Many odd zeta values are irrational

Building upon ideas of the second and third authors, we prove that at least $2^{(1-\varepsilon)\frac{\log s}{\log\log s}}$ values of the Riemann zeta function at odd integers between 3 and $s$ are irrational, where $\varepsilon$ is any positive real number and $s$ is large enough in terms of $\varepsilon$. This lower bound is asymptotically larger than any power of $\log s$; it improves on the bound $\frac{1-\varepsilon}{1+\log2}\log s$ that follows from the Ball--Rivoal theorem. The proof is based on construction of several linear forms in odd zeta values with related coefficients.


Introduction
When s ≥ 2 is an even integer, the value ζ(s) of the Riemann zeta function is a non-zero rational multiple of π s and, therefore, a transcendental number. On the other hand, no such relation is expected to hold for ζ(s) when s ≥ 3 is odd; a folklore conjecture states that the numbers π, ζ(3), ζ(5), ζ(7), . . . are algebraically independent over the rationals. This conjecture is predicted by Grothendieck's period conjecture for mixed Tate motives. But both conjectures are far out of reach and we do not even know the transcendence of a single odd zeta value.
Their corresponding result for small s has been refined several times [26,8], but the question whether ζ (5) is irrational remains open. The proof of Theorem 1 involves the well-poised hypergeometric series n! s−2r ∞ t=1 (2r+1)n j=0 (t − rn + j) n j=0 (t + j) s+1 , (0.1) which happens to be a Q-linear combination of 1 and odd zeta values when s is odd and n is even, and Nesterenko's linear independence criterion [17]. The bound 1−ε 1+log 2 log s follows from comparison of how small the linear combination is with respect to the size of its coefficients, after multiplying by a common denominator to make them integers. To improve on this bound using the same strategy, one has to find linear combinations that are considerably smaller, with not too large coefficients, -it comes out to be a rather difficult task. This may be viewed as an informal explanation of why the lower bound in Theorem 1 has never been improved for large values of s, whereas the theorem itself has been generalized to several other families of numbers.
Using (with s = 20) the series which is a Q-linear combination of 1 and odd zeta values starting from ζ(5), Rivoal has proved [21] that among the numbers ζ(5), ζ (7), . . . , ζ(21), at least one is irrational. This result has been improved by the third author [25]: among the four numbers ζ(5), ζ(7), ζ(9), ζ(11), at least one is irrational; and he also showed [26] that, for any odd ℓ ≥ 1, there is an irrational number among ζ(ℓ + 2), ζ(ℓ + 4), . . . , ζ(8ℓ − 1). Proofs of these results do not require use of linear independence criteria: if a sequence of Z-linear combinations of real numbers from a given (fixed) collection tends to 0, and is non-zero infinitely often, then at least one of these numbers is irrational. A drawback of this approach is that it only allows one to prove that one number in a family is irrational.
The situation has drastically changed when the third author introduced [27] a new method (see also [13]). He casts (with s = 25) the rational function in the form and proves that both series are Q-linear combinations of 1, ζ(3), ζ(5), . . . , ζ(s) with related coefficients. This allows him to eliminate one odd zeta value, and to prove that at least two zeta values among ζ(3), ζ(5), . . . , ζ(25) are irrational. In view of Apéry's Theorem, the result means that one number among ζ(5), . . . , ζ(25) is irrational -nothing really novel, but the method of proof is new and more elementary than the ones in [21] and [25] as it avoids use of the saddle point method. More importantly, the method allows to prove the irrationality of at least two zeta values in a family without having to produce very small linear forms. The same strategy has been adopted by Rivoal and the third author [23] to prove that among ζ(5), ζ(7), . . . , ζ(69), at least two numbers are irrational.
The method in [27] has been generalized by the second author [24], who introduces another integer parameter D > 1 and considers the rational function He proves that for any divisor d of D the series is a Q-linear combination of 1, ζ(3), ζ(5), . . . , ζ(s). The crucial point of this construction is that each ζ(i) appears in this Q-linear combination with a coefficient that depends on d in a very simple way. This makes it possible to eliminate from the entire collection of these linear combinations as many odd zeta values as the number of divisors of D. Finally, taking D equal to a power of 2 and s sufficiently large with respect to D, the second author proves that at least log D log 2 numbers are irrational among ζ(3), ζ(5), . . . , ζ(s). This strategy represents a new proof that ζ(i) is irrational for infinitely many odd integers i.
Building upon the approach in [27] and [24] we prove the following result. In this result, the lower bound is asymptotically greater than exp( √ log s), and than any power of log s; "to put it roughly, [it is] much more like a power of s than a power of log s" [11, Chapter XVIII, §1].
In comparison, Theorem 1 gives only 1−ε 1+log 2 log s irrational odd zeta values, but they are linearly independent over the rationals, whereas Theorem 2 ends up only with their irrationality.
Our proof of Theorem 2 follows the above-mentioned strategy of the second and third authors. The main new ingredient, compared to the proof in [24], is taking D large (about s 1−2ε ) and equal to the product of the first prime numbers (the so-called primorial) -such a number has asymptotically the largest possible number of divisors with respect to its size (see [11,Chapter XVIII,§1]). To perform the required elimination of a prescribed set of odd zeta values, we need to establish that a certain auxiliary matrix is invertible. Whereas the second author's choice of D in [24] allows him to deal with elementary properties of a Vandermonde matrix, we use at this step a generalization of the corresponding result. We give three different proofs of the latter, based on arguments from combinatorics of partitions, from linear algebra accompanied with a lemma of Fekete, and from analysis using Rolle's theorem.
The structure of this paper is as follows. In §1 we construct linear forms in values of the Hurwitz zeta function. Denominators of the coefficients are studied in §2; and the asymptotics of the linear forms are dealt with in §3. Section 4 is devoted to the proof that an auxiliary matrix is invertible. Finally, we establish Theorem 2 in §5.

Construction of linear forms
From now on we let s, D be positive integers such that s ≥ 3D; we assume that s is odd. Let n be a positive integer, such that Dn is even. Consider the following rational function: which, of course, depends also on s and D. Notice that the difference of the function from the corresponding one in [24] is in the factor D 3Dn instead of D 6(D−1)n (see Eq. (0.2)). Similar rational functions have already been considered, see [22] for the case D = 2 and [16,18,6] for general D. However the "central" factors t − n + j D with Dn < j < 2Dn are missing, and (as the second author noticed [24]) they play a central role in the arithmetic estimates (see Lemma 2 below).
Remark 1. Though one can implement an additional parameter r in the definition of the rational function R n (t), in a way similar to the one for the Ball-Rivoal series (0.1), we have verified that this does not bring any improvement to the result of Theorem 2.
The rational function R n (t) has a partial fraction expansion For any j ∈ {1, . . . , D}, take We recall that the Lerch and Hurwitz zeta functions are defined by where α > 0 and also i ≥ 2 for the latter.
The following is precisely [24,Lemma 1.5]; the change of the normalizing factor D 3Dn does not affect the statement.
does not depend on j, and Proof . We follow the strategy of proofs in [27, Lemma 3] and [24, Lemma 1.5]. Let z be a real number such that 0 < z < 1. We have Now we let z tend to 1 in the equality we have obtained; the left-hand side tends to r n,j . On the right-hand side, the term involving the Lerch function with i = 1 has coefficient has only a logarithmic divergence as z → 1 and n k=0 a 1,k = lim t→∞ tR n (t) = 0, this term tends to 0 as z → 1. All other terms have finite limits as z → 1, so that where ρ 0,j is given by Eq. (1.2), and ρ i = n k=0 a i,k for any i ∈ {2, . . . , s}.
To complete the proof, we apply the symmetry phenomenon of [2,20]. Since s is odd and Dn is even we have R n (−n − t) = −R n (t). Now the partial fraction expansion (1.1) is unique, so that a i,n−k = (−1) i+1 a i,k for any i and k. This implies that ρ i = 0 when i is even, and Lemma 1 follows.
where A α,k is an integer in view of the explicit formulas We also consider From this expression we compute the partial fraction expansion of R n (t) using the rules A denominator appears each time the second rule is applied, and the denominator is always a divisor of d n (see [4] or [27, Lemma 1]). This happens s + 1 − i times in each term that contributes to a i,k because there are s + 1 factors in the product (2.3) (apart from t − n). Therefore, d s+1−i n a i,k ∈ Z for any i and k, implying (2.1).
We now proceed with the second part of Lemma 2, that is, with demonstrating the inclusions (2.2). Recall from Lemma 1 that are integers for any k, ℓ and i, so that d s+1 n+1 ρ 0,j ∈ Z. From now on, we assume that 1 ≤ j ≤ D − 1 and we prove that for any k and any ℓ the internal sum over i in Eq. (2.4) is an integer. With this aim in mind, fix integers k 0 and ℓ 0 , with 0 ≤ ℓ 0 ≤ k 0 ≤ n, and assume that the corresponding sum is not an integer.
This rational number is not an integer: it has negative p-adic valuation for at least one prime number p. Therefore, on either side of (2.5) there is at least one term with negative p-adic valuation: there exist i 0 , i 1 ∈ {1, . . . , s} and k 1 ∈ {0, . . . , n}, Since d s+1−i n+1 a i,k ∈ Z for any i and k, this leads to .
, which is impossible in view of the inequality 0 < |k 0 − k 1 | ≤ n. The contradiction completes the proof of Lemma 2.

Remark 2.
It is made explicit in [23], for a particular situation considered there, that the inclusions in Lemma 2 can be sharpened as follows: . . , s, and Φ −1 n d s+1 n+1 ρ 0,j ∈ Z for any j ∈ {1, . . . , D}, where Φ n = Φ n (D) is a certain product over primes in the range 2 ≤ p ≤ n, whose asymptotic behavior φ = φ(D) = lim n→∞ log Φ n n can be controlled by means of the prime number theorem. It is possible to show that the quantity φ(D)/D increases to ∞ and at the same time φ(D)/(D log ε D) → 0 as D → ∞, for any choice of ε > 0. Later, we choose D such that D log D < s, implying that the arithmetic gain coming from the factors Φ −1 n is asymptotically negligible as s → ∞.

Asymptotic estimates of the linear forms
The following lemma is proved along the same lines as [24, Lemma 2.1] (see also [27, Lemma 4] and the second proof of [2, Lemme 3]). The difference is that here we only assume s D log D to be sufficiently large, whereas in [24] parameter D is fixed and s → ∞. where and x 0 is the unique positive root of the polynomial Proof . For j ∈ {1, . . . , D} and k ≥ 0, let implying that, for any j, the quotient c k+1,j c k,j tends to f (κ) as n → ∞ assuming k ∼ κn for κ > 0 fixed, where For the logarithmic derivative of this function we have with a = s + 1 − 3D > 0 and c = −6D < 0, hence the derivative f ′ (x) vanishes exactly at one positive real number x 1 . This means that the function f (x) decreases on (0, x 1 ] and increases on [x 1 , +∞). Since lim x→0 + f (x) = +∞ and lim x→+∞ f (x) = 1, we deduce that there exists a unique positive real number x 0 such that f (x 0 ) = 1.
Let us now prove (3.2). As in [3, §3.4] we wish to demonstrate that the asymptotic behaviour of r n,j is governed by the terms c k,j with k close to x 0 n (see Eq. (3.8) below). To begin with, notice that Denoting by k 0 (n) the integer part of x 0 n and applying the Stirling formula to the factorial and gamma factors we obtain, as n → ∞, We shall now give details that the asymptotic behavior of r n,j as n → ∞ is determined by the terms c k,j with k close to x 0 n. Given D and s, we take ε > 0 sufficiently small to accommodate the condition terms of D, s and ε), so that taking k 1 = ⌊(x 0 + 2ε)n⌋ and k 2 = ⌊(x 0 + 3ε)n⌋ we obtain for all n sufficiently large. In the same way, we get the estimate for all n large (in terms of D, s and ε). At last, choosing ε small we can assume that A(ε) is sufficiently large (in terms of D and s), so that for k ≥ (A(ε) − ε)n we have for n large. Using hypothesis (3.1) and the Stirling formula, the latter estimate implies provided n is sufficiently large. Combining Eqs. Now for any k in the range (x 0 − 3ε)n ≤ k ≤ (x 0 + 3ε)n it follows from the proof of Eq. (3.4) that for n large (in terms of D, s and ε), where h is a positive function of ε such that lim ε→0 + h(ε) = 0. This implies for n sufficiently large, and finishes the proof of lim n→∞ r 1/n n,j = g(x 0 ) for any j.
To establish lim n→∞ r n,j ′ r n,j = 1 for any j, j ′ ∈ {1, . . . , D}, we can assume that 1 ≤ j ≤ D − 1 and j ′ = j + 1. For any k we have It follows from the Stirling formula that Γ(x + 1 D ) ∼ x 1/D Γ(x) as x → ∞, so that for k = ⌊x 0 n⌋ we have, as n → ∞, More generally, for k in the range (x 0 − 3ε)n ≤ k ≤ (x 0 + 3ε)n and n sufficiently large we have with lim ε→0 +h(ε) = 0. Using Eq. (3.8) this concludes the proof of (3.2), except for the upper bound on g(x 0 ) that we shall verify now.
By taking s D log D sufficiently large, we may ensure that the first two terms are sufficiently small in comparison with s and that a is sufficiently close to 0, so that log g(x 0 ) < −(s + 1) log 3.
This completes our proof of Lemma 3. In particular, we obtain that two out of {ζ (5), ζ (7), . . . , ζ(77)} are irrational. This result is slightly weaker than the result of Rivoal and the third author [23], but it drops out as a byproduct of the construction above. The arithmetic gain given by Φ n (4) for Φ n (D) defined in Remark 2 can be used to slightly reduce the bound of 77 to 73, still weaker than the one in [23].

A non-vanishing determinant
The following lemma is used to eliminate irrational zeta values in §5 below.
Lemma 4. For t ≥ 1, let x 1 < . . . < x t be positive real numbers and α 1 < . . . < α t non-negative integers. Then the generalized Vandermonde matrix [x αi j ] 1≤i,j≤t has positive determinant. We remark that, subject to the hypothesis that x 1 , . . . , x t are real and positive, Lemma 4 is a stronger version of [14, Lemme 1] and, therefore, has potential applications to the zero estimates for linear forms in two logarithms.
The above result is quite classical and known to many people. While writing this paper we have found various proofs of rather different nature, three given below. We leave it to the readers to choose their favorite proof.
Linear algebra proof of Lemma 4. Write A J,K for the minor of an n × m-matrix A, where n ≤ m, determined by ordered index sets J and K. A classical result due to Fekete [5] asserts that if all (n − 1)-minors Analytical proof of Lemma 4 (see [10, p. 76-77]). By induction on t one proves the following claim: A non-zero function with c i , α i ∈ R, has at most t − 1 positive zeros. Indeed, if f has t positive zeros then Rolle's theorem provides t − 1 positive zeros of the derivative d dx (x −α1 f (x)). The non-vanishing of the determinant in Lemma 4 is an immediate consequence of this claim. Since the determinant depends continuously on the parameters α i , we deduce the required positivity from the positivity of the Vandermonde determinant.

Elimination of odd zeta values
Let 0 < ε < 1 3 , and let s be odd and sufficiently large with respect to ε. We take D to be the product of all primes less than or equal to (1 − 2ε) log s (such a product has asymptotically the largest possible number of divisors with respect to its size, see [11, Chapter XVIII, §1]). We have by the prime number theorem, that is, D ≤ s 1−ε . Then D log D ≤ s 1−ε log s: the assumption of Lemma 3 holds.
Assume that the number of irrational odd zeta values between ζ(3) and ζ(s) is less than δ. Let 3 = i 1 < i 2 < . . . < i δ−1 ≤ s be odd integers such that if ζ(i) ∈ Q and i is odd, 3 ≤ i ≤ s, then i = i j for some j. We set i 0 = 1, and consider the set D of all divisors of D, so that Card D = δ. Lemma 4 implies that the matrix [d ij ] d∈D,0≤j≤δ−1 is invertible. Therefore, there exist integers w d ∈ Z, where d ∈ D, such that We shall use now the integers w d to eliminate the odd zeta values ζ(i j ) for j = 1, . . . , δ − 1, including all irrational ones, as in [27] and [24]. For that, consider r n = d∈D w d r n,d . Now all ζ(i), i ∈ I, are assumed to be rational. Denoting by A their common denominator, we deduce from Lemma 2 that Ad s+1 n+1 r n is an integer. From the prime number theorem we have lim n→∞ d 1/n n+1 = e, hence the sequence of integers satisfies 0 < lim n→∞ |Ad s+1 n+1 r n | 1/n = e s+1 g(x 0 ) < e 3 s+1 < 1.
This contradiction concludes the proof of Theorem 2.