One of the Odd Zeta Values from ζ(5) to ζ(25) Is Irrational. By Elementary Means
SSymmetry, Integrability and Geometry: Methods and Applications SIGMA (2018), 028, 8 pages One of the Odd Zeta Valuesfrom ζ (5) to ζ (25) Is Irrational.By Elementary Means Wadim ZUDILINDepartment of Mathematics, IMAPP, Radboud University,PO Box 9010, 6500 GL Nijmegen, The Netherlands
E-mail: [email protected]
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Received January 31, 2018, in final form March 26, 2018; Published online March 29, 2018https://doi.org/10.3842/SIGMA.2018.028
Abstract.
Available proofs of result of the type ‘at least one of the odd zeta values ζ (5) , ζ (7) , . . . , ζ ( s ) is irrational’ make use of the saddle-point method or of linear indepen-dence criteria, or both. These two remarkable techniques are however counted as highly non-elementary, therefore leaving the partial irrationality result inaccessible to general mathe-matics audience in all its glory. Here we modify the original construction of linear formsin odd zeta values to produce, for the first time, an elementary proof of such a result —a proof whose technical ingredients are limited to the prime number theorem and Stirling’sapproximation formula for the factorial. Key words: irrationality; zeta value; hypergeometric series
Without touching deeply a history of the question (see [3] for an excellent account of this), wenotice that the irrationality of the zeta values — values of Riemann’s zeta function ζ ( s ) = ∞ (cid:88) n =1 n s at integers s = 2 , , . . . , is known for even s and also for s = 3, while there are only partialresults in this direction for odd s ≥
5. A starting point here has been set in the work [1] withfurther development, particularly focusing on ζ (5), in [7] and [10].We fix an odd integer s ≥
7. Our strategy is constructing two sequences of linear forms r n and ˆ r n living in the Q -space Q + Q ζ (3) + Q ζ (5) + · · · + Q ζ ( s ), for which we have a control of thecommon denominators λ n of rational coefficients and an elementary access to their asymptoticbehaviour as n → ∞ ; more importantly, the two coefficients of ζ (3) in these forms are propor-tional (with factor 7), so that 7 r n − ˆ r n belongs to the space Q + Q ζ (5) + · · · + Q ζ ( s ). Finally,using 7 r n − ˆ r n > λ n (7 r n − ˆ r n ) → n → ∞ of the linear forms λ n (7 r n − ˆ r n ) ∈ Z + Z ζ (5) + Z ζ (7) + · · · + Z ζ ( s )when s = 25, we conclude that it cannot happen that all the quantities ζ (5) , ζ (7) , . . . , ζ (25) arerational. a r X i v : . [ m a t h . N T ] M a r W. ZudilinThe original idea of using the so-called well-poised hypergeometric series to construct linearforms in zeta values of a given parity is due to Ball and Rivoal [1]; our new ingredient hereis using simultaneously such series and their ‘twists by half’, for an appropriate choice of theparameters. More precisely, our hypergeometric series assume the form r n = ∞ (cid:88) ν =1 R n ( ν ) and ˆ r n = ∞ (cid:88) ν =1 R n (cid:0) ν − (cid:1) , (1)where the rational-function summand R n ( t ) is defined as follows: R ( t ) = R n ( t ) = n ! s − (cid:81) nj =1 ( t − j ) · (cid:81) nj =1 ( t + n + j ) · n (cid:81) nj =1 (cid:0) t − n − + j (cid:1)(cid:81) nj =0 ( t + j ) s = 2 n n ! s − (cid:81) nj =0 (cid:0) t − n + j (cid:1)(cid:81) nj =0 ( t + j ) s +1 . (2)The following Sections 2 and 3 discuss, respectively, the arithmetic and analysis of theforms (1). In Section 4 we use this information to conclude with the proof of the claimedresult and make some relevant comments. The notation d n will be used for the least common multiple of 1 , , . . . , n . Recall that the primenumber theorem is equivalent to the asymptoticslim n →∞ d /nn = e. (3)A rational function S ( t ) of the form S ( t ) = P ( t )( t − t ) s ( t − t ) s · · · ( t − t q ) s q , whose denominator has degree larger than its numerator, possesses a unique partial-fractiondecomposition S ( t ) = q (cid:88) j =1 s j (cid:88) i =1 b i,j ( t − t j ) i . The coefficients here can be computed on the basis of explicit formula b i,j = 1( s j − i )! (cid:0) S ( t )( t − t j ) s j (cid:1) ( s j − i ) (cid:12)(cid:12)(cid:12) t = t j for all i , j in question. This procedure can be illustrated on the following examples, in whichall the exponents s j are equal to 1: n ! (cid:81) nj =0 ( t + j ) = n (cid:88) k =0 ( − k (cid:0) nk (cid:1) t + k , (cid:81) nj =1 ( t − j ) (cid:81) nj =0 ( t + j ) = n (cid:88) k =0 ( − n + k (cid:0) n + kn (cid:1)(cid:0) nk (cid:1) t + k , (cid:81) nj =1 ( t + n + j ) (cid:81) nj =0 ( t + j ) = n (cid:88) k =0 ( − k (cid:0) n − kn (cid:1)(cid:0) nk (cid:1) t + k , ne of the Odd Zeta Values from ζ (5) to ζ (25) Is Irrational 32 n (cid:81) nj =1 ( t + − j ) (cid:81) nj =0 ( t + j ) = n (cid:88) k =0 ( − n + k (cid:0) n +2 k n (cid:1)(cid:0) nn + k (cid:1) t + k , n (cid:81) nj =1 ( t − + j ) (cid:81) nj =0 ( t + j ) = n (cid:88) k =0 (cid:0) kk (cid:1)(cid:0) n − kn − k (cid:1) t + k , n (cid:81) nj =1 ( t + n − + j ) (cid:81) nj =0 ( t + j ) = n (cid:88) k =0 ( − k (cid:0) n − k n (cid:1)(cid:0) nk (cid:1) t + k . It also means that the function R ( t ) in (2) can be written as R ( t ) = s (cid:88) i =1 n (cid:88) k =0 a i,k ( t + k ) i (4)with the recipe to compute the coefficients a i,k in its partial-fraction decomposition. At thesame time, the function R ( t ) is a product of ‘simpler’ rational functions given above, with allcoefficients of their partial fractions being integral. Lemma 1.
Let k , . . . , k q be pairwise distinct numbers from the set { , , . . . , n } and s , . . . , s q positive integers. Then the coefficients in the expansion (cid:81) qj =1 ( t + k j ) s j = q (cid:88) j =1 s j (cid:88) i =1 b i,j ( t + k j ) i satisfy d s − in b i,j ∈ Z , where i = 1 , . . . , s j and j = 1 , . . . , q, (5) where s = s + · · · + s q .In particular, d s − in a i,k ∈ Z , where i = 1 , . . . , s and k = 0 , , . . . , n, (6) for the coefficients in (4) . Proof .
Denote the rational function in question by S ( t ). The statement is trivially true when q = 1, therefore we assume that q ≥
2. In view of the symmetry of the data, it is sufficientto demonstrate the inclusions (5) for j = 1. Differentiating a related product m times, for any m ≥
0, we obtain1 m ! (cid:0) S ( t )( t + k ) s (cid:1) ( m ) = 1 m ! q (cid:89) j =2 ( t + k j ) − s j ( m ) = (cid:88) (cid:96) ,...,(cid:96) q ≥ (cid:96) + ··· + (cid:96) q = m q (cid:89) j =2 (cid:96) j ! (cid:0) ( t + k j ) − s j (cid:1) ( (cid:96) j ) = (cid:88) (cid:96) ,...,(cid:96) q ≥ (cid:96) + ··· + (cid:96) q = m q (cid:89) j =2 ( − (cid:96) j (cid:18) s j + (cid:96) j − (cid:96) j (cid:19) ( t + k j ) − ( s j + (cid:96) j ) . This implies that b i, = (cid:88) (cid:96) ,...,(cid:96) q ≥ (cid:96) + ··· + (cid:96) q = s − i q (cid:89) j =2 ( − (cid:96) j (cid:18) s j + (cid:96) j − (cid:96) j (cid:19) k j − k ) s j + (cid:96) j W. Zudilinfor i = 1 , . . . , s . Using d n / ( k j − k ) ∈ Z for j = 2 , . . . , q and (cid:80) qj =2 ( s j + (cid:96) j ) = s − i for eachindividual summand, we deduce the desired inclusion in (5) for j = 1, hence for any j .The second claim in the lemma follows from considering R ( t ) as a product of the ‘simpler’rational functions. (cid:4) Lemma 2.
For the coefficients a i,k in (4) , we have a i,k = ( − i − a i,n − k for k = 0 , , . . . , n and i = 1 , . . . , s, so that n (cid:88) k =0 a i,k = 0 for i even . Proof .
Since s is odd, the function (2) possesses the following (well-poised) symmetry: R ( − t − n ) = − R ( t ). Substitution of the relation into (4) results in − s (cid:88) i =1 n (cid:88) k =0 a i,k ( t + k ) i = s (cid:88) i =1 n (cid:88) k =0 a i,k ( − t − n + k ) i = s (cid:88) i =1 ( − i n (cid:88) k =0 a i,k ( t + n − k ) i = s (cid:88) i =1 ( − i n (cid:88) k =0 a i,n − k ( t + k ) i , and the identities in the lemma follow from the uniqueness of decomposition into partial fractions.The second statement follows from n (cid:88) k =0 a i,k = ( − i − n (cid:88) k =0 a i,n − k = ( − i − n (cid:88) k =0 a i,k . (cid:4) Lemma 3.
For each n , r n = s (cid:88) i =2 i odd a i ζ ( i ) + a and ˆ r n = s (cid:88) i =2 i odd a i (cid:0) i − (cid:1) ζ ( i ) + ˆ a , with the following inclusions available : d s − in a i ∈ Z for i = 3 , , . . . , s, and d sn a , d sn ˆ a ∈ Z . Notice that (cid:0) i − (cid:1) ζ ( i ) = ∞ (cid:88) (cid:96) =1 (cid:0) (cid:96) − (cid:1) i for i ≥ Proof .
Our strategy here is to write the series in (1) using the partial-fraction decomposition (4)of R ( t ). To treat the first sum r n we additionally introduce an auxiliary parameter z >
0, whichwe later specialise to z = 1: r n ( z ) = ∞ (cid:88) ν =1 R n ( ν ) z ν = ∞ (cid:88) ν =1 s (cid:88) i =1 n (cid:88) k =0 a i,k z ν ( ν + k ) i = s (cid:88) i =1 n (cid:88) k =0 a i,k z − k ∞ (cid:88) ν =1 z ν + k ( ν + k ) i = s (cid:88) i =1 n (cid:88) k =0 a i,k z − k (cid:32) Li i ( z ) − k (cid:88) (cid:96) =1 z (cid:96) (cid:96) i (cid:33) = s (cid:88) i =1 Li i ( z ) n (cid:88) k =0 a i,k z − k − s (cid:88) i =1 n (cid:88) k =0 k (cid:88) (cid:96) =1 a i,k z − ( k − (cid:96) ) (cid:96) i , ne of the Odd Zeta Values from ζ (5) to ζ (25) Is Irrational 5whereLi i ( z ) = ∞ (cid:88) (cid:96) =1 z (cid:96) (cid:96) i for i = 1 , . . . , s are the polylogarithmic functions. The latter are well defined at z = 1 for i ≥ i (1) = ζ ( i ), while Li ( z ) = − log(1 − z ) does not have a limit as z → − . By taking thelimit as z → − in the above derivation and using R n ( ν ) = O (cid:0) ν − (cid:1) as ν → ∞ , we conclude that n (cid:88) k =0 a ,k = lim z → − n (cid:88) k =0 a ,k z − k = 0 , and r n = s (cid:88) i =2 ζ ( i ) n (cid:88) k =0 a i,k − s (cid:88) i =1 n (cid:88) k =0 a i,k k (cid:88) (cid:96) =1 (cid:96) i . (7)We proceed similarly for ˆ r n , omitting introduction of the auxiliary parameter z . Since R ( t )in (2) vanishes at t = − , − , . . . , − n + , we can shift the starting point of summation for ˆ r n to t = − m − , where m = (cid:4) n − (cid:5) , so thatˆ r n = ∞ (cid:88) ν = − m R n (cid:0) ν − (cid:1) = ∞ (cid:88) ν = − m s (cid:88) i =1 n (cid:88) k =0 a i,k (cid:0) ν + k − (cid:1) i = s (cid:88) i =1 n (cid:88) k =0 a i,k ∞ (cid:88) ν = − m (cid:0) ν + k − (cid:1) i = s (cid:88) i =1 m (cid:88) k =0 a i,k ∞ (cid:88) ν = − m (cid:0) ν + k − (cid:1) i + s (cid:88) i =1 n (cid:88) k = m +1 a i,k ∞ (cid:88) ν = − m (cid:0) ν + k − (cid:1) i = s (cid:88) i =1 m (cid:88) k =0 a i,k (cid:32) (cid:88) (cid:96) = k − m (cid:0) (cid:96) − (cid:1) i + ∞ (cid:88) (cid:96) =1 (cid:0) (cid:96) − (cid:1) i (cid:33) + s (cid:88) i =1 n (cid:88) k = m +1 a i,k (cid:32) ∞ (cid:88) (cid:96) =1 (cid:0) (cid:96) − (cid:1) i − k − m − (cid:88) (cid:96) =1 (cid:0) (cid:96) − (cid:1) i (cid:33) = s (cid:88) i =2 (2 i − ζ ( i ) n (cid:88) k =0 a i,k + s (cid:88) i =1 m (cid:88) k =0 a i,k m − k (cid:88) (cid:96) =0 ( − i (cid:0) (cid:96) + (cid:1) i − s (cid:88) i =1 n (cid:88) k = m +1 a i,k k − m − (cid:88) (cid:96) =1 (cid:0) (cid:96) − (cid:1) i . (8)Now the statement of the lemma follows from the representations in (7) and (8), Lemma 2, theinclusions (6) of Lemma 1 and d in k (cid:88) (cid:96) =1 (cid:96) i ∈ Z for 0 ≤ k ≤ n and i ≥ ,d in m − k (cid:88) (cid:96) =0 ( − i ( (cid:96) + ) i ∈ Z for 0 ≤ k ≤ m and i ≥ ,d in − k − m − (cid:88) (cid:96) =1 (cid:0) (cid:96) − (cid:1) i ∈ Z for m + 1 ≤ k ≤ n and i ≥ . (cid:4) W. Zudilin
In this section we make frequent use of Stirling’s asymptotic formula n ! ∼ √ πn (cid:16) ne (cid:17) n as n → ∞ , and its corollary (cid:18) nn (cid:19) ∼ n √ πn as n → ∞ for the central binomial coefficients. (One may also use somewhat weaker but ‘more elementary’lower and upper bounds (cid:90) n log x d x ≤ log( n !) ≤ (cid:90) n +12 log x d x for the factorial coming out from estimating integral sums of the logarithm function, witha nemesis of running into more sophisticated versions for the asymptotics and inequalities below.)Because the rational function R n ( t ) in (2) vanishes at 1 , , . . . , n and at , , . . . , n − , thehypergeometric series (1) can be alternatively written as r n = ∞ (cid:88) ν = n +1 R n ( ν ) = ∞ (cid:88) k =0 c k and ˆ r n = ∞ (cid:88) ν = n +1 R n ( ν − ) = ∞ (cid:88) k =0 ˆ c k , with the involved summands c k = R n ( n + 1 + k ) = 2 n n ! s − (cid:81) nj =0 (cid:0) k + 1 + j (cid:1)(cid:81) nj =0 ( n + k + 1 + j ) s +1 = n ! s − (6 n + 2 k + 2)!( n + k )! s +1 k + 1)!(2 n + k + 1)! s +1 (9)and ˆ c k = R n (cid:0) n + + k (cid:1) = 2 n n ! s − (cid:81) nj =0 (cid:0) k + + j (cid:1)(cid:81) nj =0 (cid:0) n + k + + j (cid:1) s +1 strictly positive . Observe that c k ˆ c k = (cid:81) nj =0 (2 k + 2 + j ) (cid:81) nj =0 (2 k + 1 + j ) · n (cid:89) j =0 n + k + + jn + k + 1 + j s +1 = 6 n + 2 k + 22 k + 1 · (cid:32) − n +1) (cid:0) n +2 k +22 n + k +1 (cid:1)(cid:0) n +2 kn + k (cid:1) (cid:33) s +1 ∼ n + 2 k + 22 k + 1 (cid:18) n + k n + k + 1 (cid:19) ( s +1) / as n + k → ∞ . (10) Lemma 4.
For s ≥ odd, lim n →∞ r /nn = lim n →∞ ˆ r /nn = g ( x ) and lim n →∞ r n ˆ r n = 1 where g ( x ) = 2 ( x + 3) ( x + 1) s +1 ( x + 2) s +1) and x is the unique positive zero of the polynomial x ( x + 2) ( s +1) / − ( x + 3)( x + 1) ( s +1) / . ne of the Odd Zeta Values from ζ (5) to ζ (25) Is Irrational 7 Proof .
We have c k +1 c k = (cid:0) k + 3 n + (cid:1) ( k + 3 n + 2)( k + 1) (cid:0) k + (cid:1) (cid:18) k + n + 1 k + 2 n + 2 (cid:19) s +1 ∼ f (cid:18) kn (cid:19) as n + k → ∞ , (11)where f ( x ) = x + 3 x (cid:18) x + 1 x + 2 (cid:19) ( s +1) / . For an ease of notation write q = ( s + 1) / ≥
4. Since f (cid:48) ( x ) f ( x ) = 1 x + 3 − x + q (cid:18) x + 1 − x + 2 (cid:19) = ( q − x + 3( q − x − x ( x + 1)( x + 2)( x + 3)and the quadratic polynomial in the latter numerator has a unique positive zero x , the func-tion f ( x ) monotone decreases from + ∞ to f ( x ) when x ranges from 0 to x and then monotoneincreases from f ( x ) to f (+ ∞ ) = 1 (not attaining the value!) when x ranges from x to + ∞ .In particular, there is exactly one positive solution x to f ( x ) = 1. Notice that 0 < x < f (1) = 4 · (2 / q < c k +1 /c k > k < x n − γ √ n and c k +1 /c k < k > x n + γ √ n for an appropriate choice of γ > c k in (9) (see [2, Section 3.4] aswell as the second proof of Lemma 3 in [1]). This means that the asymptotic behaviour ofthe sum r n = (cid:80) ∞ k =0 c k is determined by the asymptotics of c k and its neighbours c k , where k = k ( n ) ∼ x n and | k − k | ≤ γ √ n , so thatlim n →∞ r /nn = lim n →∞ c /nk ( n ) = lim n →∞ (cid:32)(cid:16) ne (cid:17) ( s − n (cid:18) n + 2 k + 2 e (cid:19) n +2 k +2 (cid:18) e k + 1 (cid:19) k +1 × (cid:18) n + k e (cid:19) ( s +1)( n + k ) (cid:18) e n + k + 1 (cid:19) ( s +1)(2 n + k +1) (cid:33) /n = (2 x + 6) x +6 ( x + 1) ( s +1)( x +1) (2 x ) x ( x + 2) ( s +1)( x +2) = 2 ( x + 3) ( x + 1) s +1 ( x + 2) s +1) · f ( x ) x = g ( x ) . It now follows from (10) thatˆ c k +1 ˆ c k ∼ c k +1 c k as n + k → ∞ , (12)so that the above analysis applies to the sum ˆ r n = (cid:80) ∞ k =0 ˆ c k as well, and its asymptotic behaviouris determined by the asymptotics of ˆ c k and its neighbours ˆ c k , where k = k ( n ) ∼ x n and | k − k | ≤ ˆ γ √ n . From (12) we deduce that the limits of ˆ c /nk ( n ) and c /nk ( n ) as n → ∞ coincide,hence ˆ r /nn → g ( x ) as n → ∞ . In addition to this, we also getlim n →∞ r n ˆ r n = lim n →∞ c k ( n ) ˆ c k ( n ) = lim n →∞ n + 2 k + 22 k + 1 (cid:18) n + k n + k + 1 (cid:19) ( s +1) / = f ( x ) , which leads to the remaining limiting relation. (cid:4) W. Zudilin
We choose s = 25 and apply Lemma 4 to find out that 7 r n − ˆ r n > n sufficiently large, andlim n →∞ (7 r n − ˆ r n ) /n = g ( x ) = exp( − . . . . ) , where x = 0 . . . . . Assuming that the odd zeta values from ζ (5) to ζ (25) are all rationaland denoting by a their common denominator, we use Lemma 3 and the asymptotics (3) toconclude that the sequence of positive integers ad n (7 r n − ˆ r n )tends to 0 as n → ∞ ; contradiction. Thus, at least one of the numbers ζ (5) , ζ (7) , . . . , ζ (25) isirrational.Those who count the prime number theorem as insufficiently elementary may use weakerversions of (3), for example, d n < n from [5] and the choice s = 33 instead, to arrive at thesame conclusion (for the larger value of s , of course).Finally, we remark that the novelty of eliminating an ‘unwanted’ term of ζ (3) in linear formsin odd zeta values can be further used with the arithmetic method in [10] to significantly reducethe size of s . Since this does not let s be down to s = 9, hence leaving the achievement ‘atleast one of the four numbers ζ (5), ζ (7), ζ (9), ζ (11) is irrational’ unchanged, we do not discussthis generalisation in greater details. We point out, however, that there are other applicationsof the hypergeometric ‘twist-by-half’ idea, some discussed in the joint papers [6, 8], and thata far-going extension to general ‘twists’ introduced by J. Sprang in [9] leads to an elementaryproof of a version of the Ball–Rivoal theorem from [1] as well as to a significant improvement ofthe latter — see [4] for details. Acknowledgements
I thank St´ephane Fischler, Tanguy Rivoal, Johannes Sprang and the anonymous referees fortheir feedback on the manuscript.
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