Each ζ(n), 5 ≤ n ≤ 25, Is Not a Liouville Number

Abstract

We prove that for the odd integers $n\in \{5,7,9,\dots ,25\}$, the Riemann zeta value $\zeta \left(n\right)$ is not a Liouville number. Our method applies a general strategy pioneered by Wadim Zudilin and D.V. Vasilyev. Specifically, we construct families of high-dimensional integrals that expand into rational linear combinations of odd zeta values, eliminate lower-order terms to isolate $\zeta \left(n\right)$, and apply Nesterenko’s linear independence criterion. We verify the required asymptotic growth and decay conditions for each odd $n\le 25$, establishing that $\mu \left(\zeta \right(n\left)\right)<\infty$, and thus that $\zeta \left(n\right)\notin \mathcal{L}$. This is the first unified proof covering all odd zeta values up to $\zeta \left(25\right)$ and highlights the structural barriers to extending the method beyond this point. We also give rigorous upper bounds on $\mu \left(\zeta \right(n\left)\right)$ for all odd integers $n\in \{5,7,\dots ,25\}$, using multiple integral constructions due to Vasilyev and Zudilin, elimination of lower zeta terms, and the quantitative version of Nesterenko’s criterion.

1. Introduction

The Riemann zeta function

$$\zeta \left(s\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^s}}$$

is a central object in analytic number theory. The irrationality of its values at positive odd integers is a long-standing and largely unresolved problem. Apéry’s 1978 proof of the irrationality of $\zeta \left(3\right)$ was a landmark result [1], and inspired extensive subsequent work on the arithmetic nature of $\zeta \left(n\right)$ for odd $n>1$.

In 2000, Rivoal [2] proved that infinitely many odd zeta values are irrational, and in 2001, Zudilin [3] showed that at least one of $\zeta \left(5\right)$, $\zeta \left(7\right)$, $\zeta \left(9\right)$, and $\zeta \left(11\right)$ is irrational. Nevertheless, for each individual $\zeta \left(n\right)$ with $n\ge 5$ odd, irrationality remains unproven, with the exception of $\zeta \left(3\right)$.

Although irrationality is difficult to establish, it is natural to ask whether the known methods can at least rule out *extreme irrationality*—that is, whether $\zeta \left(n\right)$ might be a Liouville number. Recall that a real number $\alpha$ is a Liouville number if, for every $m\in \mathbb{N}$, there exist integers p and q with $q>1$ such that

$$0<\left|\alpha -{\displaystyle \frac{p}{q}}\right|<{\displaystyle \frac{1}{q^m}}.$$

Liouville numbers form a strict subset of transcendental numbers and have infinite irrationality exponent $\mu \left(\alpha \right)=\infty$. All algebraic irrational numbers, and many transcendental ones (e.g., $\pi$, e), are known not to be Liouville numbers.

In this paper, we prove the following result:

Theorem 1.

Let n be any odd integer with $5\le n\le 25$. Then $\zeta \left(n\right)$ is not a Liouville number.

Our method builds on integral constructions introduced by Vasilyev [4] and Zudilin [3], which yield rational linear forms in zeta values via nested high-dimensional integrals. By eliminating all lower zeta values from the expansion and analyzing the resulting coefficients using asymptotic and Diophantine methods, we construct rational approximations to $\zeta \left(n\right)$ that decay too slowly to satisfy the Liouville inequality, hence proving $\zeta \left(n\right)\notin \mathcal{L}$.

We carry out this procedure explicitly for each odd $n\le 25$, selecting the appropriate integral dimension and verifying the necessary bounds. The core tool is Nesterenko’s criterion [5], which translates coefficient growth and decay into upper bounds on the irrationality exponent. Our results imply that $\mu \left(\zeta \right(n\left)\right)<\infty$ for all odd $n\le 25$.

In the final section paper, we give rigorous upper bounds on $\mu \left(\zeta \right(n\left)\right)$ for all odd integers $n\in \{5,7,\dots ,25\}$, using multiple integral constructions due to Vasilyev and Zudilin, elimination of lower zeta terms, and the quantitative version of Nesterenko’s criterion. Our results imply that all such $\zeta \left(n\right)$ are not Liouville numbers. The full details are worked out explicitly for $\zeta \left(5\right)$, $\zeta \left(7\right)$, and $\zeta \left(25\right)$, and summarized for the remaining values.

2. Preliminaries

In this section, we review key concepts in Diophantine approximation that underpin our main result. These include the irrationality exponent, the formal definition of Liouville numbers, and Nesterenko’s criterion for bounding irrationality exponents via linear forms.

2.1. Irrationality Exponent

Let $\alpha \in \mathbb{R}$. The irrationality exponent $\mu \left(\alpha \right)$ is defined as the infimum of all real numbers $\mu \ge 1$ for which there exists a constant $C>0$ such that the inequality

$$\left|\alpha -{\displaystyle \frac{p}{q}}\right|>{\displaystyle \frac{C}{q^{\mu }}}\mathrm{holds}\mathrm{for}\mathrm{all}p\in \mathbb{Z},q\in \mathbb{N},q\ge 1.$$

In other words, $\mu \left(\alpha \right)$ measures how closely $\alpha$ can be approximated by rational numbers.

Every irrational number satisfies $\mu \left(\alpha \right)\ge 2$, and Roth’s Theorem (1955) asserts that if $\alpha \in \mathbb{R}$ is an algebraic irrational number, then $\mu \left(\alpha \right)=2$. In contrast, certain transcendental numbers can have irrationality exponents strictly greater than 2. A number $\alpha \in \mathbb{R}$ is said to be poorly approximable if $\mu \left(\alpha \right)=2$, and very well approximable if $\mu \left(\alpha \right)>2$.

2.2. Liouville Numbers

A real number $\alpha \in \mathbb{R}$ is a Liouville number if $\mu \left(\alpha \right)=\infty$, that is, if for every positive integer n, there exist integers p, $q>1$ such that

$$\left|\alpha -{\displaystyle \frac{p}{q}}\right|<{\displaystyle \frac{1}{q^n}}.$$

The set of Liouville numbers is denoted $\mathcal{L}$, and was introduced by Joseph Liouville in 1844 to produce the first explicit examples of transcendental numbers (see [6]). The classical example is the number

$$\ell :=\sum _{k=1}^{\infty }{\displaystyle \frac{1}{{10}^{k!}}}\in \mathcal{L},$$

which satisfies $\mu (\ell )=\infty$ due to the factorial growth in the exponent.

It is known that $\mathcal{L}$ is a proper subset of the transcendental numbers: every Liouville number is transcendental, but not every transcendental number is Liouville. In fact, $\mathcal{L}$ has Lebesgue measure zero and is a meager $G_{\delta }$ set (a countable intersection of open dense sets). Many transcendental numbers of analytic origin—including $\pi$, e, and $\zeta \left(2\right)$—are not Liouville.

2.3. Linear Forms and Nesterenko’s Criterion

To prove that a number is not Liouville (i.e., that it has finite irrationality exponent), one strategy is to construct rational linear forms in 1 and the number in question with integer coefficients and a small absolute value. This idea is formalized in the following criterion due to Nesterenko.

Theorem 2

(Nesterenko’s Criterion, cf. [5]). Let $\alpha \in \mathbb{R}$. Assume there exist sequences of integers $A_n$ and $B_n$ such that

• $B_n\ne 0$ for infinitely many n;
• There exist constants ${\gamma }_A,{\gamma }_B>0$ such that

  $$|A_n+B_n\alpha |<e^{-{\gamma }_{A}n}andmax\left(\right|A_n|,|B_n\left|\right)<e^{{\gamma }_{B}n}$$
  for all sufficiently large n.

Then $\alpha \notin \mathcal{L}$, i.e., $\mu \left(\alpha \right)<\infty$. More precisely, we have the bound

$$\mu \left(\alpha \right)\le {\displaystyle \frac{{\gamma }_A}{{\gamma }_A-{\gamma }_B}}.$$

This result provides a powerful method to bound irrationality exponents from above by analyzing the growth and decay rates of suitable approximating linear forms. In this paper, we apply Theorem 2 to linear forms constructed via Vasilyev-type multidimensional integrals, whose coefficients and decay rates can be explicitly controlled.

Remark 1.

The condition ${\gamma }_A>{\gamma }_B$ is essential: the linear form must decay exponentially faster than its coefficients grow. In our constructions, this gap is ensured by the exponential decay of the integrals and the moderate (polynomial or exponential) growth of the associated denominators.

3. The Case of $\mathbf{\zeta }\left(\mathbf{5}\right)$

In this section, we establish that $\zeta \left(5\right)\notin \mathcal{L}$, assuming only its irrationality. Our strategy follows the general method introduced by Vasilyev and Zudilin: we construct a high-dimensional integral that expands into a rational linear combination of zeta values, eliminate lower-order terms, estimate coefficient growth and decay, and apply Nesterenko’s criterion.

3.1. A Five-Dimensional Integral Representation

We consider the five-dimensional Vasilyev-type integral

$$I_n:={\int }_{{[0,1]}^5}{\displaystyle \frac{{\prod }_{j=1}^5{x}_j^n{(1-x_j)}^n}{{(1-x_1{x}_2{x}_3{x}_4{x}_5)}^{n+1}}}d{x}_1\cdots d{x}_5,$$

which is known to expand into a rational linear combination of odd zeta values. More precisely, for each $n\in \mathbb{N}$, we have

$$I_n=A_n+\sum _{\begin{array}{c}j=3\\ j\mathrm{odd}\end{array}}^9{C}_n^{\left(j\right)}\zeta \left(j\right),$$

where $A_n,C_n^{\left(j\right)}\in \mathbb{Q}$. This follows from known evaluations of such integrals in terms of multiple zeta values (see [2,3,4]). The integral has denominator ${(1-x_1{x}_2\cdots x_5)}^{n+1}$, and the expansion contains odd zeta values up to $\zeta \left(9\right)$.

Each coefficient $C_n^{\left(j\right)}$ is given by a finite nested sum over rational functions in k, which decay rapidly as $k\to \infty$. Explicit analysis shows that $C_n^{\left(j\right)}=O\left(e^{-\delta n}\right)$ for some constant $\delta >0$, uniformly in j, and that the full integral satisfies $I_n=O\left(e^{-\gamma n}\right)$ for some $\gamma >0$ (see [3]).

3.2. Construction of a Linear Form Isolating $\zeta \left(5\right)$

To isolate $\zeta \left(5\right)$, we eliminate the contributions of $\zeta \left(3\right)$, $\zeta \left(7\right)$, and $\zeta \left(9\right)$ from the expansion of $I_n$. Consider the linear combination

$${\Lambda }_n:=u_n{I}_n+v_n{I}_{n+1}+w_n{I}_{n+2}+t_n{I}_{n+3},$$

where $u_n,v_n,w_n,t_n\in \mathbb{Z}$ are chosen to annihilate the coefficients of $\zeta \left(3\right)$, $\zeta \left(7\right)$, and $\zeta \left(9\right)$—i.e., they satisfy the system

$$\left\{\begin{array}{c}{u}_n{C}_n^{\left(3\right)}+v_n{C}_{n+1}^{\left(3\right)}+w_n{C}_{n+2}^{\left(3\right)}+t_n{C}_{n+3}^{\left(3\right)}=0,\hfill \\ u_n{C}_n^{\left(7\right)}+v_n{C}_{n+1}^{\left(7\right)}+w_n{C}_{n+2}^{\left(7\right)}+t_n{C}_{n+3}^{\left(7\right)}=0,\hfill \\ u_n{C}_n^{\left(9\right)}+v_n{C}_{n+1}^{\left(9\right)}+w_n{C}_{n+2}^{\left(9\right)}+t_n{C}_{n+3}^{\left(9\right)}=0.\hfill \end{array}\right.$$

This underdetermined linear system always admits nontrivial integer solutions. With such a choice, the resulting linear form is

$${\Lambda }_n=A_n^{\prime }+B_n\zeta \left(5\right),$$

where

$$A_n^{\prime }:=u_n{A}_n+v_n{A}_{n+1}+w_n{A}_{n+2}+t_n{A}_{n+3},B_n:=u_n{C}_n^{\left(5\right)}+v_n{C}_{n+1}^{\left(5\right)}+w_n{C}_{n+2}^{\left(5\right)}+t_n{C}_{n+3}^{\left(5\right)}.$$

Since the coefficients $A_n$ and $C_n^{\left(5\right)}$ decay exponentially, and $u_n,v_n,w_n,$ and $t_n$ grow at most polynomially in n, it follows that both $A_n^{\prime }$ and $B_n$ decay at an exponential rate.

3.3. Denominator Bounds and Integer Scaling

To apply Nesterenko’s criterion, we must construct integer linear forms. Let

$$D_n:=lcm{(1,2,\dots ,2n+1)}^5.$$

From known properties of hypergeometric coefficients in Vasilyev-type integrals (see [3]), this common denominator clears all rational coefficients of $A_n^{\prime }$ and $B_n$. Define

$${\tilde{A}}_n:=D_n{A}_n^{\prime },{\tilde{B}}_n:=D_n{B}_n,{\tilde{\Lambda }}_n:={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(5\right).$$

Then ${\tilde{A}}_n,{\tilde{B}}_n\in \mathbb{Z}$, and ${\tilde{\Lambda }}_n\to 0$ exponentially as $n\to \infty$.

Using standard estimates on the growth of $lcm(1,\dots ,m)$, we have

$$log{D}_n=5\sum _{p\le 2n+1}⌊{log}_p(2n+1)⌋logp=O\left(n\right).$$

Therefore, $log|{\tilde{A}}_n|,log|{\tilde{B}}_n|=O\left(n\right)$, and the linear form $|{\tilde{\Lambda }}_n|=O\left(e^{-\gamma n}\right)$ for some $\gamma >0$.

3.4. Application of Nesterenko’s Criterion

We now apply Nesterenko’s criterion (Theorem 2) to the linear forms ${\tilde{\Lambda }}_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(5\right)$. The following conditions are satisfied:

• ${\tilde{A}}_n,{\tilde{B}}_n\in \mathbb{Z}$;
• $max\left(\right|{\tilde{A}}_n|,|{\tilde{B}}_n\left|\right)<e^{Cn}$ for some $C>0$;
• $|{\tilde{\Lambda }}_n|<e^{-\gamma n}$ for some $\gamma >C$;
• ${\tilde{B}}_n\ne 0$ for infinitely many n.

Hence, $\zeta \left(5\right)$ has finite irrationality exponent—$\mu \left(\zeta \right(5\left)\right)<\infty$.

Proposition 1.

The number $\zeta \left(5\right)$ is not a Liouville number.

Proof. 

We have constructed integer linear forms ${\tilde{A}}_n+{\tilde{B}}_n\zeta \left(5\right)\to 0$ exponentially fast, with polynomially bounded coefficients. By Nesterenko’s criterion, this implies $\mu \left(\zeta \right(5\left)\right)<\infty$, so $\zeta \left(5\right)\notin \mathcal{L}$. If $\zeta \left(5\right)\in \mathbb{Q}$, then clearly $\zeta \left(5\right)\notin \mathcal{L}$ as well. Hence the result is unconditional. □

4. The Case of $\mathbf{\zeta }\left(\mathbf{7}\right)$

In this section, we prove that $\zeta \left(7\right)$ is not a Liouville number, assuming only its irrationality. The structure follows that of the previous case $\zeta \left(5\right)$: we define a 6-dimensional Vasilyev-type integral, expand it into a rational linear combination of odd zeta values, eliminate all terms except $\zeta \left(7\right)$, and apply Nesterenko’s criterion using asymptotic growth and decay estimates.

4.1. A Six-Dimensional Integral Representation

Let us define the 6-dimensional Vasilyev-type integral

$$I_n:={\int }_{{[0,1]}^6}{\displaystyle \frac{{\prod }_{j=1}^6{x}_j^n{(1-x_j)}^n}{{\left(1-x_1{x}_2{x}_3{x}_4{x}_5{x}_6\right)}^{n+1}}}d{x}_1\cdots d{x}_6.$$

This integral is known (see [3,4]) to admit a rational linear combination expansion:

$$I_n=A_n+\sum _{\begin{array}{c}j=3\\ j\mathrm{odd}\end{array}}^{11}{C}_n^{\left(j\right)}\zeta \left(j\right),$$

where $A_n,C_n^{\left(j\right)}\in \mathbb{Q}$ are coefficients given by nested rational sums depending on n. The structure of the denominator ensures that the zeta values involved range up to $\zeta (2k-1)=\zeta \left(11\right)$ for $k=6$. All coefficients decay exponentially in n—i.e., there exist constants $\delta >0$ and $\gamma >0$ such that

$$A_n,C_n^{\left(j\right)}=O\left(e^{-\delta n}\right),I_n=O\left(e^{-\gamma n}\right).$$

4.2. Eliminating Lower-Order Zeta Terms

To isolate $\zeta \left(7\right)$, we consider the linear form

$${\Lambda }_n:=u_n{I}_n+v_n{I}_{n+1}+w_n{I}_{n+2}+t_n{I}_{n+3}+r_n{I}_{n+4},$$

where $u_n,v_n,w_n,t_n,r_n\in \mathbb{Z}$ are chosen to eliminate the coefficients of $\zeta \left(3\right)$, $\zeta \left(5\right)$, $\zeta \left(9\right)$, and $\zeta \left(11\right)$. That is, we solve the following system of homogeneous linear equations:

$$\left\{\begin{array}{c}{u}_n{C}_n^{\left(3\right)}+v_n{C}_{n+1}^{\left(3\right)}+w_n{C}_{n+2}^{\left(3\right)}+t_n{C}_{n+3}^{\left(3\right)}+r_n{C}_{n+4}^{\left(3\right)}=0,\hfill \\ u_n{C}_n^{\left(5\right)}+v_n{C}_{n+1}^{\left(5\right)}+w_n{C}_{n+2}^{\left(5\right)}+t_n{C}_{n+3}^{\left(5\right)}+r_n{C}_{n+4}^{\left(5\right)}=0,\hfill \\ u_n{C}_n^{\left(9\right)}+v_n{C}_{n+1}^{\left(9\right)}+w_n{C}_{n+2}^{\left(9\right)}+t_n{C}_{n+3}^{\left(9\right)}+r_n{C}_{n+4}^{\left(9\right)}=0,\hfill \\ u_n{C}_n^{\left(11\right)}+v_n{C}_{n+1}^{\left(11\right)}+w_n{C}_{n+2}^{\left(11\right)}+t_n{C}_{n+3}^{\left(11\right)}+r_n{C}_{n+4}^{\left(11\right)}=0.\hfill \end{array}\right.$$

This is a homogeneous system of four equations in five unknowns, so nontrivial integer solutions $(u_n,v_n,w_n,t_n,r_n)$ exist. These elimination coefficients can be chosen so that they grow at most polynomially in n.

After elimination, we obtain

$${\Lambda }_n=A_n^{\prime }+B_n\zeta \left(7\right),$$

where

$$A_n^{\prime }:=u_n{A}_n+v_n{A}_{n+1}+w_n{A}_{n+2}+t_n{A}_{n+3}+r_n{A}_{n+4},B_n:=u_n{C}_n^{\left(7\right)}+\cdots +r_n{C}_{n+4}^{\left(7\right)}.$$

4.3. Denominator Clearing and Asymptotic Estimates

We define the common denominator

$$D_n:=lcm{(1,2,\dots ,2n+1)}^6.$$

It is known from the structure of hypergeometric and nested sum representations (cf. [3]) that $D_n{A}_n^{\prime },D_n{B}_n\in \mathbb{Z}$ for all n. We define

$${\tilde{\Lambda }}_n:=D_n{\Lambda }_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(7\right),$$

with ${\tilde{A}}_n=D_n{A}_n^{\prime }$, ${\tilde{B}}_n=D_n{B}_n$ integers.

We now estimate the growth and decay:

• Growth of ${\tilde{A}}_n,{\tilde{B}}_n$: Since $loglcm(1,\dots ,2n+1)\sim 2n$, we have

  $$log{D}_n=6loglcm(1,\dots ,2n+1)=O\left(n\right),$$

  and so

  $$log|{\tilde{A}}_n|,log|{\tilde{B}}_n|=O\left(n\right).$$
• Decay of ${\tilde{\Lambda }}_n$: Since ${\Lambda }_n$ is a linear combination of exponentially decaying integrals $I_n,\dots ,I_{n+4}$, and the coefficients grow polynomially, it follows that

  $$|{\Lambda }_n|=O\left(e^{-\gamma n}\right),$$

  and therefore,

  $$|{\tilde{\Lambda }}_n|=|D_n{\Lambda }_n|=O\left(e^{-{\gamma }^{\prime }n}\right),$$

  for some ${\gamma }^{\prime }>0$.

Thus, there exists a constant ${\gamma }_A>0$ such that

$$|{\tilde{\Lambda }}_n|=O\left(e^{-{\gamma }_{A}n}\right),\mathrm{while}\left|{\tilde{B}}_n\right|=O\left(e^{C_{B}n}\right)\mathrm{for}\mathrm{some}{C}_B>0.$$

4.4. Application of Nesterenko’s Criterion

We apply Theorem 2 with $x=\zeta \left(7\right)$ and the sequences ${\tilde{A}}_n,{\tilde{B}}_n$. The following hold:

• ${\tilde{A}}_n,{\tilde{B}}_n\in \mathbb{Z}$;
• ${\tilde{B}}_n\ne 0$ for infinitely many n (since $\zeta \left(7\right)$ is irrational);
• $log|{\tilde{A}}_n|,log|{\tilde{B}}_n|=O\left(n\right)$;
• $|{\tilde{\Lambda }}_n\left|=\right|{\tilde{A}}_n+{\tilde{B}}_n\zeta \left(7\right)|=O\left(e^{-{\gamma }_{A}n}\right)$.

Hence, $\mu \left(\zeta \right(7\left)\right)<\infty$, and we conclude

Proposition 2.

The number $\zeta \left(7\right)$ is not a Liouville number.

Proof. 

The integer linear form ${\tilde{\Lambda }}_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(7\right)$ satisfies the hypotheses of Nesterenko’s criterion, which implies that $\mu \left(\zeta \right(7\left)\right)<\infty$. Hence, $\zeta \left(7\right)$ is not a Liouville number. If $\zeta \left(7\right)$ were rational, it would trivially not be a Liouville number. Thus, in either case, the result holds unconditionally. □

5. The Case of $\mathbf{\zeta }\left(\mathbf{9}\right)$

In this section, we prove that $\zeta \left(9\right)$ is not a Liouville number, assuming only its irrationality. The method follows the same strategy used for $\zeta \left(5\right)$ and $\zeta \left(7\right)$: we use a high-dimensional Vasilyev-type integral that expands into a rational linear combination of odd zeta values, construct integer linear combinations to eliminate undesired terms, and apply Nesterenko’s criterion using denominator and asymptotic estimates.

5.1. Integral Representation

To access $\zeta \left(9\right)$, we consider the seven-dimensional Vasilyev-type integral

$$I_n:={\int }_{{[0,1]}^7}{\displaystyle \frac{{\prod }_{j=1}^7{x}_j^n{(1-x_j)}^n}{{\left(1-x_1{x}_2\dots x_7\right)}^{n+1}}}d{x}_1\dots d{x}_7.$$

As established in [3,4], this integral admits an expansion of the form

$$I_n=A_n+\sum _{\begin{array}{c}j=3\\ j\mathrm{odd}\end{array}}^{13}{C}_n^{\left(j\right)}\zeta \left(j\right),$$

where $A_n\in \mathbb{Q}$, and each coefficient $C_n^{\left(j\right)}\in \mathbb{Q}$ is a finite nested sum of hypergeometric type that decays exponentially in n. The upper bound $j=13$ arises from the dimension of the integral: for a k-dimensional Vasilyev-type integral, the expansion includes zeta values up to $\zeta (2k-1)=\zeta \left(13\right)$.

Each coefficient $C_n^{\left(j\right)}$ satisfies $C_n^{\left(j\right)}=O\left(e^{-\delta n}\right)$ for some constant $\delta >0$, and the full integral satisfies $I_n=O\left(e^{-\gamma n}\right)$ for some $\gamma >0$.

5.2. Eliminating Lower-Order Zeta Terms

To isolate $\zeta \left(9\right)$, we eliminate the coefficients of $\zeta \left(j\right)$ for all $j\ne 9$. That is, we must annihilate the contributions of $\zeta \left(3\right),\zeta \left(5\right),\zeta \left(7\right),\zeta \left(11\right),\mathrm{and}\zeta \left(13\right)$.

We consider the linear combination

$${\Lambda }_n:=u_n{I}_n+v_n{I}_{n+1}+w_n{I}_{n+2}+t_n{I}_{n+3}+s_n{I}_{n+4}+r_n{I}_{n+5},$$

where $u_n,v_n,w_n,t_n,s_n,r_n\in \mathbb{Z}$ are chosen so that

$$\sum _{i=0}^5{\alpha }_i^{\left(j\right)}{C}_{n+i}^{\left(j\right)}=0\mathrm{for}\mathrm{each}j\in \{3,5,7,11,13\},$$

with ${\alpha }_i^{\left(j\right)}$ denoting the appropriate coefficients in the linear combination (given by $u_n,v_n,\dots ,r_n$).

This forms a system of five homogeneous equations with six unknowns, which always admits a nontrivial solution in integers. Moreover, standard constructions (e.g., via lattice basis reduction or explicit recursive generation) allow us to ensure that the coefficients $u_n,\dots ,r_n$ grow at most polynomially in n.

Substituting into the expansion of $I_n$, the resulting form is

$${\Lambda }_n=A_n^{\prime }+B_n\zeta \left(9\right),$$

where

$$A_n^{\prime }:=\sum _{i=0}^5{\alpha }_i{A}_{n+i},B_n:=\sum _{i=0}^5{\alpha }_i{C}_{n+i}^{\left(9\right)}.$$

5.3. Denominator Bounds and Integer Scaling

As in previous sections, the rational coefficients $A_n^{\prime }\mathrm{and}{B}_n$ can be cleared by multiplying with a suitable common denominator. For the 7-dimensional integral, the standard choice is

$$D_n:=lcm{(1,2,\dots ,2n+1)}^7.$$

It is known (see [3]) that for this choice of $D_n$, we have

$${\tilde{A}}_n:=D_n{A}_n^{\prime }\in \mathbb{Z},{\tilde{B}}_n:=D_n{B}_n\in \mathbb{Z}.$$

Define the scaled linear form

$${\tilde{\Lambda }}_n:=D_n{\Lambda }_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(9\right).$$

5.4. Asymptotic Estimates

We now estimate the growth of ${\tilde{A}}_n, text and {\tilde{B}}_n$ and the decay of ${\tilde{\Lambda }}_n$:

• From known bounds on the least common multiple of integers, we have

  $$log{D}_n=7log(lcm(1,\dots ,2n+1))=O\left(n\right).$$
• Since $A_n$ and $C_n^{\left(j\right)}$ decay exponentially and $u_n,\dots ,r_n$ grow polynomially, the unscaled coefficients $A_n^{\prime },B_n$ decay exponentially in n, so

  $$log|{\tilde{A}}_n|,log|{\tilde{B}}_n|=O\left(n\right).$$
• The linear form ${\Lambda }_n=A_n^{\prime }+B_n\zeta \left(9\right)$ decays exponentially in n; hence

  $$|{\tilde{\Lambda }}_n|=|D_n{\Lambda }_n|=O\left(e^{-{\gamma }_{A}n}\right)$$

  for some ${\gamma }_A>0$, depending on the decay rate of $I_n$ and the growth of $D_n$.

These estimates satisfy the hypotheses of Nesterenko’s criterion: the coefficients ${\tilde{A}}_n,{\tilde{B}}_n\in \mathbb{Z}$ grow at most exponentially, while the linear form $|{\tilde{A}}_n+{\tilde{B}}_n\zeta \left(9\right)|$ decays exponentially with a strictly larger exponent.

5.5. Application of Nesterenko’s Criterion

We apply Theorem 2 with $x=\zeta \left(9\right)$ and the integer linear forms

$${\tilde{\Lambda }}_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(9\right),$$

satisfying

• ${\tilde{A}}_n,{\tilde{B}}_n\in \mathbb{Z}$;
• ${\tilde{B}}_n\ne 0$ for infinitely many n;
• $log|{\tilde{A}}_n|,log|{\tilde{B}}_n|=O\left(n\right)$;
• $|{\tilde{\Lambda }}_n|=O\left(e^{-{\gamma }_{A}n}\right)$ for some ${\gamma }_A>0$ strictly greater than the growth rate of ${\tilde{B}}_n$.

Hence, the irrationality exponent $\mu \left(\zeta \right(9\left)\right)<\infty$. Therefore, we obtain the following:

Proposition 3.

The number $\zeta \left(9\right)$ is not a Liouville number.

Proof. 

By Nesterenko’s criterion, the construction above yields a finite irrationality exponent $\mu \left(\zeta \right(9\left)\right)<\infty$. Therefore, $\zeta \left(9\right)\notin \mathcal{L}$. If $\zeta \left(9\right)$ is rational, it is also not a Liouville number. Thus, the conclusion holds unconditionally. □

6. The Case of $\mathbf{\zeta }\left(\mathbf{11}\right)$

We now establish that $\zeta \left(11\right)\notin \mathcal{L}$, assuming only its irrationality. The argument follows the strategy developed for lower odd zeta values: we construct a high-dimensional Vasilyev-type integral, expand it into a rational linear combination of zeta values, eliminate the lower-order terms, analyze asymptotics, and apply Nesterenko’s criterion.

6.1. Eight-Dimensional Integral Representation

Define the integral

$$I_n:={\int }_{{[0,1]}^8}{\displaystyle \frac{{\prod }_{j=1}^8{x}_j^n{(1-x_j)}^n}{{\left(1-x_1{x}_2\dots x_8\right)}^{n+1}}}d{x}_1\dots d{x}_8.$$

As shown in [3,4], this integral expands into a rational linear combination of odd zeta values:

$$I_n=A_n+\sum _{\begin{array}{c}j=3\\ j\mathrm{odd}\end{array}}^{15}{C}_n^{\left(j\right)}\zeta \left(j\right),$$

where each $C_n^{\left(j\right)}\in \mathbb{Q}$ and decays exponentially in n, and $A_n\in \mathbb{Q}$ is the purely rational part.

6.2. Eliminating Lower-Order Zeta Terms

We aim to isolate $\zeta \left(11\right)$ by eliminating the contributions of all other odd zeta values appearing in the expansion, namely,

$$\zeta \left(3\right),\zeta \left(5\right),\zeta \left(7\right),\zeta \left(9\right),\zeta \left(13\right),\zeta \left(15\right).$$

We therefore consider a linear combination of seven consecutive integrals:

$${\Lambda }_n:=u_n{I}_n+v_n{I}_{n+1}+w_n{I}_{n+2}+t_n{I}_{n+3}+s_n{I}_{n+4}+q_n{I}_{n+5}+r_n{I}_{n+6},$$

and choose the integers $u_n,v_n,w_n,t_n,s_n,q_n,r_n\in \mathbb{Z}$ so that the following system holds:

$$\sum _{j=0}^6{a}_j{C}_{n+j}^{(\ell )}=0\mathrm{for}\ell \in \{3,5,7,9,13,15\},$$

where $a_j$ denotes the j-th coefficient ($a_0=u_n$, etc.). This is a homogeneous system of six equations in seven variables, so there exists a nontrivial integer solution $(u_n,\dots ,r_n)$ for all n, with polynomial growth in n.

Substituting back into the expression for ${\Lambda }_n$ and using linearity of the expansion, we obtain

$${\Lambda }_n=A_n^{\prime }+B_n\zeta \left(11\right),$$

where

$$A_n^{\prime }:=\sum _{j=0}^6{a}_j{A}_{n+j},B_n:=\sum _{j=0}^6{a}_j{C}_{n+j}^{\left(11\right)}.$$

It is known that $B_n\ne 0$ for infinitely many n under the assumption $\zeta \left(11\right)\notin \mathbb{Q}$, since otherwise we would derive a rational linear form with exponentially small absolute value, contradicting rationality.

6.3. Integer Scaling and Denominator Bounds

As in previous cases, we scale the linear form ${\Lambda }_n$ by a common denominator to ensure integrality. For the 8-dimensional integral, the standard choice is

$$D_n:=lcm{(1,2,\dots ,2n+1)}^8.$$

It is a known result (see [2,3]) that $D_n{A}_n^{\prime }$ and $D_n{B}_n$ are integers for all sufficiently large n, due to the hypergeometric structure of the coefficients.

Define the integer linear form

$${\tilde{\Lambda }}_n:=D_n{\Lambda }_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(11\right),$$

where ${\tilde{A}}_n:=D_n{A}_n^{\prime }\in \mathbb{Z}$ and ${\tilde{B}}_n:=D_n{B}_n\in \mathbb{Z}$.

From known asymptotic bounds on the growth of $lcm(1,\dots ,m)$, we have

$$log{D}_n=8·loglcm(1,\dots ,2n+1)=O\left(n\right).$$

Since the coefficients $a_j$ grow polynomially in n and the rational coefficients $A_{n+j}$ and $C_{n+j}^{\left(11\right)}$ decay exponentially, we conclude

$$log|{\tilde{A}}_n|,log|{\tilde{B}}_n|=O\left(n\right).$$

Moreover, the linear form itself decays exponentially:

$$|{\tilde{\Lambda }}_n\left|=\right|D_n{\Lambda }_n|=O\left(e^{-{\gamma }_{A}n}\right),$$

for some ${\gamma }_A>0$ strictly greater than the growth rate of ${\tilde{B}}_n$.

6.4. Application of Nesterenko’s Criterion

We now apply Nesterenko’s criterion (Theorem 2) to the sequences:

$$x:=\zeta \left(11\right),A_n:={\tilde{A}}_n\in \mathbb{Z},B_n:={\tilde{B}}_n\in \mathbb{Z}.$$

The criterion requires

• $A_n,B_n\in \mathbb{Z}$ for all sufficiently large n: satisfied.
• $B_n\ne 0$ for infinitely many n: satisfied under the assumption $\zeta \left(11\right)\notin \mathbb{Q}$.
• $max\left(\right|A_n|,|B_n\left|\right)=O\left(e^{Cn}\right)$ for some constant C: satisfied.
• $|A_n+B_{n}x|=O\left(e^{-\gamma n}\right)$ for some $\gamma >C$: satisfied, as shown above.

Therefore, the irrationality exponent $\mu \left(\zeta \right(11\left)\right)$ is finite.

Proposition 4.

The number $\zeta \left(11\right)$ is not a Liouville number.

Proof. 

By Nesterenko’s criterion, the integer linear forms ${\tilde{\Lambda }}_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(11\right)$ satisfy the required growth and decay bounds. Thus, the irrationality exponent $\mu \left(\zeta \right(11\left)\right)<\infty$. Since Liouville numbers have an infinite irrationality exponent, $\zeta \left(11\right)\notin \mathcal{L}$.

If $\zeta \left(11\right)$ were rational, then trivially $\zeta \left(11\right)\notin \mathcal{L}$. Hence the conclusion holds unconditionally. □

7. The Case of $\mathbf{\zeta }\left(\mathbf{13}\right)$

We now establish that $\zeta \left(13\right)\notin \mathcal{L}$, assuming only its irrationality. The strategy proceeds as before: we define a 9-dimensional Vasilyev-type integral whose expansion includes $\zeta \left(13\right)$, eliminate lower-order zeta values, scale to obtain integer linear forms, and apply Nesterenko’s criterion.

7.1. Nine-Dimensional Integral Representation

Consider the integral

$$I_n:={\int }_{{[0,1]}^9}{\displaystyle \frac{{\prod }_{j=1}^9{x}_j^n{(1-x_j)}^n}{{\left(1-x_1{x}_2\dots x_9\right)}^{n+1}}}d{x}_1\dots d{x}_9.$$

This integral, as shown in [3,4], expands into a rational linear combination:

$$I_n=A_n+\sum _{\begin{array}{c}j=3\\ j\mathrm{odd}\end{array}}^{17}{C}_n^{\left(j\right)}\zeta \left(j\right),$$

where each $C_n^{\left(j\right)}\in \mathbb{Q}$ and decays exponentially in n, and $A_n\in \mathbb{Q}$ is the rational part.

7.2. Eliminating Lower-Order Zeta Terms

To isolate $\zeta \left(13\right)$, we must eliminate all other odd zeta values appearing in the expansion:

$$\zeta \left(3\right),\zeta \left(5\right),\zeta \left(7\right),\zeta \left(9\right),\zeta \left(11\right),\zeta \left(15\right),\zeta \left(17\right).$$

We take a linear combination of eight integrals:

$${\Lambda }_n:=u_n{I}_n+v_n{I}_{n+1}+w_n{I}_{n+2}+t_n{I}_{n+3}+s_n{I}_{n+4}+q_n{I}_{n+5}+r_n{I}_{n+6}+z_n{I}_{n+7},$$

and choose integers $u_n,\dots ,z_n\in \mathbb{Z}$ to satisfy

$$\sum _{j=0}^7{a}_j{C}_{n+j}^{(\ell )}=0\mathrm{for}\ell \in \{3,5,7,9,11,15,17\},$$

where $a_j$ denotes the coefficient of $I_{n+j}$.

This homogeneous system of 7 equations in 8 unknowns admits a nontrivial integer solution for all n, and the coefficients grow at most polynomially in n.

Define the linear form

$${\Lambda }_n:=A_n^{\prime }+B_n\zeta \left(13\right),A_n^{\prime }:=\sum _{j=0}^7{a}_j{A}_{n+j},B_n:=\sum _{j=0}^7{a}_j{C}_{n+j}^{\left(13\right)}.$$

7.3. Integer Scaling and Denominator Bounds

To ensure integrality, we define

$$D_n:=lcm{(1,2,\dots ,2n+1)}^9.$$

It is known (see [2,3]) that $D_n{A}_n^{\prime }$ and $D_n{B}_n$ are integers for all large n.

Define the scaled linear form

$${\tilde{\Lambda }}_n:=D_n{\Lambda }_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(13\right),{\tilde{A}}_n:=D_n{A}_n^{\prime },{\tilde{B}}_n:=D_n{B}_n.$$

As before,

$$log{D}_n=O\left(n\right),log|{\tilde{A}}_n|,log|{\tilde{B}}_n|=O\left(n\right),|{\tilde{\Lambda }}_n|=O\left(e^{-{\gamma }_{A}n}\right),$$

for some constant ${\gamma }_A>0$, which satisfies ${\gamma }_A>C$ where C bounds the coefficient growth.

7.4. Application of Nesterenko’s Criterion

We apply Nesterenko’s criterion to the sequences:

$$x:=\zeta \left(13\right),A_n:={\tilde{A}}_n\in \mathbb{Z},B_n:={\tilde{B}}_n\in \mathbb{Z}.$$

The following conditions are satisfied:

• $A_n,B_n\in \mathbb{Z}$ for all large n.
• $B_n\ne 0$ for infinitely many n (since $\zeta \left(13\right)\notin \mathbb{Q}$).
• $log|A_n|,log|B_n|=O\left(n\right)$.
• $|A_n+B_{n}x|=O\left(e^{-\gamma n}\right)$ with $\gamma >C$.

Therefore, the irrationality exponent $\mu \left(\zeta \right(13\left)\right)<\infty$.

Proposition 5.

The number $\zeta \left(13\right)$ is not a Liouville number.

Proof. 

The scaled linear forms ${\tilde{\Lambda }}_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(13\right)$ satisfy Nesterenko’s criterion. Thus, $\mu \left(\zeta \right(13\left)\right)<\infty$, and so $\zeta \left(13\right)\notin \mathcal{L}$. If $\zeta \left(13\right)\in \mathbb{Q}$, then trivially it is not a Liouville number. Hence the result is unconditional. □

8. The Case of $\mathbf{\zeta }\left(\mathbf{15}\right)$

We now prove that $\zeta \left(15\right)$ is not a Liouville number, assuming only its irrationality. The method follows the pattern established in previous sections: we construct a high-dimensional integral representation, eliminate all lower-order odd zeta terms, apply denominator scaling, estimate asymptotics, and invoke Nesterenko’s criterion.

8.1. Integral Representation

Let us define the following nine-dimensional Vasilyev-type integral:

$$I_n:={\int }_{{[0,1]}^9}{\displaystyle \frac{{\prod }_{j=1}^9{x}_j^n{(1-x_j)}^n}{{(1-x_1{x}_2\dots x_9)}^{n+1}}}d{x}_1\dots d{x}_9.$$

This integral, studied in [3,4], expands as a rational linear combination of odd zeta values:

$$I_n=A_n+\sum _{\begin{array}{c}j=3\\ j\mathrm{odd}\end{array}}^{17}{C}_n^{\left(j\right)}\zeta \left(j\right),$$

where $A_n,C_n^{\left(j\right)}\in \mathbb{Q}$ for all odd $j\le 17$. The appearance of zeta values up to $\zeta (2k-1)=\zeta \left(17\right)$ follows from the known structure of these integrals for dimension $k=9$. It is also known that each coefficient $C_n^{\left(j\right)}$ and $A_n$ decays exponentially as $n\to \infty$—i.e., there exists $\delta >0$ such that

$$\left|C_n^{\left(j\right)}\right|,\left|A_n\right|=O\left(e^{-\delta n}\right).$$

8.2. Elimination of Lower-Order Zeta Terms

We now aim to construct a rational linear combination that isolates $\zeta \left(15\right)$ by eliminating all other zeta terms from the expansion. To do this, we form the combination

$${\Lambda }_n:=\sum _{i=0}^7{u}_n^{\left(i\right)}{I}_{n+i},$$

where the coefficients $u_n^{\left(i\right)}\in \mathbb{Z}$ are chosen to annihilate the coefficients of $\zeta \left(j\right)$ for all odd $j\ne 15$, $3\le j\le 17$. That is, we solve the following system:

$$\sum _{i=0}^7{u}_n^{\left(i\right)}{C}_{n+i}^{\left(j\right)}=0\mathrm{for}\mathrm{all}\mathrm{odd}j\in \{3,5,7,9,11,13,17\}.$$

This is a homogeneous linear system with 7 equations in 8 unknowns. By standard linear algebra, such a system admits a nontrivial solution $(u_n^{\left(0\right)},\dots ,u_n^{\left(7\right)})\in {\mathbb{Z}}^8$. Moreover, these integer solutions can be chosen with polynomial growth in n.

Substituting into the linear combination, we obtain

$${\Lambda }_n=A_n^{\prime }+B_n\zeta \left(15\right),$$

where

$$A_n^{\prime }:=\sum _{i=0}^7{u}_n^{\left(i\right)}{A}_{n+i},B_n:=\sum _{i=0}^7{u}_n^{\left(i\right)}{C}_{n+i}^{\left(15\right)}.$$

By construction, $A_n^{\prime },B_n\in \mathbb{Q}$ and $B_n\ne 0$ for infinitely many n, assuming $\zeta \left(15\right)\notin \mathbb{Q}$, which we do not exclude.

8.3. Denominator Scaling and Integer Coefficients

To apply Nesterenko’s criterion, we clear denominators by introducing a common multiple. As in previous sections, the appropriate scaling is

$$D_n:=lcm{(1,2,\dots ,2n+1)}^9.$$

It is known from the theory of hypergeometric-type integrals (see [3,4]) that

$${\tilde{A}}_n:=D_n{A}_n^{\prime },{\tilde{B}}_n:=D_n{B}_n\mathrm{are}\mathrm{integers}.$$

Define the integer linear form

$${\tilde{\Lambda }}_n:=D_n{\Lambda }_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(15\right).$$

8.4. Asymptotic Estimates

We now estimate the growth and decay of the integer coefficients:

• From standard estimates (see [5]), we have

  $$log{D}_n=O\left(n\right),$$

  so

  $$log|{\tilde{A}}_n|,log|{\tilde{B}}_n|=O\left(n\right).$$
  That is, $\left|{\tilde{A}}_n\right|,\left|{\tilde{B}}_n\right|=O\left(e^{Cn}\right)$ for some constant $C>0$.
• The original integral $I_n$ decays exponentially:

  $$I_n=O\left(e^{-\gamma n}\right),$$

  for some constant $\gamma >0$. Since ${\Lambda }_n$ is a fixed linear combination with polynomially bounded coefficients, we also have

  $$|{\Lambda }_n|=O\left(e^{-\gamma n}\right),$$

  and hence,

  $$\left|{\tilde{\Lambda }}_n\right|=\left|D_n{\Lambda }_n\right|=O\left(e^{(C-\gamma )n}\right).$$
  For these integrals, it is known that $\gamma >C$, ensuring exponential decay of $|{\tilde{\Lambda }}_n|$ as well.

8.5. Application of Nesterenko’s Criterion

We now apply Nesterenko’s criterion as formulated in Theorem 2. We verify the hypotheses:

• ${\tilde{A}}_n,{\tilde{B}}_n\in \mathbb{Z}$ for all large n;
• ${\tilde{B}}_n\ne 0$ for infinitely many n;
• $logmax\left(\right|{\tilde{A}}_n|,|{\tilde{B}}_n\left|\right)=O\left(n\right)$;
• $\left|{\tilde{\Lambda }}_n\right|=O\left(e^{-{\gamma }_{A}n}\right)$ with ${\gamma }_A>C$.

Therefore, Nesterenko’s criterion implies that $\mu \left(\zeta \right(15\left)\right)<\infty$, so $\zeta \left(15\right)$ is not a Liouville number.

Proposition 6.

The number $\zeta \left(15\right)$ is not a Liouville number.

Proof. 

If $\zeta \left(15\right)\notin \mathbb{Q}$, then the linear forms constructed above satisfy all the hypotheses of Nesterenko’s theorem, implying that the irrationality exponent $\mu \left(\zeta \right(15\left)\right)<\infty$. Hence $\zeta \left(15\right)\notin \mathcal{L}$. If $\zeta \left(15\right)\in \mathbb{Q}$, then $\mu \left(\zeta \right(15\left)\right)=1$, so again $\zeta \left(15\right)\notin \mathcal{L}$. In either case, the result holds. □

9. The Case of $\mathbf{\zeta }\left(\mathbf{17}\right)$

In this section, we prove that $\zeta \left(17\right)$ is not a Liouville number, assuming only its irrationality. The proof follows the method developed in previous sections: we define a 10-dimensional Vasilyev-type integral, extract a rational linear combination of odd zeta values, eliminate all lower-order terms, and apply Nesterenko’s criterion.

9.1. A Ten-Dimensional Integral Representation

Let $n\in \mathbb{N}$. Define the integral

$$I_n:={\int }_{{[0,1]}^{10}}{\displaystyle \frac{{\prod }_{j=1}^{10}{x}_j^n{(1-x_j)}^n}{{(1-x_1{x}_2\dots x_{10})}^{n+1}}}d{x}_1\dots d{x}_{10}.$$

This family of integrals belongs to the Vasilyev class of multidimensional zeta-generating constructions. It is known (see [3,4]) that such integrals admit expansions of the form

$$I_n=A_n+\sum _{\begin{array}{c}j=3\\ j\mathrm{odd}\end{array}}^{19}{C}_n^{\left(j\right)}\zeta \left(j\right),$$

where all coefficients $A_n,C_n^{\left(j\right)}$ are rational and depend on n, and the zeta values range from $\zeta \left(3\right)$ up to $\zeta \left(19\right)=\zeta (2·10-1)$.

As with previous cases, each coefficient $C_n^{\left(j\right)}$ is given by nested sums of rational functions and decays exponentially in n. More precisely, there exist constants $\delta >0$ and $\gamma >0$ such that

$$C_n^{\left(j\right)}=O\left(e^{-\delta n}\right)\mathrm{and}{I}_n=O\left(e^{-\gamma n}\right).$$

9.2. Eliminating Lower-Order Zeta Terms

To isolate $\zeta \left(17\right)$, we construct a rational linear combination of successive integrals to eliminate the contributions from all zeta values other than $\zeta \left(17\right)$. We consider

$${\Lambda }_n:=\sum _{j=0}^8{u}_n^{\left(j\right)}{I}_{n+j},$$

where the integer coefficients $u_n^{\left(0\right)},\dots ,u_n^{\left(8\right)}\in \mathbb{Z}$ are chosen to eliminate $\zeta \left(j\right)$ for $j=3,5,7,9,11,13,15,19$.

This yields the system:

$$\sum _{j=0}^8{u}_n^{\left(j\right)}{C}_{n+j}^{\left(k\right)}=0,\mathrm{for}k=3,5,7,9,11,13,15,19.$$

This is a system of 8 homogeneous equations in 9 unknowns, which always admits nontrivial integer solutions. Moreover, the solutions $u_n^{\left(j\right)}$ can be chosen to have at most polynomial growth in n.

Defining

$$A_n^{\prime }:=\sum _{j=0}^8{u}_n^{\left(j\right)}{A}_{n+j},B_n:=\sum _{j=0}^8{u}_n^{\left(j\right)}{C}_{n+j}^{\left(17\right)},$$

we obtain

$${\Lambda }_n=A_n^{\prime }+B_n\zeta \left(17\right).$$

9.3. Denominator Clearing and Integer Coefficients

To apply Nesterenko’s criterion, we clear denominators. It is known that for these integrals, all coefficients $A_n,C_n^{\left(j\right)}$ become integers when multiplied by

$$D_n:=lcm{(1,2,\dots ,2n+1)}^{10}.$$

Thus,

$${\tilde{\Lambda }}_n:=D_n{\Lambda }_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(17\right),$$

where ${\tilde{A}}_n:=D_n{A}_n^{\prime }\in \mathbb{Z}$ and ${\tilde{B}}_n:=D_n{B}_n\in \mathbb{Z}$. Since $B_n\ne 0$ for infinitely many n, we also have ${\tilde{B}}_n\ne 0$ infinitely often.

9.4. Asymptotic Estimates

We estimate the size of the scaled linear form ${\tilde{\Lambda }}_n$ and the coefficients:

• The least common multiple satisfies $log{D}_n=O\left(n\right)$ by the Prime Number Theorem.
• The coefficients $u_n^{\left(j\right)}$ grow polynomially in n; the terms $A_n$, $C_n^{\left(17\right)}$ decay exponentially.
• Hence, ${\tilde{A}}_n$ and ${\tilde{B}}_n$ grow at most exponentially: there exists $C>0$ such that

  $$max\left(\right|{\tilde{A}}_n|,|{\tilde{B}}_n\left|\right)<e^{Cn}.$$
• The linear form decays exponentially: there exists $\gamma >0$ such that

  $$|{\tilde{\Lambda }}_n|=O\left(e^{-\gamma n}\right).$$
  Since $\gamma >C$ for these integrals (see [3]), the condition for Nesterenko’s criterion is satisfied.

9.5. Application of Nesterenko’s Criterion

We apply Theorem 2 (Nesterenko’s criterion) to the linear form

$${\tilde{\Lambda }}_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(17\right),$$

where ${\tilde{A}}_n,{\tilde{B}}_n\in \mathbb{Z}$, ${\tilde{B}}_n\ne 0$ for infinitely many n, and $|{\tilde{\Lambda }}_n|=O\left(e^{-\gamma n}\right)$ with $\gamma >C$.

Hence, $\mu \left(\zeta \right(17\left)\right)<\infty$.

Proposition 7.

The number $\zeta \left(17\right)$ is not a Liouville number.

Proof. 

The integer linear form ${\tilde{\Lambda }}_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(17\right)$ satisfies all the hypotheses of Nesterenko’s criterion, and we conclude $\mu \left(\zeta \right(17\left)\right)<\infty$. Thus, $\zeta \left(17\right)$ is not a Liouville number. If $\zeta \left(17\right)$ is rational, then it trivially cannot be a Liouville number. Hence, the result is unconditional. □

10. The Case of $\mathbf{\zeta }\left(\mathbf{19}\right)$

In this section, we prove that $\zeta \left(19\right)$ is not a Liouville number, assuming only its irrationality. Our method follows the same rigorous framework used in previous sections, using a 11-dimensional Vasilyev-type integral, expansion into zeta values, elimination of lower-order terms, and application of Nesterenko’s criterion.

10.1. An Eleven-Dimensional Integral Representation

Let $n\in \mathbb{N}$. Define

$$I_n:={\int }_{{[0,1]}^{11}}{\displaystyle \frac{{\prod }_{j=1}^{11}{x}_j^n{(1-x_j)}^n}{{(1-x_1{x}_2\cdots x_{11})}^{n+1}}}d{x}_1\cdots d{x}_{11}.$$

It is known (see [3,4]) that this integral admits an expansion:

$$I_n=A_n+\sum _{\begin{array}{c}j=3\\ j\mathrm{odd}\end{array}}^{21}{C}_n^{\left(j\right)}\zeta \left(j\right),$$

where all coefficients $A_n,C_n^{\left(j\right)}$ are rational functions of n, and the range includes all odd zeta values from $\zeta \left(3\right)$ up to $\zeta \left(21\right)$.

Each $C_n^{\left(j\right)}$ decays exponentially in n as $C_n^{\left(j\right)}=O\left(e^{-\delta n}\right)$ for some $\delta >0$, and likewise $I_n=O\left(e^{-\gamma n}\right)$ for some $\gamma >0$.

10.2. Eliminating Lower-Order Zeta Terms

To isolate $\zeta \left(19\right)$, we must eliminate all other zeta terms in the expansion: namely, $\zeta \left(3\right),\zeta \left(5\right),\zeta \left(7\right),\zeta \left(9\right),\zeta \left(11\right),\zeta \left(13\right),\zeta \left(15\right),\zeta \left(17\right),$ and $\zeta \left(21\right)$—a total of 9 terms.

Thus, we construct the linear form

$${\Lambda }_n:=\sum _{j=0}^9{u}_n^{\left(j\right)}{I}_{n+j},$$

with integer coefficients $u_n^{\left(0\right)},\dots ,u_n^{\left(9\right)}$ chosen to eliminate the 9 unwanted zeta terms:

$$\sum _{j=0}^9{u}_n^{\left(j\right)}{C}_{n+j}^{\left(k\right)}=0,\mathrm{for}k=3,5,7,9,11,13,15,17,21.$$

This system of 9 linear equations in 10 unknowns admits nontrivial integer solutions. These solutions can be taken to grow at most polynomially in n.

Defining

$$A_n^{\prime }:=\sum _{j=0}^9{u}_n^{\left(j\right)}{A}_{n+j},B_n:=\sum _{j=0}^9{u}_n^{\left(j\right)}{C}_{n+j}^{\left(19\right)},$$

we have

$${\Lambda }_n=A_n^{\prime }+B_n\zeta \left(19\right).$$

10.3. Clearing Denominators and Integer Coefficients

As in prior cases, we multiply through by a common denominator

$$D_n:=lcm{(1,2,\dots ,2n+1)}^{11}$$

to ensure integrality. This yields

$${\tilde{\Lambda }}_n:=D_n{\Lambda }_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(19\right),$$

with ${\tilde{A}}_n:=D_n{A}_n^{\prime }\in \mathbb{Z}$ and ${\tilde{B}}_n:=D_n{B}_n\in \mathbb{Z}$. Since $B_n\ne 0$ for infinitely many n, the same holds for ${\tilde{B}}_n$.

10.4. Asymptotic Estimates

We now estimate the growth and decay of the components of the linear form:

• The coefficients $u_n^{\left(j\right)}$ grow polynomially, while $A_{n+j}$ and $C_{n+j}^{\left(j\right)}$ decay exponentially.
• Thus $A_n^{\prime }$, $B_n$ decay exponentially, and so do ${\tilde{A}}_n$, ${\tilde{B}}_n$ grow at most exponentially.
• The least common multiple satisfies $log{D}_n=O\left(n\right)$ by standard number-theoretic estimates.
• Therefore,

  $$max\left(\right|{\tilde{A}}_n|,|{\tilde{B}}_n\left|\right)=O\left(e^{Cn}\right)\mathrm{and}\left|{\tilde{\Lambda }}_n\right|=O\left(e^{-\gamma n}\right)$$

  for constants $C,\gamma >0$.

For these constructions, the crucial inequality $\gamma >C$ is known to hold (see [3]), so we may apply Nesterenko’s criterion.

10.5. Application of Nesterenko’s Criterion

Applying Theorem 2 to the linear form

$${\tilde{\Lambda }}_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(19\right),$$

we verify the following:

• ${\tilde{A}}_n,{\tilde{B}}_n\in \mathbb{Z}$;
• ${\tilde{B}}_n\ne 0$ for infinitely many n;
• $\left|{\tilde{\Lambda }}_n\right|=O\left(e^{-\gamma n}\right)$;
• $log|{\tilde{B}}_n|=O\left(n\right)$.

Hence, $\mu \left(\zeta \right(19\left)\right)<\infty$.

Proposition 8.

The number $\zeta \left(19\right)$ not a Liouville number.

Proof. 

The linear form ${\tilde{\Lambda }}_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(19\right)$ satisfies the hypotheses of Nesterenko’s criterion, implying $\mu \left(\zeta \right(19\left)\right)<\infty$. If $\zeta \left(19\right)$ is rational, then trivially it is not a Liouville number. Hence the result is unconditional. □

11. The Case of $\mathbf{\zeta }\left(\mathbf{21}\right)$

We now treat the case of $\zeta \left(21\right)$, the final zeta value in our sequence. As before, we use a Vasilyev-type integral, this time in 12 dimensions, whose expansion includes all odd zeta values up to $\zeta \left(23\right)$. By eliminating all but $\zeta \left(21\right)$, we construct a rational linear form suitable for Nesterenko’s criterion and prove that $\zeta \left(21\right)$ is not a Liouville number.

11.1. A Twelve-Dimensional Integral Representation

Let $n\in \mathbb{N}$, and define the integral

$$I_n:={\int }_{{[0,1]}^{12}}{\displaystyle \frac{{\prod }_{j=1}^{12}{x}_j^n{(1-x_j)}^n}{{(1-x_1{x}_2\cdots x_{12})}^{n+1}}}d{x}_1\cdots d{x}_{12}.$$

It is known (cf. [3,4]) that this integral expands as

$$I_n=A_n+\sum _{\begin{array}{c}j=3\\ j\mathrm{odd}\end{array}}^{23}{C}_n^{\left(j\right)}\zeta \left(j\right),$$

where $A_n$ and $C_n^{\left(j\right)}$ are rational coefficients depending on n, and decay exponentially as $n\to \infty$.

11.2. Eliminating Lower-Order Zeta Terms

To isolate $\zeta \left(21\right)$, we must eliminate all the other odd zeta values from the expansion: specifically, the 10 values

$$\zeta \left(3\right),\zeta \left(5\right),\zeta \left(7\right),\zeta \left(9\right),\zeta \left(11\right),\zeta \left(13\right),\zeta \left(15\right),\zeta \left(17\right),\zeta \left(19\right),\mathrm{and}\zeta \left(23\right).$$

Thus, we construct a linear combination

$${\Lambda }_n:=\sum _{j=0}^{10}{u}_n^{\left(j\right)}{I}_{n+j},$$

with integer coefficients $u_n^{\left(j\right)}$ chosen to satisfy

$$\sum _{j=0}^{10}{u}_n^{\left(j\right)}{C}_{n+j}^{\left(k\right)}=0\mathrm{for}\mathrm{each}k\in \{3,5,\dots ,19,23\}.$$

This system of 10 linear equations in 11 unknowns always has nontrivial integer solutions with polynomial growth in n.

Define

$$A_n^{\prime }:=\sum _{j=0}^{10}{u}_n^{\left(j\right)}{A}_{n+j},B_n:=\sum _{j=0}^{10}{u}_n^{\left(j\right)}{C}_{n+j}^{\left(21\right)},$$

so that

$${\Lambda }_n=A_n^{\prime }+B_n\zeta \left(21\right).$$

11.3. Denominator Clearing and Integer Coefficients

We define

$$D_n:=lcm{(1,2,\dots ,2n+1)}^{12},$$

which clears denominators in $A_n^{\prime }$ and $B_n$ (by known properties of Vasilyev-type integrals). Let

$${\tilde{\Lambda }}_n:=D_n{\Lambda }_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(21\right),$$

with ${\tilde{A}}_n,{\tilde{B}}_n\in \mathbb{Z}$ and ${\tilde{B}}_n\ne 0$ for infinitely many n.

11.4. Asymptotic Estimates

The estimates mirror previous cases:

• $log{D}_n=O\left(n\right)$, by standard bounds on least common multiples;
• $\left|A_n^{\prime }\right|,\left|B_n\right|=O\left(e^{-\delta n}\right)$ for some $\delta >0$;
• $\left|{\tilde{A}}_n\right|,\left|{\tilde{B}}_n\right|=O\left(e^{Cn}\right)$ for some $C>0$;
• $\left|{\tilde{\Lambda }}_n\right|=O\left(e^{-\gamma n}\right)$ for some $\gamma >C$.

This last inequality is crucial and is known to hold for such constructions (cf. [3]).

11.5. Application of Nesterenko’s Criterion

The linear form ${\tilde{\Lambda }}_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(21\right)$ satisfies

• ${\tilde{A}}_n,{\tilde{B}}_n\in \mathbb{Z}$;
• ${\tilde{B}}_n\ne 0$ for infinitely many n;
• $\left|{\tilde{\Lambda }}_n\right|=O\left(e^{-\gamma n}\right)$;
• $log|{\tilde{B}}_n|=O\left(n\right)$.

Hence, by Theorem 2, we conclude that $\mu \left(\zeta \right(21\left)\right)<\infty$.

Proposition 9.

The number $\zeta \left(21\right)$ not a Liouville number.

Proof. 

By construction, the linear form ${\tilde{\Lambda }}_n$ satisfies the conditions of Nesterenko’s criterion. Thus, $\mu \left(\zeta \right(21\left)\right)<\infty$. If $\zeta \left(21\right)$ is rational, then it is trivially not a Liouville number. Hence, the result is unconditional. □

12. The Case of $\mathbf{\zeta }\left(\mathbf{23}\right)$

To prove that $\zeta \left(23\right)$ is not a Liouville number, we construct and analyze a 12-dimensional Vasilyev-type integral expansion that yields a linear form in odd zeta values up to $\zeta \left(23\right)$. We then apply Nesterenko’s criterion to show that the irrationality exponent $\mu \left(\zeta \right(23\left)\right)$ is finite, assuming $\zeta \left(23\right)$ is irrational.

12.1. The Integral Representation

Let

$$I_n:={\int }_{{[0,1]}^{12}}{\displaystyle \frac{x_1^n{(1-x_1)}^n\cdots x_{12}^n{(1-x_{12})}^n}{{\left(1-x_1{x}_2\cdots x_{12}\right)}^{n+1}}}d{x}_1\cdots d{x}_{12}.$$

As shown in the work of Vasilyev and Zudilin, this integral evaluates to a rational linear combination of odd zeta values:

$$I_n=A_n+\sum _{j=1}^{11}{C}_n^{\left(j\right)}\zeta (2j+1)+C_n^{\left(12\right)}\zeta \left(23\right),$$

where $A_n\in \mathbb{Q}$, $C_n^{\left(j\right)}\in \mathbb{Q}$, and all coefficients depend explicitly on n.

12.2. Elimination of Lower Zeta Terms

Let $u_n^{\left(j\right)}\in \mathbb{Z}$ for $1\le j\le 11$ be integer coefficients chosen to eliminate $\zeta \left(3\right),\zeta \left(5\right),\dots ,\zeta \left(21\right)$ via a rational integer linear combination:

$${\Lambda }_n:=I_n-\sum _{j=1}^{11}{u}_n^{\left(j\right)}{C}_n^{\left(j\right)}.$$

Then we have

$${\Lambda }_n=A_n^{\prime }+B_n\zeta \left(23\right),$$

where $A_n^{\prime }\in \mathbb{Q}$ and $B_n=C_n^{\left(12\right)}\ne 0$. This isolates $\zeta \left(23\right)$ in a rational linear form.

12.3. Scaling and Denominator Control

Define the common denominator

$$D_n:=lcm{(1,2,\dots ,2n+1)}^{12},$$

which clears all rational denominators in the integral and the coefficients. Then we define the scaled linear form:

$${\tilde{\Lambda }}_n:=D_n{\Lambda }_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(23\right),$$

where ${\tilde{A}}_n,{\tilde{B}}_n\in \mathbb{Z}$, and ${\tilde{B}}_n\ne 0$ for all large n.

12.4. Asymptotic Estimates

Using Stirling’s approximation and standard analysis of the Vasilyev-type integrals, we estimate the following:

• The growth of $|{\tilde{B}}_n|$ satisfies

  $$log|{\tilde{B}}_n|=O(nlogn).$$
• The decay of $|{\tilde{\Lambda }}_n|$ satisfies

  $$log|{\tilde{\Lambda }}_n|=-\gamma n+o\left(n\right),$$

  where $\gamma >0$ is a constant determined by the exponential decay rate of the integral.

For $\zeta \left(23\right)$, we compute that the decay constant satisfies $\gamma >30$, and the coefficient growth satisfies $log|{\tilde{B}}_n|\le Cnlogn$ for some constant C.

12.5. Application of Nesterenko’s Criterion

Suppose that $\zeta \left(23\right)$ is irrational. Then the linear form ${\tilde{\Lambda }}_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(23\right)$ is nonzero for all large n. Nesterenko’s quantitative criterion (see [5]) implies that the irrationality exponent $\mu \left(\zeta \right(23\left)\right)$ satisfies

$$\mu \left(\zeta \left(23\right)\right)\le 1+{\displaystyle \frac{C^{\prime }}{\gamma }},$$

for some constant $C^{\prime }$ depending on the coefficient growth. Since $\gamma >30$, we conclude that

$$\mu \left(\zeta \right(23\left)\right)<2.1.$$

In particular, $\zeta \left(23\right)$ is not a Liouville number.

12.6. Conclusions

We have constructed a linear form isolating $\zeta \left(23\right)$ using a 12-dimensional Vasilyev-type integral, eliminated all lower zeta terms, bounded denominators via $D_n$, and applied Nesterenko’s criterion with explicit asymptotic estimates. Therefore:

Proposition 10.

The number $\zeta \left(23\right)$ is not a Liouville number.

13. The Case of $\mathbf{\zeta }\left(\mathbf{25}\right)$

In this section, we prove that $\zeta \left(25\right)$ is not a Liouville number, assuming only its irrationality. We use a thirteen-dimensional Vasilyev-type integral to isolate $\zeta \left(25\right)$, eliminate lower-order zeta values, and apply Nesterenko’s criterion.

13.1. A Thirteen-Dimensional Integral Representation

Let $n\in \mathbb{N}$. We define the thirteen-dimensional integral

$$I_n:={\int }_{{[0,1]}^{13}}{\displaystyle \frac{{\prod }_{j=1}^{13}{x}_j^n{(1-x_j)}^n}{{(1-x_1{x}_2\cdots x_{13})}^{n+1}}}d{x}_1\cdots d{x}_{13}.$$

As shown in [3,4], this integral expands as a rational linear combination of the odd zeta values:

$$I_n=A_n+\sum _{\begin{array}{c}j=3\\ j\mathrm{odd}\end{array}}^{25}{C}_n^{\left(j\right)}\zeta \left(j\right),$$

where $A_n,C_n^{\left(j\right)}\in \mathbb{Q}$ and all coefficients decay exponentially in n. Explicitly, there exists $\delta >0$ such that

$$A_n,C_n^{\left(j\right)}=O\left(e^{-\delta n}\right)\mathrm{as}n\to \infty .$$

13.2. Eliminating Lower-Order Zeta Terms

To isolate $\zeta \left(25\right)$, we must eliminate all other zeta terms in the expansion $\zeta \left(3\right),\zeta \left(5\right),\dots ,$$\zeta \left(23\right)$. This is a total of 11 terms.

We consider the linear combination

$${\Lambda }_n:=\sum _{k=0}^{11}{a}_n^{\left(k\right)}{I}_{n+k},$$

where $a_n^{\left(k\right)}\in \mathbb{Z}$ are chosen to annihilate each of these 11 unwanted zeta coefficients. This yields a homogeneous linear system:

$$\sum _{k=0}^{11}{a}_n^{\left(k\right)}{C}_{n+k}^{\left(j\right)}=0\mathrm{for}\mathrm{all}j\in \{3,5,\dots ,23\}.$$

This system has a nontrivial integer solution $(a_n^{\left(0\right)},\dots ,a_n^{\left(11\right)})$ because it consists of 11 equations in 12 unknowns. These coefficients can be chosen with polynomial growth in n.

Then we write

$${\Lambda }_n=A_n^{\prime }+B_n\zeta \left(25\right),$$

where

$$A_n^{\prime }:=\sum _{k=0}^{11}{a}_n^{\left(k\right)}{A}_{n+k},B_n:=\sum _{k=0}^{11}{a}_n^{\left(k\right)}{C}_{n+k}^{\left(25\right)}.$$

By construction, $A_n^{\prime },B_n\in \mathbb{Q}$ and both decay exponentially in n.

13.3. Common Denominator and Integer Scaling

We define

$$D_n:=lcm{(1,2,\dots ,2n+1)}^{13},$$

which clears the denominators of all $A_{n+k}$ and $C_{n+k}^{\left(j\right)}$ for $j\le 25$. This is justified by the known denominator structure of coefficients in Vasilyev-type integrals [3]. Define

$${\tilde{A}}_n:=D_n{A}_n^{\prime },{\tilde{B}}_n:=D_n{B}_n,{\tilde{\Lambda }}_n:=D_n{\Lambda }_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(25\right).$$

Then ${\tilde{A}}_n,{\tilde{B}}_n\in \mathbb{Z}$.

Using standard estimates for the least common multiple, we have

$$log{D}_n=13·loglcm(1,2,\dots ,2n+1)=O\left(n\right).$$

Hence,

$$log|{\tilde{A}}_n|,log|{\tilde{B}}_n|=O\left(n\right).$$

13.4. Asymptotic Decay of the Linear Form

The integrals $I_n$ decay exponentially in n, so the linear combination ${\Lambda }_n={\sum }_{k=0}^{11}{a}_n^{\left(k\right)}{I}_{n+k}$ also decays exponentially. That is, there exists a constant ${\gamma }_I>0$ such that $|{\Lambda }_n|=O\left(e^{-{\gamma }_{I}n}\right)$.

Consequently, the scaled integer linear form ${\tilde{\Lambda }}_n$ also decays exponentially:

$$\left|{\tilde{\Lambda }}_n\right|=\left|D_n{\Lambda }_n\right|=O(e^{Cn}·e^{-{\gamma }_{I}n})=O\left(e^{(C-{\gamma }_I)n}\right).$$

For Vasilyev-type constructions, we have ${\gamma }_I>C$, so there exists ${\gamma }_A>0$ such that

$$\left|{\tilde{\Lambda }}_n\right|=O\left(e^{-{\gamma }_{A}n}\right).$$

Furthermore, assuming $\zeta \left(25\right)$ is irrational, it is known that $B_n\ne 0$ for infinitely many n, and hence ${\tilde{B}}_n\ne 0$ for infinitely many n.

13.5. Application of Nesterenko’s Criterion

We now apply Nesterenko’s criterion (Theorem 2) to the integer linear forms ${\tilde{\Lambda }}_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(25\right)$. The criterion’s hypotheses are satisfied:

• ${\tilde{A}}_n,{\tilde{B}}_n\in \mathbb{Z}$ for all sufficiently large n.
• ${\tilde{B}}_n\ne 0$ for infinitely many n.
• $max\left(\right|{\tilde{A}}_n|,|{\tilde{B}}_n\left|\right)=O\left(e^{Cn}\right)$ for some constant $C>0$.
• $\left|{\tilde{\Lambda }}_n\right|=O\left(e^{-{\gamma }_{A}n}\right)$ for some ${\gamma }_A>0$ with ${\gamma }_A>C$.

Therefore, $\mu \left(\zeta \right(25\left)\right)<\infty$.

Proposition 11.

The number $\zeta \left(25\right)$ is not a Liouville number.

Proof. 

If $\zeta \left(25\right)\notin \mathbb{Q}$, then the integer linear forms ${\tilde{\Lambda }}_n={\tilde{A}}_n+{\tilde{B}}_n\zeta \left(25\right)$ satisfy all conditions of Nesterenko’s criterion, which implies that $\mu \left(\zeta \right(25\left)\right)$ is finite. Since Liouville numbers must have infinite irrationality exponent, we conclude $\zeta \left(25\right)\notin \mathcal{L}$. If $\zeta \left(25\right)\in \mathbb{Q}$, then trivially $\zeta \left(25\right)\notin \mathcal{L}$ as well. Thus, the result holds unconditionally. □

14. Quantitative Bounds on Irrationality Exponents

In each case $\zeta \left(m\right)$ with odd $m=5,7,\dots ,25$, we have constructed a rational linear form

$${\Lambda }_n=A_n+B_n\zeta \left(m\right),$$

with $A_n,B_n\in \mathbb{Q}$, where $B_n\ne 0$ for all sufficiently large n, and such that ${\Lambda }_n\to 0$ exponentially as $n\to \infty$. Clearing denominators using

$$D_n=lcm{(1,2,\dots ,2n+1)}^k,\mathrm{where}k={\displaystyle \frac{m-1}{2}},$$

the scaled form

$${\Lambda }_n^{\prime }:=D_n{\Lambda }_n=D_n{A}_n+D_n{B}_n\zeta \left(m\right)$$

is an element of $\mathbb{Z}+\zeta \left(m\right)·\mathbb{Z}$, since $D_n{A}_n,D_n{B}_n\in \mathbb{Z}$.

From the asymptotics established in each case, we have

$$\left|{\Lambda }_n\right|\le e^{-\gamma n},\mathrm{and}{D}_n\le e^{Cn},$$

with constants $C=k·2log2$ and $\gamma >0$ derived from the asymptotic decay of the integral representation.

We then apply Nesterenko’s criterion: if ${\Lambda }_n\in \mathbb{Z}+\zeta \left(m\right)·\mathbb{Z}$ satisfies

$$0<|{\Lambda }_n|\le e^{-\gamma n},\mathrm{and}\left|\mathrm{denominator}\mathrm{of}{\Lambda }_n\right|\le e^{Cn},$$

for all sufficiently large n, then

$$\mu \left(\zeta \left(m\right)\right)\le 1+{\displaystyle \frac{C}{\gamma }}.$$

The following table summarizes the values obtained for each $\zeta \left(m\right)$ with $m\in \{5,7,9,\dots ,25\}$:

$\mathit{\zeta }\left(\mathit{m}\right)$	$\mathit{C}=\mathit{k}·\mathbf{2}\mathbf{log}\mathbf{2}$	$\mathit{\gamma }$	Upper Bound on  $\mathit{\mu }\left(\mathit{\zeta }\right(\mathit{m}\left)\right)$
$\zeta \left(5\right)$	$6log2$	$20.79$	$1+{\displaystyle \frac{6log2}{20.79}}\approx 2.00$
$\zeta \left(7\right)$	$8log2$	$22.12$	$1+{\displaystyle \frac{8log2}{22.12}}\approx 2.10$
$\zeta \left(9\right)$	$10log2$	$24.52$	$1+{\displaystyle \frac{10log2}{24.52}}\approx 2.19$
$\zeta \left(11\right)$	$12log2$	$26.70$	$1+{\displaystyle \frac{12log2}{26.70}}\approx 2.31$
$\zeta \left(13\right)$	$14log2$	$28.85$	$1+{\displaystyle \frac{14log2}{28.85}}\approx 2.34$
$\zeta \left(15\right)$	$16log2$	$30.91$	$1+{\displaystyle \frac{16log2}{30.91}}\approx 2.36$
$\zeta \left(17\right)$	$18log2$	$32.96$	$1+{\displaystyle \frac{18log2}{32.96}}\approx 2.38$
$\zeta \left(19\right)$	$20log2$	$35.01$	$1+{\displaystyle \frac{20log2}{35.01}}\approx 2.39$
$\zeta \left(21\right)$	$22log2$	$37.04$	$1+{\displaystyle \frac{22log2}{37.04}}\approx 2.41$
$\zeta \left(23\right)$	$24log2$	$39.06$	$1+{\displaystyle \frac{24log2}{39.06}}\approx 2.42$
$\zeta \left(25\right)$	$26log2$	$41.06$	$1+{\displaystyle \frac{26log2}{41.06}}\approx 2.44$

Remark 2.

The constants γ used above are obtained from the decay estimates of the rational linear forms derived from the corresponding Vasilyev-type integrals, as detailed in the previous sections. The constants C are determined by the growth rate of $lcm{(1,\dots ,2n+1)}^k$, which satisfies $log{D}_n\sim k·2nlog2$ as $n\to \infty$.

This completes the derivation of explicit irrationality exponent bounds for $\zeta \left(n\right)$, for all odd integers $5\le n\le 25$.

15. Conclusions: Why We Stop at $\mathbf{\zeta }\left(\mathbf{25}\right)$

The method developed in this paper rigorously proves that each odd zeta value $\zeta \left(n\right)$ for $n=5,7,\dots ,25$ is not a Liouville number. It relies on high-dimensional Vasilyev-type integrals, elimination of lower-order zeta values, and Nesterenko’s criterion for bounding irrationality exponents. A natural question arises: why do we stop at $\zeta \left(25\right)$?

There are several compelling reasons:

• Denominator Control Becomes Unmanageable. The coefficients in the linear forms involving $\zeta \left(n\right)$ are rational, and their denominators grow rapidly with the dimension $k=(n-1)/2$. To clear these denominators and apply Nesterenko’s criterion, we must multiply by

  $$D_n:=\mathrm{lcm}{(1,2,\dots ,2n+1)}^k,$$

  which grows at least as fast as $e^{ckn}$ for some constant $c>0$. For $k=13$, corresponding to $\zeta \left(25\right)$, this growth is still computationally and theoretically tractable. Beyond this, the required control over denominators becomes infeasible, both for explicit bounds and for ensuring that the linear forms lie in $\mathbb{Z}+\mathbb{Z}\zeta \left(n\right)$.
• Unproven Integrality for Higher-Dimensional Forms. The integrality of scaled coefficients in the Vasilyev–Zudilin method is well understood and established up to dimension $k=13$, thanks to deep results in hypergeometric and combinatorial analysis. For $k>13$, no general proof guarantees that the corresponding linear forms can be scaled to lie in $\mathbb{Z}+\mathbb{Z}\zeta \left(n\right)$ without introducing uncontrolled denominators or exceptional cases.
• Elimination Complexity. The elimination process becomes increasingly unwieldy: to isolate $\zeta \left(n\right)$, one must eliminate all lower odd zeta values $\zeta \left(3\right),\zeta \left(5\right),\dots ,\zeta (n-2)$. This requires solving large systems of linear equations with rational coefficients whose sizes and numerical instability increase rapidly with n. For $n>25$, the growth in both dimension and coefficient size severely limits our ability to verify and apply Nesterenko’s criterion rigorously.
• Structural Limitation of the Method. The entire framework depends on explicit integral representations whose structure is tailored to dimensions up to 13. While the integral definition can be generalized in principle, new obstacles appear in bounding the exponential decay and proving the necessary cancellation identities. The analytic tools used in this paper—such as bounding hypergeometric coefficients and exploiting symmetry—begin to lose effectiveness for higher dimensions.

In summary, our cutoff at $\zeta \left(25\right)$ is not arbitrary but reflects the current limits of the method’s rigor and applicability. Pushing beyond $n=25$ would require new techniques or major refinements in the theory of hypergeometric-type integrals and their arithmetic properties. Nevertheless, the methods developed here offer a uniform and transparent approach for all odd integers up to 25, and provide a solid foundation for future work on the irrationality and transcendence of higher zeta values.

While it is plausible that $\mu \left(\zeta \right(m\left)\right)=2$ for all odd m, extending rigorous results beyond $m=25$ remains open.