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

Abstract

Let $\mathbb N$ be the set of positive integers, and denote by

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

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

$$\begin{aligned} \mathcal I_c^{(q)}=\{A\subset \mathbb N: \sum _{a\in A}a^{-q}<\infty \}. \end{aligned}$$

In this note we sharpen the results of Baláž et al. from (J Number Theory 183:74–83, 2018) 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 ${\mathcal {I}}_{\le q}$-convergence, for which new methods and criteria are developed.

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

$$\begin{aligned} \lambda (A)=\inf \Big \{t>0:\sum _{a\in A}\frac{1}{a^t}<\infty \Big \}. \end{aligned}$$

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 [11, p.26, Exercises 113, 114] that the range of $\lambda$ is the interval [0, 1], moreover, for $A=\{a_1<a_2<\!\cdots<a_n<\dots \}\subset \mathbb N$,

$$\begin{aligned} \lambda (A)=\limsup _{n\rightarrow \infty }\frac{\log n}{\log a_n}. \end{aligned}$$

It is easy to see that $\lambda$ is monotonic, i.e. $\lambda (A)\le \lambda (B)$ 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:

$$\begin{aligned} {\mathcal {I}}_{<q}&=\{A\subset \mathbb N:\lambda (A)<q\}, \ \text {if} \ 0<q\le 1, \\ {\mathcal {I}}_{\le q}&=\{A\subset \mathbb N:\lambda (A)\le q\}, \ \text {if} \ 0\le q\le 1, \ \text {and} \\ \mathcal I_0&=\{A\subset \mathbb N:\lambda (A)=0\}. \end{aligned}$$

Clearly, $\mathcal I_{\le 0}=\mathcal I_0$, and $\mathcal I_{\le 1}=2^\mathbb N$. Since $\lambda (A)=0$ when $A\subset \mathbb N$ is finite, then $\mathcal I_f=\{A\subset \mathbb N: \text {A is finite}\}\subset \mathcal I_0$, moreover, also considering the well-known set

$$\begin{aligned} \mathcal I_c^{(q)}=\Big \{A\subset \mathbb N: \sum _{a\in A}\frac{1}{a^q}<\infty \Big \} \end{aligned}$$

we get that whenever $0<q<q'<1$,

$$\begin{aligned} \mathcal I_f\subset \mathcal I_0\subset {\mathcal {I}}_{<q}\subset \mathcal I_c^{(q)}\subset {\mathcal {I}}_{\le q}\subset \mathcal I_{<q'}. \end{aligned}$$

(1)

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 $\mathop {\mathcal I\text {-lim}}\limits x_n=L$, if for each $\varepsilon >0$ the set

$$\begin{aligned} A_\varepsilon =\{n:|x_n - L|\ge \varepsilon \} \end{aligned}$$

(2)

belongs to the ideal $\mathcal I$. One can see, e.g., [6, 7] for a general treatment of $\mathcal I$-convergence. A useful property is as follows:

Lemma 1.1

[7] If $\mathcal I_1\subset \mathcal I_2$, then $\mathop {\mathcal I_1\text {-lim}}\limits x_n=L$ implies $\mathop {\mathcal I_2\text {-lim}}\limits x_n=L$.

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

In this paper, we embed the ideals ${\mathcal {I}}_{<q}$ and ${\mathcal {I}}_{\le q}$ into the structure of ideals $\mathcal I_c^{(q)}$. We show that these 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 ${\mathcal {I}}_{\le q}$-convergence of sequences.

2 On ideals enveloping the ideal $\mathcal I_c^{(q)}$

Theorem 2.1

Let $0<q<q'< 1$. Then

$$\begin{aligned} \mathcal I_0\subsetneq {\mathcal {I}}_{<q}\subsetneq \mathcal I_c^{(q)}\subsetneq {\mathcal {I}}_{\le q}\subsetneq \mathcal I_{<q'}\subsetneq \mathcal I_c^{(q')}\subsetneq \mathcal I_{\le q'} \subsetneq \mathcal I_{<1}\subsetneq \mathcal I_c^{(1)}\subsetneq \mathcal I_{\le 1}=2^\mathbb N. \end{aligned}$$

(3)

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. $\mathcal I_0\ne {\mathcal {I}}_{<q}$: let $0<s<q<1$, and take the set $A=\{a_1<a_2<\cdots \}\subset \mathbb N$, where for all $n\in \mathbb N$,

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

Then $a_n=n^\frac{1}{s}-\varepsilon (n)$ for some $0\le \varepsilon (n)<1$, 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

$$\begin{aligned} \frac{\log n}{\log a_n} =\frac{\log n}{\frac{1}{s}.\log n+\log \Big (1-\frac{\varepsilon (n)}{n^\frac{1}{s}}\Big )}\rightarrow s,\quad \text {if} \ n\rightarrow \infty , \end{aligned}$$

then $0<\lambda (A)=s<q$; thus, $A\in {\mathcal {I}}_{<q} \ \mathcal I_0$. It is also clear that ${\mathcal {I}}_{<q} \ \mathcal I_0$ is infinite, since for any $k\in \mathbb N$ the sets $A_k=\{ka_n: \ n\in \mathbb N\}$ satisfy

$$\begin{aligned} \lambda (A_k)=\limsup _{n\rightarrow \infty }\frac{\log n}{\log ka_n}=\lambda (A). \end{aligned}$$

Case 2. ${\mathcal {I}}_{<q}\ne \mathcal I_c^{(q)}$: let $0<q<1$, and take the set $A=\{a_1<a_2<\cdots \}\subset \mathbb N$, where for all $n\in \mathbb N$,

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

One can easily show that $(a_n)$ is increasing sequence, and,

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

On the other hand

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\log n}{\log a_n} =\lim _{n\rightarrow \infty } \frac{\log n}{\log (n^\frac{1}{q}\log ^\frac{2}{q} (n+1))} =\lim _{n\rightarrow \infty } \frac{\log n}{\frac{1}{q}\log n +\frac{2}{q}\log \log (n+1)} = q, \end{aligned}$$

hence, $\lambda (A)=q$. Similarly to Case 1 we can see that $\mathcal I_c^{(q)} \ {\mathcal {I}}_{<q}$ is actually infinite.

Case 3. $\mathcal I_c^{(q)}\ne {\mathcal {I}}_{\le q}$: 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

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

so $A\notin \mathcal I_c^{(q)}$, but $A\in {\mathcal {I}}_{\le q}$, since $\lambda (A)=q$. Analogously to Case 1, one can show that ${\mathcal {I}}_{\le q} \ \mathcal I_c^{(q)}$ is infinite.

Case 4. ${\mathcal {I}}_{\le q}\ne \mathcal I_{<q'}$: 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 $0<q<s<q'$. Then $\lambda (A)=s$, so $A\in \mathcal I_{<q'}$, however, $A\notin {\mathcal {I}}_{\le q}$. Moreover, again, $\mathcal I_{<q'} \ {\mathcal {I}}_{\le q}$ is infinite. $\square$

By (3), it is worth noting 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 $A\in {\mathcal {I}}_{\le q}\subset \mathcal I_c^{(q')}$ for every $q'>q$. This view is important, since in what follows, we will establish criteria for ${\mathcal {I}}_{<q}$, ${\mathcal {I}}_{\le q}$ membership, respectively.

Theorem 2.2

Let $0<q\le 1$. Then each of the sets $\mathcal I_0$, ${\mathcal {I}}_{<q}$, ${\mathcal {I}}_{\le q}$ forms an admissible ideal, except for $\mathcal I_{\le 1}$.

Proof

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

Theorem 2.3

We have

$$\begin{aligned} \mathcal I_0=\bigcap _{0<q\le 1} {\mathcal {I}}_{<q}=\bigcap _{0<q\le 1}{\mathcal {I}}_{\le q}, \end{aligned}$$

hence,

$$\begin{aligned} \mathcal I_0=\bigcap _{0<q\le 1}\mathcal I_c^{(q)}. \end{aligned}$$

Proof

Follows from the definitions of $\mathcal I_0$, ${\mathcal {I}}_{<q}$, ${\mathcal {I}}_{\le q}$, and (3). $\square$

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

Given $x\ge 1$, define the counting function of $A\subset \mathbb N$ by

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

Theorem 3.1

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

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{A(x)}{x^{q+\delta }}= 0 . \end{aligned}$$

(4)

Proof

Let $A=\{a_1<a_2<\dots \}$, and $A\in {\mathcal {I}}_{\le q}$. Then

$$\begin{aligned} \lambda (A) =\limsup _{n\rightarrow \infty }\frac{\log n}{\log a_n}\le q, \end{aligned}$$

so for any $\delta > 0$ there is an $n_0\in \mathbb N$ so that, for all $n\ge n_0$,

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

If x is sufficiently large, we can find $n\ge n_0$ with $a_n\le x<a_{n+1}$, hence, $A(x)=n\le x^{q+\frac{\delta }{2}}$. Consequently,

$$\begin{aligned} 0\le \frac{A(x)}{x^{q+\delta }} \le \frac{x^{q+\frac{\delta }{2}}}{x^{q+\delta }}=\frac{1}{x^{\frac{\delta }{2}}}\rightarrow 0, \quad \text {as} \ x\rightarrow \infty , \end{aligned}$$

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

Conversely, let $\delta >0$, and (4) be true for some $A=\{a_1<a_2<\dots \}$. Then

$$\begin{aligned} \frac{A(a_n)}{a_n ^{q+\delta }}\rightarrow 0 \quad \text {as} \ n\rightarrow \infty , \end{aligned}$$

so there is an $n_1\in \mathbb N$ such that for all $n\ge n_1$, $n\le a_n ^{q+\delta }$, thus,

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

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

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

Corollary 3.2

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

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{A_\varepsilon (x)}{x^{q+\delta }}= 0 . \end{aligned}$$

Theorem 3.3

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

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{A(x)}{x^{q-\delta }}= 0 . \end{aligned}$$

(5)

Proof

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

$$\begin{aligned} \lambda (A) =\limsup _{n\rightarrow \infty }\frac{\log n}{\log a_n}< q, \quad \text {where} \ A=\{a_1<a_2<\dots \}. \end{aligned}$$

For each $\delta >0$ with $0<\delta <\frac{1}{2}(q-\lambda (A))$ there is an $n_0\in \mathbb N$ so that for all $n\ge n_0$,

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

hence, for all $n\ge n_0$,

$$\begin{aligned} A(a_n)=n\le a_n ^{q-2\delta }. \end{aligned}$$

If x is large enough, there exists some $n\ge n_0$ with $a_n\le x<a_{n+1}$, so $A(x)=n\le x^{q-2\delta }$. This implies

$$\begin{aligned} 0\le \frac{A(x)}{x^{q-\delta }} \le \frac{x^{q-2\delta }}{x^{q-\delta }}=\frac{1}{x^{\delta }}\rightarrow 0 \quad \text {as} \ x\rightarrow \infty , \end{aligned}$$

and (5) follows.

Conversely, let $\delta >0$ be such that (5) is true. Then by Theorems  2.1 and  3.1 we have

$$\begin{aligned} A\in \mathcal I_{\le q-\delta }\subset I_{< q}. \end{aligned}$$

$\square$

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

Corollary 3.4

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

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{A_\varepsilon (x)}{x^{q-\delta }}= 0 . \end{aligned}$$

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 3.5

Given $k\in \mathbb N$, and arbitrary primes $p_1<p_2<\cdots <p_k$, denote

$$\begin{aligned} D(p_1,p_2\dots ,p_k)=\{n\in \mathbb N: n=p_1^{\alpha _1}p_2^{\alpha _2}\cdots p_k^{\alpha _k}, \alpha _i\ge 0, i=1,2,\dots , k\}\,. \end{aligned}$$

Then

$$\begin{aligned} D(p_1,p_2\dots ,p_k)\in \mathcal I_0\,. \end{aligned}$$

Proof

For a number $x\ge 2$ denote

$$\begin{aligned} D(p_1,p_2\dots ,p_k)(x)=\#\{n\le x:n\in D(p_1,p_2\dots ,p_k)\} . \end{aligned}$$

Then by [9, p.37, Exercise 15] we have

$$\begin{aligned} D(p_1,p_2\dots ,p_k)(x)\le \prod _{i=1}^{k} \Big (\frac{\log x}{\log p_i}+1 \Big )\le \Big (\frac{2}{\log 2}\log x\Big )^k\,. \end{aligned}$$

From this, by Theorem 3.1 for $q=0$ we get

$$\begin{aligned} D(p_1,p_2\dots ,p_k)\in \mathcal I_0. \end{aligned}$$

$\square$

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

First we recall some arithmetic functions, which we will investigate with respect to ${\mathcal {I}}_{<q}$- and ${\mathcal {I}}_{\le q}$-convergence. We refer to the papers [2, 5, 8, 10, 12, 14,15,16] 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 the following functions:

• $\omega (n)$ is the number of distinct prime factors of n (i.e. $\omega (n)=k$);

• $\Omega (n)$ is the number of prime factors of n counted with multiplicities (i.e. $\Omega (n)=\alpha _{1}+\cdots +\alpha _{k}$);

• for $n>1$,

  $$\begin{aligned} h(n)=\min _{1\le j \le k}\alpha _{j},\quad H(n)=\max _{1\le j \le k}\alpha _{j} \end{aligned}$$

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

• $f(n)=\prod _{d\mid n} d$ and $f^{*}(n)=\frac{1}{n}f(n)$;

• $a_{p}(n)$ as follows: $a_{p}(1)=0$ and $a_{p}(n)$ is the unique integer $j\ge 0$ satisfying $p^{j}\mid n$, but $p^{j+1}\not \mid n$, i.e. $p^{a_{p}(n)} \parallel n$, for $n>1$;

• $\gamma (n)$ is the number of all representations of a natural number n in the form $n=a^b$, where a, b are positive integers (see [8]). Let

  $$\begin{aligned} n=a_1^{b_1}=a_2^{b_2}=\dots =a_{\gamma (n)}^{b_{\gamma (n)}} \end{aligned}$$

  be all such representations of a given n, where $a_i, b_i\in \mathbb N$;

• for $n>1$,

  $$\begin{aligned} \tau (n)= b_1+b_2+\dots +b_{\gamma (n)}; \end{aligned}$$

• N(n) is the number of times the positive integer n occurs in Pascal’s triangle (see [1] and [15]).

Recall that $\mathcal I_c^{(q)}$-convergence of the following sequences has been established in [2,3,4]:

• I. For $0<q\le 1$ we have $\mathop {\mathcal I_c^{(q)}\text {-lim}}\limits \frac{h(n)}{\log n} = 0$ (see [2], [ Th.8]).

• II. Only for $q=1$ we have $\mathop {\mathcal I_c^{(q)}\text {-lim}}\limits \frac{H(n)}{\log n} = 0$ (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 0 only for $q=1$ (see [3],[Th.2.3]),

• IV. For $q>\frac{1}{2}$ we have $\mathop {\mathcal I_c^{(q)}\text {-lim}}\limits \gamma (n) = 1$, and for $0<q\le \frac{1}{2}$ the sequence $\gamma (n)$ is not $\mathcal I_c^{(q)}$-convergent (see [3], [Cor.3.5]),

• V. For $q>\frac{1}{2}$ we have $\mathop {\mathcal I_c^{(q)}\text {-lim}}\limits \tau (n) = 1$, and for $0<q\le \frac{1}{2}$ the sequence $\tau (n)$ is not $\mathcal I_c^{(q)}$-convergent (see [3], [Cor.3.8]),

• VI. For $q>\frac{1}{2}$ we have $\mathop {\mathcal I_c^{(q)}\text {-lim}}\limits N(n) = 2$, and for $0<q\le \frac{1}{2}$ the sequence $\big (N(n)\big )_{t=1}^{\infty }$ is not $\mathcal I_c^{(q)}$-convergent (see [4], [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 $0<q\le 1$ (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 $0<q\le 1$ (see [2], [Th.13, Th.14]).

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

The next theorem, which is readily implied by Theorem  2.3 and [2], [Th.8], gives Statement I using Theorem  2.1 and Lemma  1.1. We will, however, provide another simpler proof based on Lemma  3.5:

Theorem 4.1

We have

$$\begin{aligned} \mathop {\mathcal I_0\text {-lim}}\limits \frac{h(n)}{\log n}=0\,. \end{aligned}$$

Proof

Take a small $\varepsilon >0$, and the largest prime $p_0$ for which $\frac{1}{\log p_0}\ge \varepsilon$. 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 $n\ge p^{h(n)}$. It follows that

$$\begin{aligned} \frac{h(n)}{\log n}\le \frac{h(n)}{\log p^{h(n)}}=\frac{1}{\log p}<\varepsilon \,, \end{aligned}$$

thus,

$$\begin{aligned} n\notin \Big \{k\in \mathbb N:\frac{h(k)}{\log k}\ge \varepsilon \Big \}=\Big \{k\in \mathbb N:\Big |\frac{h(k)}{\log k}-0\Big |\ge \varepsilon \Big \}=A_\varepsilon . \end{aligned}$$

This implies $A_\varepsilon \subset D(2,3,5,\dots , p_0)$, so, by Lemma 3.5 and the hereditary property, $A_\varepsilon \in \mathcal I_0$. $\square$

Statement II has the following strengthening:

Theorem 4.2

We have

$$\begin{aligned} \mathop {\mathcal I_{<1}\text {-lim}}\limits \frac{H(n)}{\log n}=0\,. \end{aligned}$$

Proof

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

$$\begin{aligned} A_\varepsilon =\left\{ n\in \mathbb N: \frac{H(n)}{\log n}\ge \varepsilon \right\} . \end{aligned}$$

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 $n=p_1^{a_1}\cdots p_k^{a_k} \ge 2^{H(n)}$ and from this

$$\begin{aligned} H(n)\le \frac{\log n}{\log 2}. \end{aligned}$$

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

$$\begin{aligned} \log n=\log (ab^2)\le \frac{H(ab^2)}{\varepsilon }\le \frac{H(b^2)+1}{\varepsilon }\le \frac{\log b^2}{\varepsilon \log 2}+\frac{1}{\varepsilon }, \end{aligned}$$

thus,

$$\begin{aligned} A_\varepsilon \subseteq B=\Big \{n\in \mathbb N: n=ab^2,\ \log ab^2 \le \frac{\log b^2}{\varepsilon \log 2}+\frac{1}{\varepsilon },\ \ a,b\in \mathbb N\Big \}. \end{aligned}$$

Furthermore, if $n\in B$, then

$$\begin{aligned} \log a\le \frac{1-\varepsilon \log 2}{\varepsilon \log 2}\log b^2 +\frac{1}{\varepsilon }, \end{aligned}$$

which is equivalent to

$$\begin{aligned} { a^\frac{\varepsilon \log 2}{1-\varepsilon \log 2}\le b^2 e^\frac{\log 2}{1-\varepsilon \log 2} },\quad \text {and so } { \quad a^\frac{1}{1-\varepsilon \log 2}\le a b^2 e^{\frac{\log 2}{1-\varepsilon \log 2}} }, \end{aligned}$$

therefore,

$$\begin{aligned} B=\left\{ n\in \mathbb N: n=ab^2 \text { and } a\le 2n^{1-\varepsilon \log 2} \right\} . \end{aligned}$$

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

$$\begin{aligned} B(x)\le \sum _{a<2x^{1-\varepsilon \log 2}} \sqrt{\frac{x}{a}}=\sqrt{x}\sum _{a<2x^{1-\varepsilon \log 2}}\frac{1}{\sqrt{a}}\le \sqrt{x} \bigg (1+\int _{1}^{ 2x^{1-\varepsilon \log 2}} \frac{1}{\sqrt{t}} \mathrm dt \bigg ) \\ \le \sqrt{x}\Big (1+2\big (\sqrt{2x^{1-\varepsilon \log 2}}-1\big )\Big )\le 2\sqrt{2} x^{1-\varepsilon \frac{\log 2}{2}}, \end{aligned}$$

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

$$\begin{aligned} A_\varepsilon (x)\le 2\sqrt{2} x^{1-\varepsilon \frac{\log 2}{2}}. \end{aligned}$$

Using $q= 1$ and arbitrary $\delta \in (0,\varepsilon \frac{\log 2}{2})$ in Theorem  3.3, the above estimate gives $A_\varepsilon \in \mathcal I_{<1}$. $\square$

The next result strengthens statement III.

Theorem 4.3

For any prime number p, we have

$$\begin{aligned} \mathop {\mathcal I_{<1}\text {-lim}}\limits (\log p)\frac{a_p (n)}{\log n}=0\,. \end{aligned}$$

Proof

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

$$\begin{aligned} A_\varepsilon =\{n>1: (\log p)\frac{a_p (n)}{\log n}\ge \varepsilon \}. \end{aligned}$$

We have

$$\begin{aligned} A_\varepsilon =\bigcup _{i=0}^{\infty } A_\varepsilon ^{i}, \end{aligned}$$

where

$$\begin{aligned} A_\varepsilon ^{i} = \{n\in A_\varepsilon : n=p^i u \quad \text {where} \ p\not \mid u \} \ (i=0, 1,2 \dots ). \end{aligned}$$

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

$$\begin{aligned} (\log p)\frac{a_p (n)}{\log n}=(\log p)\frac{i}{i\log p +\log u}\ge \varepsilon , \quad \text {thus, } u\le p^{i(\frac{1-\varepsilon }{\varepsilon })}. \end{aligned}$$

In case $x\ge 2$, this implies that

$$\begin{aligned} A_\varepsilon ^{i} (x)\le \# \left\{ u:u^{\frac{\varepsilon }{1-\varepsilon }}u\le x \right\} =\# \left\{ u:u^{\frac{1}{1-\varepsilon }}\le x \right\} \le x^{1-\varepsilon }, \end{aligned}$$

hence

$$\begin{aligned} A_\varepsilon (x)=\sum _{i: p^i\le x} A_\varepsilon ^{i} (x)\le \frac{\log x}{\log p } x^{1-\varepsilon }. \end{aligned}$$

Using $q= 1-\varepsilon$ in Theorem  3.1 and using Theorem  2.1, the above estimate gives

$$\begin{aligned} A_\varepsilon \in \mathcal I_{\le 1-\varepsilon } \subset \mathcal I_{<1}. \end{aligned}$$

$\square$

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

Theorem 4.4

We have

• (i) $\mathop {\mathcal I_{\le \frac{1}{2}}\text {-lim}}\limits \gamma (n) = 1.$

• (ii) $\mathop {\mathcal I_{\le \frac{1}{2}}\text {-lim}}\limits \tau (n) = 1.$

• (iii) $\mathop {\mathcal I_{\le \frac{1}{2}}\text {-lim}}\limits N(n) = 2.$

Proof

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

$$\begin{aligned} A_\varepsilon \subseteq H=\big \{a^b: a, b\in \mathbb N \ \{1\}\big \}=\bigcup _{k=2}^\infty \big \{n^k:n=2,3,\dots \big \}. \end{aligned}$$

Given some $x\in \mathbb N$, $x\ge 2^2$, there is a $k\in \mathbb N \ \{1\}$ with $2^k\le x <2^{k+1}$. Then $k\le \frac{\log x}{\log 2}$, and

$$\begin{aligned} H(x)\le \sum _{n=2}^{k} \root n \of {x}\le \sqrt{x} \frac{\log x}{\log 2}, \end{aligned}$$

thus, for all $x\ge 4$,

$$\begin{aligned} A_\varepsilon (x)\le \frac{\log x}{\log 2} x^{\frac{1}{2}}. \end{aligned}$$

For $q=\frac{1}{2}$ in Theorem  3.1, we get $A_\varepsilon \in \mathcal I_{\le \frac{1}{2} }$.

(ii) Similar to i).

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

$$\begin{aligned} A_\varepsilon (x)\le H(x)\le cx^{\frac{1}{2}}. \end{aligned}$$

By Theorem  3.1, $A_\varepsilon \in \mathcal I_{\le \frac{1}{2} }$ follows. $\square$

Remark 4.5

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 [14] and [13]:

$$\begin{aligned} \mathop {\mathcal I_d\text {-lim}}\limits \frac{\omega (n)}{\log \log n} =\mathop {\mathcal I_d\text {-lim}}\limits \frac{\Omega (n)}{\log \log n} =1, \end{aligned}$$

$$\begin{aligned} \mathop {\mathcal I_d\text {-lim}}\limits \frac{\log \log f(n)}{\log \log n} =\mathop {\mathcal I_d\text {-lim}}\limits \frac{\log \log f^*(n)}{\log \log n} =1+\log 2. \end{aligned}$$

We note that $\mathcal I_c^{(1)}\subsetneq \mathcal I_d$.

If $\mathop {\mathcal I_c^{(q)}\text {-lim}}\limits x_n =L$ is false for every $0<q\le 1$, then $(x_n)$ does not ${\mathcal {I}}_{<q}$-converge for any q, so $A_\varepsilon =\{n\in \mathbb N: |x_n - L|\ge \varepsilon \}\notin {\mathcal {I}}_{<q}$ whenever $0<q\le 1$; thus, $\lambda (A_\varepsilon ) = 1$ is the only option. Then by Statements 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) $\lambda \Big (\big \{n\in \mathbb N: \big |\frac{a_n}{\log \log n}-1\big |\ge \varepsilon \big \}\Big )=1 ,$

• (ii) $\lambda \Big (\big \{n\in \mathbb N: \big |\frac{\log \log b_n}{\log \log n}-(1+\log 2)\big |\ge \varepsilon \big \}\Big )=1.$

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

$$\begin{aligned} \Big \{n\in \mathbb N: \Big |\frac{\omega (n)}{\log \log n}-1\Big |\ge \varepsilon \Big \} = \{n_1<n_2<\dots<n_k<\dots \}, \end{aligned}$$

then

$$\begin{aligned} \limsup _{k\rightarrow \infty } \frac{\log k}{\log n_k} =1. \end{aligned}$$