fourierlargesymbols147 \addtokomafontlabelinglabel

On Robin’s Inequality and the Kaneko-Lagarias Inequality

Idris Assani Department of Mathematics, The University of North Carolina at Chapel Hill, 120 E Cameron Avenue, CB 3250 Chapel Hill, NC 27599-3250, USA [email protected] https://idrisassani.web.unc.edu/ , Aiden Chester The University of North Carolina at Chapel Hill, 120 E Cameron Avenue, CB 3250 Chapel Hill, NC 27599-3250, USA [email protected] and Alex Paschal The University of North Carolina at Chapel Hill, 120 E Cameron Avenue, CB 3250 Chapel Hill, NC 27599-3250, USA [email protected]

Abstract.

We prove that Robin’s inequality and the Lagarias inequality hold for almost every number, including all numbers not divisible by one of the prime numbers $\{2,3,5\}$ , primorials, sufficiently big numbers of the form $2^{k}n$ for odd $n$ and $21$ -free integers. We also prove that the Kaneko-Lagarias inequality holds for all numbers if and only if it holds for all superabundant numbers.

Key words and phrases:

Robin’s inequality, Kaneko-Lagarias inequality

2020 Mathematics Subject Classification:

11N56, 11M26

1. Introduction

We denote by $\sigma(n)$ and $\varphi(n)$ the sum of divisors function and Euler’s totient function respectively. Robin’s inequality ([8]) states that the Riemann hypothesis is equivalent to the assertion that

(1)

\sigma(n)<e^{\gamma}n\log(\log(n))

for all $n>5040$ , where $\gamma$ denotes the Euler-Mascheroni constant. Similarly, the Lagarias inequality ([6]) states that the Riemann hypothesis is equivalent to the assertion that

(2)

\sigma(n)<H_{n}+\exp(H_{n})\log(H_{n})

for all $n>1$ , where $H_{n}$ denotes the $n$ -th harmonic number. Lagarias also published an equality that we call the Kaneko-Lagarias inequality (see tthe acknowledgements in [6]), which states that the Riemann hypothesis is equivalent to the assertion that

(3)

\sigma(n)<\exp(H_{n})\log(H_{n})

for all $n>60$ .

2. Robin’s Inequality

2.1. Sufficiently big numbers not divisible by one of the prime numbers 2,3,5

Let $p_{1}=2$ , $p_{2}=3$ , etc. be an enumeration of the prime numbers which we denote by $\mathbb{P}$ . Fix $j\in\mathbb{N}$ and let $q_{1}<q_{2}<\cdots<q_{k}$ be some prime numbers distinct from $p_{j}$ . Given $\alpha_{1},\alpha_{2},\dots,\alpha_{k}\in\mathbb{N}$ , let $n=q_{1}^{\alpha_{1}}q_{2}^{\alpha_{2}}\cdots q_{k}^{\alpha_{k}}$ .

Lemma 2.1.

We have

(4)

\frac{\sigma(n)}{n}<\prod_{\ell=1}^{k}\frac{q_{\ell}}{q_{\ell}-1}\leq\prod_{\vbox{\Let@\restore@math@cr\default@tag\halign{\hfil$\m@th\scriptstyle#$&$\m@th\scriptstyle{}#$\hfil\cr\ell=&1,\dots,j-1\\ &j+1,\ldots,k+1\crcr}}}\frac{p_{\ell}}{p_{\ell}-1}=\frac{n}{\varphi(n)}.

Proof.

The first inequality follows from the fact that for any $p\in\mathbb{P}$ and $\alpha\in\mathbb{N}$

(5)

\frac{\sigma(p^{\alpha})}{p^{\alpha}}=\frac{p-\frac{1}{p^{\alpha}}}{p-1}\nearrow\frac{p}{p-1}\text{ as }\alpha\to\infty.

The second inequality follows from the fact that $p_{i}\leq q_{i}$ for all $1\leq i\leq k$ . ∎

Note that

(6)

\frac{n}{\varphi(n)}=\left(\prod_{\vbox{\Let@\restore@math@cr\default@tag\halign{\hfil$\m@th\scriptstyle#$&$\m@th\scriptstyle{}#$\hfil\cr\ell=&1,\dots,j-1\\ &j+1,\ldots,k+1\crcr}}}\frac{p_{\ell}+1}{p_{\ell}}\right)\left(\prod_{\vbox{\Let@\restore@math@cr\default@tag\halign{\hfil$\m@th\scriptstyle#$&$\m@th\scriptstyle{}#$\hfil\cr\ell=&1,\dots,j-1\\ &j+1,\ldots,k+1\crcr}}}\frac{p_{\ell}^{2}}{p_{\ell}^{2}-1}\right)=:A(k)B(k).

We can bound $A(k)$ as follows:

(7)

\log(A(k))=\sum_{\vbox{\Let@\restore@math@cr\default@tag\halign{\hfil$\m@th\scriptstyle#$&$\m@th\scriptstyle{}#$\hfil\cr\ell=&1,\dots,j-1\\ &j+1,\ldots,k+1\crcr}}}\log\left(1+\frac{1}{p_{\ell}}\right)\leq\sum_{\vbox{\Let@\restore@math@cr\default@tag\halign{\hfil$\m@th\scriptstyle#$&$\m@th\scriptstyle{}#$\hfil\cr\ell=&1,\dots,j-1\\ &j+1,\ldots,k+1\crcr}}}\frac{1}{p_{\ell}}=\left(\sum_{\ell=1}^{k+1}\frac{1}{p_{\ell}}\right)-\frac{1}{p_{j}}

and

(8)

\sum_{\ell=1}^{k+1}\frac{1}{p_{\ell}}\leq\log(\log(p_{k+1}))+c_{1}+\frac{5}{\log(p_{k+1})},

where $c_{1}\approx.261497$ by Theorem 1.10 in [11]. Thus we obtain

Lemma 2.2.

For all $k\in\mathbb{N}$ ,

(9)

A(k)\leq\log(p_{k+1})\exp\left(c_{1}-\frac{1}{p_{j}}+\frac{5}{\log(p_{k+1})}\right).

Combining [4] and Theorem 3 from [9], we obtain the following:

Theorem 2.3.

For $k\geq 6$ ,

(10)

k(\log(k)+\log(\log(k))-1)<p_{k}<k(\log(k)+\log(\log(k))).

Furthermore, combining 2.2 and 2.3, we see that

Lemma 2.4.

For $k\geq 6$ , $A(k)<C(k)$ where

(11)		$\displaystyle C(k)$	$\displaystyle=\log((k+1)(\log(k+1)+\log(\log(k+1))))$
(11)			$\displaystyle\qquad\exp\left(c_{1}-\frac{1}{p_{j}}+\frac{5}{\log((k+1)(\log(k+1)+\log(\log(k+1))-1))}\right).$

Now, put $m=p_{k+1}\#/p_{j}$ . Our goal is to show the following, since it implies that Robin’s inequality for $n$ as above:

Theorem 2.5.

For any $j\in\{1,2,3\}$ , there exists a $K_{j}\in\mathbb{N}$ such that $k\geq K_{j}$ implies

(12)

C(k)B(k)<e^{\gamma}\log(\log(m)).

Corollary 2.6.

Suppose 2.5 holds. Then Robin’s inequality holds for $n$ as above.

Proof.

We calculate

(13)

\frac{\sigma(n)}{n}<\frac{n}{\varphi(n)}\leq A(k)B(k)<C(k)B(k)<e^{\gamma}\log(\log(m))\leq e^{\gamma}\log(\log(n)),

where the last inequality follows from the fact that $m\leq n$ . ∎

Definition 2.7.

The Chebyshev function is defined as follows:

(14)

\theta(x)=\sum_{p\in\mathbb{P},p\leq x}\log(p)=\log\left(\prod_{p\in\mathbb{P},p\leq x}p\right).

Theorem 2.8.

For $x\geq 529$ ,

(15)

\prod_{\begin{subarray}{c}p\in\mathbb{P}\\ p\leq x\end{subarray}}p=e^{\theta(x)}>e^{x\left(1-\frac{1}{2\log x}\right)}\geq(2.51)^{x}.

Proof.

The first inequality is given by (3.14) in [9] and the second follows from computations since the function $f(x)=1-\frac{1}{2\log x}$ increases for $x>1$ . ∎

Lemma 2.9.

For $k\geq 99$ ,

\log(\log(m))>\log((k+1)(\log(k+1)+\log(\log(k+1))-1)\log(2.51)-\log(p_{j}))=:D(k).

Proof.

Noting that $k\geq 99$ implies $p_{k+1}>529$ , we calculate

(16)	$\displaystyle\log(\log(m))$	$\displaystyle=\log\left(\log\left(\frac{p_{k+1}\#}{p_{j}}\right)\right)>\log\left(\log\left(\frac{(2.51)^{p_{k+1}}}{p_{j}}\right)\right)$
		$\displaystyle=\log(p_{k+1}\log(2.51)-\log(p_{j}))$
		$\displaystyle>\log((k+1)(\log(k+1)+\log(\log(k+1))-1)\log(2.51)-\log(p_{j})),$

where the last inequality uses 2.3. ∎

The following implies 2.5:

Proposition 2.10.

For $j\in\{1,2,3\}$ , there exists a $K_{j}\in\mathbb{N}$ such that $k\geq K_{j}$ implies

(17)

C(k)B(k)<e^{\gamma}D(k).

Proof.

Denote

(18)		$\displaystyle\widetilde{C}(k)=e^{-c_{1}+\frac{1}{p_{j}}}C(k)$	$\displaystyle=\log((k+1)(\log(k+1)+\log(\log(k+1))))$
(18)			$\displaystyle\qquad\exp\left(\frac{5}{\log((k+1)(\log(k+1)+\log(\log(k+1))-1))}\right)$

and

(19)

\widehat{C}(k)=\exp\left(\frac{5}{\log((k+1)(\log(k+1)+\log(\log(k+1))-1))}\right).

Multiplying both sides of (17) by $e^{-c_{1}+\frac{1}{p_{j}}}p_{j}^{2}/(p_{j}^{2}-1)$ , we obtain

(20)

\widetilde{C}(k)\prod_{\ell=1}^{k+1}\frac{p_{\ell}^{2}}{p_{\ell}^{2}-1}<\frac{e^{\gamma-c_{1}+\frac{1}{p_{j}}p_{j}^{2}}}{p_{j}^{2}-1}D(k).

Noting that

(21)

\prod_{\ell=1}^{k+1}\frac{p_{\ell}^{2}}{p_{\ell}^{2}-1}\nearrow\frac{\pi^{2}}{6}\text{ as }k\to\infty,

we see that (20) is implied by

(22)

\widetilde{C}(k)<\frac{6p_{j}^{2}e^{\gamma-c_{1}+\frac{1}{p_{j}}}}{\pi^{2}(p_{j}^{2}-1)}D(k)=:E_{j}D(k).

Raising both sides to the power of $e$ , we see that (22) is implied by

(23)			$\displaystyle[(k+1)(\log(k+1)+\log(\log(k+1))]^{\widehat{C}(k)}$
(23)			$\displaystyle<[(k+1)(\log(k+1)+\log(\log(k+1))-1)\log(2.51)-\log(p_{j})]^{E_{j}}.$

(23) is equivalent to

(24)		$\displaystyle 1$	$\displaystyle<[(k+1)(\log(k+1)+\log(\log(k+1)))]^{-\widehat{C}(k)+E_{j}}$
(24)			$\displaystyle\qquad\left[1-\frac{(k+1)\log(2.51)+\log(p_{j})}{(k+1)(\log(k+1)+\log(\log(k+1)))}\right]^{E_{j}}.$

Noting that $E_{j}>1$ for $j\in\{1,2,3\}$ , we see that there exists a $K_{j}\in\mathbb{N}$ such that $k\geq K_{j}$ implies $-\widehat{C}(k)+E_{j}>\epsilon$ for some $\epsilon\in(0,1)$ . If needed, we can increase $K_{j}$ so that $k\geq K_{j}$ implies

(25)

\left[1-\frac{(k+1)\log(2.51)+\log(p_{j})}{(k+1)(\log(k+1)+\log(\log(k+1)))}\right]^{E_{j}}>\epsilon,

and also so that $k\geq K_{j}$ implies

(26)

1<\epsilon[(k+1)(\log(k+1)+\log(\log(k+1))]^{\epsilon},

which implies (24). ∎

2.2. All numbers not divisible by one of the prime numbers 2,3 5

Letting $j=1$ in (22), we seek to show that

(27)

\widetilde{C}(k)<\frac{8e^{\gamma-c_{1}+.5}}{\pi^{2}}D(k).

Lemma 2.11.

For $k\geq 13042$ , $\widehat{C}(k)<1.525$ .

Proof.

$\widehat{C}(k)$ is decreasing, so the result follows from computation. ∎

Denote $f(k)=(k+1)(\log(k+1)+\log(\log(k+1))$ . Applying 2.11to (27) and performing some algebraic manipulations, our goal reduces to showing that

(28)

\log(f(k))<\frac{8e^{\gamma-c_{1}+.5}}{\pi^{2}(1.525)}\log((f(k)-1)\log(2.51)-\log(2)).

Raising both sides to the power of $e$ , this becomes

(29)

1<f(k)^{2.0166}\left[1-\frac{(k+1)\log(2.51)-\log(2)}{f(k)}\right]^{1.20166}.

The RHS of (29) is increasing, and a computation reveals that it holds for $k\geq 13042$ . Additionally, using 2.1, one can check that

(30)

\frac{\sigma(n)}{n}<\frac{n}{\varphi(n)}<e^{\gamma}\log(\log(m))

for $k\geq 3$ . Finally, when $k\in\{1,2\}$ , we check that

(31)

\frac{\sigma(n)}{n}<\frac{n}{\varphi(n)}\leq\frac{15}{8}<e^{\gamma}\log(\log(n))

for $n\geq 680$ . This confirms the following for $j=1$ :

Theorem 2.12.

For $j\in\{1,2,3\}$ , Robin’s inequality holds for every natural number $>5040$ which is not divisible by $p_{j}$ .

To confirm 2.12 when $j\in\{2,3\}$ , one can repeat the above process to see that, for sufficiently big $k$ , (22) is satisfied. The cases with smaller $k$ have been verified in [7].

2.3. Primorials and sufficiently big even numbers

We consider numbers of the form $2^{k}n$ for odd $n$ .

Fix $k\in\mathbb{N}$ and let $n$ be odd. We calculate

(32)

\frac{\sigma(2^{k}n)}{2^{k}n}=\frac{\sigma(2^{k})}{2^{k}}\frac{\sigma(n)}{n}<\frac{\sigma(2^{k})}{2^{k}}\frac{n}{\varphi(n)}=\frac{\sigma(2^{k})}{2^{k}}\frac{\varphi(2^{k})}{2^{k}}\frac{2^{k}n}{\varphi(2^{k}n)}=\left(1-\frac{1}{2^{k+1}}\right)\frac{2^{k}n}{\varphi(2^{k}n)}.

Applying Theorem 15 from [9], we know

(33)

\left(1-\frac{1}{2^{k+1}}\right)\frac{2^{k}n}{\varphi(2^{k}n)}<\left(1-\frac{1}{2^{k+1}}\right)\left(e^{\gamma}\log(\log(2^{k}n))+\frac{2.51}{\log(\log(2^{k}n))}\right).

We ask which $n$ satisfy

(34)

\left(1-\frac{1}{2^{k+1}}\right)\left(e^{\gamma}\log(\log(2^{k}n))+\frac{2.51}{\log(\log(2^{k}n))}\right)<e^{\gamma}(\log(\log(2^{k}n))).

This is equivalent to asking when

(35)

\frac{2.51(2^{k+1}-1)}{e^{\gamma}}<(\log(\log(2^{k}n)))^{2}

holds, which is when

(36)

n>\frac{e^{e^{\sqrt{\frac{2.51(2^{k+1}-1)}{e^{\gamma}}}}}}{2^{k}}=:b(k).

Thus, we obtain the following:

Theorem 2.13.

Given any $k\in\mathbb{N}$ , Robin’s inequality holds for all numbers of the form $2^{k}n$ when $n$ is odd and satisfies (36).

In particular, we have

Corollary 2.14.

If $n\geq 620$ is odd, then Robin’s inequality holds for $2n$ . Furthermore, Robin’s inequality holds for all primorials $>30$ .

Proof.

The first statement follows immediately from 2.13 and the second follows from the computation of primorials $<1240$ .3 ∎

2.4. All 21-free numbers

The results of the previous subsection, are based on the inequality in Theorem 15 from [9]. This inequality can be improved by using a better bound stated in [2]:

(37)

\frac{m}{\varphi(m)}<e^{\gamma}\log(\log(m))+\frac{.0168}{(\log(\log(m)))^{2}}.

for $m\geq 10^{10^{13.11485}}=C.$ Using the same reasoning as before, we derive the following result.

Theorem 2.15.

Given any $k$ natural number, Robin’s inequality holds for all numbers of the form $2^{k}n$ when $n$ is odd and satisfies

(38)

2^{k}n>e^{e^{\left(\frac{.0168(2^{k+1}-1)}{e^{\gamma}}\right)^{\frac{1}{3}}}}=:2^{k}\tilde{b}(k)

Furthermore, it was shown in [7] that Robin’s inequality holds for all natural numbers $5041<m\leq C.$ . We can thus conclude the following:

Theorem 2.16.

Robin’s inequality holds for all natural numbers of the form $2^{k}n$ with $n$ odd as long as $2^{k}\tilde{b}(k)<C.$ In particular, Robin’s inequality holds for all 21-free numbers.

Proof.

Let $k$ be a natural number and $n$ be an odd natural number.

•

if $5041<2^{k}n\leq 2^{k}\tilde{b}(k)<C$ then $2^{k}n$ satisfies Robin’s inequality by [7]
•

Alternatively, if $2^{k}n>2^{k}\tilde{b}(k)$ then $2^{k}n$ satisfies Robin’s inequality by 2.15.

Recalling that a $\ell$ -free number is a natural number not divisible by any $\ell$ power of a prime number greater than or equal to $2$ , we can see that if $2^{k}\tilde{b}(k)<C$ then all $(k+1)$ -free numbers satisfy Robin’s inequality. Since $\log(2^{20}\tilde{b}(20))<6(10^{11})<2.3(10^{13.11485}<\log(C)$ , we can conclude that Robin’s inequality holds for all 21-free numbers. ∎

Remark The validity of Robin’s inequality for $\ell$ -free numbers was proved for $\ell=7$ in [10], for $\ell=11$ in [3] and for $\ell=20$ in [7].

2.5. Almost every number

Definition 2.17.

The natural density of a set $E$ is

(39)

d(E)=\lim_{n\to\infty}\frac{\#E\cap\{1,2,\dots,n\}}{n}

when the limit exists.

Theorem 2.18.

Denote by $\mathcal{R}$ the set of numbers satisfying Robin’s inequality. Then the natural density of $\mathcal{R}$ is $1$ .

Proof.

We will prove that the natural density of $\mathcal{R}^{c}$ is $0$ . Fix $\epsilon>0$ . Let $E_{k}=\{2^{k}n:n\in\mathbb{N}_{\text{odd}},n\leq b(k)\}$ and note that $\mathcal{R}^{c}\subseteq\bigcup_{k\geq 1}E_{k}$ by 2.12 and 2.13.¹¹1Here $\mathbb{N}_{\text{odd}}:=\{1,3,\dots\}$ . Pick $M$ so that $\sum_{k=M+1}^{\infty}\frac{1}{2^{k}}<\frac{\epsilon}{2}$ . For $n\in\mathbb{N}$ we calculate

(40)		$\displaystyle\frac{\#\mathcal{R}^{c}\cap\{1,2,\dots,n\}}{n}$	$\displaystyle\leq\frac{\#\bigcup_{k\geq 1}E_{k}\cap\{1,2,\dots,n\}}{n}=\frac{\sum_{k\geq 1}\#E_{k}\cap\{1,2,\dots,n\}}{n}$
(40)			$\displaystyle=\frac{\sum_{k=1}^{M}\#E_{k}\cap\{1,2,\dots,n\}+\sum_{k=M+1}^{\infty}\#E_{k}\cap\{1,2,\dots,n\}}{n},$

where the first equality follows from the fact that the $E_{k}$ ’s are disjoint. Noting that $\sum_{k=1}^{M}\#E_{k}\cap\{1,2,\dots,n\}<\infty$ for all $n\in\mathbb{N}$ , we see that we can pick $N$ so that $n\geq N$ implies that the RHS of (40) is $<\epsilon$ , completing our proof. ∎

3. The Lagarias and Kaneko-Lagarias Inequalities

3.1. Superabundant numbers

Let $\Gamma(x)$ denote the gamma function. We define two functions:

(41)		$\displaystyle H(x)$	$\displaystyle=\int_{0}^{1}\frac{t^{x}-1}{t-1}\mathop{}\!\mathrm{d}t,$
(41)		$\displaystyle\psi(x)$	$\displaystyle=\frac{\Gamma^{\prime}(x)}{\Gamma(x)}.$

$\psi$ is known as the digamma function. One can verify that $H$ is smooth for $x\geq 1$ and that $H(n)=H_{n}$ for all $n\in\mathbb{N}$ . It’s also easy to see that $\psi$ , known as the digamma function, satisfies

(42)

H(x)=\psi(x+1)+\gamma.

Lemma 3.1.

For all $x\geq 1$ ,

(43)

H(x)<\log(x)+\gamma+\frac{1}{2x}.

Proof.

By (2.2) from [1],

(44)

\psi(x)<\log(x)-\frac{1}{2x}

for all $x\geq 1$ . Then we use (42) and $\psi(x+1)=\psi(x)+\frac{1}{x}$ to finish. ∎

Lemma 3.2.

For all $x\geq 4$ ,

(45)

H(x)<\frac{2\log(x)}{1+\frac{6}{\pi^{2}x}}.

Proof.

By 3.1, it suffices to show that

(46)

\log(x)+\gamma+\frac{1}{2x}<\frac{2\log(x)}{1+\frac{6}{\pi^{2}x}}

for $x\geq 4$ . By arithmetic manipulations, (46) becomes

(47)

\frac{1}{\pi^{2}x-6}\left(\gamma\pi^{2}x+\frac{\pi^{2}}{2}+6\gamma+\frac{3}{x}\right)<\log(x).

Computation reveals that (47) holds for $x=4$ , and the LHS of (47) is decreasing while the RHS is increasing so we obtain the result. ∎

Lemma 3.3.

The following hold:

(1)

For all $n>1$ , $H_{n+1}\leq\frac{n}{\log(n)}$ .
(2)

For all $x\geq 4$ , $\log(H(x))\leq\frac{x}{2\log(x)}$ .

Proof.

(a) We can manually verify the inequality for $n\leq 6$ . Noting that

(48)

H_{n+1}=\sum_{k=1}^{n+1}\frac{1}{k}\leq 1+\int_{1}^{n+1}\frac{\mathop{}\!\mathrm{d}t}{t}=1+\log(n+1),

it suffices to show that

(49)

\log(x)(\log(x+1)+1)\leq x.

Put $g(t)=e^{t}-t^{2}-t-1$ . We see that $g(2)>0$ and that $g^{\prime}(t)=e^{t}-2t-1>0$ for $t\geq 2$ , so $g(t)>0$ for $t\geq 2$ . For $x\geq e^{2}-1$ we have

(50)

0<g(\log(x+1))=x+1-(\log(x+1))^{2}-\log(x+1)-1<x-\log(x)(\log(x+1)+1).

(b) For $x\geq 4$ , note that the function mapping $x\mapsto\frac{x}{\log(x)}$ is increasing. If $n\leq x<n+1$ , then

(51)

H_{n}\leq H(x)<H_{n+1}\leq\frac{n}{\log(n)}\leq\frac{x}{\log(x)}.

For $y>2$ we see that $\log(y)<\frac{y}{2}$ , so let $y=H(x)$ and apply (51) finish. ∎

Lemma 3.4.

For $x\geq 4$ ,

(52)

H(x)\log(H(x))<\frac{x^{2}}{x+\frac{6}{\pi^{2}}}.

Proof.

Apply 3.1 and 3.3. ∎

Lemma 3.5.

For $x\geq 4$ ,

(53)

H^{\prime}(x)>\frac{H(x)\log(H(x))}{x^{2}}.

Proof.

We will use (51) from [5] which states that

(54)

\frac{1}{\psi^{\prime}(x)}\leq x+\frac{6}{\pi^{2}}-1

for $x\geq 1$ . We calculate

(55)

H^{\prime}(x)=\psi^{\prime}(x+1)\geq\frac{1}{x+6\pi^{2}}>\frac{H(x)\log(H(x))}{x^{2}},

where the equality follows from taking the derivative of (42) and the second inequality follows from 3.4. ∎

Proposition 3.6.

The function

(56)

g(x)=\frac{\exp(H(x))\log(H(x))}{x}

is increasing for $x\geq 4$ .

Proof.

We start with (3.5) from [6]:

(57)			$\displaystyle H_{n}=\log(n)+\gamma+\int_{n}^{\infty}\frac{x-\lfloor x\rfloor}{x^{2}}\mathop{}\!\mathrm{d}x$
			$\displaystyle\implies\exp(H_{n})=e^{\gamma}n\exp\left(\int_{n}^{\infty}\frac{x-\lfloor x\rfloor}{x^{2}}\mathop{}\!\mathrm{d}x\right)$
			$\displaystyle\implies\frac{\exp(H_{n})\log(H_{n})}{n}=e^{\gamma}\log(H_{n})\exp\left(\int_{n}^{\infty}\frac{x-\lfloor x\rfloor}{x^{2}}\mathop{}\!\mathrm{d}x\right).$

Given $k\in\mathbb{N}$ , put

(58)

g_{k}(x)=e^{\gamma}\log(H(x))\exp\left(\int_{x}^{k}\frac{t-\lfloor t\rfloor}{t^{2}}\mathop{}\!\mathrm{d}t\right)

so that $\lim_{k\to\infty}g_{k}(x)=g(x)$ . We compute

(59)

g_{k}^{\prime}(x)=e^{\gamma}\exp\left(\int_{x}^{k}\frac{t-\lfloor t\rfloor}{t^{2}}\mathop{}\!\mathrm{d}t\right)\left(\frac{H^{\prime}(x)}{H(x)}+\log(H(x))\left(-\frac{x-\lfloor x\rfloor}{x^{2}}\right)\right),

so $g^{\prime}_{k}(x)>0$ if and only if

(60)

\frac{H^{\prime}(x)}{H(x)}+\log(H(x))\left(-\frac{x-\lfloor x\rfloor}{x^{2}}\right)\geq\frac{H^{\prime}(x)}{H(x)}-\frac{\log(H(x))}{x^{2}}>0,

which is the content of 3.5. Thus, $g(x)$ is the limit of monotonically increasing functions and is therefore monotonically increasing. ∎

Corollary 3.7.

The sequence

(61)

\left\{\frac{\exp(H_{n})\log(H_{n})}{n}\right\}_{n=1}^{\infty}

is monotonically increasing.

Proof.

3.6 gives the result for $n\geq 4$ and we can manually check the smaller cases. ∎

Definition 3.8.

A number $n$ is superabundant if $\sigma(m)/m<\sigma(n)/n$ for all $m<n$ .

Theorem 3.9.

If there are counterexamples to the Kaneko-Lagarias inequality, the smallest such counterexample is a superabundant number.

Proof.

Suppose, for sake of contradiction, that $m$ is the smallest counterexample to the Kaneko-Lagarias inequality and that $m$ is not superabundant. Let $n$ be the greatest superabundant number $<m$ . We calculate,

(62)

\frac{\sigma(n)}{n}>\frac{\sigma(m)}{n}\geq\frac{\exp(H_{m})\log(H_{m})}{m}>\frac{\exp(H_{n})\log(H_{n})}{n},

so $n<m$ violates the Kaneko-Lagarias inequality: a contradiction. ∎

3.2. Connection to Robin’s inequality

Theorem 3.10.

If Robin’s inequality holds for some $n\in\mathbb{N}$ , then the Kaneko-Lagarias inequality holds for $n$ .

Proof.

We use the approximation

(63)

H_{n}\geq\log(n)+\gamma+\frac{1}{2n+1}

to calculate

(64)

\displaystyle\frac{\exp(H_{n})\log(H_{n})}{n}\geq\frac{e^{\gamma+\frac{1}{2n+1}}n\log\left(\log(n)+\gamma+\frac{1}{2n+1}\right)}{n}>e^{\gamma}\log(\log(n)),

which implies the result. ∎

Note that we obtain the same result for the Lagarias inequality.

Acknowledgments. The first author thanks Jeff Lagarias for his comments and the references he provided. The third author thanks Keith Briggs for providing his code used to compute superabundant numbers, Perry Thompson and Owen McAllister for their help implementing it in Rust and Jean-Louis Nicolas for sharing the paper [2].

References

[1] Horst Alzer “On Some Inequalities for the Gamma and Psi Functions” In Mathematics of Computation 66.217 American Mathematical Society, 1997, pp. 373–389
[2] Christian Axler and Jean-Louis Nicolas “Large values of $\frac{n}{\varphi(n)}$ ” In Acta Arithmetica 209, 2022, pp. 357–383
[3] Kevin Boughan and Tim Trudgian “Robin’s inequality for 11-free integers” In Integers 15, 2015
[4] Pierre Dusart “The $k$ -th Prime is Greater than $k(\log(k)()+\log(\log(k))-1)$ for $k\geq 2$ ” In Mathematics of Computation 68.225 American Mathematical Society, 1999, pp. 411–415
[5] M. R. Farhangdoost and M. Kargar Dolatabadi “New Inequalities for Gamma and Digamma Functions” In Journal of Applied Mathematics 2014, 2014, pp. 1–7
[6] Jeffrey C. Lagarias “An Elementary Problem Equivalent to the Riemann Hypothesis” In The American Mathematical Monthly 109.6 Mathematical Association of America, 2002, pp. 534–543
[7] Thomas Morrill and David Platt “Robin’s inequality for 20-free integers” In Integers 21, 2020
[8] Guy Robin “Grandes valeurs de la fonction somme des diviseurs et hypothèse de Riemann” In Journal de Mathématiques Pures et Appliquées 63, 1984, pp. 187–213
[9] J. Barkley Rosser and Lowell Schoenfeld “Approximate formulas for some functions of prime numbers” In Illinois Journal of Mathematics 6.1 Duke University Press, 1962, pp. 64 –94
[10] P. Sole and M. Planat “The Robin inequality for 7-free integers” In Integers 12, 2012, pp. 301–309
[11] Gérald Tenenbaum “Introduction to analytic and probabilistic number theory” 46, Cambridge Studies in Advanced Mathematics Cambridge University Press, Cambridge, 1995, pp. xvi+448