Infinitely many odd zeta values are irrational. By elementary means
Johannes Sprang


In this small note, we provide an elementary proof of the fact that infinitely many odd zeta values are irrational. For the first time, this celebrated theorem been proven by Rivoal and Ball–Rivoal. The original proof uses highly non-elementary methods like the saddle-point method and Nesterenko's linear independence criterion. Recently, Zudilin has re-proven a slightly weaker form of his important result that at least one of the odd zeta values ζ(5),ζ(7),ζ(9) and ζ(11) is irrational, by elementary means. His new main ingredient are certain 'twists by half' of hypergeometric series. Generalizing this to 'higher twists' allows us to give a purely elementary proof of the result of Rivoal and Ball–Rivoal.


An efficient explicit full-discrete scheme for strong approximation of stochastic Allen-Cahn equationT1
    Xiaojie Wang

[label=e1]x.j.wang7@csu.edu.cn


    December 30, 2023
=======================================================================================================

The question about the values of the Riemann zeta function 

    ζ(s)=∑_n=1^∞1/n^s

at the positive integers is very old and has its origins in the 17th century. The first breakthrough has been made by Euler who showed the remarkable formulas

    ζ(2)=π^2/6,  ζ(4)=π^4/90,  ζ(6)=π^6/945, ...

More generally, he has proven

    ζ(2n)=(-1)^n+1B_2n(2π)^2n/2(2n)!,

where B_n denotes the n-th Bernoulli number. Combined with Lindemann's proof of the transcendence of π, this proves the transcendence of all even zeta values. While we have a good understanding of the structure of the even zeta values, almost nothing is known for the odd zeta values ζ(2n+1). The first non-trivial result on the structure of ζ(2n+1) has been obtained by Apéry in 1979. He was able to proof that ζ(3) is irrational. Although this is the only odd zeta value for which we know the irrationality for sure, the best known result aiming towards the irrationality of ζ(5) is due to Zudilin. He was able to prove the amazing result that at least one of the numbers

    ζ(5),ζ(7),ζ(9),ζ(11)

is irrational. Further, we know by a celebrated theorem of Rivoal and Ball–Rivoal, that infinitely many among the odd zeta values

    ζ(3),ζ(5),ζ(9),...

are irrational, <cit.>,<cit.>. While the original proofs of the results of Zudilin, Rivoal and Rivoal–Ball are highly non-elementary, Zudilin has recently presented a very elementary proof of the fact that at least one of the zeta values

    ζ(5),ζ(7),...,ζ(25)

is irrational. He explicitly constructs a sequence of non-zero linear forms

    r̂_n=a_0,n+a_5,nζ(5)+a_7,nζ(7)+...+a_25,nζ(25),   a_ij∈

tending to zero as n goes to infinity. If all involved zeta values were irrational, we could multiply by their common denominator and we would obtain a non-zero sequence of integers converging to zero - an obvious contradiction. The new key ingredient for the construction of the involved linear forms is that he introduces certain 'twists by half' of well-poised hypergeometric series. Comparing both, the twisted and the untwisted, hypergeometric series allows him to eliminate the unwanted value ζ(3). The present work builds heavily on Zudilin's beautiful and elementary proof. Without his paper <cit.>, this paper would probably not exist.In this work, we introduce 'higher twists' by integers D>1 of certain hypergeometric series. This allows us to construct certain linear-forms 

    ∑_i=3 
     i  odd^s a^(D)_iζ(i,j/D)+a^(D,j)_0.

for special values of the Hurwitz zeta function

    ζ(s,α):=∑_n=0^∞1/(n+α)^s.

By using the formula

    (D^i-1)·ζ(i)=∑_j=1^D-1ζ(i,j/D)

and a thorough analysis of the convergence properties of the involved linear forms, we can eliminate arbitrary many odd zeta values. In order to guarantee convergence, the parameter s which specifies the range of the zeta values has to be chosen appropriately. Let us finally remark, that Rivoal's result is stronger in the sense that it provides an asymptotic lower bound for the dimension of the  vector space spanned by

    {ζ(3),ζ(5),...,ζ(2n+1)}.

Nevertheless, our proof uses only very elementary techniques. As in Zudilin's proof, the technical ingredients of our proof are limited to Stirling's formula and the asymptotic formula

    lim_n→∞√({1,...,n})=e

which can be deduced from the prime number theorem.




§ INTEGRALITY RESULTS





A rational function over the complex numbers

    S(t)=P(t)/(t-α_1)^s_1(t-α_2)^s_2·...·(t-α_r)^s_r∈ℂ(t)

with P≤ s:=∑_i=0^r s_i has a unique partial-fraction decomposition

    S(t)=∑_k=1^r∑_i=1^s_kb_i,k/(t-α_k)^i.

The coefficients can be computed explicitly by the formula

    b_i,k=.1/(s_k-i)!∂^(s_k-i)/∂ t^(s_k-i)( S(t)(t-α_k)^s_k)|_t=α_k.

Let us write d_n:={1,...,n}. The following Lemma can be proven using only the above explicit formula for the coefficients of the partial-fraction decomposition and the Leibniz rule.

 Let k_1,...,k_r be pairwise distinct numbers from the set {0,...,n} and s_1,...,s_r positive integers. Then, the coefficients in the expansion
 
    1/∏_j=1^r(t+k_j)^s_j=∑_j=1^r∑_i=1^s_jb_i,j/(t+k_j)^i

 satisfy
 
    d_n^s-ib_i,j∈,  for  i=1,...,s_j  and  j=1,...,r

 where s:=s_1+...+s_r.


We refer to <cit.> for the proof.

We will also need the following

	Let D,n,k,i∈ with n,D≥ 1. Then n!/(D^n,n!) divides ∏_j=k^n+k-1(Dj+i).


First, let us observe that n!/(D^n,n!) is the maximal divisor of n! which is co-prime to D. Let p≤ n be a prime which is co-prime to D. By the Chinese remainder theorem, at least ⌊n/p⌋ of the numbers

    {Dk+i,D(k+1)+i,...,D(k+n-1)+i}

are divisible by p, at least ⌊n/p^2⌋ of them are divisible by p^2, etc. Let us write ν_p(m) for the p-adic valuation, i.e. for the exponent of the maximal p-power p^ν_p(m) dividing m. By the above observation, we obtain

    ∑_l≥ 1⌊n/p^l⌋≤ν_p(∏_j=k^n+k-1(Dj+i)).

On the other hand, we have

    ν_p(n!/(D^n,n!))=ν_p(n!)=∑_l≥ 1⌊n/p^l⌋.

Since this holds for every prime dividing n!/(D^n,n!) the lemma follows.


Let D be a positive integer. For a fixed integer s≥ 3D, let us define

    R_n^(D)(t):=D^6(D-1)n· (n!)^s-(3D-1)∏_j=0^3Dn(t-n+j/D)/∏_j=0^n(t+j)^s+1.




The rational function R_n^(2) coincides with the function studied in <cit.>. For t→∞ we have the asymptotic behaviour

    R_n^(D)(t)=O( 1/t^(s+1)(n+1)-3Dn-1).

  Let us consider for j∈{1,...,D} the following 'twisted' hypergeometric series

    r_n^(D,j):=∑_m=1^∞ R^(D)_n(m+j/D).

Let us write a_i,k^(D) for the coefficients appearing in the partial-fraction decomposition of R^(D)_n(t), i.e.

    R_n^(D)(t)=∑_i=1^s∑_k=0^n a_i,k^(D)/(t+k)^i.

Using the above lemma, we obtain:

	The coefficients a_i,k^(D) in the partial-fraction decomposition of R_n^(D)(t) satisfy
	
    d_n^s-i a_i,k^(D)∈.

	Further, we have
	
    a_i,k^(D)=(-1)^i-1(-1)^nDa_i,n-k^(D).



	In the case D=2, the proof is contained in <cit.>. The proof given there generalizes to more general integers D≥ 2: The symmetry
	
    a_i,k^(D)=(-1)^i-1(-1)^nDa_i,n-k^(D).

	is an immediate consequence of the symmetry R_n^(D)(-n-t)=(-1)^nDR_n(-t). For the integrality statement, let us consider the following decomposition of R^(D)_n into more elementary rational functions:
	
    R_n^(D)(t):   =D^6(D-1)n· (n!)^s-(3D-1)∏_j=0^3Dn(t-n+j/D)/∏_j=0^n(t+j)^s+1=
       =(n!)^s-(3D-1)·∏_j=1^n(t-j)∏_j=1^n(t+n+j)∏_i=1^D-1D^6n∏_j=1^3n(t-n+j-i/D)/∏_j=0^n(t+j)^s.

	Each of the following factors has an integral partial-fraction decomposition:
	
    n!/∏_j=0^n(t+j)   =∑_k=0^n(-1)^knk/(t+k)
    ∏_j=1^n(t-j)/∏_j=0^n(t+j)   =∑_k=0^n(-1)^n+kn+knnk/(t+k)
    ∏_j=1^n(t+n+j)/∏_j=0^n(t+j)   =∑_k=0^n(-1)^k2n-knnk/(t+k)


	
    D^2n∏_j=1^n(t-j+i/D)/∏_j=0^n(t+j)   =∑_k=0^n(-1)^n+k∏_j=k+1^n+k(Dj-i)/n!/(D^n,n!)nkD^n/(D^n,n!)/(t+k)
    D^2n∏_j=1^n(t+j-i/D)/∏_j=0^n(t+j)   =∑_k=0^n(-1)^n+k∏_j=1-k^n-k(Dj+i)/n!/(D^n,n!)nkD^n/(D^n,n!)/(t+k)
    D^2n∏_j=1^n(t+n+j-i/D)/∏_j=0^n(t+j)   =∑_k=0^n(-1)^k∏_j=n-k+1^2n-k(Dj-i)/n!/(D^n,n!)nkD^n/(D^n,n!)/(t+k)

The last three partial-fraction decompositions are integral by <ref>. The statement of the lemma follows now by applying <ref> to the product of simpler rational functions with integral partial-fraction decomposition.

For later reference, let us record the following:

The coefficients a^(D)_i,k in the partial-fraction decomposition of R^(D)_n satisfy for every k∈_≥0, j∈{1,...,D} and l∈{0,...,n} 

    d_n^s ·∑_i=1^s a_i,k^(D)1/(l+j/D)^i∈.



The rational function R_n^(D) has zeros at -2n,-2n+1/D,-2n+2/D,...,n-1/D,n. In particular,

    0=R_n^(D)(-n+l+j/D)=∑_i=1^s∑_k=0^na_i,k^(D)/(-n+k+l+j/D)^i.

This can be reformulated as follows:

    ∑_i=1^s a_i,k^(D)1/(l+j/D)^i=-∑_i=1^s∑_k=0^n-1a_i,k^(D)/(-n+k+l+j/D)^i.

Let p be a prime. Recall that the p-adic valuation ν_p is a homomorphism from ^× to  and satisfies the non-Archimedean triangle inequality

    ν_p(x+y)≥min(ν_p(x),ν_p(y)),   ∀ x,y∈^×

with equality if ν_p(x)≠ν_p(y). Assume there were some prime p with

    ν_p( ∑_i=1^s a_i,k^(D)1/(l+j/D)^i)<-s·ν_p(d_n).

From <ref> we deduce ν_p(a_i,k^(D))≥ -(s-i)ν_p(d_n). Under our assumption, this implies

    ν_p(1/(l+j/D)^i)<-iν_p(d_n)

or equivalently

    ν_p(l+j/D)>ν_p(d_n).

By the non-Archimedean triangle inequality we obtain for k∈{0,...,n-1}

    ν_p(-n+k+l+j/D)=ν_p(-n+k)≤ν_p(d_n).

This implies

    ν_p( ∑_i=1^s a_i,k^(D)1/(l+j/D)^i)   =ν_p(-∑_i=1^s∑_k=0^n-1a_i,k^(D)/(-n+k+l+j/D)^i)≥
       ≥min_i,k{ν_p(a_i,k^(D)1/(-n+kl+j/D)^i)}≥ -ν_p(d_n^s)

a contradiction to (<ref>). Thus, we have

    ν_p( ∑_i=1^s a_i,k^(D)1/(l+j/D)^i)≥ -s·ν_p(d_n)

for every prime p and deduce the desired integrality result

    d_n^s ·∑_i=1^s a_i,k^(D)1/(l+j/D)^i∈.


Let us write

    ζ(s,q):=∑_n=0^∞1/(n+q)^s,  (s)>1,q≠ 0

for the Hurwitz zeta function.

Let n,D≥1 be integers with nD even. For j∈{1,...,D}, we have	

    r^(D,j)_n=∑_i=3 
     i  odd^s a^(D)_iζ(i,j/D)+a^(D,j)_0

with 

    d_n^s-ia^(D)_i∈ for  i=3,5,...,s   and     d_n^s a_0^(D,j) for  j∈{0,...,D-1}.



	We proceed as in <cit.>: We introduce an auxiliary parameter z>0:
	
    r_n^(D,j)(z)   =∑_m=1^∞ R_n(m+j/D)z^m=∑_m=1^∞∑_i=1^s∑_k=0^n a^(D)_i,kz^m/(m+k+j/D)^i=
       = ∑_i=1^s∑_k=0^na^(D)_i,kz^-k∑_m=1^∞z^m+k/(m+k+j/D)^i=
       = ∑_i=1^s∑_k=0^na^(D)_i,kz^-k(Φ(z,j/D,i)-∑_l=0^k z^l/(l+j/D)^i)

	Here,
	
    Φ(z,α,s):=∑_n=0^∞z^n/(n+α)^s

	is the Lerch zeta function. The Lerch zeta function converges for s≥ 2 at z=1 and Φ(1,j/D,i)=ζ(i,j/D). For i=1 the limit z↗ 1 does not exist. Thus, by taking the limit z↗ 1 in the above equality, we conclude
	
    ∑_k=0^na^(D)_1,k=lim_z↗ 1∑_k=0^na^(D)_1,kz^-k=0.

	Let us define
	
    a^(D)_i:=∑_k=0^n a_i,k^(D).

	Since we assumed Dn to be even,  we have the symmetry
	
    a_i,k^(D)=(-1)^i-1a_i,n-k^(D)

	and hence  a^(D)_i=0 for i even. By <ref>  and <ref> we have 
	
    a_0^(D,j):=-∑_i=1^s∑_k=0^n∑_l=0^k a^(D)_i,kz^l/(l+j/D)^i∈ d_n^-s,   d_n^s-ia^(D)_i   ∈

	as desired.




§ ASYMPTOTIC BEHAVIOUR

Let us analyze the asymptotic behaviour of r_n^(D,j). We follow closely the argument in <cit.>.
Because R_n^(D)(t) vanishes at 1+j/D,2+j/D,...,n-1+j/D we have for j∈{1,...,D}:

    r^(D,j)_n=∑_m=1^∞ R^(D)_n(m+j/D)=∑_k=0^∞ c_k^(D,j)

with

    c_k^(D,j):=R^(D)_n(n+k+j/D).



Let s≥ 3D+1 be odd and j,j̃∈{1,...,D}. We have the asymptotic behaviour

    lim_n→∞√(r_n^(D,j))=g_D(x_0)  and  lim_n→∞r^D,j_n/r^D,j̃_n=1

where

    g_D(x)=D^6(D-1)(3+x)^3D(1+x)^s+1/(2+x)^2(s+1)

and x_0 is the unique positive zero of the polynomial

    (x+3)^D(x+1)^s+1-x^D(x+2)^s+1.


Again, we follow closely the proof in the case D=2 given in <cit.>:	The definition of c_k^(D,j) gives
	
    c_k^(D,j)=D^6(D-1)n(n!)^s-(3D-1)∏_l=0^3Dn(k+j+l/D)/∏_l=0^n(n+k+l+j/D)^(s+1)

	which allows us to deduce the asymptotics
	
    c^(D,j)_k+1/c^(D,j)_k	=(∏_l=1^D(k+3n+j+l/D)/(k+j+l-1/D))( k+n+j/D/k+2n+1+j/D)^s+1∼ f_D(k/n),   as n→∞,

	with f_D(x)=(x+3/x)^D(x+1/x+2)^s+1. The logarithmic derivative of f_D(x) is
	
    f'_D(x)/f_D(x)=D/x+3-D/x+(s+1)(1/x+1-1/x+2)=(s+1-3D)x^2+(3s+3-9D)x-6D/x(x+1)(x+2)(x+3).

	The quadratic polynomial in the numerator of the latter fraction has a unique positive zero x_1. Thus, f_D monoton decreases from +∞ to f_D(x_1) on (0,x_1) and then monoton increases on (x_1,∞) from f_D(x_1) to 1. In particular, there is a unique value x_0 with f_D(x_0)=1. Since f_D(1)=4^D·(2/3)^(s+1)<1 we have 0<x_0<1. Thus, the sequence (c_k^(D,j))_k≥ 1 attains its maximal values around x_0n. We follow the asymptotic analysis of series in de Bruijn's book <cit.>. Using Stirling's formula

    Γ(x)=√(2π/x)(x/e)^x(1+O( 1/x)),   x→∞.

	one can prove, that there is some C>0 such that
	 
    lim_n→∞R_n^(D)( n+x_0n± t·√(n))/R_n^(D)( n+x_0n)=exp( -C/t^2),   t∈ℝ.

	Recall, that c_k^(D,j) has been defined as the evaluation of R_n^(D) at n+k+j/D. We deduce the existence of γ>0 with
	
    c_k^(D,j)<1/2max_k{c_k^(D,j)},   for k∈ with  |k-x_0n|>γ√(n).

	The total contribution of the terms outside [x_0n-γ√(n), x_0n+γ√(n)] is relatively small. Thus the asymptotic behaviour of r_n^(D,j)=∑_k=0^∞ c_k^(D,j) is determined by the asymptotics of c_k for k in { x_0n-γ√(n),x_0n+γ√(n)}. Let us define k_0(n):= ⌊ x_0n ⌋. Using Stirling's formula, we compute
	
    lim_n→∞√(r_n^(D,j))   =lim_n→∞√(c_k_0(n)^(D,j))=lim_n→∞[D^3(D-2)n(n/e)^(s-3D+1)n×
         ×(3Dn+Dk_0(n)+j/e)^3Dn+Dk_0(n)+j(k_0(n)+n+j/D/e)^(s+1)(k_0(n)+n+j/D)×
         ×(e/k_0(n)+2n+j/D+1)^(s+1)(k_0(n)+2n+j/D+1)(e/Dk_0(n)+j-1)^Dk_0(n)+j-1]^1/n=
       =D^3(D-2)(3D+Dx_0)^3D(3D+Dx_0)^Dx_0(1+x_0)^(s+1)(x_0+1)/(Dx_0)^Dx_0(2+x_0)^(s+1)(2+x_0)=
       =D^3(D-2)D^3D·(3+x_0)^3D(1+x_0)^(s+1)/(2+x_0)^2(s+1)f_D(x_0)^x_0=
       =D^6(D-1)·(3+x_0)^3D(1+x_0)^(s+1)/(2+x_0)^2(s+1)=g(x_0)

	Let us compare for n→∞ the quotients

    c_k^(D,j)/c_k^(D,D)   =(∏_l=0^D-j-1k+3n+j+l+1/D/k+j+l/D)(Γ(2n+k+1+j/D)Γ(n+k+2)/Γ(2n+k+2)Γ(n+k+1+j/D))^s+1∼
       ∼(3n+k/2k)^D-j( n+k+2/2n+k+2)^D-j/D(s+1).

	This allows us to compare r_n^(D,j) for different values of j∈{1,...,D}:
	
    lim_n→∞r_n^(D,j)/r_n^(D,D)   =lim_n→∞c_k_0(n)^(D,j)/c_k_0(n)^(D,D)=lim_n→∞( 3n+k_0(n)/k_0(n))^D-j·( n+k_0(n)+2/2n+k_0(n)+2)^D-j/D(s+1)=
       =f_D(x_0)^D-j/D=1.

	Since this holds for any j∈{1,...,D} we get lim_n→∞r^D,j_n/r^D,j̃_n=1.




§ CONCLUSION

The aim of this section is to prove that there are infinitely many irrational zeta values of the form ζ(2n+1). We will use the above linear forms in the Hurwitz zeta function together with the formula

    ∑_j=1^D-1ζ(i,j/D)=∑_j=1^D-1∑_n=0^∞D^i/(Dn+j)^i= (D^i-1)ζ(i)

to eliminate certain odd zeta values. In particular, we get for each divisor d|D

    ((D/d)^i-1)ζ(i)=∑_j=1^(D/d)-1ζ(i,jd/D).

This motivates the following definition

    r̂_n^(D,d):=∑_j=1^(D/d)-1 r_n^(D,dj).

With the above observation, we obtain for every divisor d|D linear forms in odd zeta values:

    r̂_n^(D,d)=∑_i=3 
     i  odd^s a^(D)_i((D/d)^i-1)ζ(i)+â^(D,d)_0

with â^(D,d)_0=∑_j=1^(D/d)-1 a_0^(D,dj). Observe, that the coefficients a_i^(D) do not depend on d|D. Choosing D>1 with many divisors and using that the matrix

    ((D/d)^i-1)_i=3,...,s 
     d|D, d≠ D

has full rank, allows us to eliminate arbitrary many odd zeta values in such a linear form. In order to guarantee convergence, it is necessary to choose s>>1 appropriately. We obtain:

For each integer m≥ 0, there exists an integer N_m>m such that every subset of

    {ζ(3),ζ(5),...,ζ(2N_m+1)}

with N_m-m elements contains at least one irrational number.


Let us choose D=2^m+1. Let us write x_0(s) for the unique positive solution of f_D(x_0,s)=1 where

    f_D(x,s)=(x+3/x)^D( x+1/x+2)^s+1.

We have f_D(4· 2^-s+1/D,s)<1<f_D(2^-s+1/D,s) for s>>1. By the discussion in the proof of <ref> we obtain

    2^-s+1/D<x_0(s)<4· 2^-s+1/D.

In particular, we deduce that

    g_D(x_0(s),s)=O(1/4^s),   s→∞

where g_D(x,s) is the polynomial

    g_D(x,s)=D^6(D-1)(3+x)^3D(1+x)^s+1/(2+x)^2(s+1)

appearing in the statement of <ref>. Since e^-1>1/4, we can choose a large enough integer N_m with g_D(x_0(s),s)<e^-s for s=2N_m+1. The prime number theorem guaranties lim_n→∞√(d_n)=e. With the choice s:=2N_m+1, <ref> implies the convergence of

    d_n^s·r̂_n^(D,d)→ 0, as n→∞.

Thus for the above choice of s, d_n^s·r̂_n^(D,d) provides integral linear forms in odd zeta values converging to zero. Let

    J⊂{ 3,5,...,2N_m+1 }

be a subset with N_m-m elements. Let us choose some j∈ J and write

    {j}∪ ({ 3,5,...,2N_m+1 }∖ J)={ i_1,...,i_m+1}

for its complement with i_1<i_2<...<i_m+1. Let l be the index corresponding to j, i.e. j=i_l. By the choice D=2^m+1, every divisor d|D is of the form d=2^m+1-β for β=0,...,m+1. The matrix

    M=( (D/d)^i_α-1 )_α,d=(2^i_α·β-1)_α,β=1,...,m+1

is invertible. Define the following vector with integral entries

    w=(w_1,...,w_m+1):=(M)· e_l· M^-1∈^m+1

with e_l=(0,...,0,1,...,0) the l-th unit vector. Then

    d_n^sr̂_n:=(-1)^l-1d_n^s∑_k=1^m+1w_k r̂_n^(2^m+1,2^m+1-k)

is an integral linear combination of the odd zeta values in J and tends to zero as n goes to infinity. A straight-forward computation shows

    (-1)^l-1·∑_k=1^m+1w_k ·(2^k-1)>0

which implies the positivity of the involved linear forms:

    lim_n→∞r̂_n/r_n^(D,D)=lim_n→∞ (-1)^l-1∑_k=1^m+1w_k∑_j=1^2^k-1r_n^(2^m+1,2^m+1-kj)/r_n^(2^m+1,2^m+1)=(-1)^l-1∑_k=1^m+1w_k(2^k-1)>0.

Thus we obtain a sequence of non-zero integral linear combinations of 

    {ζ(j)| j∈ J }

tending to zero. If all of the zeta values ζ(j) where rational, we could multiply by there common denominator and would get a sequence of non-zero integers converging to zero - an obvious contradiction.



	For each m≥ 1 there is an integer N_m>m such that at least one of the numbers
	
    {ζ(2m+1),ζ(2m+3)...,ζ(2N_m+1)}

	is irrational. In particular, there are infinitely many odd zeta values, which are irrational.



We have chosen the rational functions R_n^(D) in a preferably simple way. For improving convergence properties, there are probably much better choices.




amsalpha 


