A Sufficient Condition for Uniform Convergence of Trigonometric Series with p-Bounded Variation Coefficients

Abstract

In this paper we consider trigonometric series with p-bounded variation coefficients. We presented a sufficient condition for uniform convergance of such series in case $p>1$. This condition is significantly weaker than these obtained in the results on this subject known in the literature.

1 Introduction

It is well known that there is a great number of interesting results in Fourier analysis established by assuming monotonicity of Fourier coefficients. The following classical convergence result can be found in many monographs (see for example [3, 18] or [1]).

Theorem 1

Suppose that $b_{n}\ge b_{n+1}$ and $b_{n}\rightarrow 0$ as $n \rightarrow \infty$. Then a necessary and sufficient condition for the uniform convergence of the series

$$\begin{aligned} \sum \limits _{n=1}^{\infty }b_{n}\sin nx \end{aligned}$$

(1.1)

is $nb_{n}\rightarrow 0$ as $n \rightarrow \infty$.

This result has been generalized by weakening the monotonicity conditions of the coefficient (see for example [2, 14]). We present below a historical outline of the generalizations of this theorem.

In 2001 Leindler defined (see [8] and [10]) a new class of sequences named as sequences of Rest Bounded Variation, briefly denoted by RBVS, i.e.,

$$\begin{aligned} RBVS=\left\{ a:=\left( a_{n}\right) \in \mathbb {C}:\sum \limits _{n=m}^{\infty }\left| \Delta _1a_n\right| \le C \left| a_{m}\right| \text { for all }m\in \mathbb {N} \right\} , \end{aligned}$$

where here and throughout the paper $C=C\left( a\right)$ always indicates a constant only depending on a and $\Delta _ra_n = a_n-a_{n+r}$ for $r \in \mathbb {N}$.

Denote by MS the class of monotone decreasing sequences, then it is clear that

$$\begin{aligned} MS\subsetneq RBVS. \end{aligned}$$

Further, Tikhonov introduced a class of General Monotone Sequences GMS defined as follows (see [16]):

$$\begin{aligned} GMS=\left\{ a:=\left( a_{n}\right) \in \mathbb {C}:\sum \limits _{n=m}^{2m-1}\left| \Delta _1a_n\right| \le C \left| a_{m}\right| \text { for all }m\in \mathbb {N} \right\} . \end{aligned}$$

It is clear that

$$\begin{aligned} RBVS\subsetneq GMS{.} \end{aligned}$$

The class of GMS was generalized by Tikhonov (see [15]) and independently by Zhou, Zhou and Yu (see [17]) to the class of Mean Value Bounded Variation Sequences (MVBVS). We say that a sequence $a:=\left( a_{n}\right)$ of complex numbers is said to be MVBVS if there exists $\lambda \ge 2$ such that

$$\begin{aligned} \sum \limits _{k=n}^{2n-1}\left| \Delta _1a_k\right| \le \frac{C}{n} \sum \limits _{k=[n/\lambda ]}^{\lambda n} \left| a_{k} \right| \end{aligned}$$

holds for $n \in \mathbb {N}$, where [x] is the integer part of x. They proved also in [17] that

$$\begin{aligned} GMS \subsetneq MVBVS. \end{aligned}$$

Theorem 1 was generalized for the class RBVS in [8], for the class GMS in [16] and for the class MVBVS in [17].

Next, Tikhonov [13, 15, 16] and Leindler [9] defined the class of $\beta$—general monotone sequences as follows:

Definition 1

Let $\beta :=\left( \beta _{n}\right)$ be a nonnegative sequence. The sequence of complex numbers $a:=\left( a_{n}\right)$ is said to be $\beta { -}$ general monotone, or $a\in GM\left( \beta \right)$, if the relation

$$\begin{aligned} \sum \limits _{n=m}^{2m-1}\left| \Delta _1a_n\right| \le C \beta _{m} \end{aligned}$$

holds for all $m \in \mathbb {N}$.

In the paper [15] Tikhonov considered i.e. the following examples of the sequences $\beta _{n}$:

• (1) $_{1}\beta _{n}=\left| a_{n}\right| ,$

• (2) $_{2}\beta _{n}=\sum \nolimits _{k=\left[ n/c \right] }^{\left[ cn\right] } \frac{\left| a_{k}\right| }{k}$ for some $c>1$.

It is clear that $GM\left( _{1}\beta \right) =GMS$. Moreover, Tikhonov showed in [15] that

$$\begin{aligned} GM\left( _{1}\beta \right) \subsetneq GM\left( _{2}\beta \right) \equiv \text {MVBVS}. \end{aligned}$$

Tikhonov proved also in [15] the following theorem:

Theorem 2

Let a sequence $(b_n) \in GM(_2\beta )$. If $n|b_n| \rightarrow 0$ as $n\rightarrow \infty$, then the series (1.1) converges uniformly.

Further, Szal defined a new class of sequences in the following way (see [11]):

Definition 2

Let $\beta :=\left( \beta _{n}\right)$ be a nonnegative sequence and r a natural number. The sequence of complex numbers $a:=\left( a_{n}\right)$ is said to be $\left( \beta ,r\right) -$general monotone, or $a\in GM\left( \beta ,r\right)$, if the relation

$$\begin{aligned} \sum \limits _{n=m}^{2m-1}\left| \Delta _ra_n\right| \le C \beta _{m} \end{aligned}$$

holds for all $m \in \mathbb {N}$.

It is clear that $GM\left( \beta ,1\right) \equiv GM\left( \beta \right)$. Moreover, it is easy to show that the sequence

$$\begin{aligned} a_n = \frac{\left( -1\right) ^n}{n} \end{aligned}$$

belongs to $GM\left( {}_1\beta ,2\right)$ and does not belong to $GM\left( {}_1\beta \right)$. This example shows that the class $GM\left( {}_1\beta \right)$ is essentially wider than the class $GM\left( {}_1\beta \right)$. In [11] Szal showed more general relations

$$\begin{aligned} GM({}_2\beta ,1) \subsetneq GM({}_2\beta ,r) \end{aligned}$$

for all $r >1$.

In the paper [11] Szal generalized Theorem 1 by proving the following theorem.

Theorem 3

[11]. Let a sequence $\left( b_{n}\right) \in GM\left( {}_2\beta ,r\right)$, where $r \in \mathbb {N}$. If $n|b_n| \rightarrow 0$ as $n \rightarrow \infty$ and

$$\begin{aligned} \sum \limits _{n=1}^{\infty }\sum \limits _{k=1}^{[r/2]}\left| b_{r\cdot n +k} - b_{r\cdot n + r - k} \right| < \infty \text { for } r\ge 3, \end{aligned}$$

then the series (1.1) converges uniformly.

In the paper [4] Kórus defined a new class of sequences in the following way:

Definition 3

The sequence of complex numbers $a:=\left( a_{n}\right)$ is in the class $SBVS_{2}$ (Supremum Bounded Variation Sequence), if the relation

$$\begin{aligned} \sum \limits _{n=m}^{2m-1}\left| \Delta _{1}a_{n}\right| \le \frac{C}{ n}\underset{m\ge b(n)}{\sup }\sum \limits _{k=m}^{2m} \left| a_{k} \right| \end{aligned}$$

holds for all $m\in \mathbb {N}$, where (b(n)) is a nonnegative sequence tending monotonically to infinity depending only on a.

In the paper [4] Kórus also proved the following theorem:

Theorem 4

Let a sequence $(b_n) \in SBVS_2$. If $n|b_n| \rightarrow 0$ as $n\rightarrow \infty$, then the series (1.1) converges uniformly.

Next Tikhonov and Liflyand defined a class of $GMS_p(\beta )$ in the following way (see [6, 7]):

Definition 4

Let $\beta =(\beta _{n})$ be a nonnegative sequence and p a positive real number. We say that a sequence of complex numbers $a=(a_{n})\in GMS_{p}(\beta )$ if the relation

$$\begin{aligned} \left( \sum \limits _{n=m}^{2m-1}\left| \Delta _{1}a_{n}\right| ^{p}\right) ^{\frac{ 1}{p}}\le C\beta _{m} \end{aligned}$$

holds for all $m\in \mathbb {N}$.

It is clear that $GMS_1(\beta )=GM(\beta )$.

The latest class of sequences was defined by Kubiak and Szal in [5] as follows:

Definition 5

Let $\beta :=\left( \beta _{n}\right)$ be a nonnegative sequence, r a natural number and p a positive real number. The sequence of complex numbers $a:=\left( a_{n}\right)$ is said to be $\left( p,\beta ,r\right) -$ general monotone, or $a\in GM\left( p, \beta ,r\right)$, if the relation

$$\begin{aligned} \left( \sum \limits _{n=m}^{2m-1}\left| \Delta _ra_n\right| ^p\right) ^{\frac{1}{p}} \le C \beta _m \end{aligned}$$

holds for all $m\in \mathbb {N}$.

It is clear that $GM(p,\beta ,1)=GMS_p(\beta )$ and $GM(1,\beta ,r) = GM(\beta ,r)$.

Further we will consider the following sequence:

$$\begin{aligned} {}_3\beta _n(q)=\frac{1}{n} \underset{m\ge b(n)}{\sup }m \left( \frac{1}{m} \sum \limits _{k=m}^{2m} \left| a_{k}\right| ^q \right) ^{\frac{1}{q}}, \end{aligned}$$

where $(a_n) \subset \mathbb {C}, a_n\rightarrow 0$ as $n\rightarrow \infty$, $q > 0$, (b(n)) is a nonnegative sequence such that $b(n) \nearrow$ and $b(n) \rightarrow \infty$ as $n \rightarrow \infty$. It is clear that $SBVS_2 = GM(1, {}_3\beta (1),1)$.

In the further part of our paper we will consider the following series:

$$\begin{aligned} \sum \limits _{n=1}^{\infty } b_n \sin (cnx), \end{aligned}$$

(1.2)

$$\begin{aligned} \sum \limits _{n=1}^{\infty } a_n \cos (cnx), \end{aligned}$$

(1.3)

$$\begin{aligned} \sum \limits _{n=1}^{\infty } c_n e^{icnx}, \end{aligned}$$

(1.4)

where $c >0.$

In the paper [5] Kubiak and Szal showed the following embedding relations:

Theorem 5

Let $q >0, r\in \mathbb {N}$ and $0 < p_1 \le p_2$. Then

$$\begin{aligned} GM(p_1,{}_3\beta (q),r) \subseteq GM(p_2,{}_3\beta (q),r). \end{aligned}$$

Theorem 6

Let $p \ge 1$, $q > 0$, $r_1,r_2 \in \mathbb {N}$, $r_1 \le r_2$. If $r_1 \mid r_2$, then

$$\begin{aligned} GM(p,{}_3\beta (q),r_1) \subseteq GM(p,{}_3\beta (q),r_2). \end{aligned}$$

Moreover they proved in [5] the following generalization of Theorem 1:

Theorem 7

Let a sequence $\left( b_{n}\right) \in GM\left( p, {}_{3}\beta (q),r\right)$, where $p,q\ge 1$, $r \in \mathbb {N}$ and $b(n)\ge n$ for $n \in \mathbb {N}$. If

$$\begin{aligned} n^{2-\frac{1}{p}}\left| b_{n}\right| \rightarrow 0\text { as } n\rightarrow \infty \end{aligned}$$

(1.5)

and

$$\begin{aligned} \sum \limits _{k=1}^{\infty } b_k \sin \left( \frac{2l \pi }{r} k\right) <\infty ,\text { for }r\ge 3, \end{aligned}$$

for all $l= 1,...,[\frac{r}{2}]-1$ when r is an even number and $l= 1,...,[\frac{r}{2}]$ when r is an odd number, then the series (1.2) is uniformly convergent.

Theorem 8

Let a sequence $\left( a_{n}\right) \in GM\left( p, {}_{3}\beta (q),r\right)$, where $p,q\ge 1$, $r \in \mathbb {N}$ and $b(n)\ge n$ for $n \in \mathbb {N}$. If

$$\begin{aligned} n^{2-\frac{1}{p}}\left| a_{n}\right| \rightarrow 0\text { as } n\rightarrow \infty \end{aligned}$$

and

$$\begin{aligned} \sum \limits _{k=1}^{\infty } a_k \cos \left( \frac{2l \pi }{r} k\right) <\infty , \end{aligned}$$

for all $l= 0,1,...,[\frac{r}{2}]$, then the series (1.3) is uniformly convergent.

Theorem 9

Let a sequence $\left( c_{n}\right) \in GM\left( p, {}_{3}\beta (q),r\right)$, where $p,q\ge 1$, $r \in \mathbb {N}$ and $b(n)\ge n$ for $n \in \mathbb {N}$. If

$$\begin{aligned} n^{2-\frac{1}{p}}\left| c_{n}\right| \rightarrow 0\text { as } n\rightarrow \infty \end{aligned}$$

and

$$\begin{aligned} \sum \limits _{k=1}^{\infty } c_k e^{\left( \frac{2l \pi }{r} k\right) i} <\infty , \end{aligned}$$

for all $l= 0,1,...,[\frac{r}{2}]$, then the series (1.4) is uniformly convergent.

In this paper we will show that Theorems 7, 8, 9 are true under weakened assumptions in case $p>1$.

2 Main Results

We have the following results:

Theorem 10

Let a sequence $\left( b_{n}\right) \in GM\left( p, {}_{3}\beta (q),r\right)$, where $q\ge 1$, $p>1$ and $r \in \mathbb {N}$. If

$$\begin{aligned} n \ln n\left| b_{n}\right| \rightarrow 0\text { as } n\rightarrow \infty \end{aligned}$$

(2.1)

and

$$\begin{aligned} \sum \limits _{k=1}^{\infty } b_k \sin \left( \frac{2l \pi }{r} k\right) <\infty ,\text { for }r\ge 3, \end{aligned}$$

(2.2)

for all $l= 1,...,[\frac{r}{2}]-1$ when r is an even number and $l= 1,...,[\frac{r}{2}]$ when r is an odd number, then the series (1.2) is uniformly convergent.

Proposition 1

There exist an $x_0 \in \mathbb {R}$ and a sequence $(b_n) \in GM(p, {}_3\beta (1), 3)$ for $p>1$ with the properties $nb_n \rightarrow 0$ as $n \rightarrow \infty$ and $(b_n) \notin GM(1, {}_3\beta (1),3)$, for which the series (1.2) is divergent in $x_0$.

Theorem 11

Let a sequence $\left( a_{n}\right) \in GM\left( p, {}_{3}\beta (q),r\right)$, where $q\ge 1$, $p>1$ and $r \in \mathbb {N}$. If

$$\begin{aligned} n \ln n\left| a_{n}\right| \rightarrow 0\text { as } n\rightarrow \infty \end{aligned}$$

and

$$\begin{aligned} \sum \limits _{k=1}^{\infty } a_k \cos \left( \frac{2l \pi }{r} k\right) <\infty , \end{aligned}$$

(2.3)

for all $l= 0,1,...,[\frac{r}{2}]$, then the series (1.3) is uniformly convergent.

Theorem 12

Let a sequence $\left( c_{n}\right) \in GM\left( p, {}_{3}\beta (q),r\right)$, where $q\ge 1$, $p>1$ and $r \in \mathbb {N}$. If

$$\begin{aligned} n \ln n\left| c_{n}\right| \rightarrow 0\text { as } n\rightarrow \infty \end{aligned}$$

and

$$\begin{aligned} \sum \limits _{k=1}^{\infty } c_k e^{\left( \frac{2l \pi }{r} k\right) i} <\infty , \end{aligned}$$

(2.4)

for all $l= 0,1,...,[\frac{r}{2}]$, then the series (1.4) is uniformly convergent.

Remark 1

It is clear that if a sequence $(b_n)$ satisfies the condition (2.1) then it fulfills the condition (1.5) with $p>1$, too. Therefore, from Theorem 10 we get Theorem 7 is case $p>1$. The same remark applies to Theorems 11, 8 and Theorems 12, 9, respectively.

3 Lemma

Denote, for $r \in \mathbb {N}$ and $k=0,1,2...$ by

$$\begin{aligned} \tilde{D}_{k,r}{(x)}=\frac{\cos \left( k+\frac{r}{2}\right) x}{2\sin \frac{rx}{ 2}}, \hspace{15mm} D_{k,r}{(x)}=\frac{\sin \left( k+\frac{r}{2}\right) x}{ 2\sin \frac{rx}{2}} \end{aligned}$$

the Dirichlet type kernels.

Lemma 1

. [11, 12] Let $r,m,n \in \mathbb {N}, l\in \mathbb {Z}$ and $(a_k) \subset \mathbb {C}$. If $x \ne \frac{2l\pi }{r}$, then for $m \ge n$

$$\begin{aligned} \sum \limits _{k=n}^{m} a_{k} sin(kx)&= - \sum \limits _{k=n}^{m} \Delta _{r} a_{k} \tilde{D}_{k,r}(x) + \sum \limits _{k=m+1}^{m+r} a_{k} \tilde{D} _{k,-r}(x) \nonumber&\\&\quad + \sum \limits _{k=n}^{n+r-1} a_{k} \tilde{D}_{k,-r}(x) \end{aligned}$$

(3.1)

and

$$\begin{aligned} \sum \limits _{k=n}^{m}a_{k}\cos kx&= \sum \limits _{k=n}^{m} \Delta _r a_k \tilde{D}_{k,r}(x) - \sum \limits _{k=m+1}^{m+r} a_k \tilde{D}_{k,-r}(x) \nonumber&\\&\quad + \sum \limits _{k=n}^{n+r-1} a_k \tilde{D}_{k,-r}(x). \end{aligned}$$

(3.2)

Lemma 2

[5]. Let $r,m,n\in \mathbb {N}, l \in \mathbb {Z}$ and $a=(a_n) \subset \mathbb {C}$. If $x \ne \frac{2l\pi }{r}$, then for $m \ge n$

$$\begin{aligned} \begin{aligned} \sum \limits _{k=n}^{m}a_{k}e^{ikx}&= \frac{-i}{2\sin \left( \frac{rx}{2}\right) } \left( \sum \limits _{k=n}^{m} \Delta _r a_k e^{-i\left( k+\frac{r}{2} \right) x} - \sum \limits _{k=m+1}^{m+r}a_{k} e^{-i\left( k-\frac{r}{2} \right) x} \right. \\ {}&\quad + \left. \sum \limits _{k=n}^{n+r-1}a_{k} e^{-i\left( k-\frac{r}{2} \right) x} \right) . \end{aligned} \end{aligned}$$

Lemma 3

Let $n, N \in \mathbb {N}$. Then for $p \ge 1$

$$\begin{aligned} \int _{n+N^{\frac{1}{p}}}^{n+N} \frac{1}{k\ln k} dk \le \ln p. \end{aligned}$$

Proof

This inequality is true for $p =1$. Consider the function

$$\begin{aligned} f(p) = \left( n+N^{\frac{1}{p}}\right) ^p \end{aligned}$$

for $p > 0$. We get:

$$\begin{aligned} \begin{aligned} f'(p) \ge \left( n+N^{\frac{1}{p}}\right) ^{p-1} \frac{1}{p}n \ln N \ge 0 \text{ for } \text{ all } p>0. \end{aligned} \end{aligned}$$

It means that the function is non-decresing with respect to p. Thus:

$$\begin{aligned} n+N = f(1) \le f(p) = \left( n+N^{\frac{1}{p}}\right) ^p \text { for } p\ge 1. \end{aligned}$$

Hence we get that:

$$\begin{aligned} \ln \left( n+N\right) \le \ln \left( n+N^{\frac{1}{p}}\right) ^p. \end{aligned}$$

(3.3)

Therefore, integrating by substitution with $\ln k = t$ and using (3.3), we get

$$\begin{aligned} \int _{n+N^{\frac{1}{p}}}^{n+N} \frac{1}{k \ln k } dk&= \int _{\ln (n+N)^{\frac{1}{p}}}^{\ln (n+N)} \frac{1}{t} dt = \ln (\ln (n+N)) - \ln (\ln (n+N^\frac{1}{p}))&\\&= \ln \left( \frac{\ln (n+N)}{\ln \left( n+N^{\frac{1}{p}}\right) }\right) = \ln \left( p \frac{\ln \left( n+N\right) }{\ln \left( n+N^{\frac{1}{p}}\right) ^p}\right) \le \ln p \end{aligned}$$

and the proof is completed. $\square$

4 Proofs of the Main Results

4.1 Proof of the Theorem 10

Let $\epsilon >0$. Then from (2.1) and (2.2) we obtain:

$$\begin{aligned} n\ln n \left| b_n \right| <\varepsilon , \end{aligned}$$

(4.1)

$$\begin{aligned} \left| \sum \limits _{k=n}^{\infty } b_{k}\sin \left( k\frac{2l\pi }{r}\right) \right| < \varepsilon , \end{aligned}$$

(4.2)

and

$$\begin{aligned} \left| \sum \limits _{k=n}^{n+N } b_{k}\sin \left( k\frac{2l\pi }{r}\right) \right| < \varepsilon , \end{aligned}$$

(4.3)

for all $n > N_\epsilon$ and $N \in \mathbb {N}$, where $l= 1,...,[\frac{r}{2} ]-1$ when r is an even number and $l= 1,...,[\frac{r}{2}]$ when r is an odd number. Denote by

$$\begin{aligned} \tau _{n}\left( x\right) =\sum \limits _{k=n}^{\infty }b_{k}\sin (ckx). \end{aligned}$$

We will show that

$$\begin{aligned} \left| \tau _{n}\left( x\right) \right| \ll \varepsilon \end{aligned}$$

(4.4)

holds for any $n\ge \max \{N_\varepsilon ,2\}$ and $x \in \mathbb {R}$. Since $\tau _{n}\left( 0 \right) =0$ and $\tau _{n}\left( \frac{\pi }{c} \right) =0$ it suffices to prove (4.4) for $0<x< \frac{\pi }{c}$.

First, we will show that (4.4) is valid for $x=\frac{2\,l\pi }{rc}$, where l is an integer number such that $0<2\,l<r.$ Using (4.2) we get

$$\begin{aligned} \left| \tau _{n}\left( \frac{2l\pi }{rc}\right) \right| < \varepsilon . \end{aligned}$$

Now, we prove that (4.4) holds for $\frac{2\,l\pi }{rc}<x\le \frac{ 2\,l\pi }{rc}+\frac{\pi }{rc}$, where $0\le 2\,l<r$.

Let $N^{\frac{1}{p}}:=N^{\frac{1}{p}}\left( x\right) \ge r$ be a natural number such that

$$\begin{aligned} \frac{2l\pi }{rc}+\frac{\pi }{c(N+1)^{\frac{1}{p}}}<x\le \frac{2l\pi }{rc}+ \frac{\pi }{cN^{\frac{1}{p}}}. \end{aligned}$$

(4.5)

Then

$$\begin{aligned} \tau _{n}\left( x\right)&= \sum \limits _{k=n}^{n+N^{\frac{1}{p} }-1}b_{k}\sin (ckx) + \sum \limits _{k=n+N^{\frac{1}{p}}}^{n+N}b_{k}\sin (ckx) + \sum \limits _{k=n+N+1}^{\infty }b_{k}\sin (ckx)\\&=\tau _{n}^{\left( 1\right) }\left( x\right) +\tau _{n}^{\left( 2\right) }\left( x\right) +\tau _{n}^{\left( 3\right) }\left( x\right) . \end{aligned}$$

Applying Lagrange’s mean value theorem to the function $f\left( x\right) =\sin (ckx)$ on the interval $\left[ \frac{2\,l\pi }{rc},x\right]$ we obtain that for each k there exists $y_k\in \left( \frac{2\,l\pi }{rc},x\right)$ such that

$$\begin{aligned} \sin (ckx)-\sin \left( k\frac{2l\pi }{r}\right) =ck\cos (cky_k)\left( x- \frac{2l\pi }{ rc}\right) . \end{aligned}$$

Using this we get

$$\begin{aligned} \begin{aligned} \tau _{n}^{\left( 1\right) }\left( x\right)&= \sum \limits _{k=n}^{n+N^{\frac{1}{ p}}-1}ckb_{k}\cos (cky_k)\left( x-\frac{2l\pi }{rc}\right) +\sum \limits _{k=n}^{n+N^{\frac{1}{p}}-1}b_{k}\sin \left( k\frac{2l\pi }{r} \right) \\ {}&= \tau _n^{(1.1)}(x) + \tau _n^{(1.2)}(x). \end{aligned} \end{aligned}$$

From (4.3) we have

$$\begin{aligned} \begin{aligned} \left| \tau _n^{(1.2)}(x) \right| <\varepsilon . \end{aligned} \end{aligned}$$

By (4.5) and (4.1)

$$\begin{aligned} \left| \tau _{n}^{\left( 1.1\right) }\left( x\right) \right|&\le \left( x-\frac{2l\pi }{rc}\right) \sum \limits _{k=n}^{n+N^{\frac{1}{p} }-1}ck\left| b_{k}\right| \le \left( x-\frac{2l\pi }{rc}\right) \sum \limits _{k=n}^{n+N^{\frac{1}{p}}-1}\frac{ck\ln k}{\ln k}\left| b_{k}\right| \\&< \left( x-\frac{2l\pi }{rc}\right) \sum \limits _{k=n}^{n+N^{\frac{1}{p}}-1} \frac{c\varepsilon }{\ln k} \le \frac{\pi \varepsilon }{\ln 2} . \end{aligned}$$

Using Lemma 3 we obtain

$$\begin{aligned} \left| \tau _{n}^{(2)}(x)\right|&= \left| \sum \limits _{k=n+N^{ \frac{1}{p}}}^{n+N}b_{k}\sin (ckx)\right| \le \sum \limits _{k=n+N^{\frac{ 1}{p}}}^{n+N}\frac{k\ln k}{k\ln k}\left| b_{k}\right| \ll \varepsilon \int _{n+N^{\frac{1}{p}}}^{n+N}\frac{1}{k\ln k}dk\\&\le \varepsilon \ln p. \end{aligned}$$

If $\left( b_{n}\right) \in GM\left( p,{}_3\beta (q),r\right)$, then using Lemma 1, we get

$$\begin{aligned} \bigg \vert \tau _{n}&{}^{\left( 3\right) } \left( x\right) \bigg \vert =\left| \sum \limits _{j=0}^{\infty }\sum \limits _{k=2^{j}\left( n+N+1\right) }^{2^{j+1}\left( n+N+1\right) -1}b_{k}\sin (ckx)\right| \\&\le \sum \limits _{j=0}^{\infty }\left| \frac{-1}{2\sin \left( crx/2\right) }\left\{ \sum \limits _{k=2^{j}\left( n+N+1\right) }^{2^{j+1}\left( n+N+1\right) -1}\left( b_{k}-b_{k+r}\right) \cos \left( k+ \frac{r}{2}\right) c x\right. \right. \\&\quad \left. \left. +\sum \limits _{k=2^{j+1}\left( n+N+1\right) }^{2^{j+1}\left( n+N+1\right) +r-1}b_{k}\cos \left( k-\frac{r}{2}\right) c x-\sum \limits _{k=2^{j}\left( n+N+1\right) }^{2^{j}\left( n+N+1\right) +r-1}b_{k}\cos \left( k-\frac{r}{2}\right) cx\right\} \right| \\&\le \frac{1}{2\left| \sin \left( crx/2\right) \right| } \sum \limits _{j=0}^{\infty }\left\{ \sum \limits _{k=2^{j}\left( n+N+1\right) }^{2^{j+1}\left( n+N+1\right) -1}\left| b_{k}-b_{k+r}\right| +\sum \limits _{k=2^{j+1}\left( n+N+1\right) }^{2^{j+1}\left( n+N+1\right) +r-1}\left| b_{k}\right| \right. \\&\quad \left. +\sum \limits _{k=2^{j}\left( n+N+1\right) }^{2^{j}\left( n+N+1\right) +r-1}\left| b_{k}\right| \right\} . \end{aligned}$$

Further applying the Hölder inequality with $p>1$, the inequality $\frac{rc}{\pi }x-2\,l\le \left| \sin \frac{rcx}{2}\right|$ $\left( x\in \left[ \frac{2\,l\pi }{rc },\frac{2\,l\pi }{rc}+\frac{\pi }{rc}\right] \text { and }0\le 2\,l<r\right)$, (4.5) and (4.1), we obtain

$$\begin{aligned} \bigg \vert \tau _{n}&{}^{\left( 3\right) } \left( x\right) \bigg \vert \le \frac{ 1}{\frac{rc}{\pi }x-2l}\sum \limits _{j=0}^{\infty }\left\{ \left( \sum \limits _{k=2^{j}\left( n+N+1\right) }^{2^{j+1}\left( n+N+1\right) -1}\left| b_{k}-b_{k+r}\right| ^p\right) ^{\frac{1}{p}} \right. \\&\quad \left. \left( \sum \limits _{k=2^{j}\left( n+N+1\right) }^{2^{j+1}\left( n+N+1\right) -1}1\right) ^{1-\frac{1}{p}} +\sum \limits _{k=2^{j}\left( n+N+1\right) }^{2^{j}\left( n+N+1\right) +r-1}\left| b_{k}\right| +\sum \limits _{k=2^{j+1}\left( n+N+1\right) }^{2^{j+1}\left( n+N+1\right) +r-1}\left| b_{k}\right| \right\} \\&\le \frac{(N+1)^\frac{1}{p}}{r}\sum \limits _{j=0}^{\infty }\left\{ \frac{ C\left( 2^j(n+N+1)\right) ^{1-\frac{1}{p}}}{2^j(n+N+1)} \underset{m\ge b\left( 2^j(n+N+1)\right) }{\sup }m \right. \\&\quad \left. \left( \frac{1}{m} \sum \limits _{k=m}^{2m-1 } \left| b_k\right| ^q\right) ^{\frac{1}{q}} +\sum \limits _{k=2^{j}\left( n+N+1\right) }^{2^{j}\left( n+N+1\right) +r-1}\left| b_{k}\right| +\sum \limits _{k=2^{j+1}\left( n+N+1\right) }^{2^{j+1}\left( n+N+1\right) +r-1}\left| b_{k}\right| \right\} \\&= \frac{(N+1)^\frac{1}{p}}{r}\sum \limits _{j=0}^{\infty }\left\{ \frac{C}{ \left( 2^j(n+N+1)\right) ^{\frac{1}{p}}} \underset{m\ge b\left( 2^j(n+N+1)\right) }{\sup }m^{1-\frac{1}{q}} \right. \\&\quad \left( \sum \limits _{k=m}^{2m-1 } \left( \frac{k \ln k\left| b_k \right| }{ k\ln k}\right) ^q\right) ^{\frac{1}{q}} +\sum \limits _{k=2^{j}\left( n+N+1\right) }^{2^{j}\left( n+N+1\right) +r-1}\frac{k\ln k\left| b_k \right| }{k\ln k} \\&\quad \left. +\sum \limits _{k=2^{j+1} \left( n+N+1\right) }^{2^{j+1}\left( n+N+1\right) +r-1}\frac{k\ln k\left| b_k \right| }{k\ln k} \right\} \\&< \frac{\varepsilon (N+1)^\frac{1}{p}}{r}\sum \limits _{j=0}^{\infty }\left\{ \frac{C}{\left( 2^j(n+N+1)\right) ^{\frac{1}{p}}} \underset{m\ge b\left( 2^j(n+N+1)\right) }{\sup }m^{1-\frac{1}{q}} \left( \sum \limits _{k=m}^{2m-1 } \left( \frac{1 }{k}\right) ^q\right) ^{\frac{1}{q}} \right. \\&\quad \left. +\sum \limits _{k=2^{j}\left( n+N+1\right) }^{2^{j}\left( n+N+1\right) +r-1}\frac{1}{k} +\sum \limits _{k=2^{j+1}\left( n+N+1\right) }^{2^{j+1}\left( n+N+1\right) +r-1}\frac{1}{k} \right\} \\&\le \frac{\varepsilon (N+1)^\frac{1}{p}}{r}\sum \limits _{j=0}^{\infty }\left\{ \frac{C}{\left( 2^j(n+N+1)\right) ^{\frac{1}{p}}} \underset{m\ge b\left( 2^j(n+N+1)\right) }{\sup }\ \left( m^{1-\frac{1}{q}} m^{-1} m^{\frac{1}{q }}\right) \right. \\&\quad \left. + \frac{3}{2}r\left( 2^j(n+N+1)\right) ^{-\frac{1}{p}} \right\} . \end{aligned}$$

Elementary calculations give:

$$\begin{aligned} \left| \tau _{n}^{\left( 3\right) }\left( x\right) \right|&< \frac{ \varepsilon (N+1)^\frac{1}{p}}{r}\sum \limits _{j=0}^{\infty }\left\{ \left( 2^j(n+N+1)\right) ^{-\frac{1}{p}} \left( C+\frac{3}{2}r\right) \right\} \\&\le \frac{\varepsilon (N+1)^\frac{1}{p}\left( C+\frac{3}{2}r\right) }{r \left( (n+N+1)\right) ^{\frac{1}{p}}}\sum \limits _{j=0}^{\infty }\left( \frac{1 }{2^\frac{1}{p}} \right) ^j \le \frac{\varepsilon \left( C+\frac{3}{2} r\right) }{r} \frac{1}{1-2^{-\frac{1}{p}}}. \end{aligned}$$

Finally, we prove that (4.4) is true for $\frac{2\,l\pi }{rc}+\frac{\pi }{rc }\le x<\frac{2\left( l+1\right) \pi }{rc}$, where $0<2\left( l+1\right) \le r.$

Let $M^{\frac{1}{p}}:=M^{\frac{1}{p}}\left( x\right) \ge r$ be a natural number such that

$$\begin{aligned} \frac{2\left( l+1\right) \pi }{rc}-\frac{\pi }{cM^{\frac{1}{p}}}\le x<\frac{ 2\left( l+1\right) \pi }{rc}-\frac{\pi }{c(M+1)^{\frac{1}{p}}}. \end{aligned}$$

(4.6)

Then

$$\begin{aligned} \tau _{n}\left( x\right)&= \sum \limits _{k=n}^{n+M^{\frac{1}{p} }-1}b_{k}\sin (ckx) + \sum \limits _{k=n+M^{\frac{1}{p}}}^{n+M}b_{k}\sin (ckx) + \sum \limits _{k=n+M+1}^{\infty }b_{k}\sin (ckx) \\&=\tau _{n}^{\left( 4\right) }\left( x\right) +\tau _{n}^{\left( 5\right) }\left( x\right) +\tau _{n}^{\left( 6\right) }\left( x\right) . \end{aligned}$$

Applying Lagrange’s mean value theorem to the function $f\left( x\right) =\sin (ckx)$ on the interval $\left[ x,\frac{2\left( l+1\right) \pi }{rc} \right]$ we obtain that for each k there exists $z_k\in \left( x,\frac{ 2\left( l+1\right) \pi }{rc }\right)$ such that

$$\begin{aligned} \sin \left( k\frac{2\left( l+1\right) \pi }{r}\right) -\sin (ckx)=ck\cos (ckz_k)\left( \frac{2\left( l+1\right) \pi }{rc}-x\right) . \end{aligned}$$

Using this we get

$$\begin{aligned} \begin{aligned} \tau _{n}^{\left( 4\right) }\left( x\right)&=\sum \limits _{k=n}^{n+M^{\frac{1}{ p}}-1}ckb_{k}\cos (ckz_k)\left( \frac{2\left( l+1\right) \pi }{rc}-x\right) \\ {}&\quad +\sum \limits _{k=n}^{n+M^{\frac{1}{p}}-1}b_{k}\sin \left( k\frac{2\left( l+1\right) \pi }{r}\right) =\tau _{n}^{\left( 4.1\right) }\left( x\right) +\tau _{n}^{\left( 4.2\right) }(x). \end{aligned} \end{aligned}$$

From (4.2) we have

$$\begin{aligned} \begin{aligned} \left| \tau _n^{(4.2)}(x) \right| <\varepsilon . \end{aligned} \end{aligned}$$

By (4.6) and (4.1)

$$\begin{aligned} \left| \tau _{n}^{\left( 4.1\right) }\left( x\right) \right|&\le \left( \frac{2\left( l+1\right) \pi }{rc}-x\right) \sum \limits _{k=n}^{n+M^{ \frac{1}{p}}-1}ck\left| b_{k}\right| \le \frac{\pi }{M^{\frac{1}{p}}} \sum \limits _{k=n}^{n+M^{\frac{1}{p}}-1}\frac{k\ln k}{\ln k}\left| b_{k}\right| \\&\le \frac{\pi \varepsilon }{\ln 2}. \end{aligned}$$

Using Lemma 3 we get

$$\begin{aligned} \left| \tau _{n}^{(5)}(x)\right|&=\left| \sum \limits _{k=n+M^{ \frac{1}{p}}}^{n+M}b_{k}\sin (ckx)\right| \le \sum \limits _{k=n+M^{\frac{ 1}{p}}}^{n+M}\frac{k\ln k}{k\ln k}\left| b_{k}\right| \ll \varepsilon \int _{n+M^{\frac{1}{p}}}^{n+M}\frac{1}{k\ln k}dk \\&\le \varepsilon \ln p. \end{aligned}$$

If $\left( b_{n}\right) \in GM\left( p,{}_{3}\beta (q),r\right)$, then by Lemma 1

$$\begin{aligned} \bigg \vert \tau _{n}&{}^{\left( 6\right) }\left( x\right) \bigg \vert =\left| \sum \limits _{j=0}^{\infty }\sum \limits _{k=2^{j}\left( n+M+1\right) }^{2^{j+1}\left( n+M+1\right) -1}b_{k}\sin (ckx)\right| \\&\le \sum \limits _{j=0}^{\infty }\left| \frac{-1}{2\sin \left( crx/2\right) }\left\{ \sum \limits _{k=2^{j}\left( n+M+1\right) }^{2^{j+1}\left( n+M+1\right) -1}\left( b_{k}-b_{k+r}\right) \cos \left( \left( k+\frac{r}{2}\right) cx\right) \right. \right. \\&\quad \left. \left. +\sum \limits _{k=2^{j+1}\left( n+M+1\right) }^{2^{j+1}\left( n+M+1\right) +r-1}b_{k}\cos \left( k-\frac{r}{2}\right) cx-\sum \limits _{k=2^{j}\left( n+M+1\right) }^{2^{j}\left( n+M+1\right) +r-1}b_{k}\cos \left( \left( k-\frac{r}{2}\right) cx\right) \right\} \right| \\&\le \frac{1}{2\left| \sin \left( crx/2\right) \right| } \sum \limits _{j=0}^{\infty }\left\{ \sum \limits _{k=2^{j}\left( n+M+1\right) }^{2^{j+1}\left( n+M+1\right) -1}\left| b_{k}-b_{k+r}\right| +\sum \limits _{k=2^{j+1}\left( n+M+1\right) }^{2^{j+1}\left( n+M+1\right) +r-1}\left| b_{k}\right| \right. \\&\quad \left. +\sum \limits _{k=2^{j}\left( n+M+1\right) }^{2^{j}\left( n+M+1\right) +r-1}\left| b_{k}\right| \right\} . \end{aligned}$$

Next, applying the Hölder inequality with $p>1$, then using Lemma 1, the inequality

$2\left( l+1\right) -\frac{rc}{\pi }x\le \left| \sin \frac{rcx}{2} \right|$ $\left( x\in \left[ \frac{2\,l\pi }{rc}+\frac{\pi }{rc},\frac{ 2\left( l+1\right) \pi }{rc}\right] \text { and }0<2\left( l+1\right) \le r\right)$, (4.6) and (4.1), we get

$$\begin{aligned} \bigg \vert \tau&{}_{n}^{\left( 6\right) }\left( x\right) \bigg \vert \le \frac{1}{2(l+1)-\frac{rc}{\pi }x}\sum \limits _{j=0}^{\infty }\left\{ \left( \sum \limits _{k=2^{j}\left( n+M+1\right) }^{2^{j+1}\left( n+M+1\right) -1}\left| b_{k}-b_{k+r}\right| ^{p}\right) ^{\frac{1}{p}}\right. \\&\quad \left. \left( \sum \limits _{k=2^{j}\left( n+M+1\right) }^{2^{j+1}\left( n+M+1\right) -1}1\right) ^{1-\frac{1}{p}} +\sum \limits _{k=2^{j}\left( n+M+1\right) }^{2^{j}\left( n+M+1\right) +r-1}\left| b_{k}\right| +\sum \limits _{k=2^{j+1}\left( n+M+1\right) }^{2^{j+1}\left( n+M+1\right) +r-1}\left| b_{k}\right| \right\} \\&\le \frac{(M+1)^{\frac{1}{p}}}{r}\sum \limits _{j=0}^{\infty }\left\{ \frac{ C\left( 2^{j}(n+M+1)\right) ^{1-\frac{1}{p}}}{2^{j}(n+M+1)}\underset{m\ge b\left( 2^{j}(n+M+1)\right) }{\sup }m\left( \frac{1}{m}\sum \limits _{k=m}^{2m-1}\left| b_{k}\right| ^{q}\right) ^{\frac{1}{q} }\right. \\&\quad \left. +\sum \limits _{k=2^{j}\left( n+M+1\right) }^{2^{j}\left( n+M+1\right) +r-1}\left| b_{k}\right| +\sum \limits _{k=2^{j+1}\left( n+M+1\right) }^{2^{j+1}\left( n+M+1\right) +r-1}\left| b_{k}\right| \right\} =\frac{(M+1)^{\frac{1}{p}}}{r} \sum \limits _{j=0}^{\infty } \\&\quad \bigg \{\frac{C}{ \left( 2^{j}(n+M+1)\right) ^{\frac{1}{p}}}\underset{m\ge b\left( 2^{j}(n+M+1)\right) }{\sup }m^{1-\frac{1}{q}}\left( \sum \limits _{k=m}^{2m-1}\left( \frac{k\ln k\left| b_{k}\right| }{ k\ln k}\right) ^{q}\right) ^{\frac{1}{q}} \\&\quad \left. +\sum \limits _{k=2^{j}\left( n+M+1\right) }^{2^{j}\left( n+M+1\right) +r-1}\frac{k\ln k\left| b_{k}\right| }{k\ln k}+\sum \limits _{k=2^{j+1}\left( n+M+1\right) }^{2^{j+1}\left( n+M+1\right) +r-1} \frac{k\ln k\left| b_{k}\right| }{k\ln k}\right\} \\&<\frac{\varepsilon (M+1)^{\frac{1}{p}}}{r}\sum \limits _{j=0}^{\infty }\left\{ \frac{C}{\left( 2^{j}(n+M+1)\right) ^{\frac{1}{p}}}\underset{m\ge b\left( 2^{j}(n+M+1)\right) }{\sup }m^{1-\frac{1}{q}}\right. \\&\quad \left. \left( \sum \limits _{k=m}^{2m-1}\left( \frac{1}{k}\right) ^{q}\right) ^{\frac{1}{q} } +\sum \limits _{k=2^{j}\left( n+M+1\right) }^{2^{j}\left( n+M+1\right) +r-1}\frac{1}{k}+\sum \limits _{k=2^{j+1}\left( n+M+1\right) }^{2^{j+1}\left( n+M+1\right) +r-1}\frac{1}{k}\right\} . \end{aligned}$$

Elementary calculations give

$$\begin{aligned} \bigg \vert \tau&{}_{n}^{\left( 6\right) }\left( x\right) \bigg \vert <\frac{ \varepsilon (M+1)^{\frac{1}{p}}}{r}\sum \limits _{j=0}^{\infty }\left\{ \frac{C }{\left( 2^{j}(n+M+1)\right) ^{\frac{1}{p}}}\underset{m\ge b\left( 2^{j}(n+M+1)\right) }{\sup } \right. \\&\quad \left. \left( m^{1-\frac{1}{q}}m^{-1}m^{\frac{1}{q} }\right) +r\left( 2^{j}(n+M+1)\right) ^{-\frac{1}{p}}+\frac{1}{2}r\left( 2^{j}(n+M+1)\right) ^{-\frac{1}{p}}\right\} \\&\le \frac{\varepsilon (M+1)^{\frac{1}{p}}}{r}\sum \limits _{j=0}^{\infty }\left\{ \left( 2^{j}(n+M+1)\right) ^{-\frac{1}{p}}\left( C+\frac{3}{2} r\right) \right\} \\&\le \frac{\varepsilon (M+1)^{\frac{1}{p}}\left( C+\frac{3}{2}r\right) }{ r\left( (n+M+1)\right) ^{\frac{1}{p}}}\sum \limits _{j=0}^{\infty }\left( \frac{1}{2^{\frac{1}{p}}}\right) ^{j}\le \frac{\varepsilon \left( C+\frac{3 }{2}r\right) }{r}\frac{1}{1-2^{-\frac{1}{p}}}. \end{aligned}$$

Joining the obtained estimates the uniform convergence of series (1.2) follows and thus the proof is complete. $\square$

4.2 Proof of Proposition 1

Let for $n \in \mathbb {N}$:

$a_n =\left\{ \begin{array}{l} \frac{3}{n\ln (n+1)},\text { when } n=1\text { (mod }3), \\ \frac{1}{n\ln (n+1)},\text { when } n=2\text { (mod }3), \\ \frac{1}{n\ln (n+1)},\text { when } n=0\text { (mod }3) \text { and } n\ne 0 \text { (mod }6), \\ \frac{1}{(n-3)\ln (n-2)} + \frac{1}{n^{1+\frac{1}{p}}\ln (n+1)},\text { when } n=0\text { (mod }6). \end{array} \right.$

First, we prove that $(a_{n})\in GM(p,{}_{3}\beta (1),3)$ for $p>1$. Let

$$\begin{aligned}&A_{n}=\{k\in \mathbb {N}:n\le k\le 2n-1\text { and }k=1\text { }(\text {mod }3)\},\\&B_{n}=\{k\in \mathbb {N}:n\le k\le 2n-1\text { and }k=2\text { }(\text {mod }3)\},\\&C_{n}=\{k\in \mathbb {N}:n\le k\le 2n-1\text { and }k=0\text { }(\text {mod }3) \text { and }k\ne 0\text { }(\text {mod }6)\}, \\&D_{n}=\{k\in \mathbb {N}:n\le k\le 2n-1\text { and }k=0\text { }(\text {mod }6)\}. \end{aligned}$$

Using elementary calculations we get

$$\begin{aligned} \begin{aligned}&\left\{ \sum _{k=n}^{2n-1} \left| a_{k}-a_{k+3}\right| ^{p}\right\} ^{ \frac{1}{p}}=\left\{ \sum _{k\in A_{n}}\left| a_{k}-a_{k+3}\right| ^{p}+\sum _{k\in B_{n}}\left| a_{k}-a_{k+3}\right| ^{p}\right. \\ {}&\qquad \left. +\sum _{k\in C_{n}}\left| a_{k}-a_{k+3}\right| ^{p} +\sum _{k\in D_{n}}\left| a_{k}-a_{k+3}\right| ^{p}\right\} ^{ \frac{1}{p}}=\left\{ \sum _{k\in A_{n}}\left| \frac{3}{k\ln (k+1)} \right. \right. \\ {}&\qquad \left. -\frac{3 }{(k+3)\ln (k+4)}\right| ^{p} +\sum _{k\in B_{n}}\left| \frac{1}{k\ln (k+1)}-\frac{1}{(k+3)\ln (k+4)} \right| ^{p} \\ {}&\qquad +\sum _{k\in C_{n}}\left| \frac{1}{k\ln (k+1)}-\frac{1}{k\ln (k+1)}-\frac{ 1}{(k+3)^{1+\frac{1}{p}}\ln (k+4)}\right| ^{p} \\ {}&\qquad \left. +\sum _{k\in D_{n}}\left| \frac{1}{(k-3)\ln (k-2)}+\frac{1}{k^{1+ \frac{1}{p}}\ln (k+1)}-\frac{1}{(k+3)\ln (k+4)}\right| ^{p}\right\} ^{ \frac{1}{p}} \\ {}&\quad =\left\{ \sum _{k\in A_{n}}3^{p}\left| \frac{1}{k\ln (k+1)}-\frac{1}{ (k+3)\ln (k+4)}\right| ^{p}+ \sum _{k\in B_{n}}\left| \frac{1}{k\ln (k+1)} \right. \right. \\ {}&\qquad \left. -\frac{1}{(k+3)\ln (k+4)}\right| ^{p} +\sum _{k\in C_{n}}\left| \frac{1}{(k+3)^{1+\frac{1}{p}}\ln (k+4)} \right| ^{p} \\ {}&\qquad \left. +\sum _{k\in D_{n}}\left( \left| \frac{1}{(k-3)\ln (k-2)}-\frac{1}{ (k+3)\ln (k+4)} +\frac{1}{k^{1+\frac{1}{p}}\ln (k+1)}\right| \right) ^{p}\right\} ^{\frac{1}{p}}. \end{aligned} \end{aligned}$$

Moreover

$$\begin{aligned} \bigg \vert \frac{1}{k\ln (k+1)}&-\frac{1}{(k+3)\ln (k+4)}\bigg \vert =\frac{ \left| (k+3)\ln (k+4)-k\ln (k+1)\right| }{k(k+3)\ln (k+1)\ln (k+4)} \\&\le \frac{\frac{3k}{k+1}+3\ln (k+4)}{k(k+3)\ln (k+1)\ln (k+4)} \\&\le \frac{ 6\ln (k+4)}{k^{2}\ln (k+1)\ln (k+4)}=\frac{6}{k^{2}\ln (k+1)} \end{aligned}$$

for $k\ge 1$ and

$$\begin{aligned} \bigg \vert \frac{1}{(k-3)\ln (k-2)}&-\frac{1}{(k+3)\ln (k+4)}\bigg \vert = \frac{\left| (k+3)\ln (k+4)-(k-3)\ln (k-2)\right| }{(k-3)(k+3)\ln (k-2)\ln (k+4)} \\&\le \frac{(k-3)\left| \ln (k+4)-\ln (k-2)\right| +6\ln (k+4)}{ (k-3)(k+3)\ln (k-2)\ln (k+4)} \\&\le \frac{\frac{6(k-3)}{k-2}+6\ln (k+4)}{(k-3)(k+3)\ln (k-2)\ln (k+4)} \\&\le \frac{12}{(k-3)(k+3)\ln (k-2)} \le \frac{48}{k^{2}\ln (k+1)} \end{aligned}$$

for $k\ge 6$. Thus

$$\begin{aligned} \bigg \{ \sum _{k=n}^{2n-1}&\left| a_{k}-a_{k+3}\right| ^{p}\bigg \} ^{ \frac{1}{p}}\le \left\{ \sum _{k\in A_{n}}3^{p}\left( \frac{6}{k^{2}\ln (k+1) }\right) ^{p}+\sum _{k\in B_{n}}\left( \frac{6}{k^{2}\ln (k+1)}\right) ^{p}\right. \\&\quad \left. +\sum _{k\in C_{n}}\left( \frac{1}{k^{1+\frac{1}{p}}\ln (k+1)}\right) ^{p}+\sum _{k\in D_{n}}\left( \frac{48}{k^{2}\ln (k+1)}+\frac{1}{k^{1+\frac{1 }{p}}\ln (k+1)}\right) ^{p}\right\} ^{\frac{1}{p}} \\&\le 49\left\{ \sum _{k=n}^{2n-1}\left( \frac{1}{k^{1+\frac{1}{p}}\ln (k+1)} \right) ^{p}\right\} ^{\frac{1}{p}}\le 49\frac{1}{n^{1+\frac{1}{p}}\ln (n+1) }n^{\frac{1}{p}} \\&=49\frac{1}{n\ln (n+1)} \le 147\frac{1}{n} \sum _{k=n}^{2n}\left| a_{k} \right| \le 147 \frac{1}{n} \underset{m\ge b(n)}{\sup } \sum \limits _{k=m}^{2m} \left| a_{k}\right| . \end{aligned}$$

Hence $(a_{n})\in GM(p,{}_{3}\beta (1),3)$. Now, we will show that $(a_{n})\notin GM(1,{}_{3}\beta (1),3)$. We have

$$\begin{aligned} \sum _{k=n}^{2n-1} \left| a_{k}-a_{k+3}\right|&\ge \sum _{k\in C_{n}}\left| a_{k}-a_{k+3}\right| =\sum _{k\in C_{n}}\left| \frac{ 1}{k\ln (k+1)}-\frac{1}{k\ln (k+1)} \right. \\&\qquad \left. -\frac{1}{(k+3)^{1+\frac{1}{p}}\ln (k+4)} \right| =\sum _{k\in C_{n}}\frac{1}{(k+3)^{1+\frac{1}{p}}\ln (k+4)} \\&\ge \frac{1}{ (n+3)^{1+\frac{1}{p}}\ln (n+4)}\frac{n}{12} \ge \frac{1}{48(n+3)^{\frac{1}{p} }\ln (n+4)}. \end{aligned}$$

On the other hand, we get

$$\begin{aligned} \begin{aligned} \frac{1}{n} \underset{m\ge b(n)}{\sup }\ \sum \limits _{k=m}^{2m} \left| a_{k}\right| \le C\frac{1}{n}. \end{aligned} \end{aligned}$$

Therefore, the inequality

$$\begin{aligned} \sum _{k=n}^{2n}\left| a_{k}-a_{k+3}\right| \le C\frac{1}{n} \underset{m\ge b(n)}{\sup } \sum \limits _{k=m}^{2m} \left| a_{k}\right| . \end{aligned}$$

can not be satisfied if $n\rightarrow \infty$.

Now, we will show that the series (1.2) is divergent in $x_0 = \frac{ 2}{3}\pi$. We have

$$\begin{aligned} \begin{aligned} \sum _{k=6}^{6N+5}&a_k \sin (kx_0) = \sum _{k=1}^{N} \sum _{l=0}^{5} a_{6k+l} \sin \left( (6k+l)\frac{2}{3}\pi \right) = \sum _{k=1}^{N} \left( a_{6k} \sin (4\pi ) \right. \\ {}&\quad + a_{6k+1} \sin \left( (6k+1)\frac{2}{3}\pi \right) + a_{6k+2} \sin \left( (6k+2)\frac{2}{3}\pi \right) + a_{6k+3} \sin \left( (6k+3)\frac{2}{3}\pi \right) \\ {}&\quad \left. + a_{6k+4} \sin \left( (6k+4)\frac{2}{3}\pi \right) + a_{6k+5} \sin \left( (6k+5)\frac{2}{3}\pi \right) \right) \\ {}&= \sum _{k=1}^{N} \left( a_{6k+1} \sin \left( \frac{2}{3}\pi \right) + a_{6k+2} \left( -\sin \left( \frac{2}{3}\pi \right) \right) + a_{6k+4} \sin \left( \frac{2}{3}\pi \right) \right. \\ {}&\quad \left. + a_{6k+5} \left( -\sin \left( \frac{2}{3}\pi \right) \right) \right) = \sin \left( \frac{2}{3}\pi \right) \sum _{k=1}^{N} \left[ \left( a_{6k+1} - a_{6k+2} \right) + \left( a_{6k+4} - a_{6k+5} \right) \right] \\ {}&= \sin \left( \frac{2}{3}\pi \right) \sum _{k=1}^{N} \left[ \left( \frac{3}{(6k+1)\ln (6k+2)} - \frac{1}{(6k+2)\ln (6k+3)} \right) \right. \\ {}&\quad \left. + \left( \frac{3}{(6k+4)\ln (6k+5)} - \frac{1}{(6k+5)\ln (6k+6)} \right) \right] \\ {}&\ge \sin \left( \frac{2}{3}\pi \right) \sum _{k=1}^{N} \left[ \left( \frac{3}{ (6k+2)\ln (6k+2)} - \frac{1}{(6k+2)\ln (6k+2)} \right) \right. \\ {}&\left. + \left( \frac{3}{(6k+5)\ln (6k+5)} - \frac{1}{(6k+5)\ln (6k+5)} \right) \right] \\ {}&= \sin \left( \frac{2}{3}\pi \right) \sum _{k=1}^{N} \left( \frac{2}{(6k+2)\ln (6k+2)} + \frac{2}{(6k+5)\ln (6k+5)} \right) \\ {}&\ge 4\sin \left( \frac{2}{3} \pi \right) \sum _{k=1}^{N} \frac{1}{(6k+5)\ln (6k+5)} \rightarrow \infty \text{ as } N\rightarrow \infty . \end{aligned}\end{aligned}$$

This ends our proof. $\square$

4.3 Proof of Theorem 11

The proof is similar to the proof of Theorem 10. The only difference is that we use (2.3) and (3.2) instead of (2.2) and (3.1), respectively. $\square$

4.4 Proof of Theorem 12

The proof is similar to the proof of Theorem 10. The only difference is that we use (2.4) and Lemma 2 instead of (3.1) and Lemma 1, respectively. $\square$