On the integer part of the reciprocal of the Riemann zeta function tail at certain rational numbers in the critical strip

TL;DR

This paper investigates the integer part of the reciprocal of the tail of the Riemann zeta function at specific rational points within the critical strip, providing explicit descriptions for these values.

Contribution

It establishes explicit formulas for the integer part of the reciprocal of the zeta tail at certain rational numbers, utilizing finiteness results on integral points of algebraic curves.

Findings

Explicit formulas for the integer part of 1/ζ(s) tail at s=1/p and s=2/p

Application of finiteness of integral points on algebraic curves

Advances understanding of zeta function behavior at rational points

Abstract

We prove that the integer part of the reciprocal of the tail of $\zeta(s)$ at a rational number $s=\frac{1}{p}$ for any integer with $p \geq 5$ or $s=\frac{2}{p}$ for any odd integer with $p \geq 5$ can be described essentially as the integer part of an explicit quantity corresponding to it. To deal with the case when $s=\frac{2}{p},$ we use a result on the finiteness of integral points of certain curves over $\mathbb{Q}$.

Equations

$$\zeta \left(\frac{1}{2}\right)=-1.46035\cdots ,\zeta \left(\frac{1}{3}\right)=-0.97336\cdots ,\zeta \left(\frac{1}{4}\right)=-0.813278\cdots ,$$

$$\zeta \left(\frac{1}{5}\right)=-0.733921\cdots ,\zeta \left(\frac{2}{3}\right)=-2.44758\cdots ,\zeta \left(\frac{2}{5}\right)=-1.1348\cdots .$$

$${\zeta }_n(s)$$

$$A_{n,s}$$

$$B_{n,s}$$

$${\zeta }_1(s)=\zeta (s)$$

$$\zeta_{n}(s)=\begin{cases}-\frac{1}{1-2^{1-s}}A_{n,s},~{}&~{}\mbox{if~{}$n$~{}is even},\\ -\frac{1}{1-2^{1-s}}B_{n,s},&~{}\mbox{if~{}$n$~{}is odd}.\end{cases}$$

$$\left[\frac{1}{1-2^{1-s}}\cdot {\zeta }_n(s{)}^{-1}\right]=\left[(-1{)}^{n+1}\cdot 2{\left(n-\frac{1}{2}\right)}^s\right]$$

$$2{\left(n-\frac{1}{2}\right)}^s<A_{n,s}^{-1}<2{\left(n-\frac{1}{4}\right)}^s$$

$$-2{\left(n-\frac{1}{4}\right)}^s<B_{n,s}^{-1}<-2{\left(n-\frac{1}{2}\right)}^s.$$

$$-2(1-2^{1-s}){\left(n-\frac{1}{2}\right)}^s<{\zeta }_n(s{)}^{-1}<-2(1-2^{1-s}){\left(n-\frac{1}{4}\right)}^s$$

$$2(1-2^{1-s}){\left(n-\frac{1}{4}\right)}^s<{\zeta }_n(s{)}^{-1}<2(1-2^{1-s}){\left(n-\frac{1}{2}\right)}^s$$

$$2{\left(n-\frac{1}{2}\right)}^s<A_{n,s}^{-1}<2{\left(n-\frac{1}{2}+ϵ\right)}^s$$

$$-2{\left(n-\frac{1}{2}+ϵ\right)}^s<B_{n,s}^{-1}\le -2{\left(n-\frac{1}{2}\right)}^s$$

$$-2(1-2^{1-s}){\left(n-\frac{1}{2}\right)}^s<{\zeta }_n(s{)}^{-1}<-2(1-2^{1-s}){\left(n-\frac{1}{2}+ϵ\right)}^s$$

$$2(1-2^{1-s}){\left(n-\frac{1}{2}+ϵ\right)}^s<{\zeta }_n(s{)}^{-1}<2(1-2^{1-s}){\left(n-\frac{1}{2}\right)}^s$$

$$\{P\in C(K)|f(P)\in R_S\}$$

$$2^{p}n-2^{p-1}<x^p<2^{p}n-2^{p-1}+\frac{1}{2},$$

$$[A_{n,s}^{-1}]=\left[2{\left(n-\frac{1}{2}\right)}^s\right]$$

$$[B_{n,s}^{-1}]=\left[-2{\left(n-\frac{1}{2}\right)}^s\right]$$

$$2{\left(n-\frac{1}{2}\right)}^s<A_{n,s}^{-1}<2{\left(n-\frac{1}{2}+\frac{1}{2^{p+1}}\right)}^s$$

$$-2{\left(n-\frac{1}{2}+\frac{1}{2^{p+1}}\right)}^s<B_{n,s}^{-1}<-2{\left(n-\frac{1}{2}\right)}^s$$

$$\left[\frac{1}{1-2^{1-s}}\cdot {\zeta }_n(s{)}^{-1}\right]=\left[(-1{)}^{n+1}\cdot 2{\left(n-\frac{1}{2}\right)}^s\right]$$

$${\left({\left(1-\frac{1}{4x}\right)}^2+\frac{1}{8x^2}\right)}^{-1/p-1}\cdot \left(1-\frac{1}{4x}\right)+{\left({\left(1+\frac{1}{4x}\right)}^2+\frac{1}{8x^2}\right)}^{-1/p-1}\cdot \left(1+\frac{1}{4x}\right)<2.$$

$$\underset{\alpha \to \infty }{\lim }\left({\left({\left(1-\frac{1}{\alpha }\right)}^2+\frac{2}{{\alpha }^2}\right)}^{-1/p-1}\cdot \left(1-\frac{1}{\alpha }\right)+{\left({\left(1+\frac{1}{\alpha }\right)}^2+\frac{2}{{\alpha }^2}\right)}^{-1/p-1}\cdot \left(1+\frac{1}{\alpha }\right)\right)=2.$$

$$\frac{d\left(h(\alpha )+h(-\alpha )\right)}{d\alpha }$$

$$=\frac{d\left({\left({\left(1-\frac{1}{\alpha }\right)}^2+\frac{2}{{\alpha }^2}\right)}^{-1/p-1}\cdot \left(1-\frac{1}{\alpha }\right)+{\left({\left(1+\frac{1}{\alpha }\right)}^2+\frac{2}{{\alpha }^2}\right)}^{-1/p-1}\cdot \left(1+\frac{1}{\alpha }\right)\right)}{d\alpha }$$

$$\displaystyle=\frac{1}{\alpha^{4}p}\Bigg{(}\alpha^{2}p\left(\left(\frac{\alpha^{2}-2\alpha+3}{\alpha^{2}}\right)^{-1-1/p}-\left(\frac{\alpha^{2}+2\alpha+3}{\alpha^{2}}\right)^{-1-1/p}\right)$$

$$\displaystyle\quad-2(p+1)\left((\alpha-1)(\alpha-3)\left(\frac{\alpha^{2}-2\alpha+3}{\alpha^{2}}\right)^{-2-1/p}-(\alpha+1)(\alpha+3)\left(\frac{\alpha^{2}+2\alpha+3}{\alpha^{2}}\right)^{-2-1/p}\right)\Bigg{)}.$$

$${\alpha }^{2}p\left({\left(\frac{{\alpha }^2-2\alpha +3}{{\alpha }^2}\right)}^{-1-1/p}-{\left(\frac{{\alpha }^2+2\alpha +3}{{\alpha }^2}\right)}^{-1-1/p}\right)$$

$$-2(p+1)\left((\alpha -1)(\alpha -3){\left(\frac{{\alpha }^2-2\alpha +3}{{\alpha }^2}\right)}^{-2-1/p}-(\alpha +1)(\alpha +3){\left(\frac{{\alpha }^2+2\alpha +3}{{\alpha }^2}\right)}^{-2-1/p}\right)>0,$$

$$({\alpha }^2-2\alpha +3{)}^{-2-1/p}\left(p({\alpha }^2-2\alpha +3)-2(p+1)(\alpha -1)(\alpha -3)\right)$$

Full text

Key words— Riemann zeta function tail, critical strip, Siegel’s theorem

Mathematics Subject Classification 2010: 11M06, 11B83, 11G30

1 Introduction

Among various kind of zeta functions in mathematics, one of the most famous and important zeta functions is the Riemann zeta function. For $s=\sigma+it\in\mathbb{C}$ with $\sigma>1,$ consider the absolutely convergent infinite series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{s}}.$ It is well-known that this series admits an analytic continuation $\zeta(s)$ to the whole complex plane $\mathbb{C}$. If we restrict our attention to some rational numbers $0<s<1,$ then we have the following list of values of the Riemann zeta function:

[TABLE]

[TABLE]

Now, for an integer $n\geq 1$ and a real number $s$ with $0<s<1,$ we let

[TABLE]

Note that we have

[TABLE]

and

[TABLE]

Along this line, D. Kim and K. Song [3] (resp. K. Song [4]) proved that we have

[TABLE]

for every integer $n\geq 1,$ at $s=\frac{1}{2},\frac{1}{3},\frac{1}{4}$ (resp. $s=\frac{2}{3}).$ In this paper, we extend the previous results to the case when either $s=\frac{1}{p}$ for any integer with $p\geq 5$ or $s=\frac{2}{p}$ for any odd integer with $p\geq 5.$

Our main result is summarized in the following

Theorem 1.1**.**

Let $s=\frac{1}{p}$ for any integer with $p\geq 5$ or $s=\frac{2}{p}$ for any odd integer with $p\geq 5.$ Then there exists an integer $N>0$ such that we have

[TABLE]

for every integer $n\geq N.$

For more details, see Corollaries 3.4 and 4.8.

This paper is organized as follows: In Section 2, we first introduce some properties of $\zeta_{n}(s)$. Afterwards, we recall a theorem of Siegel on the integral points of a smooth algebraic curve over a number field (see Theorem 2.5). In Section 3, we deal with the case of $s=\frac{1}{p}$, using Theorem 2.3 below. In Section 4, we give a proof of Theorem 1.1 for the case when $s=\frac{2}{p}$ by invoking a version of the previously introduced theorem of Siegel (see Theorem 2.6).

2 Preliminaries

2.1 Properties of $\zeta_{n}(s)$

In this section, we give some useful properties of $\zeta_{n}(s)$ in terms of the size of its reciprocal. To achieve our goal, we first need the following

Theorem 2.1**.**

Let $s$ be a real number with $0<s<1$. Then we have

[TABLE]

for every even integer $n\geq 2$, and

[TABLE]

for every odd integer $n\geq 1$.

Proof.

For a proof, see [3, Theorem 1]. ∎

In view of the equation (1.1), Theorem 2.1 has a nice consequence:

Corollary 2.2**.**

For any real number $s$ with $0<s<1$, we have

[TABLE]

for every even integer $n\geq 2$, and

[TABLE]

for every odd integer $n\geq 1$.

If we do not require the inequality of Theorem 2.1 to hold for every even or odd integer, as indicated above, then we can obtain a slightly better upper bound of the value of $\zeta_{n}(s)^{-1}$:

Theorem 2.3**.**

Let $\epsilon>0$ be given. Then for any real number $s$ with $0<s<1$, we have

[TABLE]

for every sufficiently large even integer $n$, and

[TABLE]

for every sufficiently large odd integer $n$.

Proof.

For a proof, see [3, Theorem 2]. ∎

As before, combining the equation (1.1) and Theorem 2.3 yields the following

Corollary 2.4**.**

Let $\epsilon>0$ be given. Then for any real number $s$ with $0<s<1$, we have

[TABLE]

for every sufficiently large even integer $n$, and

[TABLE]

for every sufficiently large odd integer $n$.

2.2 Siegel’s theorem on integral points

In this section, we briefly review a theorem of Siegel on the integral points of certain curves that are defined over a number field.

In the sequel, let $K$ be a number field, $S$ a finite set of places of $K,$ and $R_{S}$ the ring of $S$-integers in $K.$ Also, let $\overline{K}$ be an algebraic closure of $K.$ Then we have the following fundamental result:

Theorem 2.5**.**

Let $C$ be a smooth projective curve of genus $g$ over $K,$ and let $f\in K(C)$ be a nonconstant function. If $g\geq 1,$ then the set

[TABLE]

is finite.

Proof.

For a proof, see [2, Theorem D.9.1]. ∎

In fact, this theorem is more general than what we actually need. We need to use a version of Theorem 2.5 regarding a hyperelliptic curve, which we describe now: suppose that $S$ includes all the infinite places.

Theorem 2.6**.**

Let $f(X)\in K[X]$ be a polynomial of degree at least $3$ with distinct roots in $\overline{K}$. Then the equation $Y^{2}=f(X)$ has only finitely many solutions $X,Y\in R_{S}.$

Proof.

For a proof, see [2, Theorem D.8.3]. ∎

Example 2.7**.**

Let $K=\mathbb{Q}$, $S=\{2,\infty\}$, and let $f(X)=\frac{1}{32}X^{5}+\frac{17}{32}.$ Then the equation $Y^{2}=f(X)$ has only finitely many solutions in $R_{S}=\mathbb{Z}\left[\frac{1}{2}\right].$ This example is closely related to the case of $s=\frac{2}{5}$ and $m=1$ in Lemma 4.4 below.

3 The case of $s=\frac{1}{p}$

Throughout this section, let $p\geq 5$ be a fixed integer and let $s=\frac{1}{p}$.

Lemma 3.1**.**

Let $n\geq 1$ be an integer. Then there is no integer between $2\left(n-\frac{1}{2}\right)^{s}$ and $2\left(n-\frac{1}{2}+\frac{1}{2^{p+1}}\right)^{s}.$

Proof.

Suppose on the contrary that there is an integer $x$ with $2\left(n-\frac{1}{2}\right)^{s}<x<2\left(n-\frac{1}{2}+\frac{1}{2^{p+1}}\right)^{s}.$ It follows that

[TABLE]

which is absurd.

This completes the proof. ∎

By a similar argument, we can also have the following

Lemma 3.2**.**

Let $n\geq 1$ be an integer. Then there is no integer between $-2\left(n-\frac{1}{2}+\frac{1}{2^{p+1}}\right)^{s}$ and $-2\left(n-\frac{1}{2}\right)^{s}.$

Combining all these results, we have:

Corollary 3.3**.**

There exist integers $n_{0},n_{1}>0$ such that

[TABLE]

for every even integer $n\geq n_{0}$, and

[TABLE]

for every odd integer $n\geq n_{1}.$

Proof.

Let $\epsilon=\frac{1}{2^{p+1}}.$ By Theorem 2.3, there exist integers $n_{0},n_{1}>0$ such that

[TABLE]

for every even integer $n\geq n_{0},$ and

[TABLE]

for every odd integer $n\geq n_{1}.$ Then by Lemmas 3.1 and 3.2, it follows that $[A_{n,s}^{-1}]=\left[2\left(n-\frac{1}{2}\right)^{s}\right]$ for every even integer $n\geq n_{0},$ and $[B_{n,s}^{-1}]=\left[-2\left(n-\frac{1}{2}\right)^{s}\right]$ for every odd integer $n\geq n_{1},$ as desired.

This completes the proof. ∎

An immediate consequence of Corollary 3.3 is the following

Corollary 3.4**.**

There exists an integer $N>0$ such that we have

[TABLE]

for every integer $n\geq N.$

Proof.

This follows from the equation (1.1) and Corollary 3.3. ∎

Remark 3.5**.**

If $p\in\{2,3,4\},$ then we may take $N=1$ in view of [3, Corollary 6].

4 The case of $s=\frac{2}{p}$

Throughout this section, let $p\geq 5$ be a fixed odd integer and let $s=\frac{2}{p}$.

To begin with, we introduce one useful inequality:

Lemma 4.1**.**

For any real number $x\geq 2,$ we have

[TABLE]

Proof.

For convenience, let $\alpha=4x$ and $h(\alpha)=\left(\left(1-\frac{1}{\alpha}\right)^{2}+\frac{2}{\alpha^{2}}\right)^{-1/p-1}\cdot\left(1-\frac{1}{\alpha}\right)$. (Note that $\alpha\geq 8$.) Then it suffices to show that $h(\alpha)+h(-\alpha)$ is a strictly increasing function because we have

[TABLE]

Indeed, we will use the first derivative test, as follows: note first that we have

[TABLE]

Since $\alpha^{4}p>0$, we need to show that

[TABLE]

or equivalently, (by multiplying $\alpha^{-4-2/p}$ and rearranging),

[TABLE]

which is equivalent to saying that

[TABLE]

because $(-p-2)\alpha^{2}+(-6p-8)\alpha+(-3p-6)<0$ for any $\alpha,p$. We prove the last inequality. Indeed, we may assume that $(-p-2)\alpha^{2}+(6p+8)\alpha+(-3p-6)<0$ because if $(-p-2)\alpha^{2}+(6p+8)\alpha+(-3p-6)\geq 0$, then the desired inequality follows trivially. Hence, we have to show that

[TABLE]

Since

[TABLE]

and

[TABLE]

for every $p\geq 5$ and $\alpha\geq 8$, we have that the inequality (4.1) holds for $\alpha\geq 8$, which in turn, implies that $h(\alpha)+h(-\alpha)$ is increasing for $\alpha\geq 8$.

This completes the proof. ∎

An important consequence of the above lemma is the following

Lemma 4.2**.**

For any even integer $n\geq 4$, we have

[TABLE]

and for any odd integer $n\geq 3$, we have

[TABLE]

Proof.

Let $n\geq 4$ be an even integer. By Theorem 2.1, we have

[TABLE]

Hence, it suffices to show that

[TABLE]

for any even integer $n\geq 4$. Let $f,g:\mathbb{R}_{\geq 2}\rightarrow\mathbb{R}$ be two functions defined by

[TABLE]

and

[TABLE]

Then we have $f(x)=g(x)-g(x+\frac{1}{2})$, and hence, we only need to show that $g(x)$ is decreasing. Since

[TABLE]

we have to show that

[TABLE]

Then it follows from Lemma 4.1 that $g(x)$ is decreasing for $x\geq 2$. Also, a similar argument can be used to show that (4.2) holds for any odd integer $n\geq 3.$

This completes the proof. ∎

Remark 4.3**.**

The inequalities in the statement of Lemma 4.2 also hold when $n=1,2$. For example, we have

[TABLE]

Now, we give a result on the finiteness of the integer points of certain affine curves, which will be used later:

Lemma 4.4**.**

The affine curve $C_{m}:x^{p}-2^{p}y^{2}-2^{p}y+2^{p-2}+m=0$ defined over $\mathbb{Q}$ has only finitely many integer points for each $1\leq m\leq 2^{p-1}-1.$

Proof.

Let $1\leq m\leq 2^{p-1}-1$ be fixed. By completing the square and rearranging, the defining equation of $C_{m}$ can be written as

[TABLE]

By letting $Y=y+\frac{1}{2},$ the equation (4.3) becomes

[TABLE]

Let $S=\{2,\infty\}$ be a finite set of places of $\mathbb{Q}$, and let $f(x)=\frac{1}{2^{p}}\cdot x^{p}+\frac{2^{p-1}+m}{2^{p}}.$ Then the equation $Y^{2}=f(x)$ has only finitely many solutions in $R_{S}=\mathbb{Z}\left[\frac{1}{2}\right]$ by Theorem 2.6, which in turn, implies that $C_{m}(\mathbb{Z})$ is also finite. (Note that if $(a,b)$ is an integer point of $C_{m}$, then $\left(a,b+\frac{1}{2}\right)$ is a solution of the equation (4.4) in $R_{S}=\mathbb{Z}\left[\frac{1}{2}\right].$) Since $m$ was arbitrary, the proof is complete. ∎

In the sequel, let $C_{m}$ denote the affine curve defined as in Lemma 4.4 for $1\leq m\leq 2^{p-1}-1$. Using the above finiteness result on integral points, we have the following

Lemma 4.5**.**

There exists an integer $n_{0}>0$ with the property that there is no integer between $2\left(n^{2}-n+\frac{1}{4}\right)^{\frac{s}{2}}$ and $2\left(n^{2}-n+\frac{3}{4}\right)^{\frac{s}{2}}$ for every integer $n\geq n_{0}.$

Proof.

Let $S$ be the set of integers $n$ such that there is an integer between $2\left(n^{2}-n+\frac{1}{4}\right)^{\frac{s}{2}}$ and $2\left(n^{2}-n+\frac{3}{4}\right)^{\frac{s}{2}}.$ Let $n\in S$ and suppose that $a$ is an integer with $2\left(n^{2}-n+\frac{1}{4}\right)^{\frac{s}{2}}<a<2\left(n^{2}-n+\frac{3}{4}\right)^{\frac{s}{2}}.$ (Note that, in view of [3, page 9], there is exactly one such $a.$) It follows that

[TABLE]

We may write $a^{p}=2^{p}n^{2}-2^{p}n+2^{p-2}+l$ for some $1\leq l\leq 2^{p-1}-1$ so that we have $(a,n)\in C_{l}(\mathbb{Z}).$ (Note also that the uniqueness of $a$ guarantees the uniqueness of such $l.$) This observation gives rise to a well-defined set map $\displaystyle\varphi:S\rightarrow\bigcup_{m=1}^{2^{p-1}-1}C_{m}(\mathbb{Z})$ given by $\varphi(n)=(a,n)$, where $a$ is uniquely determined as above. It is easy to see that $\varphi$ is injective, by construction. Now, by Lemma 4.4, the set $\displaystyle\bigcup_{m=1}^{2^{p-1}-1}C_{m}(\mathbb{Z})$ is finite, and hence, it follows that $S$ is also finite. Let $n_{0}=\max S+1.$ Then for any integer $n\geq n_{0}$, we have $n\not\in S,$ which is the desired result.

This completes the proof. ∎

By a similar argument, we can also have the following

Lemma 4.6**.**

There exists an integer $n_{1}>0$ with the property that there is no integer between $-2\left(n^{2}-n+\frac{3}{4}\right)^{\frac{s}{2}}$ and $-2\left(n^{2}-n+\frac{1}{4}\right)^{\frac{s}{2}}$ for every integer $n\geq n_{1}.$

Using Lemmas 4.2, 4.5, and 4.6, we can prove the following

Theorem 4.7**.**

There exist integers $n_{0},n_{1}>0$ such that

[TABLE]

for every even integer $n\geq n_{0}$, and

[TABLE]

for every odd integer $n\geq n_{1}.$

Proof.

By Lemmas 4.2 and 4.5, there exists an integer $n_{0}>0$ such that

[TABLE]

and there is no integer between $2\left(n-\frac{1}{2}\right)^{s}$ and $2\left(n^{2}-n+\frac{3}{4}\right)^{\frac{s}{2}}$ for every even integer $n\geq n_{0}.$ Then it follows that $\left[A_{n,s}^{-1}\right]=\left[2\left(n-\frac{1}{2}\right)^{s}\right]$ for every even integer $n\geq n_{0}.$ Similarly, by Lemmas 4.2 and 4.6, there exists an integer $n_{1}>0$ such that

[TABLE]

and there is no integer between $-2\left(n^{2}-n+\frac{3}{4}\right)^{\frac{s}{2}}$ and $-2\left(n-\frac{1}{2}\right)^{s}$ for every odd integer $n\geq n_{1}.$ Then it follows that $\left[B_{n,s}^{-1}\right]=\left[-2\left(n-\frac{1}{2}\right)^{s}\right]$ for every odd integer $n\geq n_{1}.$

This completes the proof. ∎

As an immediate consequence of Theorem 4.7, we have:

Corollary 4.8**.**

There exists an integer $N>0$ such that we have

[TABLE]

for every integer $n\geq N.$

Proof.

This follows from the equation (1.1) and Theorem 4.7. ∎

Remark 4.9**.**

If $p=3,$ then we may take $N=1$ in view of [4, Theorem 3.12].

We conclude this paper with the following

Remark 4.10**.**

In light of [2, Remark D.9.5], it might be possible to find such a suitable $N>0$ in Corollaries 3.4 and 4.8 by adopting a theorem of Baker (see [1]).