On $\mathcal I(<q)$- and $\mathcal I(\leq q)$-convergence of arithmetic functions

TL;DR

This paper refines the understanding of convergence of arithmetic functions related to specific ideals defined by the convergence exponent, introducing new methods for characterizations based on $ ext{I}(<q)$ and $ ext{I}( ext{leq} q)$ convergence.

Contribution

It sharpens previous results on $ ext{I}_c^{(q)}$-convergence by developing new criteria and methods using $ ext{I}(<q)$ and $ ext{I}( ext{leq} q)$ convergence.

Findings

Characterization of $ ext{I}_c^{(q)}$-convergence in terms of new criteria.

Development of novel methods for analyzing $ ext{I}(<q)$ and $ ext{I}( ext{leq} q)$ convergence.

Sharpened results on the structure of admissible ideals related to convergence exponents.

Abstract

Let $\mathbb N$ be the set of positive integers, and denote by $\lambda(A)=\inf\{t>0:\sum_{a\in A} a^{-t}<\infty\}$ the convergence exponent of $A\subset\mathbb N$. For $0<q\le 1$, $0\le q\le 1$, respectively, the admissible ideals $\mathcal I(<q)$, $\mathcal I(\leq q)$ of all subsets $A\subset \mathbb N$ with $\lambda(A)<q$, $\lambda(A)\le q$, respectively, satisfy $\mathcal I(<q)\subsetneq\mathcal I_c^{(q)}\subsetneq \mathcal I(\leq q)$, where $\mathcal I_c^{(q)}=\{A\subset\mathbb N: \sum_{a\in A}a^{-q}<\infty\}$. In this note we sharpen the results of Bal\'az, Gogola and Visnyai from [2], and of others papers, concerning characterizations of $\mathcal I_c^{(q)}$-convergence of various arithmetic functions in terms of $q$. This is achieved by utilizing $\mathcal I(<q)$- and $\mathcal I(\leq q)$-convergence, for which new methods and criteria are developed.

Equations

$$\lambda (A)=\inf \{t>0:\sum _{a\in A}{a}^{-t}<\infty \}$$

$${\mathcal{I}}_c^{(q)}=\{A\subset \mathbb{N}:\sum _{a\in A}{a}^{-q}<\infty \}.$$

$$\lambda(A)=\inf\Big{\{}t>0:\sum_{a\in A}\frac{1}{a^{t}}<\infty\Big{\}}.$$

$$\lambda (A)=\underset{n\to \infty }{\mathrm{limsup}}\frac{\log n}{\log a_n}.$$

$${\mathcal{I}}_{<q}$$

$${\mathcal{I}}_{\le q}$$

$${\mathcal{I}}_0$$

$$\mathcal{I}_{c}^{(q)}=\Big{\{}A\subset\mathbb{N}:\sum_{a\in A}\frac{1}{a^{q}}<\infty\Big{\}}$$

$${\mathcal{I}}_f\subset {\mathcal{I}}_0\subset {\mathcal{I}}_{<q}\subset {\mathcal{I}}_c^{(q)}\subset {\mathcal{I}}_{\le q}\subset {\mathcal{I}}_{<q^{\prime }}.$$

$$A_{\epsilon }=\{n:|x_n-L|\ge \epsilon \}$$

$${\mathcal{I}}_0⊊{\mathcal{I}}_{<q}⊊{\mathcal{I}}_c^{(q)}⊊{\mathcal{I}}_{\le q}⊊{\mathcal{I}}_{<q^{\prime }}⊊{\mathcal{I}}_c^{(q^{\prime })}⊊{\mathcal{I}}_{\le q^{\prime }}⊊{\mathcal{I}}_{<1}⊊{\mathcal{I}}_c^{(1)}⊊{\mathcal{I}}_{\le 1}=2^{\mathbb{N}}.$$

$$a_n=\lfloor n^{\frac{1}{s}}\rfloor .$$

$$\frac{\log n}{\log a_{n}}=\frac{\log n}{\frac{1}{s}.\log n+\log\Big{(}1-\frac{\varepsilon(n)}{n^{\frac{1}{s}}}\Big{)}}\to s,\quad\text{if}\ n\to\infty,$$

$$\lambda (A_k)=\underset{n\to \infty }{\mathrm{limsup}}\frac{\log n}{\log k{a}_n}=\lambda (A).$$

$$a_n=\lfloor n^{\frac{1}{q}}{\log }^{\frac{2}{q}}(n+1)\rfloor +1.$$

$$\sum _{n=1}^{\infty }\frac{1}{a_n^q}<\sum _{n=2}^{\infty }\frac{1}{n{\log }^{2}n}<\infty ,\text{thus, }A\in {\mathcal{I}}_c^{(q)}.$$

$$\underset{n\to \infty }{\lim }\frac{\log n}{\log a_n}=\underset{n\to \infty }{\lim }\frac{\log n}{\log (n^{\frac{1}{q}}{\log }^{\frac{2}{q}}(n+1))}=\underset{n\to \infty }{\lim }\frac{\log n}{\frac{1}{q}\log n+\frac{2}{q}\log \log (n+1)}=q,$$

$$\sum _{n=1}^{\infty }\frac{1}{a_n^q}\ge \sum _{n=1}^{\infty }\frac{1}{n}=\infty ,$$

$${\mathcal{I}}_0=\bigcap _{0<q\le 1}{\mathcal{I}}_{<q}=\bigcap _{0<q\le 1}{\mathcal{I}}_{\le q},$$

$${\mathcal{I}}_0=\bigcap _{0<q\le 1}{\mathcal{I}}_c^{(q)}.$$

$$A(x)=\#\{a\le x:a\in A\}.$$

$$\underset{x\to \infty }{\lim }\frac{A(x)}{x^{q+\delta }}=0.$$

$$\lambda (A)=\underset{n\to \infty }{\mathrm{limsup}}\frac{\log n}{\log a_n}\le q,$$

$$\frac{\log n}{\log a_n}\le q+\frac{\delta }{2},\text{thus}A(a_n)=n\le a_n^{q+\frac{\delta }{2}}.$$

$$0\le \frac{A(x)}{x^{q+\delta }}\le \frac{x^{q+\frac{\delta }{2}}}{x^{q+\delta }}=\frac{1}{x^{\frac{\delta }{2}}}\to 0,\text{as}x\to \infty ,$$

$$\frac{A(a_n)}{a_n^{q+\delta }}\to 0,\text{as}n\to \infty ,$$

$$\frac{\log n}{\log a_n}\le \frac{(q+\delta )\log a_n}{\log a_n}=q+\delta .$$

$$\underset{x\to \infty }{\lim }\frac{A_{\epsilon }(x)}{x^{q+\delta }}=0.$$

$$\underset{x\to \infty }{\lim }\frac{A(x)}{x^{q-\delta }}=0.$$

$$\lambda (A)=\underset{n\to \infty }{\mathrm{limsup}}\frac{\log n}{\log a_n}<q,\text{where}A=\{a_1<a_2<\dots \}.$$

$$\frac{\log n}{\log a_n}\le q-2\delta ,\text{thus,}n\le a_n^{q-2\delta },$$

Full text

[TABLE]

the convergence exponent of $A\subset\mathbb{N}$. For $0<q\leq 1$, $0\leq q\leq 1$, respectively, the admissible ideals $\mathcal{I}_{<q}$, $\mathcal{I}_{\leq q}$ of all subsets $A\subset\mathbb{N}$ with $\lambda(A)<q$, $\lambda(A)\leq q$, respectively, satisfy $\mathcal{I}_{<q}\subsetneq\mathcal{I}_{c}^{(q)}\subsetneq\mathcal{I}_{\leq q}$, where

[TABLE]

In this note we sharpen the results of Baláž, Gogola and Visnyai from 2 , and other papers, concerning characterizations of $\mathcal{I}_{c}^{(q)}$-convergence of various arithmetic functions in terms of $q$. This is achieved by utilizing $\mathcal{I}_{<q}$- and $\mathcal{I}_{\leq q}$-convergence, for which new methods and criteria are developed.

††journal: Periodica Mathematica Hungarica

1 Introduction

Denote by $\mathbb{N}$ the set of positive integers, and let $\lambda$ be the convergence exponent function on the power set $2^{\mathbb{N}}$ of $\mathbb{N}$, i.e. for $A\subset\mathbb{N}$ put

[TABLE]

If $q>\lambda(A)$ then $\sum_{a\in A}\frac{1}{a^{q}}<\infty$, and $\sum_{a\in A}\frac{1}{a^{q}}=\infty$ when $q<\lambda(A)$; if $q=\lambda(A)$, the convergence of $\sum_{a\in A}\frac{1}{a^{q}}$ is inconclusive. It follows from [13, p.26, Exercises 113, 114] that the range of $\lambda$ is the interval $[0,1]$, moreover for $A=\{a_{1}<a_{2}<\cdots<a_{n}<\dots\}\subset\mathbb{N}$,

[TABLE]

It is easy to see that $\lambda$ is monotonic, i.e. $\lambda(A)\leq\lambda(B)$ whenever $A\subset B\subset\mathbb{N}$, furthermore, $\lambda(A\cup B)=\max\{\lambda(A),\lambda(B)\}$ for all $A,B\subset\mathbb{N}$. Define the following sets:

[TABLE]

Clearly, $\mathcal{I}_{\leq 0}=\mathcal{I}_{0}$, and $\mathcal{I}_{\leq 1}=2^{\mathbb{N}}$. Since $\lambda(A)=0$ when $A\subset\mathbb{N}$ is finite, then \mathcal{I}_{f}=\{A\subset\mathbb{N}:\textrm{A is finite}\}\subset\mathcal{I}_{0}, moreover, also considering the well-known set

[TABLE]

we get that whenever $0<q<q^{\prime}<1$,

[TABLE]

In what follows, we will use the following definitions.

The set $\mathcal{I}\subseteq 2^{\mathbb{N}}$ is a so-called admissible ideal, provided $\mathcal{I}$ is additive (i.e. $A,B\in\mathcal{I}$ implies $A\cup B\in\mathcal{I}$), hereditary (i.e. $A\in\mathcal{I},\ B\subset A$ implies $B\in\mathcal{I}$), it contains the singletons, and $\mathbb{N}\notin\mathcal{I}$.

Given an ideal $\mathcal{I}\subset 2^{\mathbb{N}}$, we say that a sequence $x=(x_{n})_{n=1}^{\infty}$ $\mathcal{I}$-converges to a number $L$, and write $\mathop{\mathcal{I}\text{\rm--lim}}\limits x_{n}=L$, if for each $\varepsilon>0$ the set

[TABLE]

belongs to the ideal $\mathcal{I}$. One can see 7 , 9 for a general treatment of $\mathcal{I}$-convergence; a useful property is as follows:

Lemma 1** (9 )**

If $\mathcal{I}_{1}\subset\mathcal{I}_{2}$, then $\mathop{\mathcal{I}_{1}\text{\rm--lim}}\limits x_{n}=L$ implies $\mathop{\mathcal{I}_{2}\text{\rm--lim}}\limits x_{n}=L$.

We will study $\mathcal{I}$-convergence in the case when $\mathcal{I}$ stands for $\mathcal{I}_{<q}$, $\mathcal{I}_{c}^{(q)}$, $\mathcal{I}_{\leq q}$, respectively. We will establish necessary and sufficient conditions for a set $A\subset\mathbb{N}$ to belong to $\mathcal{I}_{<q}$, $\mathcal{I}_{\leq q}$, respectively; as well as for the set $A_{\varepsilon}=\{n:|x_{n}-L|\geq\varepsilon\}$ so that $\mathop{\mathcal{I}_{<q}\text{\rm--lim}}\limits x_{n}=L$, resp. $\mathop{\mathcal{I}_{\leq q}\text{\rm--lim}}\limits x_{n}=L$ hold. Note that analogous criteria were not known for $\mathcal{I}_{c}^{(q)}$.

In this paper, we embed the ideals $\mathcal{I}_{<q}$ and $\mathcal{I}_{\leq q}$ into the structure of ideals $\mathcal{I}_{c}^{(q)}$. We show that theses ideals are essentially distinct. Then we refine a known statement concerning the $\mathcal{I}_{c}^{(q)}$-convergence of some arithmetic functions. A new method is introduced and can be applied widely for consideration of $\mathcal{I}_{<q}$ and $\mathcal{I}_{\leq q}$-convergence of sequences.

2 On ideals enveloping the ideal $\mathcal{I}_{c}^{(q)}$

Theorem 1

Let $0<q<q^{\prime}<1$. Then

[TABLE]

Proof

The inclusions follow from the definitions of the sets. We can show that the difference of successive sets in (3) is infinite, so equality does not hold in any of the inclusions, by considering the following four cases (as usual, $\lfloor x\rfloor$ is the integer part of the real $x$):

Case 1. $\mathcal{I}_{0}\neq\mathcal{I}_{<q}$: let $0<s<q<1$, and take the set $A=\{a_{1}<a_{2}<\cdots\}\subset\mathbb{N}$, where for all $n\in\mathbb{N}$,

[TABLE]

Then $a_{n}=n^{\frac{1}{s}}-\varepsilon(n)$ for some $0\leq\varepsilon(n)<1$, and by Lagrange’s Mean Value Theorem for $f(x)=x^{\frac{1}{s}}$ on $[n,n+1]$ we get that $a_{n}<a_{n+1}$ for all $n$. Since

[TABLE]

then $0<\lambda(A)=s<q$; thus, A\in\mathcal{I}_{<q}\mathbin{\mathchoice{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{\displaystyle\smallsetminus}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{\textstyle\smallsetminus}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.6pt}{\rotatebox[origin={c}]{-20.0}{\scriptstyle\smallsetminus}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.45pt}{\rotatebox[origin={c}]{-20.0}{\scriptscriptstyle\smallsetminus}}\mspace{-4.0mu}}}\mathcal{I}_{0}. It is also clear that \mathcal{I}_{<q}\mathbin{\mathchoice{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{\displaystyle\smallsetminus}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{\textstyle\smallsetminus}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.6pt}{\rotatebox[origin={c}]{-20.0}{\scriptstyle\smallsetminus}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.45pt}{\rotatebox[origin={c}]{-20.0}{\scriptscriptstyle\smallsetminus}}\mspace{-4.0mu}}}\mathcal{I}_{0} is infinite, since for any $k\in\mathbb{N}$ the sets $A_{k}=\{ka_{n}:\ n\in\mathbb{N}\}$ satisfy

[TABLE]

Case 2. $\mathcal{I}_{<q}\neq\mathcal{I}_{c}^{(q)}$: let $0<q<1$, and take the set $A=\{a_{1}<a_{2}<\cdots\}\subset\mathbb{N}$, where for all $n\in\mathbb{N}$,

[TABLE]

One can easily show that $(a_{n})$ is increasing sequence, and,

[TABLE]

On the other hand

[TABLE]

hence, $\lambda(A)=q$. Similarly to Case 1 we can see that \mathcal{I}_{c}^{(q)}\mathbin{\mathchoice{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{\displaystyle\smallsetminus}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{\textstyle\smallsetminus}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.6pt}{\rotatebox[origin={c}]{-20.0}{\scriptstyle\smallsetminus}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.45pt}{\rotatebox[origin={c}]{-20.0}{\scriptscriptstyle\smallsetminus}}\mspace{-4.0mu}}}\mathcal{I}_{<q} is actually infinite.

Case 3. $\mathcal{I}_{c}^{(q)}\neq\mathcal{I}_{\leq q}$: let $0<q<1$, define $A=\{a_{1}<a_{2}<\cdots\}\subset\mathbb{N}$, where $a_{n}=\lfloor n^{\frac{1}{q}}\rfloor$ for all $n\in\mathbb{N}$. Then

[TABLE]

so $A\notin\mathcal{I}_{c}^{(q)}$, but $A\in\mathcal{I}_{\leq q}$, since $\lambda(A)=q$. Analogously to Case 1, one can show that \mathcal{I}_{\leq q}\mathbin{\mathchoice{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{\displaystyle\smallsetminus}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{\textstyle\smallsetminus}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.6pt}{\rotatebox[origin={c}]{-20.0}{\scriptstyle\smallsetminus}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.45pt}{\rotatebox[origin={c}]{-20.0}{\scriptscriptstyle\smallsetminus}}\mspace{-4.0mu}}}\mathcal{I}_{c}^{(q)} is infinite.

Case 4. $\mathcal{I}_{\leq q}\neq\mathcal{I}_{<q^{\prime}}$: it suffices to choose the set $A=\{a_{1}<a_{2}<\cdots\}\subset\mathbb{N}$ such that $a_{n}=\lfloor n^{\frac{1}{s}}\rfloor$ for all $n$, where $0<q<s<q^{\prime}$. Then $\lambda(A)=s$, so $A\in\mathcal{I}_{<q^{\prime}}$, however, $A\notin\mathcal{I}_{\leq q}$. Moreover, again, \mathcal{I}_{<q^{\prime}}\mathbin{\mathchoice{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{\displaystyle\smallsetminus}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{\textstyle\smallsetminus}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.6pt}{\rotatebox[origin={c}]{-20.0}{\scriptstyle\smallsetminus}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.45pt}{\rotatebox[origin={c}]{-20.0}{\scriptscriptstyle\smallsetminus}}\mspace{-4.0mu}}}\mathcal{I}_{\leq q} is infinite. ∎

It is worth noting by (3), that in order to decide if a given $A\subset\mathbb{N}$ belongs to $\mathcal{I}_{c}^{(q)}$, it may be easier, or more advantageous to first determine the convergence exponent of $A$. Indeed, if $\lambda(A)<q$, then $A\in\mathcal{I}_{<q}\subset\mathcal{I}_{c}^{(q)}$, or, if $\lambda(A)=q$ then $A\in\mathcal{I}_{\leq q}\subset\mathcal{I}_{c}^{(q^{\prime})}$ for every $q^{\prime}>q$. This view is important, since in what follows, we will establish criteria for $\mathcal{I}_{<q}$, $\mathcal{I}_{\leq q}$ membership, respectively.

Theorem 2

Let $0<q\leq 1$. Then each of the sets $\mathcal{I}_{0}$, $\mathcal{I}_{<q}$, $\mathcal{I}_{\leq q}$ forms an admissible ideal, except for $\mathcal{I}_{\leq 1}$.

Proof

Follows from properties of $\lambda$ listed in the Introduction, along with (3). ∎

Theorem 3

We have

[TABLE]

hence,

[TABLE]

Proof

Follows from the definitions of $\mathcal{I}_{0}$, $\mathcal{I}_{<q}$, $\mathcal{I}_{\leq q}$, and (3). ∎

3 Conditions for a set $A$ to belong to $\mathcal{I}_{<q}$, $\mathcal{I}_{\leq q}$

Given $x\geq 1$, define the counting function of $A\subset\mathbb{N}$ as

[TABLE]

We have

Theorem 4

Let $0\leq q<1$ be a real number, and $A\subset\mathbb{N}$. Then $A\in\mathcal{I}_{\leq q}$ if and only if for every $\delta>0$

[TABLE]

Proof

Let $A=\{a_{1}<a_{2}<\dots\}$, and $A\in\mathcal{I}_{\leq q}$. Then

[TABLE]

so for any $\delta>0$ there is an $n_{0}\in\mathbb{N}$ so that for all $n\geq n_{0}$

[TABLE]

If $x$ is sufficiently large, we can find $n\geq n_{0}$ with $a_{n}\leq x<a_{n+1}$, hence, $A(x)=n\leq x^{q+\frac{\delta}{2}}$. Consequently,

[TABLE]

which implies (4) for every $\delta>0$.

Conversely, let $\delta>0$, and (4) be true for some $A=\{a_{1}<a_{2}<\dots\}$. Then

[TABLE]

so there is $n_{1}\in\mathbb{N}$ such that for all $n\geq n_{1}$, $n\leq a_{n}^{q+\delta}$, thus,

[TABLE]

Then for all $\delta>0$, $\lambda(A)\leq q+\delta$, hence, letting $\delta\to 0$, we get $\lambda(A)\leq q$, so, $A\in\mathcal{I}_{\leq q}$. ∎

The definition of $\mathcal{I}_{\leq q}$-convergence immediately yields

Corollary 1

Let $0\leq q<1$, $\varepsilon>0$, $L$ and $x_{n}$ be real numbers for all $n\in\mathbb{N}$, and $A_{\varepsilon}=\{n:|x_{n}-L|\geq\varepsilon\}$. Then $\mathop{\mathcal{I}_{\leq q}\text{\rm--lim}}\limits x_{n}=L$ if and only if for every $\varepsilon>0$ and $\delta>0$

[TABLE]

Theorem 5

Let $0<q\leq 1$ be a real number, and $A\subset\mathbb{N}$. Then $A\in\mathcal{I}_{<q}$ if and only if there exists $\delta>0$ such that

[TABLE]

Proof

Let $A\in\mathcal{I}_{<q}$. Then

[TABLE]

For each $\delta>0$ with $0<\delta<\frac{1}{2}(q-\lambda(A))$ there is $n_{0}\in\mathbb{N}$ so that for all $n\geq n_{0}$,

[TABLE]

hence, for all $n\geq n_{0}$,

[TABLE]

If $x$ is large enough, there exists some $n\geq n_{0}$ with $a_{n}\leq x<a_{n+1}$, so $A(x)=n\leq x^{q-2\delta}$. This implies

[TABLE]

and (5) follows.

Conversely, let $\delta>0$ be such that (5) is true. Then by Theorem 1 and Theorem 4

[TABLE]

∎

The definition of $\mathcal{I}_{<q}$-convergence immediately yields

Corollary 2

Let $0<q\leq 1$, $\varepsilon>0$, $L$ and $x_{n}$ be real numbers for all $n\in\mathbb{N}$, and $A_{\varepsilon}=\{n:|x_{n}-L|\geq\varepsilon\}$. Then $\mathop{\mathcal{I}_{<q}\text{\rm--lim}}\limits x_{n}=L$ if and only if for every $\varepsilon>0$ there exists $\delta>0$ such that

[TABLE]

As an application of the above results, we will show that an important number-theoretic set belongs to the smallest element of (3), namely $\mathcal{I}_{0}$:

Lemma 2

Given $k\in\mathbb{N}$, and arbitrary primes $p_{1}<p_{2}<\cdots<p_{k}$, denote

[TABLE]

Then

[TABLE]

Proof

For a number $x\geq 2$ denote

[TABLE]

Then by [11, p.37, Exercise 15] we have

[TABLE]

From this, by Theorem 4 for $q=0$ we get

[TABLE]

4 On $\mathcal{I}_{<q}$- and $\mathcal{I}_{\leq q}$-convergence of arithmetic functions

First we recall some arithmetic functions, which we will investigate with respect to $\mathcal{I}_{<q}$- and $\mathcal{I}_{\leq q}$-convergence. We refer to the papers 2 , 6 , 10 , 12 , 14 , 16 , 17 , 18 for definitions and properties of these functions.

Let $n=p_{1}^{\alpha_{1}}\cdot p_{2}^{\alpha_{2}}\cdots p_{k}^{\alpha_{k}}$ be the canonical representation of $n\in\mathbb{N}$. Define

• $\bullet$

$\omega(n)$ - the number of distinct prime factors of $n$ (i.e. $\omega(n)=k$),

• $\bullet$

$\Omega(n)$ - the number of prime factors of $n$ counted with multiplicities (i.e. $\Omega(n)=\alpha_{1}+\dotsb+\alpha_{k}$),

• $\bullet$

for $n>1$,

[TABLE]

and $h(1)=1$, $H(1)=1$,

• $\bullet$

$f(n)=\prod_{d\mid n}d$, and $f^{*}(n)=\frac{1}{n}f(n)$,

• $\bullet$

$a_{p}(n)$ as follows: $a_{p}(1)=0$, and if $n>0$, then $a_{p}(n)$ is the unique integer $j\geq 0$ satisfying $p^{j}\mid n$, but $p^{j+1}\nmid n$ i. e., $p^{a_{p}(n)}\|n$.

• $\bullet$

$\gamma(n)$ - the number of all representations of a natural number $n$ in the form $n=a^{b}$, where $a,b$ are positive integers (see 10 ). Let

[TABLE]

be all such representations of a given $n$, where $a_{i},b_{i}\in\mathbb{N}$.

• $\bullet$

for $n>1$,

[TABLE]

• $\bullet$

$N(n)$ - the number of times the positive integer $n$ occurs in Pascal’s triangle (see 1 and 17 ).

Recall that $\mathcal{I}_{c}^{(q)}$-convergence of the following sequences has been established in 2 , 3 , 5 :

• I.

For $0<q\leq 1$ we have $\mathop{\mathcal{I}_{c}^{(q)}\text{\rm--lim}}\limits\frac{h(n)}{\log n}=0$ (see (2, , Th.8)),

• II.

Only for $q=1$ we have $\mathop{\mathcal{I}_{c}^{(q)}\text{\rm--lim}}\limits\frac{H(n)}{\log n}=0$ (see (2, , Th.10, Th.11)),

• III.

For a prime number $p$ the sequence \big{(}(\log p)\frac{a_{p}(n)}{\log n}\big{)}_{n=2}^{\infty} is $\mathcal{I}_{c}^{(q)}$-convergent to [math] only for $q=1$ (see (3, , Th.2.3)),

• IV.

For $q>\frac{1}{2}$ we have $\mathop{\mathcal{I}_{c}^{(q)}\text{\rm--lim}}\limits\gamma(n)=1$, and for $0<q\leq\frac{1}{2}$ the sequence $\gamma(n)$ is not $\mathcal{I}_{c}^{(q)}$-convergent (see (3, , Cor.3.5)),

• V.

For $q>\frac{1}{2}$ we have $\mathop{\mathcal{I}_{c}^{(q)}\text{\rm--lim}}\limits\tau(n)=1$, and for $0<q\leq\frac{1}{2}$ the sequence $\tau(n)$ is not $\mathcal{I}_{c}^{(q)}$-convergent (see (3, , Cor.3.8)),

• VI.

For $q>\frac{1}{2}$ we have $\mathop{\mathcal{I}_{c}^{(q)}\text{\rm--lim}}\limits N(n)=2$, and for $0<q\leq\frac{1}{2}$ the sequence \big{(}N(n)\big{)}_{t=1}^{\infty} is not $\mathcal{I}_{c}^{(q)}$-convergent (see (5, , Th.2.2)),

• VII.

The sequences \big{(}\frac{\omega(n)}{\log\log n}\big{)}_{n=2}^{\infty} and \big{(}\frac{\Omega(n)}{\log\log n}\big{)}_{n=2}^{\infty} are not $\mathcal{I}_{c}^{(q)}$-convergent for all $0<q\leq 1$ (see (2, , Th.12)),

• VIII.

The sequences \big{(}\frac{\log\log f(n)}{\log\log n}\big{)} and \big{(}\frac{\log\log f^{*}(n)}{\log\log n}\big{)} are not $\mathcal{I}_{c}^{(q)}$-convergent for all $0<q\leq 1$ (see (2, , Th.13, Th.14)).

In what follows, we will improve and sharpen all the statements I–VIII via the best convergences one can obtain from the ideals in (3) that are within $\mathcal{I}_{<q}$, $\mathcal{I}_{\leq q}$.

The next theorem, which is readily implied by Theorem 3 and (2, , Th.8), gives statement I. using Theorem 1 and Lemma 1. We will, however, provide another simpler proof based on Lemma 2:

Theorem 6

We have

[TABLE]

Proof

Take a small $\varepsilon>0$, and the largest prime $p_{0}$ for which $\frac{1}{\log p_{0}}\geq\varepsilon$. Then $\frac{1}{\log p}<\varepsilon$ whenever $p>p_{0}$, so if $n\in\mathbb{N}$ is such that $p|n$ for some prime $p>p_{0}$, then $n\geq p^{h(n)}$. It follows that

[TABLE]

thus,

[TABLE]

This implies $A_{\varepsilon}\subset D(2,3,5,\dots,p_{0})$, so, by Lemma 2 and the hereditary property, $A_{\varepsilon}\in\mathcal{I}_{0}$. ∎

Statement II. has the following strengthening:

Theorem 7

We have

[TABLE]

Proof

Let $0<\varepsilon<\frac{1}{\log 2}$. Then, according to (2), we have

[TABLE]

We will show that $A_{\varepsilon}\in\mathcal{I}_{<1}$: every positive integer $n$ can be uniquely represented as $n=ab^{2}$, where $a$ is a square-free number. Hence $H(a)=1$ and $H(n)\in\{H(b^{2}),H(b^{2})+1\}$. For any $n\in\mathbb{N}$ we have $n=p_{1}^{a_{1}}\cdots p_{k}^{a_{k}}\geq 2^{H(n)}$ and from this

[TABLE]

If $n\in A_{\varepsilon}$ then for $n=ab^{2}$ we get

[TABLE]

thus,

[TABLE]

Furthermore, if $n\in B$, then

[TABLE]

which is equivalent to

[TABLE]

therefore,

[TABLE]

If $n\in B$, and $n=ab^{2}\leq x$ for $x\geq 2$, then $a\leq 2x^{1-\varepsilon\log 2}$ and $b\leq\sqrt{\frac{x}{a}}$. Consequently,

[TABLE]

hence, for $x\geq 2$, we have

[TABLE]

Using $q=1$ and arbitrary $\delta\in(0,\varepsilon\frac{\log 2}{2})$ in Theorem 5, the above estimate gives $A_{\varepsilon}\in\mathcal{I}_{<1}$. ∎

The next result strengthens statement III:

Theorem 8

For any prime number $p$, we have

[TABLE]

Proof

Let $0<\varepsilon<1$. Then, according to (2), we have

[TABLE]

We have

[TABLE]

Clearly, $A_{\varepsilon}^{i}\cap A_{\varepsilon}^{j}=\emptyset$ for $i\neq j$, and if $n\in A_{\varepsilon}^{i}$, then

[TABLE]

This implies, in case $x\geq 2$, that

[TABLE]

hence,

[TABLE]

Using $q=1-\varepsilon$ in Theorem 4 and using Theorem 1, the above estimate gives

[TABLE]

∎

The statements IV., V., VI. are consequences of the following:

Theorem 9

We have

• i)

$\mathop{\mathcal{I}_{\leq\frac{1}{2}}\text{\rm--lim}}\limits\gamma(n)=1.$**

• ii)

$\mathop{\mathcal{I}_{\leq\frac{1}{2}}\text{\rm--lim}}\limits\tau(n)=1.$**

• iii)

$\mathop{\mathcal{I}_{\leq\frac{1}{2}}\text{\rm--lim}}\limits N(n)=2.$**

Proof

i) Let $0<\varepsilon<1$. Then, according to (2), we have $A_{\varepsilon}=\{n\in\mathbb{N}:|\gamma(n)-1|\geq\varepsilon\}$. Clearly,

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Given some $x\in\mathbb{N}$, $x\geq 2^{2}$, there is a k\in\mathbb{N}\mathbin{\mathchoice{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{\displaystyle\smallsetminus}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.8pt}{\rotatebox[origin={c}]{-20.0}{\textstyle\smallsetminus}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.6pt}{\rotatebox[origin={c}]{-20.0}{\scriptstyle\smallsetminus}}\mspace{-4.0mu}}{\mspace{-4.0mu}\raisebox{0.45pt}{\rotatebox[origin={c}]{-20.0}{\scriptscriptstyle\smallsetminus}}\mspace{-4.0mu}}}\{1\} with $2^{k}\leq x<2^{k+1}$. Then $k\leq\frac{\log x}{\log 2}$, and

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thus, for all $x\geq 4$,

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For $q=\frac{1}{2}$ in Theorem 4, we get $A_{\varepsilon}\in\mathcal{I}_{\leq\frac{1}{2}}$.

ii) Similar to i).

iii) Let $0<\varepsilon<1$. Then, according to (2), we have $A_{\varepsilon}=\{n\in\mathbb{N}:|N(n)-2|\geq\varepsilon\}$. If we take $H=\{1,2\}\cup M$, where $M=\{n\in\mathbb{N}:N(n)>2\}$, then $A_{\varepsilon}\subset H$. It has been proved in 1 , that $M(x)=O(\sqrt{x})$, thus, there is a $c>0$ so that for all $x\geq 2$,

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It now follows, by Theorem 4, that $A_{\varepsilon}\in\mathcal{I}_{\leq\frac{1}{2}}$. ∎

Remark 1

We note, that the set $\mathcal{I}_{d}$ containing all subsets of $\mathbb{N}$ with zero asymptotic density forms an admissible ideal. The corresponding $\mathcal{I}_{d}$-convergence is the wellknown statistical convergence. The following results were proved in 16 and 15 :

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We note, that $\mathcal{I}_{c}^{(1)}\subsetneq\mathcal{I}_{d}$.

If $\mathop{\mathcal{I}_{c}^{(q)}\text{\rm--lim}}\limits x_{n}=L$ is false for every $0<q\leq 1$, then $(x_{n})$ does not $\mathcal{I}_{<q}$-converge for any $q$, so $A_{\varepsilon}=\{n\in\mathbb{N}:|x_{n}-L|\geq\varepsilon\}\notin\mathcal{I}_{<q}$ whenever $0<q\leq 1$; thus, $\lambda(A_{\varepsilon})=1$ is the only option. Then by VII. and VIII. it follows that for all $\varepsilon>0$ and for every $n$, $a_{n}\in\{\omega(n),\Omega(n)\}$, and $b_{n}\in\{f(n),f^{*}(n)\}$ we have

• i)

\lambda\Big{(}\big{\{}n\in\mathbb{N}:\big{|}\frac{a_{n}}{\log\log n}-1\big{|}\geq\varepsilon\big{\}}\Big{)}=1,**

• ii)

\lambda\Big{(}\big{\{}n\in\mathbb{N}:\big{|}\frac{\log\log b_{n}}{\log\log n}-(1+\log 2)\big{|}\geq\varepsilon\big{\}}\Big{)}=1.**

As a consequence, say of i) for $a_{n}=\omega(n)$, we have that if

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then

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