\frac{cj^{p/q+1}\ln^{2}k}{A^{j}\ln^{2}(n+1)k^{1/q}}<\frac{cj^{p/q+1}}{A^{j}\ln^{2}(n+1)}$

and

$\displaystyle\sum_{k=i}^{n}\frac{1}{k^{1+1/q}}\sum_{j=2}^{\infty}\frac{cj^{p/q+1}}{A^{j}\ln^{2}(n+1)}<\frac{c}{A^{2}\ln^{2}(n+1)}\sum_{k=i}^{n}\frac{1}{k^{1+1/q}}<\frac{c}{A^{2}\ln^{2}(n+1)i^{1/q}}.$

Consequently for $i\geq 2$

(6.10) $\displaystyle L_{3}(i\geq
2)=\frac{q}{i^{1/q}}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}i}{A^{j}\ln^{2j}(n+1)}+\frac{1}{i^{1/q}}O\left(\frac{1}{A^{2}\ln^{2}(n+1)}\right)+O\left(\frac{1}{A^{2}n^{1/q}}\right).$

Since $S(1)=0$ we have for $i=1$

$\displaystyle L_{3}(i=1)$
$\displaystyle=\sum_{k=1}^{n}\frac{1}{k^{p}}\left(k^{1/q}-\frac{1}{p}\right)^{p/q}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}S^{j}(k)$
$\displaystyle=\sum_{k=2}^{n}\frac{1}{k^{p}}\left(k^{1/q}-\frac{1}{p}\right)^{p/q}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}S^{j}(k)$
$\displaystyle=\frac{q}{2^{1/q}}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}2}{A^{j}\ln^{2j}(n+1)}+O\left(\frac{1}{A^{2}\ln^{2}(n+1)}\right)+O\left(\frac{1}{A^{2}n^{1/q}}\right)$
(6.11)
$\displaystyle=\frac{q}{2^{1/q}}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}2}{A^{j}\ln^{2j}(n+1)}+O\left(\frac{1}{A^{2}\ln^{2}(n+1)}\right)$

We consider the cases $i=1$ and $i\geq 2$ separately.

Case $i=1$.
From (6.5), (6) and (6) we obtain

$\displaystyle(Aq)^{-p/q}M_{1}^{*}$
$\displaystyle<q-\frac{p}{A\ln^{2}(n+1)}\left[\frac{\ln^{2}2+2q^{2}}{2^{1/q}}+\frac{2}{3}\frac{q}{2^{3/q}}-2q\sum_{k=2}^{n}\frac{1}{k^{1+2/q}}-\frac{\ln^{2}2-q\ln
2+3q^{2}/2}{q2^{1+2/q}}\right]$
$\displaystyle\,\,\,\,\,\,\,\,\,+\frac{q}{2^{1/q}}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}2}{A^{j}\ln^{2j}(n+1)}+O\left(\frac{1}{A^{2}\ln^{2}(n+1)}\right)-T$
$\displaystyle=q\sum_{j=0}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}2}{A^{j}\ln^{2j}(n+1)}-q\left(1-\frac{1}{2^{1/q}}\right)\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}2}{A^{j}\ln^{2j}(n+1)}$
$\displaystyle-\frac{p}{A\ln^{2}(n+1)}\left[\frac{\ln^{2}2+2q^{2}}{2^{1/q}}+\frac{2}{3}\frac{q}{2^{3/q}}-\ln^{2}2-2q\sum_{k=2}^{n}\frac{1}{k^{1+2/q}}-\frac{\ln^{2}2-q\ln
2+3/2q^{2}}{q2^{1+2/q}}\right]$
$\displaystyle+O\left(\frac{1}{A^{2}\ln^{2}(n+1)}\right)-T$
$\displaystyle=q\sum_{j=0}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}2}{A^{j}\ln^{2j}(n+1)}+O\left(\frac{1}{A^{2}\ln^{2}(n+1)}\right)-T$
$\displaystyle-\frac{p}{A\ln^{2}(n+1)}\left[\frac{\ln^{2}2+2q^{2}}{2^{1/q}}+\frac{2}{3}\frac{q}{2^{3/q}}-\ln^{2}2-2q\sum_{k=2}^{n}\frac{1}{k^{1+2/q}}-\frac{\ln^{2}2-q\ln
2+3/2q^{2}}{q2^{1+2/q}}\right]$

where

$T=\frac{q}{(n+1)^{1/q}}-\frac{p}{A\ln^{2}(n+1)}\left[\frac{\ln^{2}n+2q^{2}}{n^{1/q}}+\frac{2q}{3n^{3/q}}\right].$

It is obvious that by taking $A=A(p)$ big enough we can make $T$ positive. By
Lemma 2.13 there is a constant $c(p)$ such that

$\displaystyle\frac{\ln^{2}2+2q^{2}}{2^{1/q}}+\frac{2}{3}\frac{q}{2^{3/q}}-\ln^{2}2-\sum_{k=2}^{n}\frac{2q}{k^{1+2/q}}-\frac{\ln^{2}2-q\ln
2+3/2q^{2}}{q2^{1+2/q}}>c(p).$

Then

$\displaystyle(Aq)^{-p/q}M_{1}^{*}<q\sum_{j=0}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}2}{A^{j}\ln^{2j}(n+1)}+O\left(\frac{1}{A^{2}\ln^{2}(n+1)}\right)-\frac{pc(p)}{A\ln^{2}(n+1)}.$

Again, by taking $A=A(p)$ big enough we have

$\displaystyle(Aq)^{-p/q}M_{1}^{*}<q\sum_{j=0}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}2}{A^{j}\ln^{2j}(n+1)}-\frac{c(p)}{A\ln^{2}(n+1)}$

and from (6) and (6) we get

$M_{1}<\left(\frac{p}{p-1}\right)^{p}-\frac{c(A,p)}{\ln^{2}(n+1)}=\left(\frac{p}{p-1}\right)^{p}-\frac{c(p)}{\ln^{2}(n+1)}.$

Case $i\geq 2$.
From (6.5), (6), (6.10) we obtain

$\displaystyle(Aq)^{-p/q}M_{i}^{*}<q\left[\frac{1}{i^{1/q}}-\frac{1}{(n+1)^{1/q}}\right]$
$\displaystyle-\frac{p}{qA\ln^{2}(n+1)}\Big{[}\frac{q\left(\ln^{2}i+2q^{2}\right)}{i^{1/q}}-\frac{q\left(\ln^{2}n+2q^{2}\right)}{n^{1/q}}-2q^{2}\sum_{k=i}^{n}\frac{1}{k^{1+2/q}}$
$\displaystyle-\frac{\ln^{2}i-q\ln
i+3/2q^{2}}{2i^{2/q}}+\frac{2q^{2}}{3i^{3/q}}-\frac{2q^{2}}{3n^{3/q}}\Big{]}+O\left(\frac{1}{A^{2}n^{1/q}}\right)$
$\displaystyle+\frac{q}{i^{1/q}}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}i}{A^{j}\ln^{2j}(n+1)}+\frac{1}{i^{1/q}}O\left(\frac{1}{A^{2}\ln^{2}(n+1)}\right)$
$\displaystyle=\frac{q}{i^{1/q}}\sum_{j=0}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}i}{A^{j}\ln^{2j}(n+1)}-\frac{q}{(n+1)^{1/q}}$
$\displaystyle-\frac{p}{A\ln^{2}(n+1)}\Big{[}\frac{2q^{2}}{i^{1/q}}-\frac{\ln^{2}n+2q^{2}}{n^{1/q}}-2q\sum_{k=i}^{n}\frac{1}{k^{1+2/q}}+\frac{2q}{3i^{3/q}}-\frac{2q}{3n^{3/q}}$
$\displaystyle-\frac{\ln^{2}i-q\ln
i+3/2q^{2}}{2qi^{2/q}}\Big{]}+\frac{1}{i^{1/q}}O\left(\frac{1}{A^{2}\ln^{2}(n+1)}\right)+O\left(\frac{1}{A^{2}n^{1/q}}\right)$
$\displaystyle=\frac{q}{i^{1/q}}\sum_{j=0}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}(i+1)}{A^{j}\ln^{2j}(n+1)}-\frac{q}{i^{1/q}}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}(i+1)-\ln^{2j}i}{A^{j}\ln^{2j}(n+1)}$
$\displaystyle-\frac{p}{A\ln^{2}(n+1)}\Big{[}\frac{2q^{2}-\ln^{2}(i+1)+\ln^{2}i}{i^{1/q}}-2q\sum_{k=i}^{n}\frac{1}{k^{1+2/q}}$
$\displaystyle-\frac{\ln^{2}i-q\ln
i+3/2q^{2}}{2qi^{2/q}}+\frac{2q}{3i^{3/q}}\Big{]}+\frac{1}{i^{1/q}}O\left(\frac{1}{A^{2}\ln^{2}(n+1)}\right)-T$

where

$T=\frac{q}{(n+1)^{1/q}}-\frac{p}{A\ln^{2}(n+1)}\left[\frac{\ln^{2}n+2q^{2}}{n^{1/q}}+\frac{2q}{3n^{3/q}}\right]+O\left(\frac{1}{A^{2}n^{1/q}}\right).$

By taking $A=A(p)$ big enough we can make $T$ positive. Now

$\displaystyle\left|\frac{q}{i^{1/q}}\sum_{j=2}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}(i+1)-\ln^{2j}i}{A^{j}\ln^{2j}(n+1)}\right|=\frac{1}{i^{1/q}}O\left(\frac{1}{A^{2}\ln^{2}(n+1)}\right).$

Also we have

$\ln^{2}(i+1)-\ln^{2}i<\frac{3}{4}$

and

$\sum_{k=i}^{n}\frac{1}{k^{1+2/q}}\leq\frac{1}{i^{1+2/q}}+\int_{i}^{n}\frac{dx}{x^{1+2/q}}=\frac{1}{i^{1+2/q}}+\frac{q}{2i^{2/q}}$

and consequently

$\displaystyle\frac{2q^{2}-\ln^{2}(i+1)+\ln^{2}i}{i^{1/q}}-2q\sum_{k=i}^{n}\frac{1}{k^{1+2/q}}-\frac{\ln^{2}i-q\ln
i+3/2q^{2}}{2qi^{2/q}}+\frac{2q}{3i^{3/q}}$
$\displaystyle>\frac{1}{i^{1/q}}\left[2q^{2}-\frac{3}{4}+\frac{2q}{3i^{2/q}}-\frac{2q}{i^{1+1/q}}-\frac{q^{2}}{i^{1/q}}-\frac{\ln^{2}i-q\ln
i+3/2q^{2}}{2qi^{1/q}}\right].$

By Lemma 2.12 there is a constant $c(p)>0$ such that

$\displaystyle\frac{2q^{2}-\ln^{2}(i+1)+\ln^{2}i}{i^{1/q}}-2q\sum_{k=i}^{n}\frac{1}{k^{1+2/q}}-\frac{\ln^{2}i-q\ln
i+3/2q^{2}}{2qi^{2/q}}+\frac{2q}{3i^{3/q}}>\frac{c(p)}{i^{1/q}}.$

Then

$\displaystyle(Aq)^{-p/q}M_{i}^{*}$
$\displaystyle<\frac{q}{i^{1/q}}\sum_{j=0}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}(i+1)}{A^{j}\ln^{2j}(n+1)}-\frac{pc(p)}{A\ln^{2}(n+1)i^{1/q}}+\frac{1}{i^{1/q}}O\left(\frac{1}{A^{2}\ln^{2}(n+1)}\right).$

By taking $A=A(p)$ big enough we obtain

$M_{i}^{*}<\frac{A^{p/q}q^{p}}{i^{1/q}}\sum_{j=0}^{\infty}(-1)^{j}\binom{p/q}{j}\frac{\ln^{2j}(i+1)}{A^{j}\ln^{2j}(n+1)}-\frac{c(A,p)}{i^{1/q}\ln^{2}(n+1)}$

and from (6) and (6) we get

$M_{i}<q^{p}-\frac{c(A,p)}{\ln^{2}(n+1)}=\left(\frac{p}{p-1}\right)^{p}-\frac{c(A,p)}{\ln^{2}(n+1)}=\left(\frac{p}{p-1}\right)^{p}-\frac{c(p)}{\ln^{2}(n+1)}.$

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