On the Irrationality of the Odd Zeta Values

1. Introduction

1.1. Background and Motivation

The study of the arithmetic properties of special values of the Riemann zeta function occupies a central place in analytic number theory. Euler’s classical formulae [10]

$$\zeta \left(2n\right)={\displaystyle \frac{{(-1)}^{n+1}{B}_{2n}{\left(2\pi \right)}^{2n}}{2\left(2n\right)!}},n\in \mathbb{N}$$

express the even zeta values in terms of Bernoulli numbers and powers of $\pi$, thereby establishing both their transcendence and algebraic relations to $\pi$. In striking contrast, the arithmetic nature of the odd zeta values $\zeta (2n+1)$ remains largely mysterious. The pioneering breakthrough was Apéry’s proof [1,11] of the irrationality of $\zeta \left(3\right)$, which not only settled a longstanding problem of Euler but also introduced a new paradigm for irrationality proofs through the construction of highly nontrivial linear recurrences. Subsequent refinements by Beukers [3] via integrals, and the landmark results of Rivoal [7] and Zudilin [12] showed that infinitely many odd zeta values must be irrational, and that at least one of $\zeta \left(5\right),\zeta \left(7\right),\zeta \left(9\right),\zeta \left(11\right)$ is irrational. Nevertheless, no specific irrationality result beyond Apéry’s $\zeta \left(3\right)$ has been obtained, and the question of whether $\zeta (2n+1)$ is irrational for $n\ge 2$ remains open.

The goal of this paper is to prove general irrationality results for the odd zeta values, by applying a refined Diophantine approximation argument.

1.2. Main Theorem

We begin by stating the principal result.

Theorem 1.1 

(Main theorem). For every $n\in \mathbb{N}$, the odd zeta value $\zeta (2n+1)$ is irrational.

1.3. Strategy of the Proof (Sketch)

We outline the logical structure of the proof before entering the technical details.

1.

Construction of integer linear forms. We construct for each $m\in \mathbb{N}$ an asymmetric beta kernel $W_m^{\left(q\right)}\left(x\right)$ and its cumulative distribution function $S_m^{\left(q\right)}\left(x\right)$, enabling an exact integral identity linking $\zeta (2n+1)$ with a sequence of weighted moments $L_m^{\left(q\right)}$. Truncating the Maclaurin expansion of the polylogarithm ${Li}_{2n+1}\left(x\right)$ yields integer linear forms ${\Lambda }_m^{\left(q\right)}=A_m^{\left(q\right)}\zeta (2n+1)-B_m^{\left(q\right)}$, whose coefficients are made integral by a suitable least common multiple of denominators.

2.

Asymptotic decay and growth estimates. Exponential decay of the kernel tail and remainder terms is established using Stirling asymptotics and $L^p$ norms, while precise denominator growth is controlled through Legendre’s formula and the Prime Number Theorem.

3.

Application of an irrationality criterion. The balance of these estimates ensures that $|{\Lambda }_m^{\left(q\right)}|\to 0$ as $m\to \infty$, permitting application of a refined Dirichlet-type irrationality criterion to prove that each $\zeta (2n+1)$ is irrational.

2. An Integral Representation of $\zeta (2n+1)$

We shall exploit the following integral representation of $\zeta (2n+1)$:

$$\zeta (2n+1)={\int }_0^1{Li}_{2n+1}^{\prime }\left(x\right)dx,$$

(1)

where ${\displaystyle {Li}_s\left(x\right)=\sum _{k=1}^{\infty }{\displaystyle \frac{x^k}{k^s}}}$ denotes the polylogarithm [5].

3. Construction of Integer Linear Forms

3.1. The Beta Kernel and the Exact Integral Identity

Define the asymmetric beta kernel [4]

$$W_m^{\left(q\right)}\left(x\right):=\left({\displaystyle \genfrac{}{}{0pt}{}{(q+1)m}{m}}\right)x^m{(1-x)}^{qm},m,q\in \mathbb{N}.$$

(2)

The corresponding probability density function on $[0,1]$ is

$${\tilde{W}}_m^{\left(q\right)}\left(x\right):={\Omega }_m^{\left(q\right)}{W}_m^{\left(q\right)}\left(x\right),$$

(3)

where the normalization ${\Omega }_m^{\left(q\right)}:=(q+1)m+1$. The cumulative distribution function is

$$S_m^{\left(q\right)}\left(x\right):={\int }_0^x{\tilde{W}}_m^{\left(q\right)}\left(t\right)dt.$$

(4)

Set

$$L_m^{\left(q\right)}:={\int }_0^1{W}_m^{\left(q\right)}\left(x\right){Li}_{2n+1}\left(x\right)dx.$$

(5)

Lemma 3.1 

(Exact integral identity). We have

$${\Omega }_m^{\left(q\right)}{L}_m^{\left(q\right)}=\zeta (2n+1)-{\int }_0^1{Li}_{2n+1}^{\prime }\left(x\right)S_m^{\left(q\right)}\left(x\right)dx.$$

(6)

Proof. 

Substitute ${\displaystyle {Li}_{2n+1}\left(x\right)={\int }_0^x{Li}_{2n+1}^{\prime }\left(t\right)dt}$ and swap the integrals:

$$\begin{array}{cc}\hfill L_m^{\left(q\right)}& ={\int }_0^1{W}_m^{\left(q\right)}\left(x\right){\int }_0^x{Li}_{2n+1}^{\prime }\left(t\right)dtdx={\int }_0^1{Li}_{2n+1}^{\prime }\left(t\right){\int }_t^1{W}_m^{\left(q\right)}\left(x\right)dxdt\hfill \\ & ={\displaystyle \frac{1}{{\Omega }_m^{\left(q\right)}}}{\int }_0^1{Li}_{2n+1}^{\prime }\left(t\right)[1-S_m^{\left(q\right)}\left(t\right)]dt.\hfill \end{array}$$

Multiply by ${\Omega }_m^{\left(q\right)}$ and use (1). □

3.2. Truncation and Denominator Control

By definition,

$${Li}_{2n+1}\left(x\right)=\sum _{k=1}^{\infty }{\displaystyle \frac{x^k}{k^{2n+1}}}.$$

Multiplying by $W_m^{\left(q\right)}$ and integrating termwise (justified by uniform convergence on $[0,1]$) gives

$$\begin{array}{cc}\hfill L_m^{\left(q\right)}& =\sum _{k=1}^{\infty }{\displaystyle \frac{1}{k^{2n+1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{(q+1)m}{m}}\right)B(m+k+1,qm+1)=\sum _{k=1}^{\infty }{\displaystyle \frac{1}{k^{2n+1}}}{\displaystyle \frac{\left(\right(q+1\left)m\right)!(m+k)!}{m!\left(\right(q+1)m+k+1)!}}\hfill \end{array}$$

$$\begin{array}{c}\hfill =\sum _{k=1}^{\infty }{\displaystyle \frac{F_{m,k}^{\left(q\right)}}{k^{2n+1}}},\end{array}$$

(7)

where

$$F_{m,k}^{\left(q\right)}:={\displaystyle \frac{\left(\right(q+1\left)m\right)!(m+k)!}{m!\left(\right(q+1)m+k+1)!}},$$

(8)

and we used $B(a,b)={\displaystyle \frac{\Gamma \left(a\right)\Gamma \left(b\right)}{\Gamma (a+b)}}$ to simplify.

Fix a truncation parameter $K=K\left(m\right)$ and split

$$L_m^{\left(q\right)}=L_m^{\left(q\right)(\le K)}+L_m^{\left(q\right)(>K)},$$

(9)

where ${\displaystyle L_m^{\left(q\right)(\le K)}:=\sum _{k=1}^K{\displaystyle \frac{F_{m,k}^{\left(q\right)}}{k^{2n+1}}}}$ and ${\displaystyle L_m^{\left(q\right)(>K)}:=\sum _{k>K}{\displaystyle \frac{F_{m,k}^{\left(q\right)}}{k^{2n+1}}}}$. Let $den\left(x\right)$ be the denominator of the rational number x written in lowest terms. Define

$$D_m^{\left(q\right)}:=\underset{1\le k\le K}{lcm}den\left({\displaystyle \frac{F_{m,k}^{\left(q\right)}}{k^{2n+1}}}\right).$$

(10)

Lemma 3.2 

(Integer linear form). With

$$\begin{array}{cc}\hfill A_m^{\left(q\right)}& :=D_m^{\left(q\right)}\in \mathbb{Z},\hfill \end{array}$$

(11)

$$\begin{array}{cc}\hfill B_m^{\left(q\right)}& :={\Omega }_m^{\left(q\right)}{D}_m^{\left(q\right)}{L}_m^{\left(q\right)(\le K)}\in \mathbb{Z},\hfill \end{array}$$

(12)

we have

$${\Lambda }_m^{\left(q\right)}:=A_m^{\left(q\right)}\zeta (2n+1)-B_m^{\left(q\right)}={\Omega }_m^{\left(q\right)}{D}_m^{\left(q\right)}{L}_m^{\left(q\right)(>K)}+D_m^{\left(q\right)}{\int }_0^1{Li}_{2n+1}^{\prime }\left(x\right)S_m^{\left(q\right)}\left(x\right)dx.$$

(13)

Proof. 

Multiply (6) by $D_m^{\left(q\right)}$ and substitute $L_m^{\left(q\right)}=L_m^{\left(q\right)(\le K)}+L_m^{\left(q\right)(>K)}$. The choice (10) guarantees every $B_m^{\left(q\right)}$ is an integer (the least common multiple clears all denominators $k^{2n+1}$ for $k\le K$ and all factorial denominators up to $(q+1)m+K+1$). □

Thus, ${\Lambda }_m^{\left(q\right)}=A_m^{\left(q\right)}\zeta (2n+1)-B_m^{\left(q\right)}$ is our integer linear form in $\zeta (2n+1)$. Equation (13) guarantees that ${\Lambda }_m^{\left(q\right)}>0$.

4. Analytic Estimates for the Linear Forms

This section proves that the constructed linear forms tend to zero as $m\to \infty$.

4.1. Exponential Decay of the Kernel Tail

Using Stirling asymptotics and large-deviation estimates, we show that the truncated tail contributes an exponentially small error.

Lemma 4.1 

(Kernel tail). There exist parameters $\alpha ,{\gamma }^{\left(q\right)}\left(\alpha \right),C_1(\alpha ,m,q),C_2(\alpha ,q)>0$, such that for $m>C_2(\alpha ,q)$,

$$|{\Omega }_m^{\left(q\right)}{L}_m^{\left(q\right)(>K)}|\le C_1(\alpha ,m,q){\left[1-{\displaystyle \frac{C_2(\alpha ,q)}{m}}\right]}^{-1}{e}^{-{\gamma }^{\left(q\right)}\left(\alpha \right)m}.$$

(14)

Proof. 

Set $k=\sigma m$, where $\sigma >0$ is a scaling parameter. Using the Stirling estimate

$$logN!=NlogN-N+O(logN),$$

we get

$$\begin{array}{cc}\hfill log{F}_{m,k}^{\left(q\right)}& =log\left(\right(q+1\left)m\right)!+log(m+k)!-logm!-log\left(\right(q+1)m+k+1)!\hfill \\ & =m{\varphi }^{\left(q\right)}\left(\sigma \right)+O(logm),\hfill \end{array}$$

where

$${\varphi }^{\left(q\right)}\left(\sigma \right):=(1+q)log(1+q)+(1+\sigma )log(1+\sigma )-(1+q+\sigma )log(1+q+\sigma )$$

(15)

is a rate function, which is strictly decreasing on $(0,\infty )$. Since ${\varphi }^{\left(q\right)}\left(0\right)=0$, this implies ${\varphi }^{\left(q\right)}\left(\sigma \right)<0$ for all $\sigma >0$. Thus,

$$\left|{\displaystyle \frac{F_{m,k}^{\left(q\right)}}{k^{2n+1}}}\right|\le m^{O\left(1\right)}{e}^{m{\varphi }^{\left(q\right)}\left(\sigma \right)}.$$

Let $K:=\lceil \alpha m\rceil$, for some fixed real $\alpha >0$. Then,

$$|L_m^{\left(q\right)(>K)}|\le m^{O\left(1\right)}\sum _{k>\alpha m}{e}^{m{\varphi }^{\left(q\right)}\left(\sigma \right)}\le m^{O\left(1\right)}{\int }_{\alpha m}^{\infty }{e}^{m{\varphi }^{\left(q\right)}\left({\displaystyle \frac{x}{m}}\right)}dx=m^{O\left(1\right)}{\int }_{\alpha }^{\infty }{e}^{m{\varphi }^{\left(q\right)}\left(\sigma \right)}d\sigma .$$

Since

$${\displaystyle \frac{d}{d\sigma }}{e}^{m{\varphi }^{\left(q\right)}\left(\sigma \right)}=m{\varphi }^{\prime \left(q\right)}\left(\sigma \right)e^{m{\varphi }^{\left(q\right)}\left(\sigma \right)},$$

an integration by parts gives

$${\int }_{\alpha }^{\infty }{e}^{m{\varphi }^{\left(q\right)}\left(\sigma \right)}d\sigma =-{\displaystyle \frac{e^{m{\varphi }^{\left(q\right)}\left(\alpha \right)}}{m{\varphi }^{\prime \left(q\right)}\left(\alpha \right)}}+{\displaystyle \frac{1}{m}}{\int }_{\alpha }^{\infty }{\displaystyle \frac{{\varphi }^{\prime \prime \left(q\right)}\left(\sigma \right)}{{\varphi }^{\prime \left(q\right)}{\left(\sigma \right)}^2}}{e}^{m{\varphi }^{\left(q\right)}\left(\sigma \right)}d\sigma .$$

The remainder integral may be bounded as

$$\left|{\int }_{\alpha }^{\infty }{\displaystyle \frac{{\varphi }^{\prime \prime \left(q\right)}\left(\sigma \right)}{{\varphi }^{\prime \left(q\right)}{\left(\sigma \right)}^2}}{e}^{m{\varphi }^{\left(q\right)}\left(\sigma \right)}d\sigma \right|\le \underset{\sigma \ge \alpha }{sup}\left|{\displaystyle \frac{{\varphi }^{\prime \prime \left(q\right)}\left(\sigma \right)}{{\varphi }^{\prime \left(q\right)}{\left(\sigma \right)}^2}}\right|{\int }_{\alpha }^{\infty }{e}^{m{\varphi }^{\left(q\right)}\left(\sigma \right)}d\sigma =C_2(\alpha ,q){\int }_{\alpha }^{\infty }{e}^{m{\varphi }^{\left(q\right)}\left(\sigma \right)}d\sigma ,$$

where

$${\displaystyle C_2(\alpha ,q):=\underset{\sigma \ge \alpha }{sup}\left|{\displaystyle \frac{{\varphi }^{\prime \prime \left(q\right)}\left(\sigma \right)}{{{\varphi }^{\prime \left(q\right)}\left(\sigma \right)}^2}}\right|.}$$

Note that $C_2(\alpha ,q)$ is finite, since for $\sigma \ge \alpha >0$,

$${\varphi }^{\prime \left(q\right)}\left(\sigma \right)=log{\displaystyle \frac{1+\sigma }{1+q+\sigma }}<0,{\varphi }^{\prime \prime \left(q\right)}\left(\sigma \right)={\displaystyle \frac{q}{(1+\sigma )(1+q+\sigma )}}>0;$$

both are continuous on $[\alpha ,\infty )$. Moreover, both ${\varphi }^{\prime \left(q\right)}\left(\sigma \right)$ and ${\varphi }^{\prime \prime }\left(q\right)\left(\sigma \right)$ tend to zero as $\sigma \to \infty$; an application of L’Hôpital’s rule shows that this limit is ${\displaystyle \frac{1}{q}}$. So, ${\displaystyle \frac{{\varphi }^{\prime \prime \left(q\right)}}{{{\varphi }^{\prime \left(q\right)}}^2}}$ is continuous and bounded on $[\alpha ,\infty )$. Therefore, we get

$${\int }_{\alpha }^{\infty }{e}^{m{\varphi }^{\left(q\right)}\left(\sigma \right)}d\sigma \le -{\displaystyle \frac{e^{m{\varphi }^{\left(q\right)}\left(\alpha \right)}}{m{\varphi }^{\prime \left(q\right)}\left(\alpha \right)}}+{\displaystyle \frac{C_2(\alpha ,q)}{m}}{\int }_{\alpha }^{\infty }{e}^{m{\varphi }^{\left(q\right)}\left(\sigma \right)}d\sigma ,$$

or,

$$\left[1-{\displaystyle \frac{C_2(\alpha ,q)}{m}}\right]{\int }_{\alpha }^{\infty }{e}^{m{\varphi }^{\left(q\right)}\left(\sigma \right)}d\sigma \le -{\displaystyle \frac{e^{m{\varphi }^{\left(q\right)}\left(\alpha \right)}}{m{\varphi }^{\prime \left(q\right)}\left(\alpha \right)}}.$$

Since $1-{\displaystyle \frac{C_2(\alpha ,q)}{m}}>0$ for all $m>C_2(\alpha ,q)$, we get

$${\int }_{\alpha }^{\infty }{e}^{m{\varphi }^{\left(q\right)}\left(\sigma \right)}d\sigma \le -{\displaystyle \frac{e^{m{\varphi }^{\left(q\right)}\left(\alpha \right)}}{m{\varphi }^{\prime \left(q\right)}\left(\alpha \right)}}{\left[1-{\displaystyle \frac{C_2(\alpha ,q)}{m}}\right]}^{-1}={\displaystyle \frac{e^{m{\varphi }^{\left(q\right)}\left(\alpha \right)}}{mlog{\displaystyle \frac{1+q+\alpha }{1+\alpha }}}}{\left[1-{\displaystyle \frac{C_2(\alpha ,q)}{m}}\right]}^{-1}.$$

Therefore,

$$|L_m^{\left(q\right)(>K)}|\le m^{O\left(1\right)}{\displaystyle \frac{e^{m{\varphi }^{\left(q\right)}\left(\alpha \right)}}{log{\displaystyle \frac{1+q+\alpha }{1+\alpha }}}}{\left[1-{\displaystyle \frac{C_2(\alpha ,q)}{m}}\right]}^{-1}.$$

Defining

$$C_1(\alpha ,m,q):={\displaystyle \frac{{\Omega }_m^{\left(q\right)}{m}^{O\left(1\right)}}{log{\displaystyle \frac{1+q+\alpha }{1+\alpha }}}}$$

and ${\gamma }^{\left(q\right)}\left(\sigma \right):=-{\varphi }^{\left(q\right)}\left(\sigma \right)$ (so ${\gamma }^{\left(q\right)}\left(\sigma \right)$ is positive and increasing on $(0,\infty )$), we get

$$|{\Omega }_m^{\left(q\right)}{L}_m^{\left(q\right)(>K)}|\le C_1(\alpha ,m,q){\left[1-{\displaystyle \frac{C_2(\alpha ,q)}{m}}\right]}^{-1}{e}^{-{\gamma }^{\left(q\right)}\left(\alpha \right)m}.$$

□

4.2. Control of the Remainder Term

Using finite difference operators and $L^1/L^{\infty }$ bounds, we prove that the remainder integral also decays exponentially.

Lemma 4.2 

(Remainder term). There exist parameters $C_3(m,q)>0$, $0<\delta <1$ such that

$$\left|{\int }_0^1{Li}_{2n+1}^{\prime }\left(x\right)S_m^{\left(q\right)}\left(x\right)dx\right|\le C_3(m,q)\left[{\left({\displaystyle \frac{\delta }{m}}\right)}^m+{\delta }^m\right].$$

(16)

Proof. 

We adopt the convention that all $L^1/L^{\infty }$ norms are defined on $[0,1]$. Then, we have

$$\left|{\int }_0^1{Li}_{2n+1}^{\prime }\left(x\right)S_m^{\left(q\right)}\left(x\right)dx\right|\le \parallel {Li}_{2n+1}^{\prime }{\parallel }_{\infty }{\parallel S_m^{\left(q\right)}\parallel }_1,$$

(17)

where

$${Li}_{2n+1}^{\prime }\left(x\right)=\sum _{k=1}^{\infty }{\displaystyle \frac{x^{k-1}}{k^{2n}}}.$$

For $0\le x\le 1$, $\left|{\displaystyle \frac{x^{k-1}}{k^{2n}}}\right|\le {\displaystyle \frac{1}{k^{2n}}}$; therefore,

$$|{Li}_{2n+1}^{\prime }\left(x\right)|\le \sum _{k=1}^{\infty }{\displaystyle \frac{1}{k^{2n}}}=\zeta \left(2n\right),$$

which particularly implies

$$\parallel {Li}_{2n+1}^{\prime }{\parallel }_{\infty }\le \zeta \left(2n\right).$$

(18)

Now consider the finite mth backward difference with step h[6], applied to $S_m^{\left(q\right)}$:

$${\nabla }_h^m{S}_m^{\left(q\right)}\left(x\right):=\sum _{k=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){(-1)}^k{S}_m^{\left(q\right)}(x-kh).$$

(19)

Appendix A proves the following properties of ${\nabla }_h^m{S}_m^{\left(q\right)}$:

$$\parallel {\nabla }_h^m{S}_m^{\left(q\right)}{\parallel }_1\le h^m\parallel S_m^{\left(q\right)\left(m\right)}{\parallel }_1,\parallel S_m^{\left(q\right)}-{\nabla }_h^m{S}_m^{\left(q\right)}{\parallel }_1\le {\left(hm\right)}^m{\parallel S_m^{\left(q\right)\left(m\right)}\parallel }_1.$$

By definition, $S_m^{\left(q\right)\left(m\right)}={\tilde{W}}_m^{\left(q\right)(m-1)}$. For the asymmetric beta density, using the Leibniz rule,

$${\displaystyle \frac{d^{m-1}}{d{x}^{m-1}}}\left[x^m{(1-x)}^{qm}\right]=\sum _{k=0}^{m-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m-1}{k}}\right){\left(m\right)}_k{\left(qm\right)}_{m-1-k}{(-1)}^{m-1-k}{x}^{m-k}{(1-x)}^{qm-(m-1-k)},$$

where ${\left(m\right)}_k$ is the falling factorial. Taking absolute values and integrating:

$$\begin{array}{cc}\hfill {\int }_0^1\left|{\displaystyle \frac{d^{m-1}}{d{x}^{m-1}}}\left[x^m{(1-x)}^{qm}\right]\right|dx& \le \sum _{k=0}^{m-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m-1}{k}}\right){\left(m\right)}_k{\left(qm\right)}_{m-1-k}B(m-k+1,(q-1)m+k+2)\hfill \\ \hfill & \le \sum _{k=0}^{m-1}\left({\displaystyle \genfrac{}{}{0pt}{}{m-1}{k}}\right){\left(m\right)}_k{\left(qm\right)}_{m-1-k}{\displaystyle \frac{\Gamma (m-k+1)\Gamma \left(\right(q-1)m+k+2)}{\Gamma (qm+3)}}\hfill \\ \hfill & =2^{m-1}{\displaystyle \frac{\Gamma (m+1)\Gamma (qm+1)}{\Gamma (qm+3)}}.\hfill \end{array}$$

Multiplying by the normalization ${\displaystyle {\Omega }_m^{\left(q\right)}\left(\genfrac{}{}{0pt}{}{(q+1)m}{m}\right)}$ gives us ${\tilde{W}}_m^{\left(q\right)}$:

$$\parallel {\tilde{W}}_m^{\left(q\right)(m-1)}{\parallel }_1\le {\Omega }_m^{\left(q\right)}\left({\displaystyle \genfrac{}{}{0pt}{}{(q+1)m}{m}}\right)2^{m-1}{\displaystyle \frac{m!\left(qm\right)!}{(qm+2)!}}={\Omega }_m^{\left(q\right)}{2}^{m-1}{\displaystyle \frac{\left(\right(q+1\left)m\right)!}{(qm+2)!}}.$$

Hence,

$$\parallel S_m^{\left(q\right)\left(m\right)}{\parallel }_1\le {\Omega }_m^{\left(q\right)}{2}^m{\displaystyle \frac{\left(\right(q+1\left)m\right)!}{\left(qm\right)!}}.$$

From [8], we obtain the Stirling–Robbins bounds

$$\sqrt{2\pi }{N}^{N+{\displaystyle \frac{1}{2}}}{e}^{-N+{\displaystyle \frac{1}{12N+1}}}\le N!\le \sqrt{2\pi }{N}^{N+{\displaystyle \frac{1}{2}}}{e}^{-N+{\displaystyle \frac{1}{12N}}}.$$

Therefore,

$$\parallel S_m^{\left(q\right)\left(m\right)}{\parallel }_1\le c_1(m,q){\left(1+{\displaystyle \frac{1}{q}}\right)}^{qm}{\left[{\displaystyle \frac{2(q+1)m}{e}}\right]}^m\le c_1(m,q){\left[2(q+1)m\right]}^m,$$

(20)

where

$$c_1(m,q):={\Omega }_m^{\left(q\right)}\sqrt{{\displaystyle \frac{q+1}{q}}}exp\left({\displaystyle \frac{1}{12(q+1)m}}-{\displaystyle \frac{1}{12qm+1}}\right).$$

From the trivial rearrangement $S_m^{\left(q\right)}\left(x\right)={\nabla }_h^m{S}_m^{\left(q\right)}\left(x\right)+[S_m^{\left(q\right)}\left(x\right)-{\nabla }_h^m{S}_m^{\left(q\right)}\left(x\right)]$, we obtain the norm inequality

$$\parallel S_m^{\left(q\right)}{\parallel }_1\le \parallel {\nabla }_h^m{S}_m^{\left(q\right)}{\parallel }_1+{\parallel S_m^{\left(q\right)}-{\nabla }_h^m{S}_m^{\left(q\right)}\parallel }_1.$$

Applying the properties of ${\nabla }_h^m{S}_m^{\left(q\right)}$ gives

$$\parallel S_m^{\left(q\right)}{\parallel }_1\le c_1(m,q)[{\left\{2(q+1)hm\right\}}^m+{\left\{2(q+1)h{m}^2\right\}}^m].$$

Set $\delta :=2(q+1)h{m}^2$ for a fixed real $\delta \in (0,1)$. Then,

$$\parallel S_m^{\left(q\right)}{\parallel }_1\le c_1(m,q)\left[{\left({\displaystyle \frac{\delta }{m}}\right)}^m+{\delta }^m\right].$$

Combining (17), (18) and the inequality above, we get

$$\left|{\int }_0^1{Li}_{2n+1}^{\prime }\left(x\right)S_m^{\left(q\right)}\left(x\right)dx\right|\le C_3(m,q)\left[{\left({\displaystyle \frac{\delta }{m}}\right)}^m+{\delta }^m\right],$$

(21)

where $C_3(m,q):=\zeta \left(2n\right)c_1(m,q)$. □

Finally, combining (13), (14), (16), we obtain

$$|{\Lambda }_m^{\left(q\right)}|\le D_m^{\left(q\right)}{C}_1(\alpha ,m,q){\left[1-{\displaystyle \frac{C_2(\alpha ,q)}{m}}\right]}^{-1}{e}^{-{\gamma }^{\left(q\right)}\left(\alpha \right)m}+D_m^{\left(q\right)}{C}_3(m,q)\left[{\left({\displaystyle \frac{\delta }{m}}\right)}^m+{\delta }^m\right].$$

(22)

4.3. Growth of Denominators

Recall Legendre’s formula [2], which gives the exponent of a prime p in $N!$:

$${\nu }_p(N!)=\sum _{j=1}^{\lfloor {log}_{p}N\rfloor }⌊{\displaystyle \frac{N}{p^j}}⌋.$$

For any integers $N,a\ge 0$, we have for each j,

$$⌊{\displaystyle \frac{a}{p^j}}⌋\le ⌊{\displaystyle \frac{N+a}{p^j}}⌋-⌊{\displaystyle \frac{N}{p^j}}⌋\le ⌊{\displaystyle \frac{a}{p^j}}⌋+1.$$

Compute

$$\begin{array}{cc}\hfill -{\nu }_p\left(F_{m,k}^{\left(q\right)}\right)& =-{\nu }_p(\left((q+1)m\right)!)-{\nu }_p((m+k)!)+{\nu }_p(m!)+{\nu }_p(((q+1)m+k+1)!)\hfill \\ \hfill & =[{\nu }_p(((q+1)m+k+1)!)-{\nu }_p(\left((q+1)m\right)!)]-[{\nu }_p((m+k)!)-{\nu }_p(m!)]\hfill \\ \hfill & \le [{\nu }_p((k+1)!)+\lfloor {log}_p((q+1)m+k+1)\rfloor ]-{\nu }_p(k!)\hfill \\ \hfill & ={\nu }_p(k+1)+\lfloor {log}_p((q+1)m+k+1)\rfloor .\hfill \end{array}$$

Since ${\displaystyle den\left(F_{m,k}^{\left(q\right)}\right)=\prod _p{p}^{max(0,-{\nu }_p\left(F_{m,k}^{\left(q\right)}\right))}}$, the above inequality results in

$$den\left(F_{m,k}^{\left(q\right)}\right)|(k+1)\prod _p{p}^{\lfloor {log}_p((q+1)m+k+1)\rfloor }=(k+1)lcm(1,2,\dots ,(q+1)m+k+1).$$

Taking the least common multiple over $1\le k\le K$, one gets a bound of the form

$$D_m^{\left(q\right)}{|lcm(1,2,\dots ,K+1)}^{\kappa +1}lcm(1,2,\dots ,(q+1)m+K+1),$$

where $\kappa :=2n+1$. With $K=\lceil \alpha m\rceil$ and $lcm(1,2,\dots ,N)=e^{\psi \left(N\right)}$ (where $\psi \left(N\right)$ is the Chebyshev psi function), we obtain the Rosser–Schoenfeld bound [9]

$$D_m^{\left(q\right)}\le exp\left([(\kappa +1)(\alpha m+2)+(1+q+\alpha )m+2][1+o\left(1\right)]\right)=c_{2}exp\left((1+q+\lambda \alpha )m[1+o\left(1\right)]\right),$$

(23)

where $c_2:=exp\left((2\kappa +4)[1+o\left(1\right)]\right)$ and $\lambda :=\kappa +2$.

5. Proof of the Main Theorem

5.1. Parameter Selection

From (22), for exponential decay to zero, we must ensure that

$$1+q+\lambda \alpha <min({\gamma }^{\left(q\right)}\left(\alpha \right),-log\delta ).$$

(24)

Lemma 5.1 

(Existence of admissible parameters). With $\lambda >0$, there exist parameters $q\in \mathbb{N},\delta \in (0,1),\alpha >0$ which satisfy (24).

Proof. 

We first solve the inequality

$$1+q+\lambda \alpha <{\gamma }^{\left(q\right)}\left(\alpha \right)=-(1+q)log(1+q)-(1+\alpha )log(1+\alpha )+(1+q+\alpha )log(1+q+\alpha ).$$

Let

$$g\left(\alpha \right):=1+q+\lambda \alpha +(1+q)log(1+q)+(1+\alpha )log(1+\alpha )-(1+q+\alpha )log(1+q+\alpha ).$$

We want the set of $\alpha >0$ with $g\left(\alpha \right)<0$. We have

$$g^{\prime }\left(\alpha \right)=\lambda -log{\displaystyle \frac{1+q+\alpha }{1+\alpha }},g^{\prime \prime }\left(\alpha \right)={\displaystyle \frac{q}{(1+\alpha )(1+q+\alpha )}}>0.$$

So, $g^{\prime }$ is strictly increasing and g is strictly convex on $(0,\infty )$. Since $g\left(0\right)=1+q>0$ and the initial slope is $g^{\prime }\left(0\right)=\lambda -log(1+q)$, we have the following two cases:

1.

If $\lambda \ge log(1+q)$, then $g^{\prime }\left(0\right)\ge 0$. Since g is convex and $g\left(0\right)>0$, we then have $g\left(\alpha \right)>0$ for all $\alpha >0$. Hence, no positive $\alpha$ satisfies the inequality.

2.

If $\lambda <log(1+q)$, then $g^{\prime }\left(0\right)<0$. By convexity, $g^{\prime }$ increases from $g^{\prime }\left(0\right)<0$ to ${\displaystyle \underset{\alpha \to \infty }{lim}{g}^{\prime }\left(\alpha \right)=\lambda >0}$; so, there is exactly one ${\alpha }^{*}>0$ with $g^{\prime }\left({\alpha }^{*}\right)=0$. The value of ${\alpha }^{*}$ is

$${\alpha }^{*}={\displaystyle \frac{q}{e^{\lambda }-1}}-1,$$

and with this,

$$\left(g\left({\alpha }^{*}\right)\right)(\lambda ,q)=(1+q)[1+log(1+q)-\lambda ]-qlog{\displaystyle \frac{q}{e^{\lambda }-1}}.$$

At small positive $\alpha$, we have $g\left(\alpha \right)>0$, while $g\left(\alpha \right)\to \infty$ as $\alpha \to \infty$; hence, for there to exist a pair of positive roots of $g\left(\alpha \right)=0$, we require that $g\left({\alpha }^{*}\right)<0$. Differentiating with respect to $\lambda$, we find that ${\partial }_{\lambda }g\left({\alpha }^{*}\right)={\displaystyle \frac{q}{e^{\lambda }-1}}-1>0$; this means that $g\left({\alpha }^{*}\right)$ is strictly increasing in $\lambda$ on $(0,log(1+q\left)\right)$. Since ${\displaystyle \underset{\lambda \to 0^{+}}{lim}g\left({\alpha }^{*}\right)=-\infty }$ and ${\displaystyle \underset{\lambda \to log{(1+q)}^{-}}{lim}g\left({\alpha }^{*}\right)=1+q>0}$, by continuity and strict monotonicity in $\lambda$, there exists a unique ${\lambda }^{*}\in (0,log(1+q))$ such that $\left(g\left({\alpha }^{*}\right)\right)({\lambda }^{*},q)=0$. Therefore, for $0<\lambda <{\lambda }^{*}$, g has exactly two positive roots ${\alpha }_1,{\alpha }_2$ (with $0<{\alpha }_1<{\alpha }^{*}<{\alpha }_2$). Thus, the solution set of the strict inequality is the interval ${\alpha }_1<\alpha <{\alpha }_2$.

From the second case, the requirement is $\lambda <log(1+q)$, or $q>e^{\lambda }-1$. Fix such a q. Then, there exists an interval $({\alpha }_1,{\alpha }_2)$ such that $1+q+\lambda \alpha <{\gamma }^{\left(q\right)}\left(\alpha \right)$ for all $\alpha \in ({\alpha }_1,{\alpha }_2)$. Fix any $\alpha$ in this interval. Then, it suffices to choose $\delta$ small enough so that $1+q+\lambda \alpha <-log\delta$; this guarantees (24). □

5.2. Irrationality Criterion and Completion of the Proof

With (24) satisfied, $|{\Lambda }_m^{\left(q\right)}|\to 0$ as $m\to \infty$. This places us precisely within the framework of the following general criterion.

Theorem 5.1 

(Irrationality criterion). Let $x\in \mathbb{R}$. Suppose there exist infinitely many integers $A_m\ne 0$ and $B_m$, such that

$$0<|{\Lambda }_m|=|A_{m}x-B_m|\le {ϵ}_m,$$

(25)

where ${ϵ}_m\to 0$ as $m\to \infty$. Then, x is irrational.

Proof. 

Assume that $x={\displaystyle \frac{p^{*}}{q^{*}}}\in \mathbb{Q}$ is in lowest terms. Then,

$$|{\Lambda }_m|={\displaystyle \frac{|A_m{p}^{*}-B_m{q}^{*}|}{|q^{*}|}}.$$

Hence, for every m, either $|{\Lambda }_m|=0$ or $|{\Lambda }_m|\ge {\displaystyle \frac{1}{|q^{*}|}}$. By (25), $|{\Lambda }_m|\le {ϵ}_m\to 0$ as $m\to \infty$. So, for all sufficiently large m, we have ${ϵ}_m<{\displaystyle \frac{1}{|q^{*}|}}$, contradicting the dichotomy above. Therefore, x cannot be rational. □

We thus conclude that each $\zeta (2n+1)$ is irrational. This completes the proof of Theorem 1.1.