Séminaire Lotharingien de Combinatoire 81 (2020), Article B81b

A NOTE ON ODD ZETA VALUES

TANGUY RIVOAL AND WADIM ZUDILIN

Dedicated to Christian Krattenthaler, on his 60th birthday

Abstract. Using a new construction of rational linear forms in odd zeta values and the saddle point method, we prove the existence of at least two irrational numbers amongst the 33 odd zeta values ζ(5), ζ(7), …, ζ(69).

1. Introduction

The arithmetic nature of the values of the Riemann zeta function at odd integers is still largely unknown. Apéry [1] proved that ζ(3) is an irrational numbers and it was proved in [2, 7] that infinitely many of the numbers ζ(2n + 1), n ≥ 1 integer, are irrational and in fact even linearly independent over ℚ. The second author proved in [10] the existence of at least one irrational number amongst ζ(5), ζ(7), ζ(9), ζ(11), and in [11] that for any integer n ≥ 0, there exists at least one irrational number amongst the odd zeta values ζ(2n + 3), ζ(2n + 5), …, ζ(16n + 7) (this result is used below with n = 5). Let us also mention that Fischler and the second author proved in [3] the existence of two distinct odd integers m, n ∈ {3, 5, …, 139} such that 1, ζ(m), ζ(n) are linearly independent over ℚ.

The goal of this note is to prove the following theorem.

Theorem 1. There exist at least two irrational numbers amongst the 33 odd zeta values ζ(5), ζ(7), …, ζ(69).

We shall in fact prove the following equivalent form of Theorem 1.

Theorem 2. For any integer m such that 1 ≤ m ≤ 33, there exists at least one irrational number amongst the 32 odd zeta values ζ(2n + 3), where n ∈ {1, 2, …, 33} \ {m}.

Let us prove the equivalence between both theorems. Applying Theorem 2 with m = 1, let ζ(2m_{0} + 3) denote one irrational number amongst ζ(7), ζ(9), …, ζ(69). We then apply again Theorem 2 with m = m_{0}: there exists an irrational number ζ(2n_{0} + 3) for some n_{0} ∈ {1, 2, …, 33} \ {m_{0}}, and Theorem 1 follows. Conversely, let ζ(2m_{0} + 3) and ζ(2n_{0} + 3) be two irrational numbers given by Theorem 1 with 1 ≤ m_{0}, n_{0} ≤ 33. Let m ∈ {1, 2, …, 33}: if m = m_{0}, we pick the irrational number ζ(2n_{0} + 3) while if m ≠ m_{0}, we pick the irrational number ζ(2m_{0} + 3), and Theorem 2 follows.

We now make some comparisons with the above mentioned results. The set {5, 7, …, 69} is much smaller than {3, 5, …, 139}, but our method cannot decide if 1 and the two irrational odd zeta values in Theorem 1 are linearly independent over ℚ. Also, there are two irrational odd zeta values ζ(m) and ζ(n) with m ∈ {5, 7, 9, 11} and n ∈ {13, 15, …, 87}: this bound is worse than in Theorem 1, but the localization is more precise.

2010 Mathematics Subject Classification. 11J72, 11M06, 33C20.

In general, results of the type “there exists an irrational number amongst (…)” are proved using the saddle point method to estimate the decay of certain sequences of linear forms in zeta values. This is the case of the proofs of the results in [8, 10, 11]. Very recently, the second author introduced a new method to prove in an “elementary” way the existence of at least one irrational number amongst ζ(5), ζ(7), …, ζ(25). In particular, he completely avoided the use of the saddle point method. Our approach is somewhat different, as we shall indeed combine this new approach with the saddle point method to obtain the proof of Theorem 2. Though one can still use an elementary strategy (as outlined in [9]) to establish a result similar to Theorem 1, this result is weaker than ours. For further applications of this new method, see [6, 9].

The paper is organized as follows. In §2, we introduce some notations and define two series S_{n} and Ŝ_{n} that enable us to construct “good” linear forms in odd zeta values in §3. In §4 we obtain integral representations for S_{n} and Ŝ_{n}, to which we apply the saddle point method in §5. We complete the proof of Theorem 2 in §6.

2. Notations

Let A denote an integer ≥ 15. For any integer n ≥ 0, we define the rational function of t

R(t) := n!^{A−15}2^{18n}(2t + n)\frac{(t − n)_{n}^{3}(t + n + 1)_{n}^{3}(t − n + \frac{1}{2})_{3n}^{3}}{(t)_{n+1}^{A}}

= n!^{A−15}2^{−3}(2t + n)\frac{(2t − 2n)_{6n+1}^{3}}{(t)_{n+1}^{A+3}} ,

where (x)_{m} := x(x + 1) ⋯ (x + m − 1).

The degree of R(t) is (15 − A)n − A + 1 ≤ −2 so that the partial fraction expansion is

R(t) = ∑_{j=1}^{A} ∑_{m=0}^{n} \frac{p_{j,m}}{(t + m)^{j}}  (1)

with

p_{j,m} := \frac{1}{(A − j)!}(R(t)(t + m)^{A})_{t=−m}^{(A−j)} ∈ ℚ  (2)

and the residue at ∞ of R(t) is 0.

For any integer n ≥ 0, we define the series which will be our main objects of study:

S_{n} := ∑_{k=1}^{∞} R′′(k),  Ŝ_{n} := ∑_{k=1}^{∞} R′′(k − \frac{1}{2}),

where the double prime stands for double differentiation.

A series similar to S_{n} with R(t) replaced by n!^{A−6}(2t + n)(t − n)_{n}^{3}(t + n + 1)_{n}^{3}(t)_{n+1}^{−A} was used in [8] to prove the existence of at least one irrational number amongst ζ(5), ζ(7), …, ζ(21). We shall follow the same approach as in [8] and combine it with [12] where the additional factor (t − n + \frac{1}{2})_{3n} is the principal innovation. Its effects are explained in the comments just after Proposition 1 below.

The arithmetic normalization n!^{A−15}2^{18n} of the rational function R(t) is not optimal in the sense that there is a large factor which can be cancelled out from the coefficients p_{j,m}. We carefully perform this analysis in Proposition 1. It is also possible that the arithmetic behavior of the coefficients of the linear forms S_{n} and Ŝ_{n} is even better, because of certain hypergeometric transformations underlying the construction. This phenomenon, known as “denominators conjecture”, is studied in [5], and the methods developed there might bring in similar improvements for this new situation. To keep our exposition at an accessible level, we do not include such considerations here.

3. Construction of two linear forms in odd zeta values

Proposition 1. Let us assume that A ≥ 16 and n ≥ 0 are both even.

(i) We have

S_{n} = q_{0,n} + ∑_{j odd^{j=5}}^{A+1} q_{j,n}ζ(j)  (3)

and

Ŝ_{n} = q̂_{0,n} + ∑_{j odd^{j=5}}^{A+1} q_{j,n}(2^{j} − 1)ζ(j),  (4)

where q_{0,n}, q̂_{0,n} and the q_{j,n}’s are rational numbers, that depend on A.

(ii) Let d_{n} := lcm{1, 2, …, n} and Φ_{n} ∈ ℕ be the product of prime powers defined in (9) below. Then Φ_{n}^{−3}d_{n}^{A+2} is a common denominator of the coefficients q_{0,n}, q̂_{0,n} and the q_{j,n} for odd j ≥ 5.

Explicit expressions for the sequences q_{0,n}, q̂_{0,n} and q_{j,n} are given in the proof. The relevance of the series S_{n} and Ŝ_{n} for Theorem 2 can now easily be shown: for any odd integer m ∈ {5, 7, …, A + 1}, we have

(2^{m} − 1)S_{n} − Ŝ_{n} = (2^{m} − 1)q_{0,n} − q̂_{0,n} + ∑_{j odd^{j=5,j≠m}}^{A+1} q_{j,n}(2^{m} − 2^{j})ζ(j),

where ζ(m) no longer appears.

4 TANGUY RIVOAL AND WADIM ZUDILIN

Proof. (i) Using the partial fraction expansion (1), we get

S_{n} = ∑_{j=1}^{A} ∑_{m=0}^{n} p_{j,m} ∑_{k=1}^{∞} \frac{d^{2}}{dk^{2}} \frac{1}{(k + m)^{j}}

= ∑_{j=1}^{A} ∑_{m=0}^{n} p_{j,m} ∑_{k=1}^{∞} \frac{j(j + 1)}{(k + m)^{j+2}}

= ∑_{j=1}^{A} ( ∑_{m=0}^{n} p_{j,m})j(j + 1)ζ(j + 2) − ∑_{j=1}^{A} ∑_{m=0}^{n} ∑_{k=1}^{m} \frac{j(j + 1)p_{j,m}}{k^{j+2}} .

Now, − ∑_{m=0}^{n} p_{j,1} is the residue at t = ∞ of R(t), hence is 0. Moreover, we can use in (1) the symmetry R(−n − t) = (−1)^{(A+1)(n+1)}R(t) = −R(t) (because A and n are even): it implies that p_{j,m} = (−1)^{j+1}p_{j,n−m}. Hence ∑_{m=0}^{n} p_{j,m} = 0 when j is even. Therefore

S_{n} = ∑_{j odd^{j=5}}^{A+1} ( ∑_{m=0}^{n}(j − 2)(j − 1)p_{j−2,m})ζ(j) − ∑_{j=1}^{A} ∑_{m=0}^{n} ∑_{k=1}^{m} \frac{j(j + 1)p_{j,m}}{k^{j+2}} .

We now let ζ̂(s) := ∑_{k=1}^{∞} \frac{1}{(k− \frac{1}{2} )^{s}} = (2^{s} − 1)ζ(s). We also have

Ŝ_{n} = ∑_{j=1}^{A} ∑_{m=0}^{n} p_{j,m} ∑_{k=1}^{∞} \frac{d^{2}}{dk^{2}} \frac{1}{(k + m − \frac{1}{2})^{j}}

= ∑_{j=1}^{A} ∑_{m=0}^{n} p_{j,m} ∑_{k=1}^{∞} \frac{j(j + 1)}{(k + m − \frac{1}{2})^{j+2}}

= ∑_{j=1}^{A} ( ∑_{m=0}^{n} j(j + 1)p_{j,m})ζ̂(j + 2) − ∑_{j=1}^{A} ∑_{m=0}^{n} ∑_{k=1}^{m} \frac{j(j + 1)p_{j,m}}{(k − \frac{1}{2})^{j+2}} .  (5)

Hence

Ŝ_{n} = ∑_{j odd^{j=5}}^{A+1} ( ∑_{m=0}^{n}(j − 2)(j − 1)p_{j−2,m})(2^{j} − 1)ζ(j) − ∑_{j=1}^{A} ∑_{m=0}^{n} ∑_{k=1}^{m} \frac{j(j + 1)p_{j,m}}{(k − \frac{1}{2})^{j+2}} .

We have thus proved (3) and (4) with

q_{j,n} := ∑_{m=0}^{n}(j − 2)(j − 1)p_{j−2,m}, j odd, j ≥ 5,  (6)
q_{0,n} := − ∑_{j=1}^{A} ∑_{m=0}^{n} ∑_{k=1}^{m} \frac{j(j + 1)p_{j,m}}{k^{j+2}} ,  (7)

and

q̂_{0,n} := − ∑_{j=1}^{A} ∑_{m=0}^{n} ∑_{k=1}^{m} \frac{j(j + 1)p_{j,m}}{(k − \frac{1}{2})^{j+2}} .  (8)

(ii) We can apply mutatis mutandis [12, Lemma 1] and get that d_{n}^{A−j}p_{j,m} ∈ ℤ for any j ∈ {1, …, A} and any m ∈ {0, …, n}.

On the other hand, observe that the rational function R(t) can be written as (2t + n) × F (t)^{3}G(t)^{A−15}, where

F (t) := 2^{6n} \frac{(t − n)_{n}(t + n + 1)_{n}(t − n + \frac{1}{2})_{3n}}{(t)_{n+1}^{5}} and G(t) := \frac{n!}{(t)_{n+1}} .

The numbers

(F (t)(t + m)^{5})_{t=−m} = \frac{(2n + 2m)! (4n − 2m)!}{m!^{6}(n − m)!^{6}}, for m = 0, 1, …, n,

have a large common factor. Indeed, by the standard formula for p-adic valuation v_{p}(n!) = ∑_{ℓ=1}^{∞}[n/p^{ℓ}] of n!, we get

v_{p}(\frac{(2n + 2m)! (4n − 2m)!}{m!^{6}(n − m)!^{6}}) = ρ(\frac{n}{p}, \frac{m}{p}) ≥ ρ_{0}(\frac{n}{p})

for an odd prime p > 2\sqrt{n}, where the function

ρ(x, y) := ⌊2x + 2y⌋ + ⌊4x − 2y⌋ − 6⌊y⌋ − 6⌊x − y⌋ = 6{y} + 6{x − y} − {2x + 2y} − {4x − 2y}

is periodic of period 1 both in x and y, and

ρ_{0}(x) := min_{y∈ℝ} ρ(x, y) = min_{0≤y<1} ρ(x, y)

is also 1-periodic. The latter can be explicitly given on the interval 0 ≤ x < 1 by

ρ_{0}(x) = \begin{cases}0, & if 0 ≤ x < \frac{1}{3}, \\ 1, & if \frac{1}{3} ≤ x < \frac{1}{2}, \\ 2, & if \frac{1}{2} ≤ x < \frac{2}{3}, \\ 3, & if \frac{2}{3} ≤ x < \frac{5}{6}, \\ 4, & if \frac{5}{6} ≤ x < 1 \end{cases}

(see, for example, [11, § 4] for similar computations). Thus, denoting

Φ_{n} := ∏_{2\sqrt{n}<p≤n} p^{ρ_{0}(n/p)}  (9)

(a product over prime numbers), we obtain

Φ_{n}^{−1}(F (t)(t + m)^{5})_{t=−m} ∈ ℤ for m = 0, 1, …, n.

Following the lines of the proof of [11, Lemma 4.2], we see that these inclusions imply

Φ_{n}^{−1}d_{n}^{k} \frac{1}{k!}(F (t)(t + m)^{5})_{t=−m}^{(k)} ∈ ℤ

for m = 0, 1, …, n and all integers k ≥ 0. For the same range of indices we also have

d_{n}^{k} \frac{1}{k!}(G(t)(t + m))_{t=−m}^{(k)} ∈ ℤ.

Combining these inclusions with the representation R(t) = (2t+n)F (t)^{3}G(t)^{A−15} and using Leibniz’s rule for differentiating products, we conclude that the numbers p_{j,m} in (2) satisfy Φ_{n}^{−3}d_{n}^{A−j}p_{j,m} ∈ ℤ for any j ∈ {1, …, A} and any m ∈ {0, …, n}.

From the expressions of q_{j,n} in (6) and (7), we then deduce that Φ_{n}^{−3}d_{n}^{A+2}q_{0,n} ∈ ℤ and Φ_{n}^{−3}d_{n}^{A+2}q_{j,n} ∈ ℤ for any odd j ≥ 5.

This argument does not work directly for q̂_{0,n} with the expression in (8) because it leads only to the weaker estimate Φ_{n}^{−3}d_{2n}^{A+2}q̂_{0,n} ∈ ℤ. We need an alternative expression for q̂_{0,n}, and for this we follow [12, p. 5]. We let ω = ⌊\frac{n−1}{2}⌋. Since R′′(k) = 0 for k = −\frac{1}{2}, −\frac{3}{2}, …, −n + \frac{1}{2}, in particular R′′(k − \frac{1}{2}) = 0 for k = −ω, …, −1, 0 when (^{1}) n ≥ 1. Hence, for any n ≥ 1,

Ŝ_{n} = ∑_{k=−ω}^{∞} R′′(k − \frac{1}{2}) = ∑_{j=1}^{A} ∑_{m=0}^{n} p_{j,m} ∑_{k=−ω}^{∞} \frac{d^{2}}{dk^{2}} \frac{1}{(k + m − \frac{1}{2})^{j}}

= ∑_{j=1}^{A} ∑_{m=0}^{ω} p_{j,m} ∑_{k=−ω}^{∞} \frac{j(j + 1)}{(k + m − \frac{1}{2})^{j+2}} + ∑_{j=1}^{A} ∑_{m=ω+1}^{n} p_{j,m} ∑_{k=−ω}^{∞} \frac{j(j + 1)}{(k + m − \frac{1}{2})^{j+2}}

= ∑_{j=1}^{A} ∑_{m=0}^{ω} p_{j,m}( ∑_{k=m−ω}^{0} \frac{j(j + 1)}{(k − \frac{1}{2})^{j+2}} + ∑_{k=1}^{∞} \frac{j(j + 1)}{(k − \frac{1}{2})^{j+2}} )

+ ∑_{j=1}^{A} ∑_{m=ω+1}^{n} p_{j,m}( ∑_{k=1}^{∞} \frac{j(j + 1)}{(k − \frac{1}{2})^{j+2}} − ∑_{k=1}^{m−ω−1} \frac{j(j + 1)}{(k − \frac{1}{2})^{j+2}} )

= ∑_{j=1}^{A} ( ∑_{m=0}^{n} j(j + 1)p_{j,m})ζ̂(j + 2) + ∑_{j=1}^{A} ∑_{m=1}^{ω}(−1)^{j}j(j + 1)p_{j,m} ∑_{k=0}^{ω−m} \frac{1}{(k + \frac{1}{2})^{j+2}}

− ∑_{j=1}^{A} ∑_{m=ω+1}^{n}(−1)^{j}j(j + 1)p_{j,m} ∑_{k=1}^{m−ω−1} \frac{1}{(k − \frac{1}{2})^{j+2}} .  (10)

Comparing (10) with (5), we see that, when n ≥ 1,

q̂_{0,n} = ∑_{j=1}^{A} ∑_{m=1}^{ω} p_{j,m} ∑_{k=0}^{ω−m} \frac{(−1)^{j}j(j + 1)}{(k + \frac{1}{2})^{j+2}} − ∑_{j=1}^{A} ∑_{m=ω+1}^{n} p_{j,m} ∑_{k=1}^{m−ω−1} \frac{(−1)^{j}j(j + 1)}{(k − \frac{1}{2})^{j+2}}.

1This is not true if n = 0, but in this case q̂_{0,n} = 0 and there is nothing to prove.

This expression is more suitable than (8) and an analysis similar to [12, p. 6] shows that Φ_{n}^{−3}d_{n}^{A+2}q̂_{0,n} ∈ ℤ as expected. □

4. Complex integral representations of S_{n} and Ŝ_{n}

In this section, the integers A ≥ 15 and n ≥ 0 are not necessarily even.

Proposition 2. Let L denote any vertical line {c + iy : y ∈ ℝ} oriented from y > 0 to y < 0, where c ∈ (\frac{1}{2}, n). We have

S_{n} = \frac{n!^{A−15}}{iπ} ∫_{L}(2t + n)\frac{Γ(t)^{A+3}Γ(2t + 4n + 1)^{3}Γ(2n − 2t + 1)^{3}}{Γ(t + n + 1)^{A+3}} cos(πt)^{4}dt (11)

and

Ŝ_{n} = −\frac{n!^{A−15}}{iπ} ∫_{L}(2t + n − 1)\frac{Γ(t − \frac{1}{2})^{A+3}Γ(2t + 4n)^{3}Γ(2n − 2t + 2)^{3}}{Γ(t + n + \frac{1}{2})^{A+3}} cos(πt)^{4}dt. (12)

Proof. We adapt the proof of [8, Lemma 4]. Using the trivial identity (x)_{m} = Γ(x + m)/ Γ(x), we first rewrite

R(t) = n!^{A−15}2^{−3}(2t + n)\frac{Γ(t)^{A+3}Γ(2t + 4n + 1)^{3}}{Γ(t + n + 1)^{A+3}Γ(2t − 2n)^{3}}  (13)
= n!^{A−15}2^{−3}(2t + n)\frac{Γ(t)^{A+3}Γ(2t + 4n + 1)^{3}Γ(2n − 2t + 1)^{3}}{Γ(t + n + 1)^{A+3}} · \frac{sin(2πt)^{3}}{π^{3}} ,  (14)

where (14) is a consequence of the reflection formula Γ(x)Γ(1 − x) = π/ sin(πx) applied to (13) with x = 2t − 2n.

Let u be such that Re(u) ≤ 0 and |Im(u)| ≤ 3π. Let c denote any real number in (\frac{1}{2}, n), and N any integer ≥ n + 1. Let C_{N} denote the rectangular contour (oriented positively) with sides [c−iN, N + \frac{1}{2} −iN], [N + \frac{1}{2} −iN, N + \frac{1}{2} +iN], [N + \frac{1}{2} +iN, c+iN], [c+iN, c−iN]. By the residue theorem, and because R^{(j)}(k) = 0 for k = 1, …, n and j ∈ {0, 1, 2}, we have

\frac{1}{2iπ} ∫_{C_{N}} R(t)\frac{π^{3}}{sin(πt)^{3}} e^{ut}dt = ∑_{k=n+1}^{N} (\frac{π^{2} + u^{2}}{2} R(k) + uR′(k) + \frac{1}{2}R′′(k)) (−e^{u})^{k}.

The conditions on u ensure that, as N → +∞, we have

\frac{1}{2iπ} ∫_{c+i∞}^{c−i∞}R(t)\frac{π^{3}}{sin(πt)^{3}} e^{ut}dt = ∑_{k=n+1}^{∞} (\frac{π^{2} + u^{2}}{2} R(k) + uR′(k) + \frac{1}{2}R′′(k)) (−e^{u})^{k}. (15)

In particular, summing the two specializations of (15) for u = iπ and u = −iπ, we get

\frac{1}{iπ} ∫_{c+i∞}^{c−i∞}R(t)\frac{π^{3} cos(πt)}{sin(πt)^{3}}dt = S_{n}.

Hence,

S_{n} = \frac{n!^{A−15}2^{−3}}{iπ} ∫_{c+i∞}^{c−i∞}(2t + n)\frac{Γ(t)^{A+3}Γ(2t + 4n + 1)^{3}Γ(2n − 2t + 1)^{3}}{Γ(t + n + 1)^{A+3}} · \frac{sin(2πt)^{3} cos(πt)}{sin(πt)^{3}}dt

= \frac{n!^{A−15}}{iπ} ∫_{c+i∞}^{c−i∞}(2t + n)\frac{Γ(t)^{A+3}Γ(2t + 4n + 1)^{3}Γ(2n − 2t + 1)^{3}}{Γ(t + n + 1)^{A+3}} cos(πt)^{4}dt.

With t − \frac{1}{2} instead of t in (14), we have

R(t − \frac{1}{2}) = −n!^{A−15}2^{−3}(2t + n − 1)\frac{Γ(t − \frac{1}{2})^{A+3}Γ(2t + 4n)^{3}Γ(2n − 2t + 2)^{3}}{Γ(t + n + \frac{1}{2})^{A+3}} · \frac{sin(2πt)^{3}}{π^{3}} .

Since R^{(j)}(k − \frac{1}{2}) = 0 for k = 1, …, n and j ∈ \{0, 1, 2\}, it follows again that

\frac{1}{iπ} ∫_{C_{N}} R(t − \frac{1}{2})\frac{π^{3} cos(πt)}{sin(πt)^{3}}dt = ∑_{k=1}^{N} R′′(k − \frac{1}{2}).

Hence, upon letting N → +∞,

Ŝ_{n} = −\frac{n!^{A−15}}{iπ} ∫_{c+i∞}^{c−i∞}(2t + n − 1)\frac{Γ(t − \frac{1}{2})^{A+3}Γ(2t + 4n)^{3}Γ(2n − 2t + 2)^{3}}{Γ(t + n + \frac{1}{2})^{A+3}} cos(πt)^{4}dt

and the proof is now complete. □

5. Asymptotic behavior of S_{n} and Ŝ_{n}

To prove Theorem 1, we have to determine A even minimal such that Φ_{n}^{−3}d_{n}^{A+2}S_{n} → 0 and Φ_{n}^{−3}d_{n}^{A+2}Ŝ_{n} → 0 along a subsequence of the even integers. It turns out that A = 68 is this minimal value.

Proposition 3. Let A = 68. There exists an increasing sequence of even integers σ(n) such that

lim_{n→+∞} |S_{σ(n)}|^{1/σ(n)} = lim_{n→+∞} |Ŝ_{σ(n)}|^{1/σ(n)} = e^{−κ}

with κ ≈ 66.1727, and

lim_{n→+∞} \frac{Ŝ_{σ(n)}}{S_{σ(n)}} = −1.

Proof. We let the integer A ≥ 15 be unspecified for the moment, and n is any integer ≥ 0. We define log(z) = ln |z| + i arg(z), with | arg(z)| < π. Stirling’s formula reads:

Γ(z) = z^{z}e^{−z}\sqrt{\frac{2π}{z}} (1 + O(\frac{1}{z}))

as z → ∞ in any angular sector | arg(z)| ≤ π − ε, and O depends on ε > 0. Let us define

φ_{0}(t) := (A + 3)t log(t) + (6t + 12) log(2t + 4)
+ (6 − 6t) log(2 − 2t) − (A + 3)(t + 1) log(t + 1),

where all the logarithms are defined with their principal determinations. This function is analytic in Ω := ℂ \ ((−∞, 0] ∪ [1, +∞)).

After changing t to tn in (11) and (12) in Proposition 2, we apply Stirling’s formula to the various Gamma functions of the integrands and get

S_{n} = \frac{2i(2π)^{(A−9)/2}}{n^{(A+9)/2}} ∫_{c−i∞}^{c+i∞}g_{n}(t)f(t)^{n} cos(πnt)^{4}dt  (16)

and

Ŝ_{n} = \frac{−2i(2π)^{(A−9)/2}}{n^{(A+9)/2}} ∫_{c−i∞}^{c+i∞}ĝ_{n}(t)f(t)^{n} cos(πnt)^{4}dt,  (17)

where c is any real number in (0, 1), and

f(t) = exp(φ_{0}(t)) = \frac{t^{t(A+3)}(2t + 4)^{3(2t+4)}(2 − 2t)^{3(2−2t)}}{(t + 1)^{(t+1)(A+3)}}

is analytic in Ω. The functions g_{n}(t) and ĝ_{n}(t) are analytic in Ω and such that as n → +∞

g_{n}(t) = g(t)(1 + O(\frac{1}{n})),
ĝ_{n}(t) = g(t)\frac{(t + 1)^{(A+3)/2}(2 − 2t)^{3}}{t^{(A+3)/2}(2t + 4)^{3}} (1 + O(\frac{1}{n})).

with

g(t) = \frac{(2t + 1)(2t + 4)^{3/2}(2 − 2t)^{3/2}}{t^{(A+3)/2}(t + 1)^{(A+3)/2}} .

For ε > 0, when t satisfies | arg(t)| ≤ π − ε and | arg(1 − t)| ≤ π − ε, the constants in the O(\frac{1}{n}) terms depend on ε and not on t.

We shall now apply the saddle point method to estimate each integral in (16) and (17). Since

cos(x)^{4} = \frac{3}{8}( cos(4x) + 4 cos(2x) + 3),

we have

S_{n} = \frac{2i(2π)^{(A−9)/2}}{n^{(A+9)/2}} ∑_{ℓ=−2}^{2} u_{ℓ}J_{ℓ,n} and Ŝ_{n} = −\frac{2i(2π)^{(A−9)/2}}{n^{(A+9)/2}} ∑_{ℓ=−2}^{2} u_{ℓ}Ĵ_{ℓ,n},  (18)

where

J_{ℓ,n} = ∫_{c−i∞}^{c+i∞}g_{n}(t)f(t)^{n}e^{2ℓiπnt}dt and Ĵ_{ℓ,n} = ∫_{c−i∞}^{c+i∞}ĝ_{n}(t)f(t)^{n}e^{2ℓiπnt}dt,

and u_{−4} = u_{4} = \frac{3}{16}, u_{−2} = u_{2} = \frac{3}{4} and u_{0} = \frac{9}{16}.

Observe that \overline{J_{ℓ,n}} = −J_{−ℓ,n} and \overline{Ĵ_{ℓ,n}} = −Ĵ_{−ℓ,n}.

We now give the details necessary to complete the proof when A = 68. For the statement of the saddle point method, see [8, Lemma 3].

• Case ℓ = 0. The equation φ′_{0}(t) = 0 admits a solution t_{0} ≈ 0.9991 ∈ (0, 1) and we have g(t_{0}) = ĝ(t_{0}) ≠ 0 and φ′′_{0}(t_{0}) ≠ 0 (see below). The real function f_{0}(y) := Re(φ_{0})(t_{0} + iy) admits a unique maximum at y = 0; this is proved by studying the sign of

f′_{0}(y) = −Im(φ′_{0})(t_{0} + iy)
= −71 arg(t_{0} + iy) − 6 arg(2t_{0} + 2iy + 4)
+ 6 arg(2 − 2t_{0} − 2iy) + 71 arg(1 + t_{0} + iy).  (19)

This is done by the same method as in [8, Lemma 4], using arg(z) = arctan(Im(z)/Re(z)) when Re(z) > 0 (which is the case of all the various values of z in (19)). Hence, shifting the line c + iℝ to the line t_{0} + iℝ (oriented from Im(t) < 0 to Im(t) > 0), we can apply the saddle point method to J_{0,n} and Ĵ_{0,n}, and get

J_{0,n} = ig(t_{0})\sqrt{\frac{2π}{nφ′′_{0}(t_{0})}}f(t_{0})^{n}(1 + o(1)) and Ĵ_{0,n} = iĝ(t_{0})\sqrt{\frac{2π}{nφ′′_{0}(t_{0})}}f(t_{0})^{n}(1 + o(1)).

Numerically,

t_{0} ≈ 0.99918, f(t_{0}) ≈ 1.8127 × 10^{−29},

g(t_{0}) = \frac{(2t_{0} + 1)(2t_{0} + 4)^{3/2}(2 − 2t_{0})^{3/2}}{t_{0}^{71/2}(t_{0} + 1)^{71/2}} ≈ 6.2647 × 10^{−14},

ĝ(t_{0}) = g(t_{0})\frac{(t_{0} + 1)^{71/2}(2 − 2t_{0})^{3}}{t_{0}^{71/2}(2t_{0} + 4)^{3}} = g(t_{0}), φ′′_{0}(t_{0}) ≈ 7373.2123.

• Case ℓ = 1. Let us define φ_{1}(t) := φ_{0}(t)+2iπt. The equation φ′_{1}(t) = 0 admits a solution t_{1} ≈ 0.9995 − i0.0007. We have g(t_{1}) = −ĝ(t_{1}) ≠ 0 and φ′′_{1}(t_{1}) = φ′′_{0}(t_{1}) ≠ 0 (see below). The real function f_{1}(y) := Re(φ_{1})(t_{1} + iy) admits a unique maximum at y = 0; again, this is proved by studying the sign of

f′_{1}(y) = −Im(φ′_{1})(t_{0} + iy) = −Im(φ′_{0})(t_{0} + iy) − 2π

as in [8, Lemma 4]. Hence, shifting the line c + iℝ to the line t_{1} + iℝ (oriented from Im(t) < 0 to Im(t) > 0), we can apply the saddle point method to J_{1,n} and Ĵ_{1,n}, and get

J_{1,n} = ig(t_{1})\sqrt{\frac{2π}{nφ′′_{0}(t_{1})}}f(t_{1})^{n}(1 + o(1)) and Ĵ_{1,n} = iĝ(t_{1})\sqrt{\frac{2π}{nφ′′_{0}(t_{1})}}f(t_{1})^{n}(1 + o(1)).

Numerically,

t_{1} ≈ 0.9995 − i0.0007, f(t_{1})e^{2iπt_{1}} ≈ 1.8171 × 10^{−29} − i7.7425 × 10^{−32},

g(t_{1}) = \frac{(2t_{1} + 1)(2t_{1} + 4)^{3/2}(2 − 2t_{1})^{3/2}}{t_{1}^{71/2}(t_{1} + 1)^{71/2}} ≈ −i1.8629 × 10^{−15} + 6.1544 × 10^{−14},

ĝ(t_{1}) = g(t_{1})\frac{(t_{1} + 1)^{71/2}(2 − 2t_{1})^{3}}{t_{1}^{71/2}(2t_{1} + 4)^{3}} = −g(t_{1}),

φ′′_{0}(t_{1}) ≈ 3724.1063 − i6320.4884.

• Case ℓ = −1. We have

J_{−1,n} = −\overline{J_{1,n}} = i\overline{g(t_{1})}\sqrt{\frac{2π}{n\overline{φ′′_{0}(t_{1})}}}\overline{f(t_{1})}^{n}(1 + o(1))

and

Ĵ_{1,n} = −\overline{Ĵ_{1,n}} = i\overline{ĝ(t_{1})}\sqrt{\frac{2π}{n\overline{φ′′_{0}(t_{1})}}}\overline{f(t_{1})}^{n}(1 + o(1)).

• Case ℓ = 2. Let us define φ_{2}(t) := φ_{0}(t) + 4iπt. The equation φ′_{2}(t) = 0 admits a solution t_{2} ≈ 1.0004 − i0.0007. We have g(t_{2}) = ĝ(t_{2}) ≠ 0 and φ′′_{2}(t_{2}) = φ′′_{0}(t_{2}) ≠ 0 (see below). The situation is different here because Re(t_{2}) > 1 and thus the straight line t_{2} + iℝ is not contained in Ω. Instead, we define the polygonal contour L := {t_{2} + y, y ≤ 0} ∪ [t_{2}, 1] ∪ {1 + iy, y ≥ 0}. One can then prove that when t varies in L, Re(φ_{2})(t) has a unique maximum at t = t_{2}; see the proof of [4, Lemma 10] or [11] for similar computations. We now replace the “corner” at t = 1 by an arc of circle centered at 1 and of small radius η > 0 (and contained in Ω), and connect it to the remaining parts of [t_{2}, 1] and {1 + iy, y ≥ 0}: we obtain a path L′ along which Re(φ_{2})(t) still has a unique maximum at t = t_{2} provided η is small enough, and L′ is at positive distance of (−∞, 0] ∪ [1, +∞). Hence, moving the line c + iℝ to L′ (oriented from Im(t) < 0 to Im(t) > 0) we can apply the saddle point method to J_{2,n} and Ĵ_{2,n}, and get

J_{2,n} = ig(t_{2})\sqrt{\frac{2π}{nφ′′_{0}(t_{2})}}f(t_{2})^{n}(1 + o(1)) and Ĵ_{2,n} = iĝ(t_{2})\sqrt{\frac{2π}{nφ′′_{0}(t_{2})}}f(t_{2})^{n}(1 + o(1)).

Numerically,

t_{2} ≈ 1.0004 − i0.0007,
f(t_{2})e^{4iπt_{2}} ≈ 1.8261 × 10^{−29} − i7.8209 × 10^{−32},

g(t_{2}) = \frac{(2t_{2} + 1)(2t_{2} + 4)^{3/2}(2 − 2t_{2})^{3/2}}{t_{2}^{71/2}(t_{2} + 1)^{71/2}} ≈ −5.9420 × 10^{−14} − i1.8161 × 10^{−15},

ĝ(t_{2}) = g(t_{2})\frac{(t_{2} + 1)^{71/2}(2 − 2t_{2})^{3}}{t_{2}^{71/2}(2t_{2} + 4)^{3}} = g(t_{2}),

φ′′_{0}(t_{2}) ≈ −3574.1082 − i6320.4861.

• Case ℓ = −2. We have

J_{−2,n} = −\overline{J_{2,n}} = i\overline{g(t_{2})}\sqrt{\frac{2π}{n\overline{φ′′_{0}(t_{2})}}}\overline{f(t_{2})}^{n}(1 + o(1))

and

Ĵ_{−2,n} = −\overline{Ĵ_{2,n}} = i\overline{ĝ(t_{2})}\sqrt{\frac{2π}{n\overline{φ′′_{0}(t_{2})}}}\overline{f(t_{2})}^{n}(1 + o(1)).

We now observe that |f(t_{2})e^{4iπt_{2}}| > |f(t_{1})e^{2iπt_{1}}| > |f(t_{0})|. By an argument similar to [8, Lemma 5], we deduce from (18) and the above asymptotic estimates for J_{2,n}, J_{−2,n}, Ĵ_{2,n} and Ĵ_{−2,n} the existence of an increasing sequence of even integers σ(n) such that

lim_{n→+∞} |S_{σ(n)}|^{1/σ(n)} = |f(t_{2})e^{4iπt_{2}}|, lim_{n→+∞} |Ŝ_{σ(n)}|^{1/σ(n)} = |f(t_{2})e^{4iπt_{2}}|

and moreover

lim_{n→+∞} \frac{Ŝ_{σ(n)}}{S_{σ(n)}} = −1.

We conclude the proof with the estimate |f(t_{2})e^{4iπt_{2}}| ≈ exp(−66.1727). □

6. Proof of Theorem 2

We first remark that the asymptotic behaviors of d_{n} and of Φ_{n} both follow from the prime number theorem:

lim_{n→∞} \frac{log(d_{n})}{n} = 1 and δ := lim_{n→∞} \frac{log(Φ_{n})}{n} = ∫_{0}^{1} ρ_{0}(t) d(ψ(t) + t^{−1}) ≈ 1.29564,

where ψ(t) := Γ′(t)/Γ(t) and the function ρ_{0}(t) is defined in the proof of Proposition 1.

We now take A = 68 in Proposition 1. For any odd integer m ∈ {5, 7, …, 69} and any even integer n, we have

Φ_{n}^{−3}d_{n}^{70}((2^{m} − 1)S_{n} − Ŝ_{n}) = Q_{0,n} + ∑_{j odd^{j=5,j≠m}}^{69} Q_{j,n}ζ(j).

with Q_{0,n} := Φ_{n}^{−3}d_{n}^{70}((2^{m} − 1)q_{0,n} − q̂_{0,n}) ∈ ℤ and Q_{j,n} = Φ_{n}^{−3}d_{n}^{70}(2^{m} − 2^{j})q_{j,n} ∈ ℤ.

By Proposition 3, we have

lim_{n→+∞} |Φ_{σ(n)}^{−3}d_{σ(n)}^{70}((2^{m} − 1)S_{σ(n)} − Ŝ_{σ(n)})|^{1/σ(n)}

= lim_{n→+∞} |Φ_{σ(n)}^{−3}d_{σ(n)}^{70}S_{σ(n)}|^{1/σ(n)} × lim_{n→+∞} |(2^{m} − 1) − \frac{Ŝ_{σ(n)}}{S_{σ(n)}} |^{1/σ(n)}

= e^{70−κ−3δ} ≈ e^{−0.0597} ∈ (0, 1).

This proves the existence of at least one irrational number amongst the odd zeta values ζ(j) with j odd in {5, 7, …, 69} \ {m}.

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Tanguy Rivoal, Institut Fourier, CNRS et Université Grenoble Alpes, CS 40700, 38058 Grenoble cedex 9, France

Email address: tanguy.rivoal [&] univ-grenoble-alpes.fr

Wadim Zudilin, Institute for Mathematics, Astrophysics and Particle Physics, Radboud Universiteit, PO Box 9010, 6500 GL Nijmegen, The Netherlands

Email address: w.zudilin [&] math.ru.nl