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1 Math. Ann. 00) 347: DOI 0.007/s y Mathematische Annalen A refinement of Nesterenko s linear independence criterion with applications to zeta values Stéphane Fischler Wadim Zudilin Received: 9 May 009 / Published online: December 009 The Authors) 009. This article is published with open access at Springerlink.com Abstract We refine and give a new proof of) Nesterenko s famous linear independence criterion from 985, by making use of the fact that some coefficients of linear forms may have large common divisors. This is a typical situation appearing in the context of hypergeometric constructions of Q-linear forms involving zeta values or their q-analogs. We apply our criterion to sharpen previously known results in this direction. Introduction. Nesterenko s criterion In this text, we refine Nesterenko s linear independence criterion by taking into account the existence of common divisors to the coefficients of the linear forms. Consider the following situation: The work of W. Zudilin was supported by the Max Planck Institute for Mathematics Bonn) and the Hausdorff Center for Mathematics Bonn). S. Fischler B) Laboratoire de Mathématiques d Orsay, Université Paris-Sud, 9405 Orsay Cedex, France Stephane.Fischler@math.u-psud.fr S. Fischler CNRS, 9405 Orsay Cedex, France W. Zudilin School of Mathematical and Physical Sciences, University of Newcastle, Callaghan, NSW 308, Australia

2 740 S. Fischler, W. Zudilin N) Let ξ 0,...,ξ r be real numbers, with r. Let 0 <α<and β>. For any n, let l 0,n,...,l r,n be integers such that r lim n i=0 /n l i,n ξ i = α and lim sup l i,n /n β for any i {0,...,r}. n Let us recall a special case of Nesterenko s criterion [4]. Theorem A Yu. Nesterenko) Assume that hypothesis N) holds. Then we have dim Q Span Q ξ 0,...,ξ r ) log α log β. Hypothesis N) implies dim Q Span Q ξ 0,...,ξ r ), because otherwise ξ 0,...,ξ r would be integer multiples of a possibly zero) real number ξ, so that all linear forms ri=0 l i,n ξ i would be integer multiples of ξ. This is impossible since these linear forms tend to 0 without vanishing for n sufficiently large). This remark shows that the first interesting case is trying to get a dimension greater than or equal to three. This special case of Theorem A reads as follows. Theorem B Assume that hypothesis N) holds. If αβ <, then dim Q Span Q ξ 0,...,ξ r ) 3. In other words, among ξ 0,...,ξ r there are at least three numbers that are linearly independent over the rationals.. A refinement We obtain the following improvement of Nesterenko s criterion, the proof of which relies on Minkowski s convex body theorem and yields a new proof of Nesterenko s result. Theorem Let ξ 0,...,ξ r be real numbers, with r. Let τ>0 and γ,...,γ r 0. For any n and any i {0,...,r}, letl i,n Z. Forn and i {,...,r}, let δ i,n be a positive divisor of l i,n such that i) δ i,n divides δ i+,n for any n and any i {,...,r }, and ii) δ j,n divides δ j,n+ for any n and any 0 i < j r, with δ 0,n =. δ i,n δ i,n+ Assume that there exists an increasing sequence Q n ) n of integers such that, as n, the following conditions are met:

3 A refinement of Nesterenko s linear independence criterion 74 Q n+ = Qn +o), max l i,n Qn +o), 0 i r r l i,n ξ i = Qn τ+o), i=0 δ i,n = Q γ i +o) n for any i {,...,r}. Let s = dim Q Span Q ξ 0,...,ξ r ). Then we have s τ + γ + +γ s. Taking Q n = β n and τ = log α)/log β) we obtain the following special case. Corollary Assume that hypothesis N) holds. For any n and any i {,...,r}, let δ i,n be a positive divisor of l i,n. Assume that i) δ i,n divides δ i+,n for any n and any i {,...,r }, and ii) δ j,n divides δ j,n+ for any n and any 0 i < j r, with δ 0,n =. δ i,n δ i,n+ Furthermore, assume that for any i {,...,r} the limit of δ /n i,n as n exists. Let s = dim Q Span Q ξ 0,...,ξ r ). Then we have s and αβ s s lim n δ/n i,n. i= The conclusion of this corollary has to be understood as a lower bound for s, namely, log α/ ) s i= lim n δ /n i,n s ; log β but is should be noted that the product contains s factors. The following special case of Corollary, in which we let d n = lcm,,...,n),is useful when studying linear independence of zeta values see, for example, the proofs of Theorems 3 and 5). Corollary Assume that hypothesis N) holds. Let e e r be non-negative integers such that d e i n divides l i,n for any n and any i {,...,r}. Let s = dim Q Span Q ξ 0,...,ξ r ). Then we have s and s e + +e s log α. log β Again the lower bound we obtain for s in this corollary actually depends on s itself.

4 74 S. Fischler, W. Zudilin It is interesting to see what happens with Theorem when we just try to prove that dim Q Span Q ξ 0,...,ξ r ) 3, as in Theorem B. Then we may assume, without loss of generality, that δ,n = = δ r,n for any n. In this case, the Assumptions i) and ii) simply mean that δ,n divides δ,n+. Actually, we obtain the following stronger improvement of Theorem B, in which this assumption is replaced with a lower bound on the greatest common divisor of δ,n and δ,n+. Theorem Assume that hypothesis N) holds. For any n, letδ n be a common positive divisor of l,n,...,l r,n. Assume that Then αβ < lim inf n gcdδ n,δ n+ )) /n. dim Q Span Q ξ 0,...,ξ r ) 3. The main interest of Theorem is actually its proof, which is simpler than that of Theorem cf. Sects.. and.3)..3 Applications Theorem is interesting whenever Nesterenko s linear independence criterion is used and the coefficients of the linear forms are known to have common divisors. Among the various situations where this phenomenon appears, we have chosen an arithmetic problem for the so-called odd zeta values the values of Riemann s zeta function ζl) = at odd integers l > ; see [,,7,,0,] for history and known results in this arithmetic direction. The following theorem improves on previous bounds i 45 and i 97) from [0, Theorem 0.3]. Theorem 3 There exist odd integers i 39 and i 96 such that the numbers are linearly independent over Q. k l, ζ3), ζi ), and ζi ) Another but related application is devoted to arithmetic properties of the following q-analog of Riemann s zeta function q < ): k l q k ζ q l) = q k = σ l k)q k,

5 A refinement of Nesterenko s linear independence criterion 743 where σ l k) = d k dl. A usual setup for a number q is to be of the form /p, where p Z\{0, ±}. Although the irrationality of ζ q ) and even the transcendence of ζ q l) for any even positive integer l are known, not so much is obtained for ζ q l) with l > odd; we refer the reader to the works [0,] for details. For example, Jouhet and Mosaki [0] show that at least one of the four numbers ζ q 3), ζ q 5), ζ q 7), ζ q 9) is irrational and give further results for the odd q-zeta values in the spirit of Theorem 3. Our next theorem sharpens slightly the corresponding bounds on i, i and i 3 proved in [0]. Theorem 4 Let q be a rational of the form /p, where p Z\{0, ±}. There exist odd integers < i 0 < i < i < i 3 such that i 0 9, i 37, i 83, i 3 45 and the numbers are linearly independent over Q., ζ q i 0 ), ζ q i ), ζ q i ), and ζ q i 3 ) Our third application also appeals to arithmetic of the odd zeta values, but this time we add log to the set. Theorem 5 There exist odd integers i 93 and i 5 such that the numbers are linearly independent over Q., log, ζi ), and ζi ) In our proof of Theorem 5 we use a seemingly) new hypergeometric construction of linear forms in, log and odd zeta values. We find rather curious that a degenerate case of our construction, when odd zeta values do not occur at all, resembles well the rational approximations [] from Apéry s proof of the irrationality of ζ3) cf., for example, [7]); this is the subject of our final Sect The linear independence criterion In this part, we prove Sect..) our main linear independence criterion namely Theorem stated in Sect. ), as well as its special case Sect..) corresponding to Theorem. We gather some remarks in Sect..3.. Proof of the criterion Let us start with two remarks. Remark The existence of arbitrarily small non-zero linear combinations of ξ 0,...,ξ r with integer coefficients implies dim Q Span Q ξ 0,...,ξ r ), that is, s.

6 744 S. Fischler, W. Zudilin Remark In the statement of Theorem and in all other linear independence criteria we prove in this text, no assumption is made on whether ξ 0 vanishes or not. Actually, we can always assume that ξ 0 = 0, because if ξ 0 = 0 then Remark provides an integer i such that ξ i = 0, and we can consider the linear forms 0ξ i + l,n ξ + +l r,n ξ r in ξ i,ξ,...,ξ r ). The proof of Theorem splits into two parts. To begin with, let us prove this result under the assumption that ξ 0,...,ξ r are linearly independent over the rationals that is, s = r). Denote by ξ the point ξ 0,...,ξ r ) R r+, and by L n the linear form l 0,n X 0 + +l r,n X r, so that L n ξ) = r i=0 l i,n ξ i. Thanks to Remark, we may assume that ξ 0 = 0 and even ξ 0 = dividing all ξ i by ξ 0 if necessary). Let n be a sufficiently large integer. In what follows, o) stands for any sequence that tends to 0 as n tends to infinity. We take R n = δ r,n L n ξ) and ε n = δ r,n 3 Ln ξ) ri= δ i,n ) /r, so that ε n = Q γ r τ+γ + +γ r )/r+o) n. Arguing by contradiction, assume that τ + γ + +γ r > r. Consider the set C n consisting in all q 0,...,q r ) R r+ such that q 0 R n δ r,n and, for any i {,...,r}, δ i,n q 0 ξ i q i δ i,n δ r,n ε n. The volume of C n is r+ R n ε r n ri= δ i,n δ r+ r,n = 3 r+ > r+. Moreover C n is convex. To prove this fact, let q 0,...,q r ) and q 0,...,q r ) be elements of C n, and let λ [0, ].Letq i = λq i + λ)q i for any i {0,...,r}. Then q 0 λ q 0 + λ) q 0 R n/δ r,n and, for any i {,...,r}, δ i,n q 0 ξ i q i λ δ i,n q 0 ξ i q i + λ) δ i,n q 0 ξ i q i ε nδ i,n /δ r,n. Therefore C n is a convex body, symmetric with respect to the origin: Minkowski s theorem shows that there is a non-zero integer point q 0,n,...,q r,n ) in C n. Then rescaling p 0,n = δ r,n q 0,n and p i,n = δ r,n δ i,n q i,n Z for any i {,...,r},

7 A refinement of Nesterenko s linear independence criterion 745 we have p 0,n,...,p r,n Z thanks to Assumption i), and also by definition of C n : p 0,n R n and p 0,n ξ i p i,n ε n for any i {,...,r}. ) Let k n denote the least positive integer such that p 0,n δ r,k n L kn ξ). By definition of R n and thanks to the upper bound )on p 0,n, k n exists and k n n; this upper bound will be used later. Since δ r,n divides l r,n,we have γ r. The assumption τ +γ + +γ r > r and the definition of ε n yield lim n ε n = 0. Thanks to ), this implies lim n p 0,n ξ p i,n ) = 0. Now,ξ,...,ξ r are assumed to be linearly independent, so that ξ Q and lim n p 0,n =+. The definition of k n then implies lim n k n =+,so that δ r,kn = Q γ r +o) k n and L kn ξ) =Q τ+o) k n. By minimality of k n,wehave with δ r,kn = Q γ r +o) k n Finally we obtain Now we can write δ r,kn L kn ξ) < p 0,n δ r,k n L kn ξ) = Q γ r +o) k n r l i,kn p i,n = p 0,n i=0 and, in the same way, L kn ξ) =Q τ+o) k n. p 0,n =Q γ r +τ+o) k n. ) r l i,kn ξ i + i=0 r l i,kn p i,n p 0,n ξ i ). 3) On the right-hand side, the first term has absolute value equal to p 0,n L kn ξ), therefore less than or equal to δ r,k n by definition of k n. Assume the second term has absolute value less than the first one. Then the absolute value of the right-hand side is less than δ r,kn. But it is equal to the left-hand side, which is an integer multiple of δ r,kn, since i=0 l i,kn p i,n = l i,kn δ r,n δ i,n q i,n is an integer multiple of δ i,kn δ r,kn δ i,kn = δ r,kn by condition ii) and n k n ): it has to be zero. But then both terms on the righthand side would have the same absolute value, in contradiction with the assumption. Therefore we have proved that the second term on the right-hand side of 3) has absolute value greater than or equal to the absolute value of the first one. Combining this

8 746 S. Fischler, W. Zudilin inequality with ) and ) we obtain r = p 0,n L kn ξ) = p r 0,n l i,kn ξ i l i,kn p i,n p 0,n ξ i ) Q+o) k n ε n. Q γ r +o) k n Now recall that k n n and γ r ; the previous estimate yields i=0 Q γ r +o) k n ε n Q γ r +o) n ε n = Q τ+γ + +γ r )/r+o) n, which contradicts the assumption τ + γ + +γ r > r for n sufficiently large and completes the proof of Theorem under the assumption that ξ 0,...,ξ r are linearly independent over the rationals. Let us now deduce Theorem from this special case. Because of Assumptions i) and ii), we cannot without loss of generality) change the ordering of ξ 0,...,ξ r in order to assume that ξ 0,...,ξ s are linearly independent. For this reason we have to focus on the integers i {,...,r} such that ξ i does not belong to the Q-vector space spanned by ξ 0,...,ξ i. Let us state this in precise terms. Thanks to Remarks and, wehaves and we may assume that ξ 0 = 0. Take i 0 = 0, and let i be the least positive integer such that ξ 0 and ξ i are linearly independent over the rationals. Define inductively i k,fork {0,...,s}, tobethe least integer such that ξ i0,ξ i,...,ξ ik are linearly independent over Q. Clearly,we have 0 = i 0 < i < < i s and, for any i {0,...,r}, we can write ξ i as a linear combination of those ξ i j with i j i. In precise terms, we write ξ i = k j=0 c i, j ξ i j with c i, j Q and k {0,...,s} defined by i k i < i k+ with i s+ = r + ). For any n, this gives i=0 r l i,n ξ i = i=0 s l j,n ξ j j=0 by letting ξ j = ξ i j and l j,n = r i=i j l i,n c i, j.letd denote a common denominator of the rational numbers c i, j ; note that d is independent of n. For any n and any j {0,...,s}, dl j,n is an integer and, moreover, a multiple of δ i j,n, since δ i j,n divides δ i,n for any i between i j and r. Since ξ 0,...,ξ s are linearly independent over the rationals, we can apply to these numbers the special case of Theorem proved at the beginning, with the linear forms sj=0 dl j,n ξ j, the same sequence Q n) n and the same τ, with divisors δ j,n = δ i j,n for j {,...,s} and exponents γ j which satisfy γ j = γ i j γ j, because δ j,n = δ i j,n δ j,n for any j and any n. This completes the proof of Theorem.. Proof of Theorem The following statement implies Theorem. Proposition Let ξ 0,...,ξ r be real numbers, with r. For any n and any i {0,...,r}, letl i,n Z. Let δ n ) n be a sequence of positive integers, such that

9 A refinement of Nesterenko s linear independence criterion 747 δ n is a common divisor of l,n,...,l r,n for any n. For any n, leth n and ε n be positive real numbers such that max 0 i r l i,n H n and r l i,n ξ i ε n. i=0 Assume that r i=0 l i,n ξ i = 0 for infinitely many n and that Then we have H n ε n+ + H n+ ε n lim n gcdδ n,δ n+ ) = 0. dim Q Span Q ξ 0,...,ξ r ) 3. Proof Thanks to Remarks and, wehavedim Q Span Q ξ 0,...,ξ r ) and we may assume that ξ 0 = 0, and even that ξ 0 =. Since ξ,...,ξ r play symmetric roles, we may assume that ξ is irrational. Let us argue by contradiction, assuming on the contrary that dim Q Span Q ξ 0,...,ξ r ) =. Then ξ,...,ξ r are rational linear combinations of ξ 0 = and ξ. As in the proof of Theorem, we obtain a positive integer d independent of n and rational numbers l 0,n and l,n such that r l i,n ξ i = l 0,n + l,n ξ i=0 with dl 0,n, dl,n Z of absolute value less than d H n for some constant d independent of n; moreover, δ n divides dl,n. Now consider the determinant n = dl 0,n dl,n dl 0,n+ dl = d l,n+ l 0,n + l,n ξ) d l,n l 0,n+ + l,n+ ξ),,n+ which satisfies n dd H n ε n+ + H n+ ε n )<gcdδ n,δ n+ ) if n N for some integer N. Now n is the determinant of a matrix in which both entries in the second column, namely, dl,n and dl,n+, are integer multiples of gcdδ n,δ n+ ). Therefore, n = 0 for any n N. Hence, for any n N the vector dl 0,n, dl,n ) is proportional to dl 0,N, dl,n ). Now the subgroup Z Qdl 0,N, dl,n ) of the finitely generated free Z-module Z is generated by a single vector, say u 0, u ) Z. Therefore for any n N there exists an integer c n such that dl 0,n = c nu 0 and dl,n = c nu. This implies l 0,n + l,n ξ = c n u 0 + u ξ ) with c n Z, in contradiction with the fact that l 0,n + l,n ξ tends to 0 without being identically equal to 0 for n sufficiently large, and Proposition follows.

10 748 S. Fischler, W. Zudilin.3 Remarks In this section, we make some comments on the proofs given above. In the case where all divisors δ j,n are equal to, the proof of Theorem gives a new proof of Theorem A, while that of Proposition yet another one in the special case of Theorem B. Nesterenko s general result in [4] is exactly Theorem in the special case δ j,n =, except for one point: Nesterenko assumes that Q τ +o) n r i=0 l i,n ξ i Q τ +o) n, whereas we treat the case τ = τ only. Our method should generalize easily to the situation where τ = τ, but we do not write it down because the equality holds in all the applications we have in mind the point is that knowing the exact size of r i=0 l i,n ξ i is not really necessary: upper and lower bounds are sufficient to make the arguments work). For the same reason, we did not try to replace Q with another number field, though Nesterenko s criterion can be generalized to this setting see [3,8]). Nesterenko s proof consists in obtaining a lower bound for the distance of ξ = ξ 0,...,ξ r ) to any linear subspace of R r+, defined over Q, of dimension t <τ+. He proceeds by induction on t, whereas we use in the proof of Theorem only the first step namely t =, see below) of his induction. Colmez [4] writes down Nesterenko s proof in another way from notes by F. Amoroso). Assume for simplicity that ξ 0,...,ξ r are Q-linearly independent the general case follows from this special case as in the proof of Theorem ). For any sufficiently large integer n 0, one constructs by an analogous induction procedure a decreasing sequence n 0 > n > > n r of positive integers such that the determinant of the matrix [l i,n j ] 0 i, j r is not zero. The easy case is when n 0,...,n r are, roughly speaking, of the same size for instance, if they are consecutive integers). Then replacing the first line with the linear. Since is a non-zero integer, this gives r τ and completes the proof in this case. The difficult part of this proof is to obtain some control upon n 0,...,n r. In Amoroso Colmez s version of Nesterenko s proof, the sequence n 0 > n > > n r is constructed, and yields the result r τ, but there might be huge gaps between n j and n j+.itwould be interesting to know whether such a sequence can always be constructed with n r nearly as large as n 0 ; for instance this would enable one to essentially replace the unnatural Assumptions i) and ii) of Theorem with: δ i,n divides δ i,n+. This is what we do in the case r = ) in the proof of Proposition Sect..): for infinitely many integers n 0, we prove that = 0 with n = n 0. This kind of method is similar to the ones used by H. Davenport and W. Schmidt see, for instance, [5,6]). This different method is the reason why the assumptions of Proposition are weaker than those of Theorem. It is interesting to point out that in Proposition we do not need to assume combination of the lines given by ξ 0,...,ξ r ), we obtain Q r τ+o) n Q n+ = Q +o) n, nor to have a lower bound for r i=0 l i,n ξ i. On the other hand, our proof of Theorem in Sect.. relies on a completely different idea. In the case when all divisors δ j,n are equal to, it can be summarized as follows see [9] for a translation in terms of exponents of Diophantine approximation). Dirichlet s box principle yields under the assumption that ξ 0,...,ξ r are linearly independent over Q) very good simultaneous approximants p /p 0,...,p r /p 0 to ξ /ξ 0,...,ξ r /ξ 0 with the same denominator p 0. This means that ξ 0,...,ξ r ) is suffi-

11 A refinement of Nesterenko s linear independence criterion 749 ciently close to the line generated by p 0,...,p r ) in R r+, and contradicts if r <τ) the lower bound proved in the first induction step of Nesterenko s proof see above). Actually, this first step is very easy to prove directly without Nesterenko s machinery for controlling the intersection of a linear subspace with a hyperplane). Indeed, for some n denoted by k n in Sect..) the hyperplane H n defined by l 0,n X l r,n X r = 0 has comparatively small height and is very close to ξ 0,...,ξ r ), hence to p 0,...,p r ), so that p 0,...,p r ) has to belong to H n. But then the distance from ξ 0,...,ξ r ) to H n is less than, or equal to, the distance of ξ 0,...,ξ r ) to p 0,...,p r ); this is too small, in contradiction with the lower bound for L n ξ). At last, let us comment briefly on the optimality of our criterion. Eventhough Assumptions i) and ii) might be refined see above), it is likely that the conclusion s τ + γ + +γ s of Theorem cannot be improved see [9,8] for related results when s = ). Now it turns out that Theorem applied to the construction of [,6] gives dim Q Span Q,ζ3), ζ5), ζ7),...,ζa)) + o) + log log a, exactly like Nesterenko s linear independence criterion, because γ,,γ s turn out to be very small and the improvement lies inside the error term). Therefore a new construction of linear forms seems to be necessary for refining this lower bound. 3 Applications of the criterion 3. First application: odd zeta values For a pair of positive integers s and t with t < s, consider the very-well-poised) hypergeometric series h n = n! s t) It is known [,6,0] that, for some a i,n Q, k + n ) tn j= k j) tn j= k + n + j) nj=0 k + j) s. 4) s h n = a 0,n + a i,n ζi + ). 5) i= Let us summarize the auxiliary results about these linear forms that can be deduced from [,0]. Denote by x 0 the maximal real zero of the polynomial x + t + ) x ) s+ x t ) x + ) s+; 6)

12 750 S. Fischler, W. Zudilin it belongs to the interval t +, + ). Introduce the function f x) = t + ) ) ) ) log x + t + + t + log x t s + ) log x + ) s + ) log x ). 7) Consider the following product over primes: t n = n t) = l= t+)n<p n {n/p} E l p l, 8) where [ ) l E l = t, l + t + / [ l E l = t + /, l ) t for l = 0,,...,t, for l =,,...,t, 9) and { }denotes the fractional part of a number. Proposition In the above notation, and log h n lim = f x 0 ) n n lim sup n log a i,n n Re f 0) = s t) log + t + ) logt + ) for i = 0,,...,s. Moreover, the rational coefficients of the forms 5) satisfy dn s n a 0,n Z and dn s i) while the asymptotic behavior of 8) is determined by log n lim n n = t )ψ) t ψ l= n a i,n Z for i =,...,s, 0) + l ) t t ψ l= + ) l, t + / where ψx) is the digamma function that is, the logarithmic derivative of the Gamma function).

13 A refinement of Nesterenko s linear independence criterion 75 Proof of Proposition This result follows from Propositions., 3., and 4. and Lemma 4.5 of [0], taking also into account the proof [] of the so-called denominator conjecture by Krattenthaler and Rivoal. They show that d s n a 0,n Z and d s i) n a i,n Z for i =,...,s, in other words, they get rid of an extra d n with respect to the analogous properties used in [0]. Proof of Theorem 3 The collection of numbers under consideration is Setting from 0) we see that ξ 0,ξ,ξ,...,ξ s ) =,ζ3), ζ5),...,ζs )). l i,n = dn s n a i,n Z for i = 0,...,s, d i+ n l i,n for i =,...,s, hence in the notation of Corollary we have r = s, δ i,n = d i+ n, and log α = f x 0 ) + s ϖ t log β = s t) log + t + ) logt + ) + s ϖ t, where ϖ t = lim n log n n. Using standard formulas for the digamma function we can write ϖ t by means of elementary functions only: ϖ t = 3t t + ) t log + t log t + l= l + π t t πl t cot t + + l= ) logt + ). l= cos 4πl log sin πl t + t + We now apply Corollary. With the choice s = 70, t = 0 we obtain hence log α 3 log β = >, dim Q Span Q,ζ3), ζ5),...,ζ39)) 3;

14 75 S. Fischler, W. Zudilin in the same way, taking s = 98, t = 65 we get yielding log α 3 + 5) log β = >3 dim Q Span Q,ζ3), ζ5),..., ζ96)) 4. This computation implies Theorem Second application: odd q-zeta values We now fix a number q of the form /p, where p Z\{0, ±}. As in the previous section, we take a pair of positive integers s and t satisfying t < s. With the help of the basic hypergeometric series h n q) = q) s t) n q k+n ) qk tn ) tn q k+n+ ) tn q k ) s n+ s q ks t)n+ks k = a 0,n q) + a i,n q)ζ q i + ), ) i= where b) n = b; q) n = n q k b) is the q-pochhammer symbol, it was shown in []seealso[0], where the q-denominator conjecture is proved) that dim Q Span Q,ζq 3), ζ q 5),...,ζ q s ) ) π + o) s π + as s. The coefficients a i,n q) are, in fact, rational functions of the variable p = /q, whose denominators involve only powers of p and of the cyclotomic polynomials j p) = j k, j)= p e π k/j ) Z[p], deg p j p) = ϕj), j =,,.... In these settings, the q-analog of the quantity d n is the least common multiple of the polynomials p, p,...,p n, which equals ) d n p) = n j p); j=

15 A refinement of Nesterenko s linear independence criterion 753 Mertens theorem asserts that, for a real number p with p >, log d n p) lim n n = 3 log p π. The following statement summarizes the analytic and arithmetic results of [0,] for the linear forms in the odd q-zeta values, sharpened using the argument of [0, Sect. 4]: one just replaces primes by cyclotomic polynomials ) cf.[, Sect. ]). Recall that the sets E l are defined in 9). Proposition 3 In the above notation, and log h n q) lim n n = ts t), log p lim sup n log a i,n q) n log p s + t 4 for i = 0,,...,s. Moreover, the coefficients of the forms ) satisfy s)!p M d n p) s n p) a 0,n q) Z[p] and s)!p M d n p) s i) n p) a i,n q) Z[p] for i =,...,s, 3) where sn + ) tntn ) M = + + s + )n 4 t n p) = n t) p) = l= t+)n< j n {n/j} E l s t)n, j p) l, 4) and the asymptotic behavior of 4) is determined by log n p) lim n n = ϖ log p where = t =t )ψ ) t ) t l= + 3 π t t ψ x) = 3 dψx) π = 3 dx π denotes the normalized) trigamma function. ψ + l ) t t l= k=0 ) l t l= ψ + l ) t +/ t ) l ψ +, t + / l= x + k)

16 754 S. Fischler, W. Zudilin Proof of Theorem 4 In the notation p = /q Z\{0, ±}, set l i,n = p M d n p) s n p) a i,n q) Z for i = 0,...,s. To these linear forms in ξ 0,ξ,ξ,...,ξ s ) =,ζ q 3), ζ q 5),...,ζ q s ) ), we apply Theorem taking Q n = β n log p and τ = log α)/log β), where and From 3) we see that log α = ts t) + s + t 4 log β = s + t + + 3s ) π ϖ t 3s ) π ϖ t. d n p) i+ l i,n for i =,...,s, hence we may take δ i,n = d n p) i+ to meet the required conditions of Theorem. The existence of an odd integer 3 i 0 9, for which ζ q i 0 ) is irrational, is already shownin[0]. The following choices of s and t and Theorem ensure the truth of Theorem 4: log α 3 3/π s = 9, t = 4 : = >, log β s = 4, t = 6 : log α 3 + 5) 3/π = >3, log β s = 73, t = 8 : log α ) 3/π = >4. log β 3.3 Third application: log and odd zeta values As in the two previous sections, we take a pair of positive integers s and t with t < s, but this time we assume s to be even. Consider the hypergeometric series h n = n n)!) s n! t k + n ) tn k j) tn j= j= k + n + j) n. 5) j=0 k + j/) s

17 A refinement of Nesterenko s linear independence criterion 755 Its k-rational summand H n k) = k + n) tn j= k j) tn ) s j= k + n + j) n n)! n! t n! t n 6) j=0 k + j/) differs from the corresponding one in 4) a little: the s products n!/ ) n j=0 k + j) in 4) are replaced by n n)!/ n j=0 k + j/) in 5), and these two have similar asymptotics as n. This similarity allows us to compute, like in [] or[0], the asymptotic behavior of 5) and of the coefficients in the zeta decomposition of 5) which we are going to describe in the next statement. Lemma In the above notation, we have where s/ h n = ã 0,n + ã,n log + ã i,n ζi ), 7) i= 4tn d s nã0,n Z and 4tn d s i+ n ã i,n Z for i =,,...,s/. 8) Proof The function 6) is the product of integer-valued polynomials k + n, nj= k ln j), l =0,,...,t, and n! nj= k+ln+ j), l =,,...,t, n! and of s copies of the rational function n n)! n j=0 k + j/) = n)! = j=0 k + j) j=0 ) j n) j k + j/. 9) It follows from the Leibniz rule for differentiating a product cf. [0, Lemmas..4] and the formula 9)) that with the property H n k) = s i= j=0 A i, j k + j/) i 4tn d s i n A i,k Z. 0)

18 756 S. Fischler, W. Zudilin For a variable x in the unit circle x <, we now perform the summation H n k)x k = where = = s i= j=0 + s x k s i= j=0 A i, j x j i= j= s Li i x ) i= s i= j=0 A i, j x j+ A i, j k + j/) i = j xk k i A i, j x j + j=0 j x k j A i, j k i s i= s s i= j=0 j xk k /) i A i, j x j ) i Li i x) Li i x ) i= j= Li i x) = j A i, j x k k i j= x k+ j k + j/) i j+ A i, j x x k j k /) i, ) denotes the ith polylogarithm function. To compute the limit x in ), we use Abel s theorem for power series, the sum residue theorem in the form A, j + A, j = j=0 j= j=0 Res k= j/ H n k) = Res k= H n k) = 0, and the identity Li x) Li x ) = Li x). Therefore, s h n = log A, j + ζi) A i, j + i ) i= j= s j A i, j j=0 i= k i + j=0 A i, j j= j= A i, j j k /) i, ) where we used the evaluations Li ) = log and Li i ) = ζi) for i =,...,s. Finally, note that the parity of s implies from 6) that H n k n) = H n k),

19 A refinement of Nesterenko s linear independence criterion 757 hence A i, j = ) i A i,n j and n j=0 A i, j = n j= A i, j = 0fori =, 4,...,s. This implies the required decomposition 7) with ã 0,n = ã,n = ã i,n = s i= j A i, j j=0 A, j, j= and k i + A i, j + i ) j=0 A i, j j= j= A i, j j k /) i, for i =,...,s/. 3) Using 0) we arrive at the inclusions 8). We are now in power to sharpen the inclusions 8) in the way we already did in Propositions and 3. Note that for an integer N > and a prime p > N we have ord p ƔN +)=ord p N!= where N p ƔN +/) N)! N and ord p =ord p Ɣ/) N N! =, p 4) x = x x 5) see the proof of Lemma 3 for another expression of x ). Lemma For the coefficients in the decomposition 7), we have the inclusions 4tn dn s n ã0,n Z and 4tn d s i+ n n ãi,n Z for i =,,...,s/, 6) where n = t) n = t+)n<p n p τn/p) 7) and the function τ ) is defined as follows: τx) = τ t x) = min {τ x, y), τ x, y)}, 8) y R τ x, y) = t + ) x + y + t + ) x y x + y x y t x, τ x, y) = t + ) x + y + t + ) x y x + y x y t x.

20 758 S. Fischler, W. Zudilin Proof In the notation of the above proof of Lemma, we can write d s i A i, j = s i)! dk s i H n k) k + j ) s ) k= j/ for i =,...,s and j = 0,,...,n. 9) Taking into account the partial fraction decomposition 9) and tn l= k l) tn l= k + n + l) n! t n! t k= j/ tn Ɣt + )n/ + j n)/ + ) Ɣt + )n/ + n j)/ + ) = ) n! t Ɣn/ + j n)/ + ) n! t Ɣn/ + n j)/ + ) for j = 0,,...,n, 30) with the help of [0, Lemma 4.] we conclude that any common multiple of the numbers in 30), involving primes p n only, can be used in sharpening the inclusions 0): tn d s i n A i, j Z for i =,...,s and j = 0,,...,n. In view of 3), it is enough to show that n defined in 7) is such a multiple. From 4) we see that Ɣt + )n/ + j n)/ + ) ord p n! t Ɣn/ + j n)/ + ) { τ n/p,j n)/p) for j even, = τ n/p,j n)/p) for j odd, Ɣt + )n/ + n j)/ + ) n! t Ɣn/ + n j)/ + ) j = 0,,...,n, and this implies the desired result. Lemma 3 The quantity 7) can be written as follows: t n = n t) = where the sets E l are given in 9). In addition, log n lim n n l= t+)n<p n {n/p} E l t = ϖ t = 4 t + ) t/ + t log t + t + l= p l, 3) l + π t cot l= πl t + ) logt + ). 3)

21 A refinement of Nesterenko s linear independence criterion 759 Proof Using the identity x = x + x + /, we see that the function 5) is equal to x + /. This implies that τ x, y) = τ x, y + ), hence τx) = min {τ x, y)}. 33) y R Furthermore, it follows from 8) that τ x +, y) = τ x, y + ), hence the function τx) is -periodic: τx) = τ{x}). Moreover, we have τ x, y + ) = τ x, y) and τ x, y) = τ x, y) implying that the minimum in 33) can be performed for y [0, [ only: τx) = min {τ x, y)}. 0 y< It remains to use the results of [0, Section 4] already taken into account in Sects. 3. and 3.): min {τ x, y)} =l for x E l, l = 0,,...,t, 0 y< where the sets E 0, E,...,E t are defined in 9); this gives us the desired form 3) of the quantity 7). To compute the asymptotics in 3) we apply [0, Lemma 4.4]: log n lim n n = = t l= t l= = ϖ t + + = ϖ t + + = ϖ t + l E l [0,/) dψ + x) + l dψ + x) + E l t l= t/ t l= t/ + t l= t/ t l= t/ + l E l l ) E l [/,) E l [/,) dψx) d ) x d ) / t x t/ /t ) x E l d d ) x t l l t + / ) ) t t l + t/ t + / l ) t ) l l t + ) t l= t/ + ) t l t,

22 760 S. Fischler, W. Zudilin where ϖ t is defined by ϖ t = t γ) t + ) t l = tψ) t l= l= ψ + l t ) + ψ t l= ψ + ) l + ψ t )) l, t + / l t + / and γ = ψ) is Euler s constant. It remains to apply identities for the digamma function, and the lemma follows. The following statement summarizes our findings in this section. Proposition 4 For positive integers s and t with s even and t < s, the linear forms 5), 7) and their coefficients admit the asymptotics and lim sup n log ã i,n n log h n lim = f x 0 ) n n Re f 0) = s t) log + t + ) logt + ) for i = 0,,...,s/, where x 0 t +, + ) is the maximal real zero of the polynomial 6) and the function f x) is defined in 7). Moreover, the coefficients in the decomposition 7) satisfy 6) and the asymptotics of the quantities 7), 3) is determined in 3). Proof of Theorem 5 This time we have Q-linear forms in whose coefficients are ξ 0,ξ,ξ,...,ξ s/ ) =, log,ζ3), ζ5),..., ζs )), )) l i,n = 4tn dn s n ãi,n Z for i = 0,,...,s/. Then 6) implies d i n l i,n for i =,,...,s/, hence in the notation of Corollary we have r = s/, δ i,n = d i n, and log α = f x 0 ) + 4t log + s ϖ t log β = s + t) log + t + ) logt + ) + s ϖ t.

23 A refinement of Nesterenko s linear independence criterion 76 Applying the theorem with the choice s = 94, t = we obtain log α log β = >, while the choice s = 5, t = 67 results in log α + 6) log β = >3. This implies the required independence result. Remark 3 Applying Nesterenko s linear independence criterion instead of Corollary yields the same bound on i, and the slightly less precise bound i Triple integrals for rational approximations to log The particular case s =, t = of our construction in Sect. 3.3 is of independent interest, since the corresponding series h n ) n)! n = k + n ) n j= k j) n j= k + n + j) n! n 34) j=0 k + j/) goes in parallel with Ball s series h n = n! k + n ) n j= k j) n j= k + n + j) nj=0 k + j) 4 for Apéry s approximations to ζ3). Switching to the classical hypergeometric notation [7] and using n n)! n! = Ɣn + ) Ɣn + /) Ɣ ) =, π we can write the series in 34) as the following very-well-poised hypergeometric series: h n = Ɣ3n + 3)Ɣn + )3 Ɣn + ) Ɣn + 3 ) πɣn + ) 3 Ɣn + 3 ) 7 F 6 3n +, 3n+ +, n +, n +, n + 3, n +, n + 3 3n+, n +, n +, n + 3, n +, n + 3 ).

24 76 S. Fischler, W. Zudilin Applying now Bailey s transformation see [7, Eq )] or [, Proposition ]), after a little reduction of the gamma factors we get the Barnes-type integral h n = n+ π πi i i Ɣn++ξ) Ɣn+ +ξ)ɣn+ 3 +ξ)ɣ ξ)ɣ ξ) Ɣn++ξ)Ɣn+ 3 +ξ) dξ, 35) where the path separates the decreasing sequence of poles ξ = n, n, n 3,... and the increasing sequence of poles ξ = 0,,,... of the integrand. The result can be expressed as a triple real integral thanks to a theorem of Nesterenko [, Proposition ]: h n = n + π [0,] 3 x n x) n y n / y) n z n+/ z) n / xy)z) n+ dx dy dz. 36) On the other hand, using the duplication formula Ɣz)Ɣz + ) = π z Ɣz) and the Barnes-type and Euler integrals for the Gauss hypergeometric function see [7, Sections.6. and 4.]) we can transform 35) further: h n = 4n + ) πi = n + ) πi i i i i Ɣn + + ξ)ɣn + + ξ)ɣ ξ) Ɣ4n ξ) Ɣn + ξ)ɣn + + ξ)ɣ ξ) Ɣ4n + + ξ) = n+) Ɣn)Ɣn+) n, n+ F Ɣ4n+) 4n + and for the latter integral the decomposition )= dξ 0 dξ x n x) n+ +x) n+ dx, h n = ã 0,n + ã,n log with d n ã 0,n Z and ã,n Z 37) is known see, for example, [9]). The arithmetic inclusions in 37) are much better than the ones we have from Proposition 4; this suggests the existence of a power denominator conjecture for the linear forms constructed in Sect In addition, a more general form of the triple integral in 36) could be of use in study of the quality of rational approximations to log ; it is due to a remarkable resemblance of such integrals with the ones used by Rhin and Viola [5] in proving the record irrationality measure for ζ3). For a different construction of rational approximations to log using the Rhin Viola method, we refer the reader to the paper [3], where Marcovecchio obtains a new irrationality measure for this constant.

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