Q-APÉRY IRRATIONALITY PROOFS BY Q-WZ PAIRS

Tewodros Amdeberhan and Doron Zeilberger

Department of Mathematics, Temple University, Philadelphia PA 19122, USA tewodros@math.temple.edu, zeilberg@math.temple.edu

Abstract. Using WZ pairs, Apéry-style proofs of the irrationality of the q-analogues of the Harmonic series and Ln(2) are given.

0. Introduction:

Let us define the following q-analogues of the Harmonic series ∑_{n=1}^{∞} \frac{1}{n} and Ln(2), respectively by:

(0.1)  h_{q}(1) : = ∑_{k=1}^{∞} \frac{1}{q^{k} − 1}  (for |q| > 1),

and

(0.2)  Ln_{q}(2) : = ∑_{n=1}^{∞} \frac{(−1)^{n}}{q^{n} − 1}  (for |q| ≠ 0, 1).

In 1948, Paul Erdös [E1] proved the irrationality of h_{2}(1). Recently, Peter Borwein used Pade approximation techniques [B1] and some complex analysis methods [B2] to prove the irrationality of both h_{q}(1) and Ln_{q}(2). Here we present a proof in the spirit of Apéry’s magnificent work [A], which was delightfully accounted by Alf van der Poorten [P]. This method of proof gives favorable irrationality measure for Ln_{q}(2) campared with those implied in [B1], [B2]. Further discussion of irrationality results for certain series is to be found in Erdös [E2].

We will assume familiarity with ref. [Z]. In particular, we use \binom{n}{k}_{q} = \frac{(q)_{n}}{(q)_{k}(q)_{n−k}}, where (q)_{n} = (1 − q) ⋯ (1 − q^{n}). N and K are forward shift operators on n and k, respectively. Δ_{n} = N − 1, Δ_{k} = K − 1.

Typeset by AMS-TEX

1. A scheme for proving the irrationality of the q-harmonic series h_{q}(1): The claims made in subsections 1.1-1.5 below were found using the Maple Package qEKHAD accom-

panying [PWZ]. The relevant script substantiating our claims can be found in this paper’s Web Pages.

1.1. The q-WZ 1-form ω is:

ω = \frac{−1}{\binom{n+k+1}{k}_{q}(q)_{n+1}} \{δk + \frac{q^{n+1}}{(q^{n+1} − 1)}δn\}.

1.2. The choice of the potential c(n, k) is:

c(n, k) = ∑_{m=1}^{n} \frac{q^{m}}{(1 − q^{m})(q)_{m}} + ∑_{m=1}^{k} \frac{1}{(q^{m} − 1)} \frac{1}{\binom{n+m}{m}_{q}(q)_{n}} .

1.3. The choice of the mollifier b(n, k) is:

b(n, k) = (−1)^{k}q^{k(k+1)/2}\binom{n + k}{k}_{q}\binom{n}{k}_{q}.

1.4. We define two sequences:

a(n) = ∑_{k=0}^{n} c(n, k)b(n, k),  and  b(n) = ∑_{k=0}^{n} b(n, k).

1.5. Introduce L = y_{2}(n)N^{2} + y_{1}(n)N + y_{0}(n) and B(n, k) = P_{q}^{1}(n, k)b(n + 1, k), where

A(n, k) = c(n, k)B(n, k) + \frac{(−1)^{k}q^{2n+3}}{q^{n+1} − 1} \binom{n + 1}{k}_{q} \frac{q^{\binom{k}{2}}}{(q)_{n+2}} P_{q}^{2}(n, k) and

P_{q}^{1}(n, k) = −qα_{n}^{2}β_{k}^{−1}(q^{2}α_{n} + 2q) + qα_{n}^{2}(q^{2}α_{n}^{3} + 2q(q + 1)α_{n}^{2} + 3qα_{n} − (q + 1) − (α_{n} + 2)β_{k})

P_{q}^{2}(n, k) = q^{2}α_{n}^{2} + qα_{n} − 2 + β_{k}(q^{2}α_{n}^{5} + q(2q + 1)α_{n}^{4} − 2α_{n}^{3} − α_{n}^{3}β_{k} − (2 − q^{−1})α_{n}^{2}β_{k}

− (3q + 5)α_{n}^{2} + 2q^{−1}α_{n}β_{k} + (q − 1 + 2q^{−1})α_{n} + (1 + 3q^{−1})),

y_{0}(n) = q(α_{n} − 1)(qα_{n} + 2), y_{2}(n) = (qα_{n} − 1)(α_{n} + 2), α_{n} = q^{n+1}, β_{k} = q^{k+1} and y_{1}(n) = q^{3}α_{n}^{5} + 2q^{2}(q + 1)α_{n}^{4} + q^{2}α_{n}^{3} − 4q(q + 1)α_{n}^{2} + (q^{2} − 4q + 1)α_{n} + 2(q + 1). Then

(∗) L(b(n, k)) = B(n, k) − B(n, k − 1) and L(b(n, k)c(n, k))= A(n, k) − A(n, k − 1).

Now, summing over k in (∗) shows that both sequences a(n) and b(n) are solutions of Lu(n) = 0.

1.6. Set b_{n} = b(n) and a_{n} = a(n). Now since b_{n+1} > b_{n} and Lb_{n} = 0, i.e. y_{2}(n)b_{n+2} + y_{1}(n)b_{n+1} + y_{0}(n)b_{n} = 0, then asymptotically we have that

\frac{b_{n+2}}{b_{n+1}} = O (\frac{y_{1}(n)}{y_{2}(n)}) = O (q^{3n+3}) .

Hence,

(1.6.1)  b_{n} = O (q^{\frac{3n^{2}}{2}} ) .

On the other hand, La_{n} = 0 and Lb_{n} = 0 lead to the system of recurrence equations,

(1.6.2) y_{2}(n)a_{n+2} + y_{1}(n)a_{n+1} + y_{0}(n)a_{n} = 0,  y_{2}(n)b_{n+2} + y_{1}(n)b_{n+1} + y_{0}(n)b_{n} = 0.

Multiplying the first and the second equations in (1.6.2), respectively by b_{n+2} and a_{n+2}, and subtracting we obtain

y_{1}(n)(a_{n+1}b_{n+2} − b_{n+1}a_{n+2}) = y_{0}(n)(a_{n+2}b_{n} − b_{n+2}a_{n}).

Rewriting this

\frac{a_{n+1}}{b_{n+1}} − \frac{a_{n+2}}{b_{n+2}} = \frac{y_{0}(n)}{y_{1}(n)} \frac{b_{n}}{b_{n+1}} (\frac{a_{n+2}}{b_{n+2}} − \frac{a_{n}}{b_{n}} ) .

Then we have the estimate

|\frac{a_{n+1}}{b_{n+1}} − \frac{a_{n+2}}{b_{n+2}} | ≤ |\frac{y_{0}(n)}{y_{1}(n)} \frac{b_{n}}{b_{n+1}} (\frac{a_{n+2}}{b_{n+2}} − \frac{a_{n+1}}{b_{n+1}} )| + |\frac{y_{0}(n)}{y_{1}(n)} \frac{b_{n}}{b_{n+1}} (\frac{a_{n+1}}{b_{n+1}} − \frac{a_{n}}{b_{n}} )| .

This yields in

(1.6.3)  \frac{a_{n+1}}{b_{n+1}} − \frac{a_{n}}{b_{n}} = O (b_{n}^{−2}) .

Therefore,

(1.6.4)  h_{q}(1) − \frac{a_{n}}{b_{n}} = O (b_{n}^{−2}) .

In particular, the sequence of rational numbers \frac{a_{n}}{b_{n}} converges moderately faster to h_{q}(1).

1.7. For a given prime p, let ord_{p}k denote the exponent of p in the prime expansion of k. Then we observe that

(1.7.1)  ord_{p}\binom{n}{m}_{q} ≤ ord_{p}(q)_{n} − ord_{p}(q)_{m}.

Note:

(1.7.2)  \binom{n + k}{k}_{q}\binom{k}{m}_{q} = \binom{n + m}{m}_{q}\binom{n + k}{k − m}_{q}.

Claim: The sequences

u_{n} = a_{n}(q)_{n+1} ∏_{s=[n/2]}^{n}(1 − q^{s}) and z_{n} = b_{n}(q)_{n+1} ∏_{s=[n/2]}^{n}(1 − q^{s})

are polynomials in q with integer coefficients, and moreover

(1.7.3)  z_{n} = O (q^{19n^{2}/8}) .

Proof: Applying (1.7.1) and (1.7.2), we estimate the denominator of u_{n}:

ord_{p} (\frac{(q^{m} − 1)(q)_{n}\binom{n+m}{m}_{q}}{\binom{n+k}{k}q} ) ≤ ord_{p} (\frac{(q^{m} − 1)(q)_{n}\binom{k}{m}_{q}}{\binom{n+k}{k−m}q} )

≤ ord_{p}(q)_{n} + ord_{p}(q^{m} − 1) + ord_{p}(q)_{k} − ord_{p}(q)_{m}

≤ ord_{p}(q)_{n} + ord_{p} ∏_{s=[n/2]}^{n}(1 − q^{s}) + ord_{p}(q)_{k} − ord_{p}(q)_{m}

≤ ord_{p} ((q)_{n} ∏_{s=[n/2]}^{n}(1 − q^{s})) ,

since m ≤ k ≤ n. This proves the claim on u_{n}. And (1.7.3) follows from (1.6.1). The rest is trivial. Claim: h_{q}(1) - \frac{u_{n}}{z_{n}} = O (\frac{1}{z_{n}^{1+δ}} ); where δ = 0.26316 … > 0. proof: From (1.6.1), (1.6.4) and (1.7.3), we gather that

h_{q}(1) − \frac{u_{n}}{z_{n}} = O (b_{n}^{−2}) = O (q^{−3n^{2}}) = O (z_{n}^{−1−(5/19)}) .

Thus, we have proved: Theorem 1: If |q| > 1 is an integer, h_{q}(1) is irrational with irrationality measure 4.80. Note: By invoking Theorem 7 ([Z], p.596) with ω as in 1.1, we obtain the series acceleration:

h_{q}(1) = ∑_{n=1}^{∞} \frac{q^{n}}{(1 − q^{n})(q)_{n}}  and  h_{q}(1) = ∑_{n=1}^{∞} \frac{1 − q^{n} − q^{2n}}{(q^{n} − 1)\binom{2n}{n}_{q}(q)_{n}} .

2. A scheme for proving the irrationality of Ln_{q}(2): The claims made in subsections 2.1-2.5 below were found using the Maple Package qEKHAD accom-

panying [PWZ]. The relevant script substantiating our claims can be found in this paper’s Web Pages.

2.1. The qWZ 1-form ω is:

ω = \frac{(−1)^{k}}{(1 − q^{k+1})} \frac{(q)_{n}}{\binom{n+k+1}{k+1}_{q}(q^{2})_{n}} \{δk + \frac{q^{n+1}}{(1 + q^{n+1})}δn\}.

2.2. The choice of the potential c(n, k) is:

c(n, k) = ∑_{m=1}^{n} \frac{q^{m}(q)_{m}}{(1 − q^{m})(q^{2})_{m}} + ∑_{m=1}^{k} \frac{(−1)^{m−1}}{(1 − q^{m})} \frac{(q)_{n}}{\binom{n+m}{m}_{q}(q^{2})_{n}} .

2.3. The choice of the mollifier b(n, k) is:

b(n, k) = q^{k(k+1)/2}\binom{n + k}{k}_{q}\binom{n}{k}_{q}.

2.4. We define two sequences:

a(n) = ∑_{k=0}^{n} c(n, k)b(n, k),  and  b(n) = ∑_{k=0}^{n} b(n, k).

2.5. Introduce L = y_{2}(n)N^{2} + y_{1}(n)N + y_{0}(n) and B(n, k) = P_{q}^{1}(n, k)b(n + 1, k), where

A(n, k) = c(n, k)B(n, k) + \frac{(−1)^{k}q^{2n+3}}{1 − q^{n+1}} \binom{n + 1}{k}_{q} \frac{q^{\binom{k}{2}}(q)_{n+1}}{(q^{2})_{n+1}} P_{q}^{2}(n, k) and

P_{q}^{1}(n, k) = qα_{n}^{2}[q^{3}α_{n}^{5} + q^{2}(1 + q)α_{n}^{4} + 2q(1 + q^{2})α_{n}^{3} − (1 − q + q^{2})α_{n} − 3(1 + q)] + qα_{n}^{2}[qβ_{k}^{−1}(q^{2}α_{n}^{3} + q(1 + q)α_{n}^{2} + (2 − q)α_{n} − 2) + (qα_{n}^{3} + (q − 1)α_{n}^{2} + (2q − 1)α_{n})α_{k} − 2]

P_{q}^{2}(n, k) = q^{2}α_{n}^{3} + q(1 + q)α_{n}^{2} + (2 + q)α_{n} + 2 − α_{n}α_{k}^{2}[α_{n}^{3} + (1 − q^{−1})α_{n}^{2} + (2 − q)α_{n} − 2q^{−1}]

− α_{k}[q^{2}α_{n}^{6} + q(1 + q)α_{n}^{5} + (2 + q + 2q^{2})α_{n}^{4} + (1 + q)α_{n}^{3} + 2α_{n}^{2} − (2 + q + q^{−1})α_{n} + (q^{−1} − 1)], y_{0}(n) = −q(α_{n} − 1)(α_{n} + 1)(q^{2}α_{n}^{2} + qα_{n} + 2), y_{2}(n) = −(qα_{n} − 1)(qα_{n} + 1)(α_{n}^{2} + α_{n} + 2),

y_{1}(n) = q^{4}α_{n}^{7} + q^{2}(1 + q)(qα_{n}^{6} + α_{n}^{4}) + q(1 + q + q^{2})(2qα_{n}^{5} + α_{n}^{3}) − (1 + 3q + 3q^{2} + q^{3})α_{n}^{2} − (1 + q^{2})(2 + α_{n}), and α_{n} = q^{n+1}, β_{k} = q^{k+1}.

Then

(∗∗) L(b(n, k)) = B(n, k) − B(n, k − 1) and L(b(n, k)c(n, k))= A(n, k) − A(n, k − 1).

Now, summing over k in (∗∗) shows that both sequences a(n) and b(n) are solutions of Lu(n) = 0.

2.6. Similar arguments and estimates as in (1.6) above lead to

(2.6.1)  Ln_{q}(2) − \frac{a_{n}}{b_{n}} = O (b_{n}^{−2}) .

In particular, the sequence of rational numbers \frac{a_{n}}{b_{n}} converges moderately faster to Ln_{q}(2).

2.7. Claim: The sequences

u_{n} = a_{n} ∏_{t=1}^{n}(1 + q^{t}) ∏_{s=[n/2]}^{n}(1 − q^{s}) and z_{n} = b_{n} ∏_{t=1}^{n}(1 + q^{t}) ∏_{s=[n/2]}^{n}(1 − q^{s})

are polynomials in q with integer coefficients, and moreover

(2.7.1)  z_{n} = O (q^{19n^{2}/8}) .

Proof: Applying (1.7.2) and (2.7.1), we estimate the denominator of u_{n}:

ord_{p} (\frac{(1 − q^{m})(q^{2})_{n}\binom{n+m}{m}_{q}}{\binom{n+k}{k}_{q}(q)_{n}} ) ≤ ord_{p} (\frac{(q^{m} − 1)(q^{2})_{n}\binom{k}{m}_{q}}{\binom{n+k}{k−m}_{q}(q)_{n}} )

≤ ord_{p} (\frac{(q^{2})_{n}}{(q)_{n}} ) + ord_{p}(q^{m} − 1) + ord_{p}(q)_{k} − ord_{p}(q)_{m}

≤ ord_{p} (\frac{(q^{2})_{n}}{(q)_{n}} ) + ord_{p} ∏_{s=[n/2]}^{n}(1 − q^{s}) + ord_{p}(q)_{k} − ord_{p}(q)_{m}

≤ ord_{p} (∏_{t=1}^{n}(1 + q^{t}) ∏_{s=[n/2]}^{n}(1 − q^{s})) ,

since m ≤ k ≤ n. This proves the claim on u_{n}. And (2.7.1) follows from (1.6.1). The rest is trivial. Claim: Ln_{q}(2) - \frac{u_{n}}{z_{n}} = O (\frac{1}{z_{n}^{1+δ}} ); where δ = 0.26316 … > 0. proof: From (1.6.1), (2.6.1) and (2.7.1), we gather that

Ln_{q}(2) − \frac{u_{n}}{z_{n}} = O (b_{n}^{−2}) = O (q^{−3n^{2}}) = O (z_{n}^{−1−(5/19)}) .

Thus, we have proved: Theorem 2: If |q| ≠ 0, 1 is an integer, Ln_{q}(2) is irrational with irrationality measure 4.80. Note: By invoking Theorem 7 ([Z], p. 596) with ω as in 2.1, we obtain the series acceleration:

Ln_{q}(2) = ∑_{n=1}^{∞} \frac{q^{n}(q)_{n}}{(1 − q^{n})(q^{2})_{n}}  and  Ln_{q}(2) = ∑_{n=1}^{∞} \frac{(−1)^{n−1}(q)_{n}(1 − q^{3n})}{(1 − q^{n})^{2}\binom{2n}{n}_{q}(q^{2})_{n}}.

References

[A] R. Apéry, Irrationalitè de ζ(2) et ζ(3), Asterisque 61 (1979), 11-13.

[B1] P. Borwein, On the irrationality of ∑ 1/(q^{n} + r), J. Number Theory 37 (1991), 253-259.

[B2] P. Borwein, On the irrationality of certain series, Proc. Camb. Phil. Soc. 112 (1992), 141-146.

[E1] P. Erdös, On arithmetical properties of Lambert Series, J. Indian Math. Soc. (N.S.) 12 (1948), 63-66.

[E2] P. Erdös, On the irrationality of certain series: problems and results, In New Advances in Transcendence Theory (Cambridge University Press), 1988, pp. 102-109.

[P] A. van der Poorten, A proof that Euler missed ..., Apéry’s proof of the irrationality of ζ(3), Math. Intel. 1 (1979), 195-203.

[PWZ] M. Petkovšek, H.S. Wilf, D. Zeilberger, “A=B”, A.K. Peters Ltd., 1996.

   The package qEKHAD is available by the www at http://www.math.temple.edu/~zeilberg/programs.html

[Z] D. Zeilberger, Closed Form (pun intended!), Contemporary Mathematics 143 (1993), 579-607.