UPPER BOUNDS FOR INFINITARY HARMONIC NUMBERS AND INFINITARY HARMONIOUS TUPLES

ELCHIN HASANALIZADE

doi:10.1017/S0004972726101324

UPPER BOUNDS FOR INFINITARY HARMONIC NUMBERS AND INFINITARY HARMONIOUS TUPLES

Part of: Elementary number theory

Published online by Cambridge University Press: 17 June 2026

ELCHIN HASANALIZADE

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ELCHIN HASANALIZADE*: Affiliation:
School of IT and Engineering, ADA University, Ahmadbey Aghaoghlu str. 61, AZ 1008 Baku, Azerbaijan
*: e-mail: ehasanalizade@ada.edu.az

Article contents

Abstract
Introduction
Preliminaries
Proofs of the theorems
Footnotes
References

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Abstract

A positive integer N is said to be an infinitary harmonic number (IHN) if the harmonic mean of its infinitary divisors is integral. It is still an open problem whether or not there exist infinitely many IHNs. Cohen and Hagis [‘Infinitary harmonic numbers’, Bull. Aust. Math. Soc. 41 (1989), 151–158] showed that there exist at most finitely many IHNs with a fixed number of I-components. We give an explicit Nielsen-type upper bound for the number of IHNs. We also consider a class of general infinitary amicable numbers and obtain the Borho-type upper bound for them.

Keywords

infinitary harmonic numbers infinitary amicable numbers infinitary harmonious tuples

MSC classification

Primary: 11A25: Arithmetic functions; related numbers; inversion formulas

Information

Type: Research Article
Information: Bulletin of the Australian Mathematical Society , Volume 114 , Issue 2 , October 2026 , pp. 221 - 227

DOI: https://doi.org/10.1017/S0004972726101324 [Opens in a new window]
Creative Commons: This is an Open Access article, distributed under the terms of the Creative Commons Attribution-NonCommercial-ShareAlike licence (https://creativecommons.org/licenses/by-nc-sa/4.0/), which permits non-commercial re-use, distribution, and reproduction in any medium, provided the same Creative Commons licence is included and the original work is properly cited. The written permission of Cambridge University Press must be obtained for commercial re-use.
Copyright: © The Author(s), 2026. Published by Cambridge University Press on behalf of Australian Mathematical Publishing Association Inc.

1 Introduction

Let $d(n)=\sum _{d\mid n}1$ and $\sigma (n)=\sum _{d\mid n}d$ denote respectively the number and the sum of the positive divisors of a positive integer n. Ore [Reference Ore7] studied the set of numbers S for which the ratio $H(n)={nd(n)}/{\sigma (n)}$ is an integer. Such numbers are called harmonic numbers. Note that the set of perfect numbers is a subset of S.

Subsequently, this notion was generalised to the other types of divisor functions. Hagis and Lord [Reference Hagis and Lord4] introduced the unitary analogue of the harmonic numbers. A positive integer d is called a unitary divisor of N if $d\mid N$ and $(d,N/d)=1$ . A positive integer N is said to be a unitary harmonic number (UHN) if the harmonic mean of the unitary divisors of N, given by

$$ \begin{align*} H^*(n)=\frac{Nd^*(N)}{\sigma^*(N)}=2^{\omega(N)}\prod_{p^a\mid\mid N}\frac{p^a}{p^a+1}, \end{align*} $$

is an integer. Here, $\omega (N)$ is the number of distinct prime factors of N.

Let

$$ \begin{align*} I=\{p^{2^\alpha}\mid p \text{ is prime and} \ \alpha\ge0\}. \end{align*} $$

From the fact that the binary representation of a natural number is unique, it follows that every $N>1$ can be written in exactly one way (up to the order of the factors) as the product of distinct elements from I. Each element of I in this product is called an I-component of N.

We denote the number of I-components of N by $J=J(N)$ . Clearly, $J(N)$ is an additive function given by $J(1)=0$ and $J(p^y)=\sum _{y_i=1}1$ if $y=\sum _{i=0}^{\infty }y_i2^i$ .

A divisor d of N is said to be an I-divisor if every I-component of d is also an I-component of N. Let $d_{\infty }(N)$ and $\sigma _{\infty }(N)$ denote respectively the number and the sum of the I-divisors of N. It is easy to see that $d_{\infty }(N)$ and $\sigma _{\infty }(N)$ are multiplicative functions defined by

$$ \begin{align*} d_{\infty}(N)=\prod\limits_{p^y\mid\mid N}2^{J(p^y)}=2^{J(N)} \quad\text{and}\quad \sigma_{\infty}(N)=\prod\limits_{p^y\mid\mid N}\prod\limits_{y_i=1}(1+p^{2^i}) \quad\text{where}\ y=\sum\limits_{i=0}^{\infty}y_i2^i.\end{align*} $$

Hagis and Cohen [Reference Hagis and Cohen3] defined N to be an infinitary harmonic number (IHN) if the infinitary harmonic mean of N, defined by

$$ \begin{align*} H_{\infty}(n)=\frac{Nd_{\infty}(N)}{\sigma_{\infty}(N)}=2^{J(N)}\prod_{p^a\mid\mid N}\prod_{y_i=1}\frac{p^{2^{i}}}{p^{2^i}+1}, \end{align*} $$

is an integer. They showed that there exist at most finitely many IHNs with a specified number of I-components and only finitely many with a specified infinitary harmonic mean. In our first theorem, we establish an explicit upper bound for the number of IHNs which is similar to Nielsen’s bound for odd perfect numbers [Reference Nielsen6]. (See also a recent paper by Stone [Reference Stone8].)

Theorem 1.1. Suppose that N is an IHN with $J=J(N)\, I$ -components and $H_{\infty }(N)=c$ . Then:

(a) $N\le c^{c^2}$ ;
(b) $N\le (2^{2^J-1})^J$ .

For an integer $k\ge 2$ , we call a tuple of k positive integers $(M_i)_{i=1}^k$ an infinitary amicable k-tuple if the equation

$$ \begin{align*} \sigma_{\infty}(M_1)=\cdots=\sigma_{\infty}(M_k)=M_1+\cdots+M_k \end{align*} $$

is satisfied. This is a generalisation of infinitary amicable pairs, defined by Cohen [Reference Cohen2]. We also introduce a new class of tuples of integers.

For an integer $k\ge 2$ , we say a tuple of positive integers $(M_i)_{i=1}^k$ is an infinitary harmonious k-tuple if the equation

(1.1) $$ \begin{align} \frac{M_1}{\sigma_{\infty}(M_1)}+\cdots+\frac{M_k}{\sigma_{\infty}(M_k)}=1 \end{align} $$ StartLayout 1st Row StartFraction upper M 1 Over sigma Subscript infinity Baseline left parenthesis upper M 1 right parenthesis EndFraction plus midline horizontal ellipsis plus StartFraction upper M Subscript k Baseline Over sigma Subscript infinity Baseline left parenthesis upper M Subscript k Baseline right parenthesis EndFraction equals 1 EndLayout

holds. Note that if $(M_i)_{i=1}^k$ is an infinitary amicable tuple, then

$$ \begin{align*} 1=\frac{M_1}{M_1+\cdots+M_k}+\cdots+\frac{M_k}{M_1+\cdots+M_k}=\frac{M_1}{\sigma_{\infty}(M_1)}+\cdots+\frac{M_k}{\sigma_{\infty}(M_k)}, \end{align*} $$

so every infinitary amicable tuple is infinitary harmonious.

Borho [Reference Borho1] proved an upper bound for amicable pairs $(M,N)$ in terms of $\Omega (MN)$ , the number of prime factors of $MN$ counted with multiplicity. In our next theorem, we give an upper bound for the number of infinitary harmonious k-tuples which is analogous to Borho’s bound.

Theorem 1.2. For any infinitary harmonious k-tuple $(M_i)_{i=1}^k$ ,

$$ \begin{align*} M_1\cdots M_k\le k^{-k}(2^{2^L}-2^{2^{L-1}}) \quad\text{where}\ L=J(M_1)+\cdots+J(M_k). \end{align*} $$

2 Preliminaries

In this section, we give some lemmas that are useful for the proof of the main results. In the first lemma due to Hagis and Cohen [Reference Hagis and Cohen3], we collect some elementary results concerning $H_{\infty }(N)$ and IHNs.

Lemma 2.1.

(a) Let $J=J(N)$ . If $N>1$ , then
(2.1) $$ \begin{align} \frac{2^{J+1}}{J+2}\le H_{\infty}(N)<2^J. \end{align} $$
(b) If N is an odd $IHN$ , then $H_{\infty }(N)$ is odd.
(c) If N is an odd $IHN$ , $p^y\mid \mid N$ and $p=4m+3$ , then y is even.

We also need an upper bound for the minimum and maximum I-components of an IHN.

Lemma 2.2. Let $p^a$ and $r^c$ respectively be the minimum and maximum I-components of an $IHN$ . Then,

$$ \begin{align*} p^a\le \frac{J\cdot H_{\infty}(N)}{2^J-H_{\infty}(N)} ,\quad r^c\ge \frac{(J-1)\cdot2^J+H_{\infty}(N)}{2^J-H_{\infty}(N)}. \end{align*} $$

Proof. We have

$$ \begin{align*} H_{\infty}(N)\ge2^J\frac{p^a}{p^a+1}\frac{p^a+1}{p^a+2}\cdots\frac{p^a+J-1}{p^a+J}=\frac{2^Jp^a}{p^a+J} \end{align*} $$

and

$$ \begin{align*} H_{\infty}(N)\le2^J\frac{r^c-J+1}{r^c-J+2}\frac{r^c-J+2}{r^c-J+3}\cdots\frac{r^c}{r^c+1}=\frac{2^J(r^c-J+1)}{r^c+1}. \\[-39pt] \end{align*} $$

The next lemma identifies IHNs with a few selected values of J and $H_{\infty }(N)$ .

Lemma 2.3. Let N be an $IHN$ with $J=J(N)\, I$ -components and let $H_{\infty }(N)=c$ .

(a) If $J\le 3$ , then $N\in \{1,6,45,60,90,15925\}$ .
(b) If $c\le 5$ , then $N\in \{1,6,45,60,90\}$ .

Proof. (a) Let $T_{k,\ell }$ denote the set of IHNs which have $k\, I$ -components and exactly $\ell $ prime factors. Clearly, $T_{k,1}$ is empty for every $k\ge 1$ . If $N\in T_{2,2}$ , then $N=p^aq^b$ , where $q>p$ are primes and $a,b$ are powers of $2$ . In this case, $H_{\infty }(N)=H^*(N)$ , where $H^*(N)$ is the harmonic mean of the unitary divisors of N and, thus, N is a unitary harmonic number (UHN). Therefore, $T_{2,2}=\{6,45\}$ .

If $N\in T_{3,2}$ , then from Lemma 2.1(a), $H_{\infty }(N)=4,5,6$ or $7$ . We consider here the cases $H_{\infty }(N)=6$ and $7$ ; the other cases can be treated in a similar way. If $H_{\infty }(N)=6$ , then N is even, $3\mid N$ and by Lemma 2.2, $p^a=4,5,7$ . If $p^a=4,5$ or $7$ , then N has three prime factors. If $H_{\infty }(N)=7$ , then $7\mid N$ . Assume first that $2\mid N$ . Then, $N=2^{2^{a^\prime }+2^a}7^{2^b}$ or $N=2^{2^a}7^{2^{b^\prime }+2^b}$ . Hence, N must satisfy either

$$ \begin{align*} (2^{2^{a^\prime}}+1)(2^{2^a}+1)(7^{2^b}+1)=2^{2^{a^\prime}+2^a+3}7^{2^b-1} \end{align*} $$

$$ \begin{align*} (2^{2^a}+1)(7^{2^b}+1)(7^{2^{b^\prime}}+1)=2^{2^a+3}7^{2^{b^\prime}+2^b-1}. \end{align*} $$

If $m\ge 1$ , then any odd prime divisor of the generalised Fermat number $7^{2^m}+1$ is of the form $k\cdot 2^{m+1}+1$ and must divide the right-hand side of the above equations, which is impossible. Now, assume that N is odd. Then, using Lemma 2.1(c) and Lemma 2.2, $7^2\mid N$ and $p^a=9,13,17$ . If $p^a=9$ or $17$ , then N has three prime factors. If $p^a=13$ , then $N=7^{2^b}13^{2^a+1}$ or $N=7^{2^c+2^d}13$ , where $a\ge 1$ , $b\ge 1$ and $d>c\ge 1$ . In the former case, $H_{\infty }(N)>H_{\infty }(13^37^2)>7$ and in the latter case, $H_{\infty }(N)>H_{\infty }(13\cdot 7^6)>7$ . By considering the remaining cases of $H_{\infty }(N)$ , one can show that no such IHN exists. Thus, $T_{3,2}$ is empty.

If $N\in T_{3,3}$ , then $N=p^{2^a}q^{2^b}r^{2^c}$ with $p<q<r$ primes and $H_{\infty }(N)=H^*(N)$ . Hence, N is a UHN and $T_{3,3}=\{60,90,15925\}$ .

(b) Note that if $J\ge 4$ , then from the inequality (2.1), we get $c>5$ . Since $H_{\infty }(15925)=7$ , the result follows from part (a).

The next two lemmas due to Goto [Reference Goto5] and Suzuki [Reference Suzuki9] respectively play an important role in the proofs of the upper bounds for IHNs and infinitary harmonious tuples.

Lemma 2.4. Let $r,a,b$ be positive integers, and let $x_1,\ldots ,x_r$ be integers such that $1\le x_1<x_2<\cdots <x_r$ and

$$ \begin{align*} 1\le\prod_{j=1}^{r-1} \bigg(1+\frac{1}{x_j}\bigg)< \frac{a}{b}\le\prod_{j=1}^{r} \bigg(1+\frac{1}{x_j}\bigg). \end{align*} $$

1 less than or equals product Underscript j equals 1 Overscript r minus 1 Endscripts left parenthesis 1 plus StartFraction 1 Over x Subscript j Baseline EndFraction right parenthesis less than StartFraction a Over b EndFraction less than or equals product Underscript j equals 1 Overscript r Endscripts left parenthesis 1 plus StartFraction 1 Over x Subscript j Baseline EndFraction right parenthesis period

Then,

$$ \begin{align*} \prod_{j=1}^r x_j\le (b+1)^{2^r-1}-(b+1)^{2^{r-1}-1} \end{align*} $$ product Underscript j equals 1 Overscript r Endscripts x Subscript j Baseline less than or equals left parenthesis b plus 1 right parenthesis Superscript 2 Super Superscript r Superscript minus 1 Baseline minus left parenthesis b plus 1 right parenthesis Superscript 2 Super Superscript r minus 1 Superscript minus 1

with equality if and only if $x_j=m_j$ , $1\le j\le r$ , where

$$ \begin{align*} m_i = \begin{cases} (b+1)^{2^{j-1}} & \text{for } j=1,\ldots,r-1,\\ (b+1)^{2^{j-1}}-1 & \text{for } j=r. \end{cases} \end{align*} $$

m Subscript i Baseline equals StartLayout Enlarged left brace 1st Row 1st Column left parenthesis b plus 1 right parenthesis Superscript 2 Super Superscript j minus 1 Baseline 2nd Column for j equals 1 comma ellipsis comma r minus 1 comma 2nd Row 1st Column left parenthesis b plus 1 right parenthesis Superscript 2 Super Superscript j minus 1 Baseline minus 1 2nd Column for j equals r period EndLayout

Lemma 2.5. Let k and R be positive integers. Consider a sequence of integers

$$ \begin{align*} \mathcal{M}=(m_j)^R_{j=1}, \quad 1<m_1\le m_2\le\cdots\le m_R, \end{align*} $$

a decomposition of the index set

$$ \begin{align*} \mathcal{J}=\{1,\ldots,R\}, \quad \mathcal{J}=\bigcup\limits_{i=1}^{k}\mathcal{J}_i \quad\text{where } \mathcal{J}_1,\ldots, \mathcal{J}_k \text{ are disjoint}, \end{align*} $$ script upper J equals StartSet 1 comma ellipsis comma upper R EndSet comma script upper J equals union Underscript i equals 1 Overscript k Endscripts script upper J Subscript i Baseline where script upper J 1 comma ellipsis comma script upper J Subscript k Baseline are disjoint comma

and a tuple of integers

$$ \begin{align*} a_1,\ldots,a_k,b_1,\ldots,b_k\ge1 \end{align*} $$

satisfying $a_i\ge b_i$ for all i. If a pair of inequalities

$$ \begin{align*} \sum_{i=1}^k\frac{b_i}{a_i}\prod\limits_{\substack{j=1 \\ j\in\mathcal{J}_i}}^R\bigg(1-\frac{1}{m_j}\bigg)\le1 \quad\text{and}\quad \sum_{i=1}^k\frac{b_i}{a_i}\prod\limits_{\substack{j=1 \\ j\in\mathcal{J}_i}}^{R-1}\bigg(1-\frac{1}{m_j}\bigg)>1 \end{align*} $$

sigma summation Underscript i equals 1 Overscript k Endscripts StartFraction b Subscript i Baseline Over a Subscript i Baseline EndFraction product Underscript StartLayout 1st Row j equals 1 2nd Row j element of script upper J Subscript i Baseline EndLayout Overscript upper R Endscripts left parenthesis 1 minus StartFraction 1 Over m Subscript j Baseline EndFraction right parenthesis less than or equals 1 and sigma summation Underscript i equals 1 Overscript k Endscripts StartFraction b Subscript i Baseline Over a Subscript i Baseline EndFraction product Underscript StartLayout 1st Row j equals 1 2nd Row j element of script upper J Subscript i Baseline EndLayout Overscript upper R minus 1 Endscripts left parenthesis 1 minus StartFraction 1 Over m Subscript j Baseline EndFraction right parenthesis greater than 1

holds, then

$$ \begin{align*} a\prod_{j=1}^Rm_j\le(a+1)^{2^R}-(a+1)^{2^{R-1}} \quad\text{where } a=a_1\cdots a_k. \end{align*} $$

a product Underscript j equals 1 Overscript upper R Endscripts m Subscript j Baseline less than or equals left parenthesis a plus 1 right parenthesis Superscript 2 Super Superscript upper R Superscript Baseline minus left parenthesis a plus 1 right parenthesis Superscript 2 Super Superscript upper R minus 1 Superscript Baseline where a equals a 1 midline horizontal ellipsis a Subscript k Baseline period

3 Proofs of the theorems

Proof of Theorem 1.1

We follow closely the argument of Goto [Reference Goto5]. Observe that

$$ \begin{align*} \frac{\sigma_{\infty}(N)}{N}=\prod_{p^y\mid\mid N}\prod_{y_i=1}\bigg(1+\frac{1}{p^{2^i}}\bigg)=\frac{d_{\infty}(N)}{H_{\infty}(N)}=\frac{2^J}{c}. \end{align*} $$

We arrange the $J\, I$ -components of N in ascending order. Applying Lemma 2.4 with $x_{\alpha }=p^{2^\alpha }$ , $a=2^J$ , $b=c$ and $r=J$ ,

(3.1) $$ \begin{align} N=\prod_{p^y\mid\mid N}\prod_{y_i=1}p^{2^i}\le (c+1)^{2^J-1}-(c+1)^{2^{J-1}-1}<(c+1)^{2^J-1}. \end{align} $$

From Lemma 2.3, the required inequalities hold if $J\le 3$ or $c\le 5$ . Thus, we may assume that $J\ge 4$ and $c\ge 6$ . From the inequality (2.1), $c+1\le 2^J$ . Combining this with inequality (3.1) gives part (b).

We now consider part (a). From $J\ge 4$ , we easily get $J+2<2^{(2J+3)/4}$ . Thus, Lemma 2.1(a) implies that

$$ \begin{align*} c\ge\frac{2^{J+1}}{J+2}>2^{(2J+1)/4}. \end{align*} $$ c greater than or equals StartFraction 2 Superscript upper J plus 1 Baseline Over upper J plus 2 EndFraction greater than 2 Superscript left parenthesis 2 upper J plus 1 right parenthesis divided by 4 Baseline period

Therefore, $2^J<c^2/\sqrt {2}$ . From $c\ge 6$ , we easily deduce that $c+1<c^{\sqrt {2}}$ . Hence, it follows that

$$ \begin{align*} N<(c+1)^{2^J}<(c+1)^{c^2/\sqrt{2}}<c^{c^2}, \end{align*} $$

as desired.

Proof of Theorem 1.2

Let $(M_i)_{i=1}^k$ be an infinitary harmonious tuple. Using

$$ \begin{align*} \frac{M}{\sigma_{\infty}(M)}=\prod_{p^y\mid\mid M}\prod_{y_i=1}\frac{p^{2^i}}{1+p^{2^i}}=\prod_{p^y\mid\mid M}\prod_{y_i=1}\bigg(1-\frac{1}{1+p^{2^i}}\bigg), \end{align*} $$

we can rewrite (1.1) as

$$ \begin{align*} \sum_{j=1}^k\prod_{p^y\mid\mid M_j}\prod_{y_i=1}\bigg(1-\frac{1}{1+p^{2^i}}\bigg)=1. \end{align*} $$ sigma summation Underscript j equals 1 Overscript k Endscripts product Underscript p Superscript y Baseline vertical bar vertical bar upper M Subscript j Baseline Endscripts product Underscript y Subscript i Baseline equals 1 Endscripts left parenthesis 1 minus StartFraction 1 Over 1 plus p Superscript 2 Super Superscript i Superscript Baseline EndFraction right parenthesis equals 1 period

sigma summation Underscript j equals 1 Overscript k Endscripts product Underscript p Superscript y Baseline vertical bar vertical bar upper M Subscript j Baseline Endscripts product Underscript y Subscript i Baseline equals 1 Endscripts left parenthesis 1 minus StartFraction 1 Over 1 plus p Superscript 2 Super Superscript i Superscript Baseline EndFraction right parenthesis equals 1 period

Arranging the I-components of $M_1,\ldots ,M_k$ in ascending order and applying Lemma 2.5 with $m_{\alpha }=1+p^{2^{\alpha }}$ , $a_i=b_i=1$ , we see that

(3.2) $$ \begin{align} \sigma_{\infty}(M_1)\cdots\sigma_{\infty}(M_k)=\prod_{j=1}^k\prod_{p^y\mid\mid M_j}\prod_{y_i=1}(1+p^{2^i})\le2^{2^L}-2^{2^{L-1}}, \end{align} $$

StartLayout 1st Row sigma Subscript infinity Baseline left parenthesis upper M 1 right parenthesis midline horizontal ellipsis sigma Subscript infinity Baseline left parenthesis upper M Subscript k Baseline right parenthesis equals product Underscript j equals 1 Overscript k Endscripts product Underscript p Superscript y Baseline vertical bar vertical bar upper M Subscript j Baseline Endscripts product Underscript y Subscript i Baseline equals 1 Endscripts left parenthesis 1 plus p Superscript 2 Super Superscript i Superscript Baseline right parenthesis less than or equals 2 Superscript 2 Super Superscript upper L Superscript Baseline minus 2 Superscript 2 Super Superscript upper L minus 1 Superscript Baseline comma EndLayout

where L is given by

$$ \begin{align*} L=\sum_{j=1}^k\sum_{p^y\mid\mid M_j}\sum_{y_i=1}1=J(M_1)+\cdots+J(M_k). \end{align*} $$

StartLayout 1st Row upper L equals sigma summation Underscript j equals 1 Overscript k Endscripts sigma summation Underscript p Superscript y Baseline vertical bar vertical bar upper M Subscript j Baseline Endscripts sigma summation Underscript y Subscript i Baseline equals 1 Endscripts 1 equals upper J left parenthesis upper M 1 right parenthesis plus midline horizontal ellipsis plus upper J left parenthesis upper M Subscript k Baseline right parenthesis period EndLayout

By using the AM-GM inequality in (1.1), we have

$$ \begin{align*} \frac{1}{k}\ge\sqrt[k]{\frac{M_1}{\sigma_{\infty}(M_1)}\cdots\frac{M_k}{\sigma_{\infty}(M_k)}} \end{align*} $$

StartLayout 1st Row StartFraction 1 Over k EndFraction greater than or equals RootIndex k StartRoot StartFraction upper M 1 Over sigma Subscript infinity Baseline left parenthesis upper M 1 right parenthesis EndFraction midline horizontal ellipsis StartFraction upper M Subscript k Baseline Over sigma Subscript infinity Baseline left parenthesis upper M Subscript k Baseline right parenthesis EndFraction EndRoot EndLayout

$$ \begin{align*} \frac{\sigma_{\infty}(M_1)\cdots\sigma_{\infty}(M_k)}{M_1\cdots M_k}\ge k^k. \end{align*} $$ StartLayout 1st Row ContinuedFraction sigma Subscript infinity Baseline left parenthesis upper M 1 right parenthesis midline horizontal ellipsis sigma Subscript infinity Baseline left parenthesis upper M Subscript k Baseline right parenthesis Over upper M 1 midline horizontal ellipsis upper M Subscript k Baseline greater than or equals k Superscript k Baseline period EndLayout

Multiplying both sides by $M_1\cdots M_k$ , we obtain

$$ \begin{align*} \sigma_{\infty}(M_1)\cdots\sigma_{\infty}(M_k)\ge k^kM_1\cdots M_k. \end{align*} $$

Combining this with (3.2), gives the required assertion.

Footnotes

This research was supported by ADA University Faculty Research and Development Funds.

References

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Article contents

UPPER BOUNDS FOR INFINITARY HARMONIC NUMBERS AND INFINITARY HARMONIOUS TUPLES

Abstract

Keywords

MSC classification

Information

1 Introduction

2 Preliminaries

3 Proofs of the theorems

Proof of Theorem 1.1

Proof of Theorem 1.2

Footnotes

References

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