Diophantine properties for q-analogues of Dirichlet’s beta function at positive integers

Frédéric Jouhet and Elie Mosaki

2000 Mathematics Subject Classification: Primary 11J72; Secondary 11M36; 33D15. Key words and phrases: q-analogues of the values of Dirichlet’s beta function at integers; modular forms; irrationality; basic hypergeometric series.

Abstract. In this paper, we define q-analogues of Dirichlet’s beta function at positive integers, which can be written as β_{q}(s) = ∑_{k≥1} ∑_{d|k} χ(k/d)d^{s−1}q^{k} for s ∈ ℕ^{∗}, where q is a complex number such that |q| < 1 and χ is the non trivial Dirichlet character modulo 4. For odd s, these expressions are connected with the automorphic world, in particular with Eisenstein series of level 4. From this, we derive through Nesterenko’s work the transcendance of the numbers β_{q}(2s + 1) for q algebraic such that 0 < |q| < 1. Our main result concerns the nature of the numbers β_{q}(2s): we give a lower bound for the dimension of the vector space over ℚ spanned by 1, β_{q}(2), β_{q}(4), …, β_{q}(A), where 1/q ∈ ℤ \ {−1; 1} and A is an even integer. As consequences, for 1/q ∈ ℤ \ {−1; 1}, on the one hand there is an infinity of irrational numbers among β_{q}(2), β_{q}(4), …, and on the other hand at least one of the numbers β_{q}(2), β_{q}(4), …, β_{q}(20) is irrational.

1 Introduction

For any complex number s with Re(s) ≥ 1, Dirichlet’s beta function at s is defined by:

β(s) = ∑_{k=0}^{∞} \frac{(−1)^{k}}{(2k + 1)^{s}} ·

Recall Euler’s identity:

β(2m + 1) = \frac{(−1)^{m}E_{2m}}{2^{2m+2}(2m)!}π^{2m+1},

where m ∈ ℕ and the rational numbers E_{2m} are Euler numbers defined by 1/ cosh(z) = ∑_{k≥0} E_{k}z^{k}/k!. Thus, as for the values of Riemann’s zeta function at odd positive integers, Lindemann’s theorem yields that for m ∈ ℕ, β(2m + 1) is a transcendental number. However, nothing similar can be said concerning the values at even positive integers; the best known result in that direction is due to Rivoal and Zudilin [13]: at least one of the numbers β(2), β(4), …, β(12) is irrational.

In this article, we define q-analogues of the values of β at positive integers. They can be written for s ∈ ℕ^{∗} and for any complex number q with |q| < 1:

β_{q}(s) := ∑_{k≥1} k^{s−1} \frac{q^{k}}{1 + q^{2k}} = ∑_{k≥1} ∑_{d|k} χ(k/d)d^{s−1}q^{k},  (1.1)

where χ is the non trivial Dirichlet character modulo 4, defined by χ(2m+1) = (−1)^{m} and χ(2m) = 0. One can justify the term of q-analogue by the following relation, valid for s ∈ ℕ^{∗} (see the last section of this paper for a proof):

lim_{q→1}(1 − q)^{s}β_{q}(s) = (s − 1)!β(s).

Similarly to the q-analogues of Riemann’s zeta function at even positive integers considered in [6, 9], our definition (1.1) is related to modular forms when s in an odd positive integer. Indeed, consider the case s = 1, for which the first expression of (1.1) can be written:

β_{q}(1) = ∑_{k≥1} ∑_{m≥0}(−1)^{m}q^{(2m+1)k} = ∑_{m≥0}(−1)^{m} \frac{q^{2m+1}}{1 − q^{2m+1}} ·  (1.2)

This yields immediately β_{q}(1) = (π_{q} − 1)/4, where π_{q} is a q-analogue of π. The series π_{q} is considered in [3, 4], where an upper bound for its irrationality exponent is given. We also have (see [3]):

β_{q}(1) = ∑_{k≥1} q^{k} ∑_{d|k} χ(d) = \frac{θ^{2}(q) − 1}{4},

where θ(q) := ∑_{n∈ℤ} q^{n^{2}} is the classical theta function. This shows that if we set q = e^{2iπz}, β_{q}(1) is, up to a rational constant, the Fourier expansion of a weight 1 modular form on Γ_{1}(4) [7, p. 138, Proposition 30]. Moreover, as remarked in [3], Nesterenko’s algebraic independence theorem from [11] shows that θ(q), and therefore β_{q}(1), is a transcendental number when q is algebraic such that 0 < |q| < 1. Concerning the other values at odd positive integers, one sees that for s ≥ 1 and q = e^{2iπz}, β_{q}(2s + 1) is also the Fourier expansion, with algebraic coefficients, of a weight 2s+1 modular form on Γ_{1}(4). More precisely, consider the level 4 Eisenstein series [7, p. 131]:

G_{2s+1}^{(1,0)}(z) := ∑_{{(m_{1},m_{2})∈ℤ^{2} \atop (m_{1},m_{2})≡(1,0) mod 4}} \frac{1}{(m_{1}z + m_{2})^{2s+1}} ·

Then we have the following Fourier expansion [7, proposition 22]:

G_{2s+1}^{(1,0)}(z) = \frac{i}{(2s)!} (\frac{π}{2} )^{2s+1} ∑_{k≥1} ∑_{d|k} χ(k/d)d^{2s}q^{k/4}.

Hence

β_{q}(2s + 1) = i(−1)^{s+1} \frac{E_{2s}}{2β(2s + 1)} G_{2s+1}^{(1,0)}(4z),

and it is not difficult to see that z ↦ G_{2s+1}^{(1,0)}(4z) is a weight 2s+1 modular form on Γ_{1}(4). Set ϕ_{s}(q) := β_{q}(2s + 1)/θ^{4s+2}(q), which is a modular function (with weight 0) on Γ_{1}(4), having a Fourier expansion with algebraic coefficients. Thus ϕ_{s} is algebraic on ℚ(J) (see [7, p. 144, Problem 7]), where J is the modular invariant. Assume first that for s ≥ 1, ϕ_{s} is not a constant. We can deduce that J(q) and ϕ_{s}(q) are algebraically dependent. Assume from now on that q is algebraic such that 0 < |q| < 1. Nesterenko’s algebraic independence theorem from [11] shows that J(q) and θ(q) are algebraically independent. Therefore β_{q}(2s + 1) and θ(q) are necessarily also algebraically independent, and in particular β_{q}(2s+1) is a transcendental number. If now ϕ_{s} is a constant (necessarily algebraic), then θ^{4s+2}(q) is a transcendental number by [11], and so β_{q}(2s + 1) is again a transcendental number. To summarize, we have the following result, which can be compared to the transcendence of the values of Dirichlet’s beta function at odd positive integers, as well as the transcendence of the values of ζ_{q} at even positive integers in [9]: for s ∈ ℕ and q algebraic such that 0 < |q| < 1, β_{q}(2s + 1) is a transcendental number.

Consider now the values at even positive integers β_{q}(2s), which do not seem to be directly related to Eisenstein series. We will prove the following theorem, which is the main result of the present paper:

Theorem 1.1. For 1/q ∈ ℤ \ {−1; 1} and any odd integer A ≥ 3, we have the following lower bound:

dim_{ℚ} (ℚ + ℚβ_{q}(2) + ⋯ + ℚβ_{q}(A − 1)) ≥ f(A),  (1.3)

where

f(A) = max_{{r∈ℕ \atop 1≤r<A/2}} f(r; A) and f(r; A) := \frac{4rA + A − 4r^{2}}{(\frac{48}{π^{2}} + 2) A + 8r^{2} − \frac{16}{π^{2}} + \frac{16r}{3}} ·

Moreover f(A) satisfies f(A) ∼ \frac{π}{2\sqrt{π^{2} + 24}} \sqrt{A} when A → +∞.

The previous asymptotic estimate for f(A) gives immediately the following:

Corollary 1.2. For 1/q ∈ ℤ \ {−1; 1}, there are infinitely many irrational numbers among β_{q}(2), β_{q}(4), β_{q}(6),…

On the other hand, the estimate f(3; 21) ≥ 1, 02... gives the following quantitative version:

Corollary 1.3. For 1/q ∈ ℤ\{−1; 1}, at least one of the numbers β_{q}(2), β_{q}(4), β_{q}(6),…, β_{q}(20) is irrational.

Now we aim to sketch the proof of Theorem 1.1, which will be more detailed later on. We need the following proposition, which is a special case of Nesterenko’s linear independence criterion from [10]:

Proposition 1.4. Let N ≥ 2 be an integer, and v_{1}, …, v_{N} be real numbers. Assume that there exist N integer sequences (p_{j,n})_{n≥0} and two real numbers α_{1} et α_{2} with α_{2} > 0 such that:

i) lim_{n→+∞} \frac{1}{n^{2}} log |p_{1,n}v_{1} + ⋯ + p_{N,n}v_{N}| = −α_{1},

ii) for all j ∈ \{1, …, N\}, we have lim sup_{n→+∞} \frac{1}{n^{2}} log |p_{j,n}| ≤ α_{2}.

Then the dimension of the ℚ-vector space spanned by v_{1}, …, v_{N} satisfies:

dim_{ℚ} (ℚv_{1} + ⋯ + ℚv_{N}) ≥ 1 + \frac{α_{1}}{α_{2}} ·

In order to use Nesterenko’s criterion in our context, we shall study the following hypergeometric series (see Section 2 for the notations):

S_{n}(q) := (q)_{n}^{A−2r} ∑_{k≥1}(−1)^{k+1}q^{(k−1/2)((A−2r)n/2+A/2−1)}

× (1 − q^{2k+n−1})\frac{(q^{k−rn}, q^{k+n})_{rn}}{(q^{k−1/2})_{n+1}^{A}} ,

where A is an integer, r ∈ ℕ^{∗} and A − 2r > 0. In a first step, we rewrite S_{n}(q^{2}) as a linear combination of some β_{q}(2m), m ∈ ℕ^{∗}:

S_{n}(q^{2}) = P̂_{0,n}(q^{2}) + ∑_{{j=2 \atop j even}}^{A−1} P̂_{j,n}(q^{2})β_{q}(j),

where |q| < 1, A and n are odd positive integers and P̂_{j,n}(q^{2}) are a priori in ℚ(q), i.e. rational fractions in the variable q (thus in the variable 1/q) with coefficients in ℚ. In a second step, we look for a common denominator D_{n}(q) to these rational fractions in the variable 1/q, satisfying:

D_{n}(q)P̂_{j,n}(q^{2}) ∈ ℤ [\frac{1}{q}] for all j ∈ \{0, 2, 4, …, A − 1\}.

Next, we shall prove the following asymptotic estimates which are true for all 0 < |q| < 1:

lim_{n→+∞} \frac{1}{n^{2}} log |S_{n}(q)| = −\frac{1}{2}r(A − 2r) log |1/q|,
lim sup_{{n→+∞ \atop n odd}} \frac{1}{n^{2}} log |P̂_{j,n}(q)| ≤ \frac{1}{8}(A + 4r^{2}) log |1/q|,

lim_{n→+∞} \frac{1}{n^{2}} log |D_{n}(q)| = (\frac{A}{4} + r^{2} + \frac{12}{π^{2}}(A − 1) + \frac{4r}{3} + \frac{8}{π^{2}} ) log |1/q|.

Assume 1/q ∈ ℤ\{−1; 1}. Then we apply Proposition 1.4 to the N = (A+1)/2 integer sequences (D_{n}(q) × P̂_{j,n}(q^{2}))_{n odd}, which yields (1.3).

The estimate f(A) ∼ π\sqrt{A}/2\sqrt{π^{2} + 24} for A → +∞, is obtained by choosing r = u\sqrt{A} and finding the maximal value of f(u\sqrt{A}; A)/\sqrt{A} in the variable u.

Remark 1.5. Corollary 1.3 can be proved direcly without Nesterenko’s criterion. Indeed, for 1/q ∈ ℤ \ {−1; 1}, it is enough to obtain an asymptotic estimate of the linear combination D_{n}(q) × S_{n}(q^{2}) in β_{q}(2), β_{q}(4), …, β_{q}(20) with integer coefficients (by choosing A = 21 and r = 3). Therefore it is not necessary to find an upper bound for the height of the coefficients of the linear form, this is only useful for linear independence.

This article is organized as follows. In Section 2 we recall a few notations on q-series, which will be useful later. The third section is concerned with the study of the series S_{n}(q) mentioned before. In particular, the structure of this series (it is a very-well-poised basic hypergeometric series) yields interesting properties. In particular, when n is an odd integer, an appropriate expansion of S_{n}(q^{2}) will give explicitely the already mentioned linear combination and its coefficients P̂_{j,n}(q^{2}). In the fourth section we find a common denominator D_{n}(q) to the coefficients P̂_{j,n}(q^{2}), by using arithmetical techniques and cyclotomic polynomials. In Section 5, we study the asymptotics of S_{n}(q), P̂_{j,n}(q) through Cauchy’s formula, and D_{n}(q) by using the Möbius inversion. Finally, in the sixth and last section, we establish some links between β_{q} and β, and we end by stating a q-denominators conjecture.

2 Notations

We recall some standard definitions and notations for q-series, which can be found in [5].

Let q be a fixed complex parameter (the “base”) with |q| ≠ 1. We define for any real number a and any k ∈ ℕ the q-shifted factorial by:

(a)_{k} ≡ (a; q)_{k} := \{ \begin{matrix}1 if k = 0 \\ (1 − a) … (1 − aq^{k−1}) if k > 0. \end{matrix}

The base q can be omitted when there is no confusion (writing (a)_{k} for (a; q)_{k}, etc). For the sake of simplicity, write for k ∈ ℕ:

(a_{1}, …, a_{m})_{k} := (a_{1})_{k} × ⋯ × (a_{m})_{k}.

Recall the definition of the q-binomial coefficient:

\left[{n \atop k}\right]_{q} := \frac{(q)_{n}}{(q)_{k}(q)_{n−k}} ,

which is a polynomial in the variable q, with integer coefficients (see for example [14]).

Further, recall the definition of the basic hypergeometric series _{s+1}ϕ_{s}:

s+1ϕ_{s}[\begin{matrix}a_{0}, a_{1}, …, a_{s} \\ b_{1}, …, b_{s} \end{matrix}; q, z] := ∑_{k=0}^{∞} \frac{(a_{0}, a_{1}, …, a_{s})_{k}}{(q, b_{1}, …, b_{s})_{k}} z^{k},

with a_{j} ∈ ℂ for 0 ≤ j ≤ s, and b_{j} q^{k} ≠ 1 for all k ∈ ℕ and 1 ≤ j ≤ s. The series converges for |q| < 1 and |z| < 1, or |q| > 1 and |z| < |b_{1} … b_{s}/a_{1} … a_{s+1}|, and we say that _{s+1}ϕ_{s} is:

• well-poised if qa_{0} = a_{1}b_{1} = ⋯ = a_{s}b_{s}

• very-well-poised if it is well-poised and a_{1} = q\sqrt{a_{0}} = −a_{2}.

3 A very-well-poised series

We consider the following basic hypergeometric series:

S_{n}(q) := (q)_{n}^{A−2r} ∑_{k≥1}(−1)^{k+1}q^{(k−1/2)((A−2r)n/2+A/2−1)}

× (1 − q^{2k+n−1})\frac{(q^{k−rn}, q^{k+n})_{rn}}{(q^{k−1/2})_{n+1}^{A}} , (3.1)

for any odd integer A, r ∈ ℕ^{∗} and A − 2r > 0. Note that this series converges for all |q| ≠ 1. The series S_{n}(q) satisfies:

S_{n}(1/q) = q^{n(r−1)}S_{n}(q),  (3.2)

which comes from the choice of the power of q inside the sum of (3.1). Thanks to (3.2), it will be possible to expand S_{n}(q^{2}) as a linear combination over ℚ(q) of the values of β_{q} at even positive integers only, although one would expect values at odd positive integers as well. Besides, notice that:

S_{n}(q) = (−1)^{rn}q^{(rn+1/2)((A−2r)n/2+A/2−1)}

× (1 − q^{(2r+1)n+1})(q)_{n}^{A−2r} \frac{(q, q^{(r+1)n+1})_{rn}}{(q^{rn+1/2})_{n+1}^{A}}

×_{A+4}ϕ_{A+3}[\begin{matrix}a, q\sqrt{a}, −q\sqrt{a}, q^{rn+1}, q^{rn+1/2} …, q^{rn+1/2} \\ \sqrt{a}, −\sqrt{a}, q^{(r+1)n+1}, q^{(r+1)n+3/2} …, q^{(r+1)n+3/2} \end{matrix}; q, z] ,

with a = q^{(2r+1)n+1} and z = −q^{(A−2r)n/2+A/2−1}. This shows that S_{n}(q) is a very-well-poised basic hypergeometric series.

3.1 Some auxiliary functions

For all |q| ≠ 1 and s ∈ ℕ^{∗}, we consider the functions:

Y_{s}(q) := ∑_{k≥0}(−1)^{k} \frac{q^{2k+1}}{(1 − q^{2k+1})^{s}} ·  (3.3)

We will need the signless Stirling numbers of the first kind (see [14]), which are integers denoted by c(s, j) (where s and j are two integers such that 1 ≤ j ≤ s) and that are defined by:

(x)_{s} := x(x + 1) … (x + s − 1) = ∑_{j=1}^{s} c(s, j)x^{j}.

The following result gives for s ≥ 2 the expansion of the functions Y_{s} in terms of values of β_{q} at positive integers:

Lemma 3.1. For all |q| < 1 and any integer s ≥ 2:

Y_{s}(q) = \frac{1}{(s − 1)!} ∑_{j=2}^{s} c(s − 1, j − 1)β_{q}(j).  (3.4)

Proof. The definition (3.3) can be expanded as follows, for |q| < 1:

Y_{s}(q) = \frac{1}{(s − 1)!} ∑_{k≥0} ∑_{ℓ≥1}(−1)^{k}(ℓ)_{s−1}q^{(2k+1)ℓ}  (3.5)
= \frac{1}{(s − 1)!} ∑_{k≥0} ∑_{ℓ≥1}(−1)^{k} ∑_{j=2}^{s} c(s − 1, j − 1)ℓ^{j−1}q^{(2k+1)ℓ}
= \frac{1}{(s − 1)!} ∑_{j=2}^{s} c(s − 1, j − 1) ∑_{ℓ≥1} \frac{q^{ℓ}}{1 + q^{2ℓ}} ,

and we conclude by using the definition (1.1).

The following lemma gives for s ≥ 2 the expansion of the functions Y_{s} in terms of values of β_{q} at even positive integers only: Lemma 3.2. For all 0 < |q| < 1 and any integer s ≥ 2:

Y_{s}(q) + Y_{s}(1/q) = \frac{2}{(s − 1)!} ∑_{{j=2 \atop jeven}}^{s} c(s − 1, j − 1)β_{q}(j).  (3.6)

Proof. Set 0 < |q| < 1. Starting from the definition (3.3), we get:

Y_{s}(1/q) = (−1)^{s} ∑_{k≥0}(−1)^{k} \frac{q^{(2k+1)(s−1)}}{(1 − q^{2k+1})^{s}}
= (−1)^{s} ∑_{k≥0} ∑_{ℓ≥1}(−1)^{k}q^{(2k+1)(s−2)} \frac{(ℓ)_{s−1}}{(s − 1)!}q^{(2k+1)ℓ}
= \frac{(−1)^{s}}{(s − 1)!} ∑_{k≥0} ∑_{ℓ≥1}(−1)^{k}(ℓ − s + 2)_{s−1}q^{(2k+1)ℓ}.  (3.7)

Then we collect expressions (3.5) and (3.7), invert summations, and sum over k:

Y_{s}(q) + Y_{s}(1/q) = \frac{1}{(s − 1)!} ∑_{ℓ≥1} [(ℓ)_{s−1} − (−ℓ)_{s−1}] \frac{q^{ℓ}}{1 + q^{2ℓ}}
= \frac{1}{(s − 1)!} ∑_{j=2}^{s} c(s − 1, j − 1)(1 + (−1)^{j}) ∑_{ℓ≥1} ℓ^{j−1} \frac{q^{ℓ}}{1 + q^{2ℓ}} ·

This shows (3.6) via the definition (1.1) of β_{q}.

3.2 Linear combination in the β_{q}(2j), j ∈ ℕ^{∗}

Define:

R_{n}(T; q) := T^{(A−2r)n/2+A/2−2}q^{−A(n^{2}−1)/2−((A−2r)n+A−2)/4}

× \frac{(q)_{n}^{A−2r}(q^{−rn}T, q^{n}T)_{rn}}{(T − q^{1/2})^{A}(T − q^{1/2−1})^{A} … (T − q^{1/2−n})^{A}} · (3.8)

Then obviously we have:

S_{n}(q) = ∑_{k≥1}(−1)^{k+1}q^{k}(1 − q^{2k+n−1})R_{n}(q^{k}; q).

Notice that when n is odd, R_{n}(T; q) is a rational fraction in the variable T with degree −n(A − 2r)/2 − A/2 − 2, which is less or equal to −3, as A > 2r ≥ 2. Assume from now on that n is a fixed odd positive integer. The partial fraction expansion of R_{n}(T; q) can be written:

R_{n}(T; q) = ∑_{s=1}^{A} ∑_{j=0}^{n} \frac{c_{s,j,n}(q)}{(T − q^{1/2−j})^{s}} = ∑_{s=1}^{A} ∑_{j=0}^{n} \frac{d_{s,j,n}(q)}{(1 − T q^{j−1/2})^{s}} ,

where

d_{s,j,n}(q) := (−1)^{s}q^{(j−1/2)s}c_{s,j,n}(q)  (3.9)

and

c_{s,j,n}(q) := \frac{1}{(A − s)!} [\frac{d^{A−s}}{dT^{A−s}} R_{n}(T; q)(T − q^{1/2−j})^{A}]_{T =q^{1/2−j}} (3.10)

= \frac{q^{(1/2−j)s}}{(A − s)!} [\frac{d^{A−s}}{du^{A−s}} R_{n}(uq^{1/2−j}; q)(u − 1)^{A}]_{u=1} . (3.11)

The definition (3.8) gives R_{n}(T q^{n−1}; 1/q) = −q^{n(r−2)+1}R_{n}(T; q), which yields for all j ∈ {0, …, n} and s ∈ {1, …, A}:

d_{s,n−j,n}(1/q) = −q^{n(r−2)+1}d_{s,j,n}(q),  (3.12)

or equivalently

c_{s,n−j,n}(1/q) = −q^{n(s+r−2)+1−s}c_{s,j,n}(q).  (3.13)

We can now prove the following lemma, which gives explicitely the expected linear combination in terms of the values of β_{q} at even positive integers:

Lemma 3.3. For all 0 < |q| < 1, any odd positive integers n and A, and r ∈ ℕ^{∗} such that A > 2r:

S_{n}(q^{2}) = P̂_{0,n}(q^{2}) + ∑_{{j=2 \atop j even}}^{A−1} P̂_{j,n}(q^{2})β_{q}(j),  (3.14)

where for j = 2, 4, …, A − 1,

P̂_{0,n}(q) := P_{0,n}(1, q) + q^{−n(r−1)}P_{0,n}(1, 1/q) − \frac{1}{2}P_{1,n}(1, q), (3.15)
P̂_{j,n}(q) := ∑_{s=j}^{A} \frac{2c(s − 1, j − 1)}{(s − 1)!}P_{s,n}(1, q),  (3.16)

and

P_{0,n}(z, q) := ∑_{s=1}^{A} ∑_{j=1}^{n} ∑_{k=1}^{j}(−1)^{j+k} \frac{q^{k−j}}{(1 − q^{k−1/2})^{s}} d_{s,j,n}(q)z^{j−k}, (3.17)

P_{s,n}(z, q) := ∑_{j=0}^{n}(−1)^{j}q^{1/2−j}d_{s,j,n}(q)z^{j}.  (3.18)

Proof. We fix an odd positive integer n and a complex number q such that 0 < |q| < 1. Define for complex z

S_{n}(z; q) := ∑_{k≥1}(−1)^{k+1}q^{k}R_{n}(q^{k}; q)z^{−k}.

It is not difficult to see that S_{n}(z; q) converges if |z| > |q|^{(A−2r)n/2+A/2−1} and S_{n}(1/z; 1/q) converges if |z| < |q|^{−(A−2r)n/2−A/2}. Thus both series S_{n}(z; q) and S_{n}(1/z; 1/q) converge if |q| < |z| ≤ 1. A direct calculation shows that:

S_{n}(1; q) + q^{−n(r−1)}S_{n}(1; 1/q) = S_{n}(q).  (3.19)

On the other hand, by using the partial fraction expansion of R_{n}(T; q), we get:

S_{n}(z; q) = P_{0,n}(z, q) + ∑_{s=1}^{A} P_{s,n}(z, q)L_{s}(z; q),  (3.20)

where P_{0,n}(z, q) and P_{s,n}(z, q) are polynomials in the variable z defined by (3.17) and (3.18), and

L_{s}(z; q) := ∑_{k≥1}(−1)^{k+1} \frac{q^{k−1/2}}{(1 − q^{k−1/2})^{s}} z^{−k}.  (3.21)

Coming back to the definition (3.18), we see that (3.12) implies that for s ≥ 1, P_{s,n}(1/z, 1/q) = z^{−n}q^{n(r−1)}P_{s,n}(z, q). This gives us the idea to study the series:

S̃_{n}(z; q) := S_{n}(z; q) + z^{n}q^{−n(r−1)}S_{n}(1/z; 1/q),  (3.22)

with the convergence condition |q| < |z| < 1. The expansion (3.20), and the previous relation between P_{s,n}(1/z, 1/q) and P_{s,n}(z, q), give for |q| < |z| < 1:

S̃_{n}(z; q) = P_{0,n}(z, q) + q^{−n(r−1)}z^{n}P_{0,n}(1/z, 1/q)

+ ∑_{s=1}^{A} P_{s,n}(z, q)(L_{s}(z; q) + L_{s}(1/z; 1/q)). (3.23)

Note that we can deduce directly from the definition (3.21):

L_{1}(1/z; 1/q) = − ∑_{k≥1} ∑_{ℓ≥0}(−1)^{k+1}q^{(k−1/2)ℓ}z^{k},

which yields L_{1}(1/z; 1/q) = −L_{1}(1/z; q) − z/(1 + z), and therefore:

lim_{z→1} P_{1,n}(z, q)(L_{1}(z; q) + L_{1}(1/z; 1/q)) = −\frac{1}{2}P_{1,n}(1, q).  (3.24)

It remains to let z tend to 1 in (3.23) to get the lemma, via the definition (3.22), the relations (3.19), (3.6) and (3.24), together with the fact that L_{s}(1; q^{2}) = Y_{s}(q) for s ≥ 2.

4 Arithmetical investigations

In this section, we find for any odd positive integer n a common denominator D_{n}(q) to the coefficients P̂_{j,n}(q^{2}) ∈ ℚ(q). Let d_{n}(x) ∈ ℤ[x] be the unitary polynomial, with lowest degree, and common multiple of 1 − x, 1 − x^{2}, …, 1 − x^{n}. Recall some standard properties on cyclotomic polynomials. For t ∈ ℕ, the t-th cyclotomic polynomial is defined by ϕ_{t}(x) := ∏_{k∧t=1,k≤t}(x − e^{2ikπ/t}). Then one can prove:

x^{n} − 1 = ∏_{d|n} ϕ_{d}(x),  (4.1)

which yields:

d_{n}(x) = ∏_{t=1}^{n} ϕ_{t}(x).  (4.2)

Besides, we have ϕ_{t}(x) ∈ ℤ[x] and ϕ_{t}(x) is irreducible in ℤ[x]. We will also need the following polynomials:

Δ_{n}(x) := ∏_{{t=1 \atop t odd}}^{2n−1} ϕ_{t}(x),  (4.3)

and

φ_{n}(x) := ϕ_{2}(x)^{n}ϕ_{4}(x)^{⌊n/2⌋} … ϕ_{2n}(x),  (4.4)

where ⌊x⌋ is the integer part of the real number x. In what follows, we denote by Ω_{t}(Q(x)) the greatest power of ϕ_{t}(x) dividing the polynomial Q(x). We have the following useful arithmetical lemma:

Lemma 4.1. Let e be an odd positive integer and w_{n}(x) := ∏_{i=1}^{n} \frac{1−x^{e+2i}}{1−x^{2i}}. Then

φ_{n}(x) w_{n}(x) ∈ ℤ [x] .

Remark 4.2. We can see that φ_{n}(x) always divides ∏_{i=1}^{n}(1 + x^{i})^{2}, so we could use this alternative polynomial later, instead of φ_{n}(x). This would give simpler manipulations, but we would have worse asymptotic properties (see Lemma 5.4).

Proof of Lemma 4.1. From (4.1), w_{n}(x) is a quotient of products of cyclotomic polynomials ϕ_{t}(x). It is enough to prove that Ω_{t}(φ_{n}(x) w_{n}(x)) ≥ 0 for all t. Assume first that t is even. In view of (4.1), as e is odd, ϕ_{t}(x) can only be a factor in the denominator of w_{n}(x). Then we have:

Ω_{t}(w_{n}(x)) = − ∑_{i=1}^{n} Ω_{t}(1 − x^{2i}) = − ∑_{i=1}^{n} ∑_{t|2i} 1 = − ∑_{i=1}^{n} ∑_{\frac{t}{2} |i} 1 = − ⌊\frac{n}{t/2}⌋ .

By the definition (4.4) we have Ω_{t}(φ_{n}(x)) = ⌊\frac{2n}{t}⌋. Thus Ω_{t}(φ_{n}(x) w_{n}(x)) = 0.

Now if t is odd, we have Ω_{t}(φ_{n}(x) w_{n}(x)) = Ω_{t}(w_{n}(x)), which is equal to

∑_{j=0}^{⌊\frac{n}{t}⌋−1} Ω_{t} ( ∏_{i=jt+1}^{(j+1)t} \frac{1 − x^{e+2i}}{1 − x^{2i}} ) + Ω_{t} ( ∏_{i=t⌊\frac{n}{t}⌋+1}^{n} \frac{1 − x^{e+2i}}{1 − x^{2i}} ) .

Now the orders of divisibility in the sum over j are all equal to 1 − 1 = 0. Indeed, as t is odd, there is exactly one integer i ∈ {jt + 1, …, (j + 1)t} for which t divides e + 2i: i is of the form kt + (t − e)/2 and it is unique for if jt + 1 ≤ i_{1} ≤ i_{2} ≤ (j + 1)t and if t divides e + 2i_{1} and e + 2i_{2}, then t divides 2(i_{1}−i_{2}) therefore divides i_{1}−i_{2}. Finally, as 0 ≤ i_{2}−i_{1} < t, we have i_{1}−i_{2} = 0. In the same vein, there is exactly one integer i ∈ {jt + 1, …, (j + 1)t} for which t divides 2i: it is i = (j + 1)t.

Moreover we now show that:

Ω_{t} ( ∏_{i=t⌊\frac{n}{t}⌋+1}^{n} \frac{1 − x^{e+2i}}{1 − x^{2i}} ) = Ω_{t} ( ∏_{i=t⌊\frac{n}{t}⌋+1}^{n}(1 − x^{e+2i})) ∈ \{0; 1\}.

Indeed, the first equality is true as there is no multiple of t lying in \{t ⌊\frac{n}{t}⌋ + 1, …, n}. Besides, as before, as there are at most t consecutive integers lying in \{t ⌊\frac{n}{t}⌋ + 1, …, n\}, there is at most one integer i ∈ \{t ⌊\frac{n}{t}⌋ + 1, …, n\} such that t divides e + 2i. This concludes the proof.

Throughout this section, we assume n and A to be odd positive integers, and r to be a positive integer such that A − 2r > 0.

Lemma 4.3. For all s ∈ {1, …, A} and j ∈ {0, …, n}, we have:

φ_{n}(1/q)^{2r} d_{n} (1/q^{2})^{A−s} c_{s,j,n}(q^{2}) ∈ ℤ [q; \frac{1}{q}] .

Proof. Rewrite (3.10) as:

c_{s,j,n}(q) = \frac{1}{(A − s)!} [\frac{d^{A−s}}{dT^{A−s}} V_{n}(T; q)]_{T =q^{1/2−j}} ,  (4.5)

with

V_{n}(T; q) := R_{n}(T; q)(T − q^{1/2−j})^{A}
= (q)_{n}^{A−2r}T^{(A−2r)n/2+A/2−2}q^{−A(n^{2}−1)/2−((A−2r)n+A−2)/4}
×\frac{(q^{−rn}T, q^{n}T)_{rn}(T − q^{1/2−j})^{A}}{(T − q^{1/2})^{A}(T − q^{1/2−1})^{A} … (T − q^{1/2−n})^{A}} · (4.6)

We collect the terms of V_{n}(T; q) as follows:

V_{n}(T; q) = q^{an^{2}+bn+c}T^{A/2−2−n/2}F (T)^{A/2−r+1/2}G(T)^{A/2−r−1/2} ∏_{ℓ=1}^{r} H_{ℓ}(T)I_{ℓ}(T),

where a, b and c are integers (or half-integers) depending only on A and r, and the functions F, G, H_{ℓ} and I_{ℓ} are defined below. First, F satisfies:

F (T) := q^{−n(n+1)/2} \frac{(q)_{n}T^{n}(T − q^{1/2−j})}{(T − q^{1/2})(T − q^{1/2−1}) … (T − q^{1/2−n})}

= (−1)^{n}(1/q; 1/q)_{n} + ∑_{{i=0 \atop i≠j}}^{n}(−1)^{n−i+1}q^{−i(i+1)/2}\left[{n \atop i}\right]_{1/q} \frac{q^{1/2−j} − q^{1/2−i}}{T − q^{1/2−i}} · (4.7)

Then G satisfies:

G(T) := q^{−n(n+1)/2} \frac{(q)_{n}(T − q^{1/2−j})}{(T − q^{1/2})(T − q^{1/2−1}) … (T − q^{1/2−n})}

= ∑_{{i=0 \atop i≠j}}^{n}(−1)^{n−i+1}q^{(i−1/2)n−i(i+1)/2}\left[{n \atop i}\right]_{1/q} \frac{q^{1/2−j} − q^{1/2−i}}{T − q^{1/2−i}} · (4.8)

For all ℓ ∈ {1, …, r}, the function H_{ℓ} satisfies:

H_{ℓ}(T) := q^{−n(n+1)/2} \frac{(q^{−ℓn}T)_{n}(T − q^{1/2−j})}{(T − q^{1/2})(T − q^{1/2−1}) … (T − q^{1/2−n})}

= (−1)^{n}q^{−ℓn^{2}−n} + ∑_{{i=0 \atop i≠j}}^{n}(−1)^{i+1}q^{−(n−i)^{2}/2−(2n+i)/2}\left[{n \atop i}\right]_{1/q}

× \frac{(q^{−(ℓ−1)n−i−1/2}; q^{−1})_{n}}{(q^{−1}; q^{−1})_{n}} \frac{q^{1/2−j} − q^{1/2−i}}{T − q^{1/2−i}} · (4.9)

Finally, I_{ℓ} satisfies:

I_{ℓ}(T) := q^{−n(n+1)/2} \frac{(q^{ℓn}T)_{n}(T − q^{1/2−j})}{(T − q^{1/2})(T − q^{1/2−1}) … (T − q^{1/2−n})}

= (−1)^{n}q^{ℓn^{2}−n} + ∑_{{i=0 \atop i≠j}}^{n}(−1)^{n+i+1}q^{ℓn^{2}−n−i(i+1)/2}\left[{n \atop i}\right]_{1/q}

× \frac{(q^{−(ℓ+1)n−i−1/2}; q^{−1})_{n}}{(q^{−1}; q^{−1})_{n}} \frac{q^{1/2−j} − q^{1/2−i}}{T − q^{1/2−i}} · (4.10)

We see that if q is replaced by q^{2} and if U denotes any of the functions F, G, or U_{0} : T ↦ T^{A/2−2−n/2}, then by using the partial fraction expansions (4.7) and (4.8):

\frac{d_{n} (1/q^{2})^{μ}}{μ!} [\frac{d^{μ}}{dT^{μ}} U(T)]_{T =q^{1−2j}} ∈ ℤ [q; \frac{1}{q}] for all μ ∈ ℕ.  (4.11)

Now if q is replaced by q^{2} and if U denotes any of the functions H_{ℓ} or I_{ℓ}, then by using the partial fraction expansions (4.9), (4.10), and Lemma 4.1, we get for all μ ∈ ℕ:

φ_{n}(1/q) \frac{d_{n} (1/q^{2})^{μ}}{μ!} [\frac{d^{μ}}{dT^{μ}} U(T)]_{T =q^{1−2j}} ∈ ℤ [q; \frac{1}{q}] .  (4.12)

If we note ∂^{μ}U := \frac{1}{μ!} [\frac{d^{μ}}{dT^{μ}} U(T)]_{T =q^{1−2j}} and apply Leibniz’s formula to the function V_{n}(T; q^{2}), we obtain via the definition (4.5):

c_{s,j,n}(q^{2}) = q^{2(an^{2}+bn+c)} ∑_{μ_{0}+···+μ_{A}=A−s}(∂^{μ_{0}}U_{0})(∂^{μ_{1}}F) ⋯ (∂^{μ_{A/2−r+1/2}}F)

× (∂^{μ_{A/2−r+3/2}}G) ⋯ (∂^{μ_{A−2r}}G)(∂^{μ_{A−2r+1}}H_{1}) ⋯ (∂^{μ_{A−r}}H_{r})

× (∂^{μ_{A−r+1}}I_{1}) ⋯ (∂^{μ_{A}}I_{r}).

Now we conclude the proof of the lemma by using (4.11) and (4.12).

Lemma 4.4. Set α = −A/4 − r^{2}. Then there exist real numbers β′ and γ′ depending only on A and r such that for all (s, j) ∈ {1, …, A}×{0, …, n}, the limit of the function q ↦ q^{αn^{2}+β′n+γ′}c_{s,j,n}(q^{2}) exists and is finite as q → +∞.

Proof. Recall the expression (4.5) from the previous proof, and let v_{n}(T; q) := \frac{d}{dT} V_{n}(T; q)/V_{n}(T; q) be the logarithmic derivative of V_{n}(T; q), in the variable T. As in [6] and [9], we use Faà di Bruno’s differentiation formula, which gives for all μ ∈ ℕ:

\frac{1}{μ!} \frac{d^{μ}}{dT^{μ}} V_{n}(T; q) = ∑_{k_{1}+···+μk_{μ}=μ} \frac{V_{n}(T; q)}{k_{1}! … k_{μ}!} ∏_{ℓ=1}^{μ} (\frac{1}{ℓ!} \frac{d^{ℓ−1}}{dT^{ℓ−1}} v_{n}(T; q))^{k_{ℓ}} . (4.13)

By (4.6) we get:

v_{n}(T; q) = \frac{d}{dT} (log V_{n}(T; q))
= \frac{(A − 2r)n/2 + A/2 − 2}{T} + ∑_{i=1}^{rn} \frac{1}{T − q^{i}}

+ ∑_{i=n}^{rn+n−1} \frac{1}{T − q^{−i}} − A ∑_{{i=0 \atop i≠j}}^{n} \frac{1}{T − q^{1/2−i}} ,

thus for all ℓ ∈ ℕ^{∗}:

\frac{(−1)^{ℓ−1}}{(ℓ − 1)!} \frac{d^{ℓ−1}}{dT^{ℓ−1}} v_{n}(T; q) = \frac{(A − 2r)n/2 + A/2 − 2}{T^{ℓ}} + ∑_{i=1}^{rn} \frac{1}{(T − q^{i})^{ℓ}}

+ ∑_{i=n}^{rn+n−1} \frac{1}{(T − q^{−i})^{ℓ}} − A ∑_{{i=0 \atop i≠j}}^{n} \frac{1}{(T − q^{1/2−i})^{ℓ}} ·

We can rewrite this as follows:

\frac{(−1)^{ℓ−1}}{(ℓ − 1)!} \frac{d^{ℓ−1}}{dT^{ℓ−1}} v_{n}(T; q) = \frac{(A − 2r)n/2 + A/2 − 2}{T^{ℓ}} + ∑_{i=1}^{rn} (\frac{q^{−i}}{T q^{−i} − 1})^{ℓ}

+ ∑_{i=n}^{rn+n−1} \frac{1}{(T − q^{−i})^{ℓ}} − A ∑_{i=0}^{j−1} (\frac{q^{i}}{T q^{i} − 1})^{ℓ} − A ∑_{i=j+1}^{n} \frac{1}{(T − q^{−i})^{ℓ}} ·

Therefore, for all j ∈ \{0, …, n\}, the limit of q ↦ q^{(1/2−j)ℓ} [\frac{d^{ℓ−1}}{dT^{ℓ−1}} v_{n}(T; q)]_{T =q^{1/2−j}}

exists and is finite as q → +∞. We deduce that for k_{1} + ⋯ + μk_{μ} = μ:

lim_{q→+∞} q^{(1/2−j)μ} [∏_{ℓ=1}^{μ} (\frac{1}{ℓ!} \frac{d^{ℓ−1}}{dT^{ℓ−1}} v_{n}(T; q))^{k_{ℓ}}]_{T =q^{1/2−j}} < ∞.  (4.14)

Besides, V_{n}(q^{1/2−j}; q) defined by (4.6) satisfies:

lim_{q→+∞} V_{n}(q^{1/2−j}; q) × q^{−j(An−A+3)/2+j^{2}A/2−rn(rn−2)/2−A/2} = 1.

It remains to choose μ = A − s in (4.13). With the help of (4.5) and (4.14) we then deduce that the limit of

q ↦ q^{(1/2−j)(A−s)−(j(An−A+3)/2−j^{2}A/2+rn(rn−2)/2+A/2)}c_{s,j,n}(q)

exists and is finite as q → +∞. Replacing q by q^{2}, we can easily conclude, for we have that for all (s, j) ∈ {1, …, A} × {0, …, n}:

(1 − 2j)(A − s) − (j(An − A + 3) − j^{2}A + rn(rn − 2) + A) ≥ αn^{2} + β′n + γ′ ,

where α = −A/4 − r^{2}, β′ = 2r − (A + 1)/2, γ′ = −1/2 − 1/(2A) (this lower bound is obtained for s = 1 and j = (n + 1)/2 + 1/(2A)).

Now we can prove the following lemma, which gives an expression for D_{n}(q), a common denominator to P̂_{j,n}(q^{2}), for j ∈ {0, 2, 4, …, A − 1}:

Lemma 4.5. Let n and A be odd positive integers, and r ∈ ℕ^{∗} such that A − 2r > 0. For α = −A/4 − r^{2}, there exist β and γ real numbers depending only on A and r such that for all j ∈ {2, 4, …, A − 1}:

(A − 1)! q^{⌊αn^{2}+βn+γ⌋}φ_{n}(1/q)^{2r} d_{n}(1/q^{2})^{A−j}P̂_{j,n}(q^{2}) ∈ ℤ [\frac{1}{q}]  (4.15)

and

q^{⌊αn^{2}+βn+γ⌋}φ_{n}(1/q)^{2r} d_{2n}(1/q)^{A−1} Δ_{n}(1/q) P̂_{0,n}(q^{2}) ∈ ℤ [\frac{1}{q}] .  (4.16)

Thus, by setting

D_{n}(q) := (A − 1)! q^{⌊αn^{2}+βn+γ⌋}φ_{n}(1/q)^{2r} d_{2n}(1/q)^{A−1} Δ_{n}(1/q),  (4.17)

we get:

D_{n}(q)P̂_{j,n}(q^{2}) ∈ ℤ [\frac{1}{q}] for all j ∈ \{0, 2, 4, …, A − 1\}.

Proof. As lim_{q→+∞} d_{n}(1/q^{2}) = d_{n}(0) = ±1 and lim_{q→+∞} ϕ_{t}(1/q) = ϕ_{t}(0) = ±1, the previous lemma implies that for all (s, j) ∈ {1, …, A} × {0, …, n}, the limit of the function

q ↦ q^{αn^{2}+β′n+γ′}φ_{n}(1/q)^{2r} d_{n}(1/q^{2})^{A−s}c_{s,j,n}(q^{2})

exists and is finite as q → +∞. By using Lemma 4.3 we then obtain that for α = −A/4 − r^{2} there exist real numbers β′ and γ′ depending only on A and r, such that for all (s, j) ∈ {1, …, A} × {0, …, n}:

q^{⌊αn^{2}+β′n+γ′⌋}φ_{n}(1/q)^{2r} d_{n} (1/q^{2})^{A−s} c_{s,j,n}(q^{2}) ∈ ℤ [\frac{1}{q}] .  (4.18)

From the expressions (3.9), (3.16) and (3.18), we deduce (4.15) (some easy computations show that the values β = β′ − A + 1 and γ = γ′ + A − 2 are convenient).

Besides, the definition (4.3) shows that Δ_{n}(x) is nothing else but the lowest common multiple (lcm) of 1 − x, 1 − x^{3},…,1 − x^{2n−1}. Recall that d_{n}(x) is the lcm of 1 − x, 1 − x^{2},…,1 − x^{n}. From the definition of lcm and equation (4.1), we deduce first

d_{n}(x^{2}) = ∏_{t=1}^{n} ϕ_{t}(x) × ∏_{{t=n+1 \atop t even}}^{2n} ϕ_{t}(x),

and then that d_{2n}(x)^{A−1}Δ_{n}(x) is the lcm of the polynomials Δ_{n}(x)^{s}d_{n}(x^{2})^{A−s} when s runs along {1, …, A}. This yields (4.16), with the help of (4.18), expressions (3.15), (3.17) and (3.18), and equation (3.13).

5 Asymptotic estimates

We now evaluate the asymptotics for S_{n}(q), the coefficients P̂_{j,n}(q) from (3.14), and finally D_{n}(q). Throughout this section, we fix an odd integer A and r ∈ ℕ^{∗} such that A − 2r > 0.

5.1 Asymptotic evaluation of S_{n}(q)

Lemma 5.1. For all 0 < |q| < 1, we have:

lim_{n→+∞} \frac{1}{n^{2}} log |S_{n}(q)| = −\frac{1}{2}r(A − 2r) log |1/q|.

Proof. Set ρ_{k}(q) := q^{k}(1−q^{2k+n−1})R_{n}(q^{k}; q), so that S_{n}(q) = ∑_{k≥1} ρ_{k}(q). By the definition (3.8) of R_{n}(T; q), it is clear that ρ_{k}(q) = 0 ⇔ k ∈ {0, …, rn}. Moreover we have for k ≥ rn + 1:

\frac{ρ_{k+1}(q)}{ρ_{k}(q)} = q^{(A−2r)n/2+A/2−1} \frac{1 − q^{2k+n+1}}{1 − q^{2k+n−1}} \frac{1 − q^{k}}{1 − q^{k−rn}}

× \frac{1 − q^{k+n+rn}}{1 − q^{k+n}} (\frac{1 − q^{k−1/2}}{1 − q^{k+n−1/2}} )^{A+1} .

Then, as A − 2r > 0, 0 < |q| < 1 and k ≥ rn + 1, we have for a sufficiently large n the following upper bound, uniformly in k:

|\frac{ρ_{k+1}(q)}{ρ_{k}(q)}| ≤ |q|^{(A−2r)n/2} (\frac{1 + |q|}{1 − |q|})^{A+3} < \frac{1}{3},

which yields, as in [6] and [9], the following inequalities:

\frac{1}{2}|ρ_{rn+1}(q)| ≤ |S_{n}(q)| ≤ \frac{3}{2}|ρ_{rn+1}(q)|.

Besides,

ρ_{rn+1}(q) = (−1)^{rn+2}q^{(rn+1/2)((A−2r)n/2+A/2−1)}

× (1 − q^{2rn+n+1})(q)_{n}^{A−2r} \frac{(q, q^{(r+1)n+1})_{rn}}{(q^{rn+1/2})_{n+1}^{A}} ,

therefore we get:

lim_{n→+∞} \frac{1}{n^{2}} log |S_{n}(q)| = lim_{n→+∞} \frac{1}{n^{2}} log |ρ_{rn+1}(q)| = −\frac{1}{2}r(A − 2r) log |1/q|.

5.2 Asymptotic evaluation of the coefficients P̂_{j,n}(q) of (3.14)

Lemma 5.2. For all j ∈ {0, 2, 4, …, A − 1} and 0 < |q| < 1, we have:

lim sup_{{n→+∞ \atop n odd}} \frac{1}{n^{2}} log |P̂_{j,n}(q)| ≤ \frac{1}{8}(A + 4r^{2}) log |1/q|.

Proof. We assume from now on that n is an odd positive integer. First note that for any complex numbers a_{i,n} (0 ≤ i ≤ n):

(∀i ∈ \{0, …, n\}, lim sup_{n→+∞} \frac{1}{n^{2}} log |a_{i,n}| ≤ c) ⇒ lim sup_{n→+∞} \frac{1}{n^{2}} log |∑_{i=0}^{n} a_{i,n}| ≤ c.

This shows, via the definition of the coefficients P̂_{j,n}(q) given by (3.15)-(3.18), that it is enough to prove the inequality of the lemma for the coefficients d_{s,j,n}(q) = (−1)^{s}q^{(j−1/2)s}c_{s,j,n}(q), uniformly in j and s. We fix the integer j ∈ {0, …, n} and η = (1 − |q|)/2 > 0. Cauchy’s formula applied to (3.10) gives:

d_{s,j,n}(q) = −\frac{1}{2iπ} ∫_{C} R_{n}(T q^{1/2−j}; q)(1 − T)^{s−1}dT,

where C is the circle of center 1 and radius η. Back to the expression (3.8), we get:

R_{n}(T q^{1/2−j}; q)(1 − T)^{s−1} = q^{−j((A−2r)n/2+A/2−2)−An−1/2}T^{(A−2r)n/2+A/2−2}

× (q)_{n}^{A−2r}(1 − T)^{s−1} \frac{(q^{−rn−j+1/2}T, q^{n−j+1/2}T)_{rn}}{(T q^{−j})_{n+1}^{A}} ·

After some elementary manipulations, we can deduce:

R_{n}(T q^{1/2−j}; q)(1−T)^{s−1} = q^{Aj^{2}/2−Anj/2+2j−(rn)^{2}/2−An−1/2}T^{A(n−2j)/2+A/2−2}

× (−1)^{Aj+rn}(q)_{n}^{A−2r}(1 − T)^{s−A−1} \frac{(q^{j+1/2}/T, q^{n−j+1/2}T)_{rn}}{(q/T)_{j}^{A}(qT)_{n−j}^{A}} ·

In order to find an upper bound to this expression for T ∈ C, we use the following inequalities from [9], valid for (a, b) ∈ ℕ^{∗} × ℕ, T ∈ C and η = (1 − |q|)/2:

0 < (|q|(1 + η); |q|)_{∞} ≤ |(q^{a}T)_{b}| ≤ (−(1 + η); |q|)_{∞},
0 < (|q|/(1 − η); |q|)_{∞} ≤ |(q^{a}/T)_{b}| ≤ (−1/(1 − η); |q|)_{∞},
|T^{A(n−2j)/2+A/2−2}| ≤ (max(1 + η; 1/(1 − η))^{An/2}(1 + η)^{A/2−2},
|(q)_{n}| ≤ (−|q|; |q|)_{∞} and |1 − T |^{s−A−1} ≤ 1/η^{A+1}.

It remains to find a lower bound for the power of q in the previous expression of R_{n}(T q^{1/2−j}; q)(1 − T)^{s−1}. This can be done by noting that the function j ↦ Aj^{2}/2 − Anj/2 + 2j − (rn)^{2}/2 − An − 1/2 is minimal at j = n/2 − 2/A, and this minimal value is equal to −An^{2}/8 − r^{2}n^{2}/2 + λn + μ, where λ and μ are real numbers depending only on A and r. All this yields to the following:

|d_{s,j,n}(q)| ≤ c_{0} × |q|^{−(A+4r^{2})n^{2}/8},

where c_{0} depends neither on j nor on s, and satisfies lim_{n→+∞} c_{0}^{1/n^{2}} = 1, and we can conclude.

5.3 Asymptotic evaluation of D_{n}(q) defined by (4.17)

We first recall Möbius function μ : ℕ^{∗} → {−1, 0, 1}, which is defined for any positive integer n by

μ(n) := \{ \begin{matrix}1 if n = 1, \\ (−1)^{s} if n is a product of s distinct primes, \\ 0 otherwise. \end{matrix}

We prove a preliminary result:

Lemma 5.3. For any positive integer n, we have:

∑_{{1≤d≤n \atop d odd}} \frac{μ(d)}{d^{2}} = \frac{8}{π^{2}} + O(1/n),  (5.1)
∑_{{1≤d≤n \atop d even}} \frac{μ(d)}{d^{2}} = −\frac{2}{π^{2}} + O(1/n).  (5.2)

Proof. Recall

\frac{6}{π^{2}} = ∑_{d≥1} \frac{μ(d)}{d^{2}} = ∑_{d≥1} \frac{μ(2d)}{4d^{2}} + ∑_{{d≥1 \atop d odd}} \frac{μ(d)}{d^{2}}.

Besides μ(2d) = −μ(d) if d is odd and μ(2d) = 0 if d is even, so we get:

∑_{{d≥1 \atop d odd}} \frac{μ(d)}{d^{2}} = \frac{8}{π^{2}} ,  (5.3)

and this implies

∑_{{d≥1 \atop d even}} \frac{μ(d)}{d^{2}} = −\frac{2}{π^{2}} ·  (5.4)

Then we immediately deduce (5.1) (resp. (5.2)) from (5.3) (resp. (5.4)).

Now we can prove the following lemma: Lemma 5.4. For all 0 < |q| < 1 we have:

lim_{n→+∞} \frac{1}{n^{2}} log |Δ_{n}(1/q)| = \frac{8}{π^{2}} log |1/q|,  (5.5)
lim_{n→+∞} \frac{1}{n^{2}} log |φ_{n}(1/q)| = \frac{2}{3} log |1/q|.  (5.6)

Proof. Recall Möbius inversion formula: if f is an arithmetic function defined on ℕ^{∗}, and if g(n) := ∑_{d|n} f(d), then f(n) = ∑_{d|n} μ(n/d)g(d) for all n ∈ ℕ^{∗}. Applying this formula to (4.1) yields for k ∈ ℕ^{∗}:

log |ϕ_{k}(x)| = ∑_{d|k} μ(d) log |x^{k/d} − 1|.  (5.7)

First recall the definition (4.3):

Δ_{n}(x) = ∏_{{k=1 \atop k odd}}^{2n−1} ϕ_{k}(x).

Thus, by using (5.7), we can write for |x| > 1:

log |Δ_{n}(x)| = ∑_{{k=1 \atop k odd}}^{2n} ∑_{d|k} μ(d) log |x^{k/d} − 1| = ∑_{{1≤d≤2n \atop d odd}} μ(d) ∑_{{1≤ℓ≤2n/d \atop ℓ odd}} log |x^{ℓ} − 1|
= ∑_{{1≤d≤2n \atop d odd}} μ(d) ∑_{{1≤ℓ≤2n/d \atop ℓ odd}}(ℓ log |x| + log |1 − x^{−ℓ}|)
= log |x| ∑_{{1≤d≤2n \atop d odd}} μ(d) ∑_{{1≤ℓ≤2n/d \atop ℓ odd}} ℓ + O (n ∑_{ℓ=1}^{2n} \frac{1}{ℓ}) .

Since ∑_{{1≤ℓ≤2n/d \atop ℓ odd}} ℓ = \frac{n^{2}}{d^{2}} + O(1) and O (n ∑_{ℓ=1}^{2n} \frac{1}{ℓ}) = O(n log n), we get:

log |Δ_{n}(x)| = n^{2} log |x| ∑_{{1≤d≤2n \atop d odd}} \frac{μ(d)}{d^{2}} + O(n log n),

which yields (5.5), by setting x = 1/q and using (5.1). To prove (5.6), recall the definition (4.4):

φ_{n}(x) = ϕ_{2}(x)^{n}ϕ_{4}(x)^{⌊n/2⌋} … ϕ_{2n}(x).

We use again (5.7), assuming |x| > 1:

log |φ_{n}(x)| = ∑_{k=1}^{n} ⌊\frac{n}{k}⌋ ∑_{d|2k} μ(d) log |x^{2k/d} − 1|
= ∑_{{d,ℓ≥1 \atop dℓ even≤2n}} μ(d) ⌊\frac{2n}{dℓ}⌋ log |x^{ℓ} − 1|
= ∑_{{1≤d≤2n \atop d even}} μ(d) ∑_{1≤ℓ≤2n/d} ⌊\frac{2n}{dℓ}⌋ log |x^{ℓ} − 1|

+ ∑_{{1≤d≤2n \atop d odd}} μ(d) ∑_{{1≤ℓ≤2n/d \atop ℓ even}} ⌊\frac{2n}{dℓ}⌋ log |x^{ℓ} − 1|.

Hence, by setting u_{n}(x) := ∑_{ℓ=1}^{n} ⌊\frac{n}{ℓ}⌋ log |x^{ℓ} − 1|, we have:

log |φ_{n}(x)| = ∑_{{1≤d≤2n \atop d even}} μ(d)u_{2n/d}(x) + ∑_{{1≤d≤2n \atop d odd}} μ(d)u_{n/d}(x^{2}).  (5.8)

Assuming |x| > 1 we can write:

u_{n}(x) = log |x| ∑_{ℓ=1}^{n} ℓ ⌊\frac{n}{ℓ}⌋ + ∑_{ℓ=1}^{n} ⌊\frac{n}{ℓ}⌋ log |1 − x^{−ℓ}|
= log |x| ∑_{ℓ=1}^{n} ℓ ∑_{1≤k≤n/ℓ} 1 + ∑_{{ℓ,k≥1 \atop ℓk≤n}} log |1 − x^{−ℓ}|
= \frac{log |x|}{2} ∑_{k=1}^{n} ⌊\frac{n}{k}⌋ (⌊\frac{n}{k}⌋ + 1) + O ( ∑_{{ℓ,k≥1 \atop ℓk≤n}} 1)
= \frac{log |x|}{2} ∑_{k=1}^{n} \frac{n^{2}}{k^{2}} + O (n log n)
= n^{2} log |x| × \frac{π^{2}}{12} + O (n log n) .

Therefore, by using (5.8), (5.2) and (5.1), we deduce:

log |φ_{n}(x)| = ∑_{{1≤d≤2n \atop d even}} μ(d) (4\frac{n^{2}}{d^{2}} log |x| × \frac{π^{2}}{12} + O (\frac{2n}{d} log \frac{2n}{d}))

+ ∑_{{1≤d≤2n \atop d odd}} μ(d) (2\frac{n^{2}}{d^{2}} log |x| × \frac{π^{2}}{12} + O (\frac{n}{d} log \frac{n}{d}))

= n^{2} log |x| × \frac{π^{2}}{3} × \frac{−2}{π^{2}} + n^{2} log |x| × \frac{π^{2}}{6} × \frac{8}{π^{2}} + O (n log^{2} n)
= \frac{2}{3}n^{2} log |x| + O (n log^{2} n) ,

which, by setting x = 1/q, shows (5.6) as desired.

Now we are able to prove the following result: Lemma 5.5. For all 0 < |q| < 1 we have:

lim_{n→+∞} \frac{1}{n^{2}} log |D_{n}(q)| = (\frac{A}{4} + r^{2} + \frac{12}{π^{2}}(A − 1) + \frac{4r}{3} + \frac{8}{π^{2}} ) log |1/q|.

Proof. For 0 < |q| < 1 recall the estimate (see [2] and [15]):

lim_{n→+∞} \frac{1}{n^{2}} log |d_{n}(1/q)| = \frac{3}{π^{2}} log |1/q|.  (5.9)

Thus, with the help of the expressions of D_{n}(q) and α = −A/4 − r^{2} given by (4.17), and by using the previous lemma, we can easily conclude.

6 Concluding remarks

6.1 Link with Dirichlet’s beta function

As mentioned in the introduction, we now justify the term q-analogues of the values of Dirichlet’s beta function at positive integers. To this aim, recall the Stirling numbers of the second kind (see [14]), which are integers denoted by S(s, j) (where s and j are integers such that 1 ≤ j ≤ s) and defined by:

x^{s} = ∑_{j=1}^{s} S(s, j)x(x − 1) … (x − j + 1).

By expanding the summand in the first expression of (1.1), we can write for all s ≥ 2 and |q| < 1:

β_{q}(s) = ∑_{k≥1} k^{s−1} ∑_{m≥0}(−1)^{m}q^{(2m+1)k}
= ∑_{k,m≥0}(k + 1)^{s−1}(−1)^{m}q^{(2m+1)(k+1)}

= ∑_{k,m≥0} ∑_{j=1}^{s−1} q^{(2m+1)(k+1)}(−1)^{m+s−1−j}S(s − 1, j)j!\binom{k + j}{j}

= ∑_{j=1}^{s−1}(−1)^{s−1−j}S(s − 1, j)j! ∑_{m≥0}(−1)^{m} \frac{q^{2m+1}}{(1 − q^{2m+1})^{j+1}}

= ∑_{j=1}^{s−1}(−1)^{s−1−j}S(s − 1, j)j! Y_{j+1}(q).  (6.1)

Note in passing that (6.1) is the inverse expansion of (3.4). Notice that our auxiliary functions Y_{s}(q) are clearly q-analogues of the function β at positive integers, since for s ≥ 1:

lim_{q→1}(1 − q)^{s}Y_{s}(q) = β(s).

This shows, by using (6.1) and (1.2), that for all s ≥ 1:

lim_{q→1}(1 − q)^{s}β_{q}(s) = (s − 1)!β(s).  (6.2)

6.2 Special emphazis on β_{q}(1) and β_{q}(2)

For s = 1, we have in fact:

β_{q}(1) = Y_{1}(q),

which, as mentioned in the introduction, is, up to constants, equal to π_{q} whose irrationality exponent was studied in [3, 4]. Now we inspect more carefully the link with Catalan’s constant, which is defined by G := ∑_{k≥0}(−1)^{k}/(2k + 1)^{2} = β(2). The q-analogue of G proposed at the end of [4] corresponds to Y_{2}(q). Although this is not really obvious from the definition of β_{q}(2), we have in fact via (6.1) the following identity, which can also be deduced from (3.4):

β_{q}(2) = Y_{2}(q) = ∑_{k≥0}(−1)^{k} \frac{q^{2k+1}}{(1 − q^{2k+1})^{2}} ·

There are many similarities between the diophantine behaviour of the values of Riemann’s zeta function at even positive integers and the values of Dirichlet’s beta function at odd positive integers. However, no analogy to Apery’s famous result [1] ζ(3) ∉ ℚ has been found for G yet. Indeed, the linear forms built in [13] do not show that G is irrational. Moreover, even the denominators conjecture formulated in [13] do not give the arithmetic nature of G. We point out that this denominators conjecture was proved in [12] through Padé approximants, and then in a simpler way by using transformation formulae for hypergeometric series in [8]. Now recall the linear combination for G studied in [8]:

n! ∑_{k≥1}(−1)^{k} (k + \frac{n − 1}{2}) \frac{(k − n)_{n}(k + n)_{n}}{(k − 1/2)_{n+1}^{3}} = a_{n}G − b_{n},  (6.3)

where the coefficients a_{n} and b_{n} are explicitely given in [8], and we recall (x)_{n} := x(x + 1) … (x + n − 1). We want to point out that our linear combination (3.14) gives a q-analogue of (6.3). Indeed, for any odd positive integer n, A = 3 and r = 1, (3.14) is:

(q)_{n} ∑_{k≥1}(−1)^{k+1}q^{(2k−1)(n+1)/4}(1 − q^{k+(n−1)/2})\frac{(q^{k−n}, q^{k+n})_{n}}{(q^{k−1/2})_{n+1}^{3}}

= A_{n}(q)β\sqrt{q}(2) + B_{n}(q), (6.4)

with

A_{n}(q) := ∑_{j=0}^{n}(−1)^{j}q^{1/2−j}(2d_{2,j,n}(q) + d_{3,j,n}(q))

and

B_{n}(q) := ∑_{s=1}^{3} ∑_{j=1}^{n} ∑_{k=1}^{j}(−1)^{j+k} (\frac{q^{k−j}d_{s,j,n}(q)}{(1 − q^{k−1/2})^{s}} + \frac{q^{−k+j}d_{s,j,n}(1/q)}{(1 − q^{−k+1/2})^{s}})

− \frac{1}{2} ∑_{j=0}^{n}(−1)^{j}q^{1/2−j}d_{1,j,n}(q).

Multiplying (6.4) by (1 − q^{1/2})^{2}, then letting q tend to 1, we get by using (6.2):

−\frac{1}{2} n! ∑_{k≥1}(−1)^{k} (k + \frac{n − 1}{2}) \frac{(k − n)_{n}(k + n)_{n}}{(k − 1/2)_{n+1}^{3}} = α_{n}G + β_{n},  (6.5)

where α_{n} := lim_{q→1} A_{n}(q) and β_{n} := lim_{q→1}(1 − q^{1/2})^{2}B_{n}(q). If the hypergeometric series on the left-hand side of (6.5) is multiplied by −2, we obtain the left-hand side of (6.3). Moreover, by using the definition (3.8) of R_{n}, the identity (3.9), and (3.11), a direct calculation gives:

α_{n} = −2 ∑_{j=0}^{n}(n − 2j)\binom{n}{j}^{3}\binom{n + j − \frac{1}{2}}{n}\binom{2n − j − \frac{1}{2}}{n}

× (\frac{1}{n − 2j} + 3H_{j} + H_{j− \frac{1}{2}} − H_{n+j− \frac{1}{2}} ) ,

where for any positive integer m, H_{m} := ∑_{j=1}^{m} \frac{1}{j} is the m-th harmonic num-

ber, whose definition is extended to half-integers by H_{m} := ∑_{j=1}^{⌊m⌋+1} \frac{1}{m−j+1}· This shows that for any odd positive integer n, α_{n} = −a_{n}/2 (see [8]). Thus our linear combination (6.4) is indeed a q-analogue of the one in [8], as we necessarily have β_{n} = b_{n}/2.

6.3 A q-denominators conjecture

As in [6], it seems possible to refine Lemma 4.5, by considering another common denominator to the P̂_{j,n}(q^{2}), having the form D̃_{n}(q) = D_{n}(q)/Δ_{n}(1/q). We formulate in this direction the following q-denominators conjecture:

Conjecture 6.1. Let n and A be odd positive integers, and set r ∈ ℕ^{∗} such that A − 2r > 0. For α = −A/4 − r^{2}, there exist real numbers β and γ only depending on A and r such that if we set:

D̃_{n}(q) := (A − 1)! q^{⌊αn^{2}+βn+γ⌋}φ_{n}(1/q)^{2r} d_{2n}(1/q)^{A−1},

then we get:

D̃_{n}(q)P̂_{j,n}(q^{2}) ∈ ℤ [\frac{1}{q}] ∀j ∈ \{0, 2, 4, …, A − 1\}.

In our opinion, Conjecture 6.1 should be solved by using transformation formulae for basic hypergeometric series, together with arithmetical techniques similar to those used in [6]. Proving this conjecture would imply for 1/q ∈ ℤ \ {−1; 1} and any odd integer A ≥ 3:

dim_{ℚ} (ℚ + ℚβ_{q}(2) + ⋯ + ℚβ_{q}(A − 1)) ≥ g(A),

where

g(A) = max_{{r∈ℕ \atop 1≤r<A/2}} g(r; A) and g(r; A) := \frac{4rA + A − 4r^{2}}{(\frac{48}{π^{2}} + 2) A + 8r^{2} − \frac{48}{π^{2}} + \frac{16r}{3}} ·

Unfortunately, and unlike the case of ζ_{q} in [6], solving this conjecture would not give a refinement of the quantitative version in Corollary 1.3. Indeed, although g(A) > f(A) for all A, we have 1 > g(19) ≃ 0.988 > f(19) ≃ 0.973 and g(21) ≃ 1.042 > f(21) ≃ 1.028 > 1.

References

[1] Apéry (R.), Irrationalité de ζ(2) et ζ(3), Astérisque 61 (1979), 11-13.

[2] Bundschuh (P.) and Väänänen (K.), Arithmetical investigations of a certain infinite product, Compositio Math. 91.2 (1994), 175-199.

[3] Bundschuh (P.) and Zudilin (W.), Rational approximation to a q-analogue of π and some other q-series, proceedings of the 70th birthday conference in honour of W. M. Schmidt(Vienna, November 2003), Vienna, Springer-Verlag (2007), 17 pages.

[4] Bundschuh (P.) and Zudilin (W.), Irrationality measures for certain qmathematical constants, Math. Scand. 101.1 (2007), 104–122.

[5] Gasper (G.) and Rahman (M.), Basic Hypergeometric Series, 2nd Edition, Encyclopedia of mathematics and its applications, Vol. 96, Cambridge University Press, Cambridge, 2004.

[6] Jouhet (F.) and Mosaki (E.), Irrationalité aux entiers impairs positifs d’un q-analogue de la fonction zêta de Riemann, Int. J. Number Theory, to appear.

[7] Koblitz (N.), Introduction to elliptic curves and modular forms, Graduate Texts in Math. No. 97, Springer-Verlag, 1984. Second edition, 1993.

[8] Krattenthaler (C.) and Rivoal (T.), On a linear form for Catalan’s constant, South East Asian J. Math. Math. Sci. 6.2 (2008), 3-15.

[9] Krattenthaler (C.), Rivoal (T.) and Zudilin (W.), Séries hypergéométriques basiques, q-analogues des valeurs de la fonction zêta et séries d’Eisenstein, J. Inst. Jussieu 5.1 (2006), 53-79.

[10] Nesterenko (Yu. V.), On the linear independance of numbers, (in russian) Vest. Mosk. Univ., Ser. I, no. 1 (1985), 46-54; english transl. in Mosc. Univ. Math. Bull. 40.1 (1985), 69-74.

[11] Nesterenko (Yu. V.), Modular functions and transcendance questions, (in russian) Math. Sb. 187.9 (1996), 65-96; english transl. in Sb. Math. 187.9 (1996), 1319-1348.

[12] Rivoal (T.), Nombres d’Euler, approximants de Padé et constante de Catalan, Ramanujan J. 11 (2006), 199-214.

[13] Rivoal (T.) and Zudilin (W.), Diophantine properties of numbers related to Catalan’s constant, Math. Annalen 326.4 (2003), 705-721.

[14] Stanley (R. P.), Enumerative Combinatorics, Vol. 1, Cambridge University Press, Cambridge, 1998.

[15] Van Assche (W.), Little q-Legendre polynomials and irrationality of certain Lambert series, Ramanujan J. 5.3 (2001), 295-310.

Frédéric Jouhet, Université de Lyon, Université Lyon I, CNRS, UMR 5208 Institut Camille Jordan, Bâtiment du Doyen Jean Braconnier, 43, bd du 11 Novembre 1918, 69622 Villeurbanne Cedex, France

jouhet@math.univ-lyon1.fr

Elie Mosaki, Université de Lyon, Université Lyon I, CNRS, UMR 5208 Institut Camille Jordan, Bâtiment du Doyen Jean Braconnier, 43, bd du 11 Novembre 1918, 69622 Villeurbanne Cedex, France

mosaki@math.univ-lyon1.fr