Arithmetic of Catalan’s constant and its relatives

Abstract

We prove that at least one of the six numbers $\beta (2i)$ for $i=1,\ldots ,6$ is irrational. Here $\beta (s)=\sum _{k=0}^{\infty }(-1)^k(2k+1)^{-s}$ denotes Dirichlet’s beta function, so that $\beta (2)$ is Catalan’s constant.

1 Introduction

In this note we discuss arithmetic properties of the values of Dirichlet’s beta function

$$\begin{aligned} \beta (s)=\sum _{n=1}^{\infty }\frac{\bigl (\frac{-4}{n}\bigr )}{n^s} =\sum _{k=0}^{\infty }\frac{(-1)^k}{(2k+1)^s} \end{aligned}$$

at positive even integers s. The very first such beta value $\beta (2)$ is famously known as Catalan’s constant; its irrationality remains an open problem, though we expect the number to be irrational and transcendental. The best known results in this direction were given by T. Rivoal and this author in [4]. Namely, we showed that at least one of the seven numbers $\beta (2),\beta (4),\ldots ,\beta (14)$ is irrational, and that there are infinitely many irrational numbers among the even beta values $\beta (2),\beta (4),\beta (6),\ldots$ . Here we use a variant of the method from [3, 8] to improve slightly on the former achievement; a significant strengthening towards the infinitude result, based on a further development of the ideas in [2, 6], is a subject of the recent preprint [1] of S. Fischler.

Theorem 1

At least one of the six numbers

$$\begin{aligned} \beta (2), \; \beta (4), \; \beta (6), \; \beta (8), \; \beta (10), \; \beta (12) \end{aligned}$$

is irrational.

In Sect. 2 we illustrate principal ingredients of the method in a particularly simple situation; this leads to a weaker version of Theorem 1, namely, to the irrationality of at least one number $\beta (2i)$ for $i=1,\ldots ,8$. The details about the general construction of approximating forms to even beta values and our proof of Theorem 1 are given in Sect. 3.

2 Outline of the construction

For an odd integer $s\ge 3$ (which we eventually set to be 17) and even $n>0$, define the rational function

$$\begin{aligned} R_n(t)=\frac{2^{6n}n!^{s-3}\,(2t+n)\,\prod _{j=1}^{3n} \left( t-n+j-\frac{1}{2}\right) }{\prod _{j=0}^n(t+j)^s} \end{aligned}$$

(1)

and assign to it the related sequence of quantities

$$\begin{aligned} r_n=\sum _{\nu =1}^{\infty } R_n\left( \nu -\tfrac{1}{2}\right) (-1)^{\nu } =\sum _{\nu =n+1}^{\infty } R_n\left( \nu -\tfrac{1}{2}\right) (-1)^{\nu }. \end{aligned}$$

(2)

The sums $r_n$ are instances of generalized hypergeometric functions, for which we can use some standard integral representations to write

$$\begin{aligned} r_n=\frac{2^{6n}(3n+1)!}{n!^3}\int \cdots \int _{[0,1]^s} \frac{(1-t_1t_2\cdots t_s)\,\prod _{j=1}^st_j^{n-1/2} (1-t_j)^n\,\hbox {d} t_j}{(1+t_1t_2\cdots t_s)^{3n+2}} \end{aligned}$$

(details are given in Lemma 2 below). This form clearly implies that $r_n>0$ and also gives access to the asymptotics

$$\begin{aligned} \lim _{n\rightarrow \infty }r_n^{1/n} =2^63^3\max _{\varvec{t}\in [0,1]^s} \frac{\prod _{j=1}^st_j(1-t_j)}{(1+t_1t_2\cdots t_s)^3} =12^3\max _{0<t<1}\frac{t^s(1-t)^s}{(1+t^s)^3}. \end{aligned}$$

(3)

An important ingredient of the construction is the following decomposition of the quantities $r_n$.

Lemma 1

For odd s and even n as above,

$$\begin{aligned} r_n=\sum _{\begin{array}{c} i=1\\ i\;\text {even} \end{array}}^sa_i\beta (i)+a_0, \end{aligned}$$

where $a_i=a_{i,n}$ satisfy the inclusions $\Phi _n^{-1}d_n^{s-i}a_i\in \mathbb {Z}$ for $i=0,1,\ldots ,s$ even. Here $d_n$ denotes the least common multiple of the numbers $1,2,\ldots ,n$, and

$$\begin{aligned} \Phi _n=\prod _{2\sqrt{n}<p\le n}p^{\varphi _0(n/p)} \qquad \text {with}\quad \varphi _0(x)={\left\{ \begin{array}{ll} 0 &{}\quad \text {if}\quad 0\le \{x\}<\tfrac{1}{3}, \\ 1 &{}\quad \text {if}\quad \tfrac{1}{3}\le \{x\}<\tfrac{1}{2}, \\ 2 &{}\quad \text {if}\quad \tfrac{1}{2}\le \{x\}<1, \end{array}\right. } \end{aligned}$$

the product taken over primes.

Note that from the prime number theorem we deduce the asymptotics

$$\begin{aligned} \lim _{n\rightarrow \infty }d_n^{1/n}=e \qquad \text {and}\qquad \lim _{n\rightarrow \infty }\Phi _n^{1/n}=e^{\varkappa }, \end{aligned}$$

where

$$\begin{aligned} \varkappa =\bigl (\psi \big (\tfrac{1}{2}\big )-\psi \big (\tfrac{1}{3}\big )-1\bigr ) +2\cdot \bigl (\psi (1)-\psi \big (\tfrac{1}{2}\big )-1\bigr ) =0.9411124762\ldots , \end{aligned}$$

the function $\psi (x)$ denotes the logarithmic derivative of the gamma function.

Remark 1

The analogous construction in [4] makes use of a slightly different rational function than (1), namely, of

$$\begin{aligned} \widetilde{R}_n(t)=\frac{2^{4n}n!^{s-2}\,(2t+n) \,\prod _{j=1}^n\left( t-n+j-\frac{1}{2}\right) \,\prod _{j=1}^n \left( t+n+j-\frac{1}{2}\right) }{\prod _{j=0}^n(t+j)^s}, \end{aligned}$$

so that

$$\begin{aligned} R_n(t)=\widetilde{R}_n(t)\cdot \frac{2^{2n} \prod _{j=1}^n\left( t+j-\frac{1}{2}\right) }{n!}. \end{aligned}$$

The analogous decomposition of a related quantity $\widetilde{r}_n$ assumes the form

$$\begin{aligned} \widetilde{r}_n=\sum _{\begin{array}{c} i=1\\ i\;\text {even} \end{array}}^s \widetilde{a}_i\beta (i)+\widetilde{a}_0, \end{aligned}$$

in which the rational coefficients $\widetilde{a}_i=\widetilde{a}_{i,n}$ satisfy $\Phi _n^{-1}d_n^{s-i}\widetilde{a}_i\in \mathbb {Z}$ for $i=1,\ldots ,s$ even, but $\Phi _n^{-1}d_{2n}^s\widetilde{a}_0\in \mathbb {Z}$. The appearance of $d_{2n}^s$ as the common denominator in place of $d_n^s$ changes the scene drastically and leads to weaker arithmetic applications.

Proof of Lemma 1

Following the strategy in [5, 8] we can write the function (1) as sum of partial fractions,

$$\begin{aligned} R_n(t)=\sum _{i=1}^s\sum _{k=0}^n\frac{a_{i,k}}{(t+k)^i}, \end{aligned}$$

in which $\Phi _n^{-1}d_n^{s-i}a_{i,k}\in \mathbb {Z}$ for all i and k. Indeed, the rational function (1) is a product of simpler ones

$$\begin{aligned} \frac{n!}{\prod _{j=0}^n(t+j)}&=\sum _{k=0}^n\frac{(-1)^k\left( {\begin{array}{c}n\\ k\end{array}}\right) }{t+k},\\ \frac{2^{2n}\prod _{j=1}^n\left( t-n+j-\frac{1}{2}\right) }{\prod _{j=0}^n(t+j)}&=\sum _{k=0}^n\frac{(-1)^{n+k}\left( {\begin{array}{c}2n+2k\\ 2n\end{array}}\right) \left( {\begin{array}{c}2n\\ n+k\end{array}}\right) }{t+k},\\ \frac{2^{2n}\prod _{j=1}^n\left( t+j-\frac{1}{2}\right) }{\prod _{j=0}^n(t+j)}&=\sum _{k=0}^n\frac{\left( {\begin{array}{c}2k\\ k\end{array}}\right) \left( {\begin{array}{c}2n-2k\\ n-k\end{array}}\right) }{t+k},\\ \frac{2^{2n}\prod _{j=1}^n\left( t+n+j-\frac{1}{2}\right) }{\prod _{j=0}^n(t+j)}&=\sum _{k=0}^n\frac{(-1)^k\left( {\begin{array}{c}4n-2k\\ 2n\end{array}}\right) \left( {\begin{array}{c}2n\\ k\end{array}}\right) }{t+k} \end{aligned}$$

and $2t+n$; the inclusions $d_n^{s-i}a_{i,k}\in \mathbb {Z}$ then follow from [8, Lemma 1]. The cancellation by the factor $\Phi _n$ originates from the p-adic analysis of the binomial factors entering

$$\begin{aligned} a_{s,k}=\frac{(2n+2k)!\,(4n-2k)!}{(n+k)!\,(2n-k)!\,k!^3(n-k)!^3} \cdot (n-2k){\left( {\begin{array}{c}n\\ k\end{array}}\right) }^{s-3} \quad \text {for} \; k=0,1,\ldots ,n, \end{aligned}$$

and the estimate $\hbox {ord}_pa_{i,k}\ge -(s-i) +\hbox {ord}_pa_{s,k}\ge -(s-i)+\varphi (n/p,k/p)$ for primes in the range $2\sqrt{n}<p\le n$, where

$$\begin{aligned} \varphi (x,y)=\lfloor 2x+2y\rfloor +\lfloor 4x-2y\rfloor -\lfloor x+y\rfloor -\lfloor 2x-y\rfloor -3\lfloor y\rfloor -3\lfloor x-y\rfloor \end{aligned}$$

is a periodic function of period 1 in both x and y, and from the inequality

$$\begin{aligned} \varphi (x,y)\ge \min _{y\in \mathbb {R}}\varphi (x,y) =\min _{0\le y<1}\varphi (x,y)=\varphi _0(x); \end{aligned}$$

the details can be borrowed from [4, Sect. 7]. Furthermore, the property $R_n(-t-n)=R_n(t)$ derived from (1) implies $a_{i,k}=(-1)^ia_{i,n-k}$ for $i=1,\ldots ,s$ and $k=0,1,\ldots ,n$.

Recall that n is even, so that $n/2=m$ is a positive integer. The summation over $\nu$ in (2) can also start from $-m-1$ (rather than 1 or $n+1$), because the function $R_n(t)$ vanishes at all half-integers between $-2n$ and n. Therefore,

$$\begin{aligned} r_n= & {} \sum _{\nu =-m-1}^{\infty } R_n\left( \nu -\tfrac{1}{2}\right) (-1)^{\nu } =\sum _{i=1}^s\sum _{k=0}^na_{i,k}\sum _{\nu =-m-1}^{\infty } \frac{(-1)^{\nu }}{\left( \nu +k-\frac{1}{2}\right) ^i} \\= & {} \sum _{i=1}^s\sum _{k=0}^n(-1)^{k-1}a_{i,k} \sum _{\ell =k-m}^{\infty }\frac{(-1)^{\ell }}{\left( \ell +\frac{1}{2}\right) ^i}\\= & {} \sum _{i=1}^s2^i\beta (i)\sum _{k=0}^n(-1)^{k-1}a_{i,k} +\sum _{i=1}^s\sum _{k=0}^{m-1}(-1)^{k-1}a_{i,k} \sum _{\ell =k-m}^{-1}\frac{(-1)^{\ell }}{\left( \ell +\frac{1}{2}\right) ^i} \\&-\sum _{i=1}^s\sum _{k=m+1}^n(-1)^{k-1}a_{i,k}\sum _{\ell =0}^{k-m-1} \frac{(-1)^{\ell }}{\left( \ell +\frac{1}{2}\right) ^i}, \end{aligned}$$

where the rules

$$\begin{aligned} \sum _{\ell =k-m}^{\infty }=\sum _{\ell =0}^{\infty }-\sum _{\ell =0}^{k-m-1} \quad \text {if}\; k>m \qquad \text {and}\qquad \sum _{\ell =k-m}^{\infty } =\sum _{\ell =0}^{\infty }+\sum _{\ell =k-m}^{-1} \quad \text {if}\; k<m \end{aligned}$$

were applied. Thus, the rational numbers

$$\begin{aligned} a_i=2^i\sum _{k=0}^n(-1)^{k-1}a_{i,k} \quad \text {for}\; i=1,\ldots ,s \end{aligned}$$

satisfy $\Phi _n^{-1}d_n^{s-i}a_i\in \mathbb {Z}$, while for the quantity

$$\begin{aligned} a_0=\sum _{i=1}^s\sum _{k=0}^{m-1}(-1)^{k-1}a_{i,k}\sum _{\ell =k-m}^{-1} \frac{(-1)^{\ell }}{\left( \ell +\frac{1}{2}\right) ^i} -\sum _{i=1}^s\sum _{k=m+1}^n(-1)^{k-1} a_{i,k}\sum _{\ell =0}^{k-m-1}\frac{(-1)^{\ell }}{\left( \ell +\frac{1}{2}\right) ^i} \end{aligned}$$

the inclusion $\Phi _n^{-1}d_n^sa_0\in \mathbb {Z}$ follows from noticing that

$$\begin{aligned} d_{n-1}^i\sum _{\ell =k-m}^{-1}\frac{(-1)^{\ell }}{\left( \ell +\frac{1}{2}\right) ^i}&\in \mathbb {Z} \quad \text {if}\; 0\le k<m, \\ d_{n-1}^i\sum _{\ell =0}^{k-m-1}\frac{(-1)^{\ell }}{\left( \ell +\frac{1}{2}\right) ^i}&\in \mathbb {Z} \quad \text {if}\; m<k\le n. \end{aligned}$$

Finally,

$$\begin{aligned} a_i=2^i\sum _{k=0}^n(-1)^{k-1}a_{i,k}&=(-1)^i2^i\sum _{k=0}^n(-1)^{k-1}a_{i,n-k}\\&=(-1)^i2^i\sum _{k=0}^n(-1)^{n-k-1}a_{i,n-k} =(-1)^ia_i, \end{aligned}$$

so that $a_i$ vanish for odd i. $\square$

Set now $s=17$, in which case we compute from (3) that

$$\begin{aligned} \lim _{n\rightarrow \infty }r_n^{1/n}=e^{-16.1123070755\ldots }, \end{aligned}$$

hence the linear forms

$$\begin{aligned} \Phi _n^{-1}d_n^{17}r_n\in \mathbb {Z}\beta (2)+\mathbb {Z}\beta (4) +\cdots +\mathbb {Z}\beta (16)+\mathbb {Z} \end{aligned}$$

are positive and tend to 0 as $n\rightarrow \infty$. This implies that the eight numbers $\beta (2),\beta (4),\ldots ,\beta (16)$ cannot be all rational.

3 General settings

A natural way to generalize the construction in Sect. 2 follows the recipe of [4] and [7].

For an odd integer $s\ge 5$, consider a collection $\varvec{\eta }=(\eta _0,\eta _1,\ldots ,\eta _s)$ of integral parameters satisfying the conditions

$$\begin{aligned} 0<\eta _j<\frac{1}{2}\eta _0 \quad \text {for}\; j=1,\ldots ,s \qquad \text {and}\qquad \eta _1+\eta _2+\cdots +\eta _s\le \frac{s-1}{2}\eta _0, \end{aligned}$$

to which we assign, for each positive integer n, the collection

$$\begin{aligned} h_0=\eta _0n+1,\quad h_j=\eta _jn +\tfrac{1}{2} \quad \text {for}\; j=1,\ldots ,s. \end{aligned}$$

In what follows, we assume that $h_0-1=\eta _0n$ is even—the condition that is automatically achieved when $\eta _0\in 2\mathbb {Z}$, otherwise by restricting to even n.

Define the rational function

$$\begin{aligned} R_n(t)=R_{n,\varvec{\eta }}(t)=\gamma _n\cdot (2t+h_0)\, \frac{(t+1)_{h_0-1}}{\prod _{j=1}^s(t+h_j)_{1+h_0-2h_j}}=R_n(-t-h_0), \end{aligned}$$

(4)

where

$$\begin{aligned} \gamma _n=4^{h_0-1}\,\frac{\prod _{j=2}^s(h_0-2h_j)!}{\left( h_1-\frac{1}{2}\right) !^2}, \end{aligned}$$

and the (very-well-poised) hypergeometric sum

$$\begin{aligned} r_n&=r_{n,\varvec{\eta }} =\sum _{\nu =0}^{\infty } R_n(\nu )(-1)^{\nu } \nonumber \\&=\gamma _n \cdot \frac{\Gamma (1+h_0)\,\prod _{j=1}^q\Gamma (h_j)}{\prod _{j=1}^q\Gamma (1+h_0-h_j)} \,{}_{s+2}F_{s+1} \biggl (\begin{matrix} h_0, \; 1+\tfrac{1}{2}h_0, \; h_1, \; \ldots ,\; h_s \\ \tfrac{1}{2}h_0, \, 1+h_0-h_1, \, \ldots , \, 1+h_0-h_s \end{matrix}\biggm |-1\biggr ). \end{aligned}$$

(5)

Then [4, Lemma 1] implies the following Euler-type integral representation of $r_n$ (see also [4, Lemma 3]).

Lemma 2

The formula

$$\begin{aligned} r_n&=\frac{4^{h_0-1}\Gamma (1+h_0)}{\Gamma \left( h_1+\frac{1}{2}\right) ^2 \,\Gamma (1+h_0-2h_1)} \\&\qquad \times \int \cdots \int _{[0,1]^s} \frac{\prod _{j=1}^st_j^{h_j-1}(1-t_j)^{h_0-2h_j}}{(1+t_1t_2\cdots t_s)^{h_0}}\, \frac{1-t_1t_2\cdots t_s}{1+t_1t_2\cdots t_s} \,\mathrm{d} t_1\,\mathrm{d} t_2\cdots \mathrm{d} t_s \end{aligned}$$

is valid. In particular, $r_n>0$ and

$$\begin{aligned} \lim _{n\rightarrow \infty }r_n^{1/n}=\frac{(4\eta _0)^{\eta _0}}{\eta _1^{2\eta _1} (\eta _0-2\eta _1)^{\eta _0-2\eta _1}} \cdot \max _{\varvec{t}\in [0,1]^s} \frac{\prod _{j=1}^st_j^{\eta _j}(1-t_j)^{\eta _0-2\eta _j}}{(1+t_1t_2\cdots t_s)^{\eta _0}}. \end{aligned}$$

Computation of the latter maximum is performed in [4, Sect. 4, Remark], and the result is as follows.

Lemma 3

Assume that $x_0$ is a unique zero of the polynomial

$$\begin{aligned} x\prod _{j=1}^s\bigl ((\eta _0-\eta _j)-\eta _jx\bigr ) -\prod _{j=1}^s\bigl (\eta _j-(\eta _0-\eta _j)x\bigr ) \end{aligned}$$

in the interval $0<x<1$, and set

$$\begin{aligned} x_j =\frac{\eta _j-(\eta _0-\eta _j)x_0}{(\eta _0-\eta _j)-\eta _jx_0} \quad \text {for}\; j=1,2,\ldots ,s. \end{aligned}$$

Then

$$\begin{aligned} \max _{\varvec{t}\in [0,1]^s}\frac{\prod _{j=1}^st_j^{\eta _j} (1-t_j)^{\eta _0-2\eta _j}}{(1+t_1t_2\cdots t_s)^{\eta _0}} =\frac{\prod _{j=1}^sx_j^{\eta _j}(1-x_j)^{\eta _0-2\eta _j}}{(1+x_1x_2\cdots x_s)^{\eta _0}}. \end{aligned}$$

Arithmetic ingredients of the construction are in line with the strategy used in the proof of Lemma 1. For simplicity we split the corresponding statement into two parts. Define

$$\begin{aligned} N=\min _{1\le j\le s}\{h_j-\tfrac{1}{2}\} \qquad \text {and}\qquad M=\max \{h_0-2N-1,\,h_1-\tfrac{1}{2}\}, \end{aligned}$$

and notice that the poles of the rational function (4) are located at the points $t=-k-\frac{1}{2}$ for integers k in the range $N\le k\le h_0-N-1$.

Lemma 4

The coefficients in the partial-fraction decomposition

$$\begin{aligned} R_n(t)=\sum _{i=1}^s\sum _{k=N}^{h_0-N-1} \frac{a_{i,k}}{\left( t+k+\tfrac{1}{2}\right) ^i} \end{aligned}$$

of (4) satisfy

$$\begin{aligned} a_{i,k}=(-1)^ia_{i,h_0-1-k} \end{aligned}$$

(6)

and

$$\begin{aligned} \Phi _n^{-1}d_M^{s-i}a_{i,k}\in \mathbb {Z} \end{aligned}$$

(7)

for $i=1,\ldots ,s$ and $N\le k\le h_0-N-1$, where the product over primes

$$\begin{aligned} \Phi _n=\prod _{\sqrt{2h_0}<p\le M}p^{\varphi _0(n/p)} \end{aligned}$$

is defined through the 1-periodic functions

$$\begin{aligned} \varphi _0(x)=\min _{0\le y<1}\varphi (x,y) \end{aligned}$$

and

$$\begin{aligned} \varphi (x,y)&=\lfloor 2(\eta _0x-y)\rfloor +\lfloor 2y\rfloor -\lfloor \eta _0x-y\rfloor -\lfloor y\rfloor -2\lfloor \eta _1x\rfloor -\lfloor (\eta _0-2\eta _1)x\rfloor \\&+\sum _{j=1}^s\bigl (\lfloor (\eta _0-2\eta _j)x\rfloor -\lfloor y -\eta _jx\rfloor -\lfloor (\eta _0-\eta _j)x-y\rfloor \bigr ). \end{aligned}$$

Proof

For this, we write the function $R_n(t-\tfrac{1}{2})$ as the product of $2t+h_0-1$, the three integer-valued polynomials

$$\begin{aligned} \frac{4^{h_1^*}\left( t+\frac{1}{2}\right) _{h_1^*}}{h_1^*!}, \quad \frac{4^{h_0-2h_1}\left( t+h_1^*+\frac{1}{2}\right) _{h_0-2h_1}}{(h_0-2h_1)!}, \quad \frac{4^{h_1^*}\left( t+h_0-h_1^*-\frac{1}{2}\right) _{h_1^*}}{h_1^*!}, \end{aligned}$$

where $h_1^*=h_1-\frac{1}{2}=\eta _1n$, and the rational functions

$$\begin{aligned} \frac{(h_0-2h_j)!}{\left( t+h_j-\frac{1}{2}\right) _{1+h_0-2h_j}} \quad \text {for}\; j=1,\ldots ,s. \end{aligned}$$

Then [4, Lemmas 4, 5, 10, 11] and the Leibniz rule for differentiating a product imply the inclusions $d_M^{s-i}a_{i,k}\in \mathbb {Z}$ and estimates

$$\begin{aligned} \hbox {ord}_pa_{i,k}\ge -(s-i)+\varphi \Bigl (\frac{n}{p},\frac{k}{p}\Bigr ) \end{aligned}$$

for the p-adic order of the coefficients. These are combined to conclude with (7).

The property (6) follows from the symmetry of the rational function (4). $\square$

Lemma 5

The decomposition

$$\begin{aligned} r_n=\sum _{\begin{array}{c} i=1\\ i\;\text {even} \end{array}}^sa_i\beta (i)+a_0 \in \mathbb {Q}\beta (2)+\mathbb {Q}\beta (4)+\cdots +\mathbb {Q}\beta (s-1)+\mathbb {Q} \end{aligned}$$

(8)

takes place, where $\Phi _n^{-1}d_M^{s-i}a_i\in \mathbb {Z}$ for $i=0,1,\ldots ,s$ even, and $\Phi _n$ is defined in Lemma 4.

Proof

Since the function (4) vanishes at $t=-1,-2,\ldots ,-h_0+2$, we can shift the summation in (5):

$$\begin{aligned} r_n =\sum _{\nu =-(h_0-1)/2}^{\infty } R_n(\nu )(-1)^{\nu } =\sum _{i=1}^s\sum _{k=N}^{h_0-N-1}(-1)^ka_{i,k}\sum _{\nu =-(h_0-1)/2}^{\infty } \frac{(-1)^{\nu +k}}{\left( \nu +k+\tfrac{1}{2}\right) ^i}. \end{aligned}$$

Now proceeding as in the proof of Lemma 1 we arrive at the desired decomposition (8) with

$$\begin{aligned} a_0= & {} \sum _{i=1}^s\sum _{k=N}^{(h_0-3)/2}(-1)^ka_{i,k} \sum _{\ell =k-(h_0-1)/2}^{-1}\frac{(-1)^{\ell }}{\left( \ell +\tfrac{1}{2}\right) ^i} \\&-\sum _{i=1}^s\sum _{k=(h_0+1)/2}^{h_0-N-1}(-1)^ka_{i,k} \sum _{\ell =0}^{k-(h_0+1)/2}\frac{(-1)^{\ell }}{\left( \ell +\tfrac{1}{2}\right) ^i} \end{aligned}$$

and

$$\begin{aligned} a_i=2^i\sum _{k=N}^{h_0-N-1}(-1)^ka_{i,k}, \end{aligned}$$

with $a_i$ vanishing for i odd in view of the property (6). The inclusions for the coefficients in (8) therefore follow from Lemma 4 and

$$\begin{aligned} d_{h_0-2N-2}^i\sum _{\ell =k-(h_0-1)/2}^{-1} \frac{(-1)^{\ell }}{\left( \ell +\tfrac{1}{2}\right) ^i}&\in \mathbb {Z} \quad \text {if}\;\; N\le k\le \frac{h_0-3}{2},\\ d_{h_0-2N-2}^i\sum _{\ell =0}^{k-(h_0+1)/2} \frac{(-1)^{\ell }}{\left( \ell +\tfrac{1}{2}\right) ^i}&\in \mathbb {Z} \quad \text {if}\;\; \frac{h_0+1}{2}\le k\le h_0-N-1. \end{aligned}$$

$\square$

Proof of Theorem 1

Take $s=13$ and

$$\begin{aligned} (\eta _0,\eta _1,\ldots ,\eta _{13}) =(31,10,10,10,10,10,11,11,11,11,12,12,12,12), \end{aligned}$$

hence $M=11n$. Then

$$\begin{aligned} \lim _{n\rightarrow \infty }r_n^{1/n}=\exp (-100.73966317\ldots ) \end{aligned}$$

and

$$\begin{aligned} \varphi _0(x)={\left\{ \begin{array}{ll} 8 &{} \text {if}\; \{x\}\in \bigl [\frac{7}{24},\frac{3}{10}\bigr ), \\ 7 &{} \text {if}\; \{x\}\in \bigl [\frac{3}{31},\frac{1}{10}\bigr ) \cup \bigl [\frac{6}{31},\frac{1}{5}\bigr ) \cup \bigl [\frac{9}{31},\frac{7}{24}\bigr ) \cup \bigl [\frac{19}{24},\frac{4}{5}\bigr ) \cup \bigl [\frac{8}{9},\frac{9}{10}\bigr ), \\ 6 &{} \text {if}\; \{x\}\in \bigl [\frac{1}{11},\frac{3}{31}\bigr ) \cup \bigl [\frac{2}{11},\frac{6}{31}\bigr ) \cup \bigl [\frac{3}{11},\frac{9}{31}\bigr ) \cup \bigl [\frac{7}{20},\frac{2}{5}\bigr ) \cup \bigl [\frac{9}{20},\frac{1}{2}\bigr ) \\ &{}\qquad \qquad \cup \bigl [\frac{11}{20},\frac{3}{5}\bigr ) \cup \bigl [\frac{13}{20},\frac{7}{10}\bigr ) \cup \bigl [\frac{3}{4},\frac{19}{24}\bigr ) \cup \bigl [\frac{6}{7},\frac{8}{9}\bigr ), \\ 5 &{} \text {if}\; \{x\}\in \bigl [\frac{7}{31},\frac{1}{4}\bigr ) \cup \bigl [\frac{7}{22},\frac{7}{20}\bigr ) \cup \bigl [\frac{3}{7},\frac{9}{20}\bigr ) \cup \bigl [\frac{17}{31},\frac{11}{20}\bigr ) \cup \bigl [\frac{19}{31},\frac{5}{8}\bigr ) \\ &{}\qquad \qquad \cup \bigl [\frac{20}{31},\frac{13}{20}\bigr ) \cup \bigl [\frac{5}{7},\frac{8}{11}\bigr ) \cup \bigl [\frac{23}{31},\frac{3}{4}\bigr ), \\ 4 &{} \text {if}\; \{x\}\in \bigl [\frac{1}{10},\frac{1}{8}\bigr ) \cup \bigl [\frac{1}{5},\frac{7}{31}\bigr ) \cup \bigl [\frac{3}{10},\frac{7}{22}\bigr ) \cup \bigl [\frac{6}{11},\frac{17}{31}\bigr ) \cup \bigl [\frac{3}{5},\frac{19}{31}\bigr ) \\ &{}\qquad \qquad \cup \bigl [\frac{7}{11},\frac{20}{31}\bigr ) \cup \bigl [\frac{8}{11},\frac{23}{31}\bigr ), \\ 2 &{} \text {if}\; \{x\}\in \bigl [\frac{1}{24},\frac{1}{11}\bigr ) \cup \bigl [\frac{3}{22},\frac{2}{11}\bigr ) \cup \bigl [\frac{1}{4},\frac{3}{11}\bigr ) \cup \bigl [\frac{13}{24},\frac{6}{11}\bigr ) \cup \bigl [\frac{5}{8},\frac{7}{11}\bigr ) \\ &{}\qquad \qquad \cup \bigl [\frac{17}{20},\frac{6}{7}\bigr ) \cup \bigl [\frac{19}{20},1\bigr ), \\ 1 &{} \text {if}\; \{x\}\in \bigl [\frac{1}{31},\frac{1}{24}\bigr ) \cup \bigl [\frac{1}{8},\frac{3}{22}\bigr ) \cup \bigl [\frac{9}{22},\frac{3}{7}\bigr ) \cup \bigl [\frac{1}{2},\frac{13}{24}\bigr ) \cup \bigl [\frac{7}{10},\frac{5}{7}\bigr )\\ &{}\qquad \qquad \cup \bigl [\frac{4}{5},\frac{9}{11}\bigr ) \cup \bigl [\frac{26}{31},\frac{17}{20}\bigr ) \cup \bigl [\frac{9}{10},\frac{10}{11}\bigr ) \cup \bigl [\frac{29}{31},\frac{19}{20}\bigr ), \\ 0 &{} \text {otherwise}, \end{array}\right. } \end{aligned}$$

so that

$$\begin{aligned} \lim _{n\rightarrow \infty }(\Phi _n^{-1}d_M^{13})^{1/n}=\exp (100.23354349\ldots ). \end{aligned}$$

This means that the positive linear forms

$$\begin{aligned} \Phi _n^{-1}d_M^{13}r_n\in \mathbb {Z}\beta (2) +\mathbb {Z}\beta (4)+\cdots +\mathbb {Z}\beta (12)+\mathbb {Z} \end{aligned}$$

tend to 0 as $n\rightarrow \infty$. Thus, at least one of the even beta values in consideration must be irrational. $\square$