%AMSTeX file of `q-Apery Irrationality Proofs ...' by T. Amdeberhan and DZ
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\title
q-Ap\'ery Irrationality Proofs by q-WZ pairs
\endtitle
\author
Tewodros Amdeberhan and Doron Zeilberger
\endauthor
\affil Department of Mathematics, Temple University,
Philadelphia PA 19122, USA \\
tewodros\@math.temple.edu, zeilberg\@math.temple.edu
\endaffil
\abstract
Using WZ pairs, Ap\'ery-style proofs of the irrationality of the q-analogues of the Harmonic series and Ln(2) are given.
\endabstract
\endtopmatter
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\bf 0. Introduction: \rm \smallskip


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

$$\alignat2
h_q(1):&=\sum_{k=1}^{\infty}\frac 1{q^k-1} &&\qquad\text{(for $|q| > 1$),} \tag0.1
\endalignat$$
and 
$$\alignat2
Ln_q(2):&=\sum_{n=1}^{\infty}\frac {(-1)^n}{q^n-1} &&\qquad\text{(for $|q| \neq 0, 1$)}.\tag0.2
\endalignat$$

In 1948, Paul Erd\"os [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 Ap\'ery'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 Erd\"os [E2]. \smallskip

We will assume familiarity with ref. [Z]. In particular, we use \par
$\binom{n}k_q = \frac {(q)_n}{(q)_k (q)_{n-k}}$, where $(q)_n = (1-q) \cdots (1-q^n)$. \par
N and K are \it forward shift operators \rm on n and k, respectively.\par
$\Delta_n = N - 1$, $\Delta_k = K - 1$. \par

\pagebreak
\midspace{.3in}

\bf 1. A scheme for proving the irrationality of the \it q-harmonic series \bf $h_q(1)$:\rm \bigskip
The claims made in subsections 1.1-1.5 below were found using the Maple Package {\tt qEKHAD} accompanying [PWZ]. The relevant script substantiating our claims can be found in this paper's Web Pages.\smallskip

1.1. The q-WZ 1-form $\omega$ is:
$$\omega = \frac {-1}{\binom {n+k+1}{k}_{q}(q)_{n+1}}
\{\delta k + \frac {q^{n+1}}{(q^{n+1}-1)}\delta n\}.$$

1.2. The choice of the potential $c(n,k)$ is:
$$c(n,k)=\sum_{m=1}^{n}\frac {q^m}{(1-q^m)(q)_{m}} + \sum_{m=1}^{k}\frac 1{(q^m-1)}\frac 1{\binom {n+m}m_{q}(q)_{n}}.$$

\smallskip

1.3. The choice of the mollifier $b(n,k)$ is:
$$b(n,k)=(-1)^k q^{k(k+1)/2}\binom {n+k}k_{q}\binom nk_{q}.$$

1.4. We define two sequences:
$$
a(n)=\sum_{k=0}^{n}c(n,k)b(n,k), \qquad\text{and} \qquad b(n)=\sum_{k=0}^{n}b(n,k).$$

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
$$
 A(n,k)=c(n,k)B(n,k)+\frac{(-1)^kq^{2n+3}}{q^{n+1}-1}\binom {n+1}k_q\frac {q^{\binom k2}}{(q)_{n+2}}P_q^2(n,k) \qquad\text{and}$$

$P_q^1(n,k)=-q\alpha_n^2\beta_k^{-1}(q^2\alpha_n+2q)+q\alpha_n^2(q^2\alpha_n^3+2q(q+1)\alpha_n^2+3q\alpha_n-(q+1)-(\alpha_n+2)\beta_k)$

$$\multline
P_q^2(n,k)=q^2\alpha_n^2+q\alpha_n-2+\beta_k(q^2\alpha_n^5+q(2q+1)\alpha_n^4-2\alpha_n^3-\alpha_n^3\beta_k-(2-q^{-1})\alpha_n^2\beta_k\\
-(3q+5)\alpha_n^2+2q^{-1}\alpha_n\beta_k+(q-1+2q^{-1})\alpha_n+(1+3q^{-1})),\endmultline$$

$y_0(n)= q(\alpha_n-1)(q\alpha_n+2)$, ${ }$ $y_2(n)=(q\alpha_n-1)(\alpha_n+2)$, $ {}$ $\alpha_n=q^{n+1}$, $ { }$ $\beta_k=q^{k+1}$ and

\bigskip
$y_1(n)=q^3\alpha_n^5+2q^2(q+1)\alpha_n^4
+q^2\alpha_n^3-4q(q+1)\alpha_n^2
+(q^2-4q+1)\alpha_n+2(q+1).$

\bigskip
Then $$\alignat3
L(b(n,k))&=B(n,k)-B(n,k-1) &\qquad\text{and} &\qquad L(b(n,k)c(n,k))&=A(n,k)-A(n,k-1).\tag{$*$}\endalignat$$
Now, summing over k in $(*)$ shows that both sequences $a(n)$ and $b(n)$ are solutions of $Lu(n) = 0$.

\pagebreak
\midspace{.4in}

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\(q^{3n+3}\).$$
Hence, 
$$b_n = O\(q^{\frac{3n^2}2}\).\tag1.6.1$$

On the other hand, $La_n = 0$ and $Lb_n = 0$ lead to the system of recurrence equations,
$$\alignat2
y_2(n)a_{n+2}+y_1(n)a_{n+1}+y_0(n)a_n = 0,&\qquad
y_2(n)b_{n+2}+y_1(n)b_{n+1}+y_0(n)b_n = 0.\tag1.6.2\endalignat$$

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

$$\left|\frac {a_{n+1}}{b_{n+1}}- \frac {a_{n+2}}{b_{n+2}}\right| \leq
\left|\frac{y_0(n)}{y_1(n)}\frac {b_n}{b_{n+1}}\(\frac{a_{n+2}}{b_{n+2}}- \frac {a_{n+1}}{b_{n+1}}\)\right|+
\left|\frac{y_0(n)}{y_1(n)}\frac {b_n}{b_{n+1}}\(\frac{a_{n+1}}{b_{n+1}}- \frac {a_{n}}{b_{n}}\)\right|.$$

This yields in
$$\frac {a_{n+1}}{b_{n+1}}- \frac {a_{n}}{b_{n}} = O\(b_n^{-2}\).\tag1.6.3$$

Therefore,
$$h_q(1)-\frac {a_n}{b_n}=O\(b_n^{-2}\).\tag1.6.4$$
In particular, the sequence of rational numbers $\frac {a_n}{b_n}$ converges moderately faster to $h_q(1)$.
\pagebreak
\midspace{.3in}

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
$$ord_p \binom nm_q \leq ord_p(q)_n - ord_p(q)_m.\tag1.7.1$$
\bf Note: \rm $$\binom{n+k}k_q\binom km_q = \binom{n+m}m_q\binom{n+k}{k-m}_q.\tag1.7.2$$

\bf Claim: \rm The sequences 
$$
u_n=a_n(q)_{n+1}\prod_{s=[n/2]}^n(1-q^s) \qquad\text{and} \qquad z_n= b_n(q)_{n+1}\prod_{s=[n/2]}^n(1-q^s)$$
are polynomials in q with integer coefficients, and moreover 
$$z_n = O\(q^{19n^2/8}\).\tag1.7.3$$
  
\bf Proof: \rm Applying (1.7.1) and (1.7.2), we estimate the denominator of $u_n$: 
$$\align
ord_p\(\frac {(q^m-1)(q)_{n}\binom{n+m}m_q}{\binom{n+k}k_q}\)
&\leq ord_p\(\frac {(q^m-1)(q)_{n}\binom km_q}{\binom{n+k}{k-m}_q}\)\\
&\leq ord_p(q)_{n} + ord_p(q^m-1) + ord_p(q)_k - ord_p(q)_m\\
&\leq ord_p(q)_{n} + ord_p\prod_{s=[n/2]}^n(1-q^s) +  ord_p(q)_k - ord_p(q)_m\\
&\leq ord_p\((q)_{n}\prod_{s=[n/2]}^n(1-q^s)\),\endalign$$
since $m \leq k \leq n$. This proves the claim on $u_n$. And (1.7.3) follows from (1.6.1). The rest is trivial. \bigskip
 
\bf Claim: \rm $h_q(1)$ - $\frac {u_n}{z_n}$ = $O\(\frac 1{z_n^{1+\delta}}\)$; where $\delta =0.26316\dots$$>$ 0.\bigskip

\bf proof: \rm 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\(z_n^{-1-(5/19)}\).$$
\smallskip
Thus, we have proved:\smallskip
\bf Theorem 1: \rm If $|q|>1$ is an integer, $h_q(1)$ is irrational with irrationality measure 4.80. \smallskip

\bf Note: \rm By invoking Theorem 7 ([Z], p.596) with $\omega$ as in 1.1, we obtain the series acceleration:
$$
h_q(1)=\sum_{n=1}^{\infty}\frac {q^n}{(1-q^n)(q)_n} \qquad\text{and}\qquad
h_q(1)=\sum_{n=1}^{\infty}\frac {1-q^n-q^{2n}}{(q^n-1)\binom {2n}n_q (q)_n}.$$

\pagebreak
\midspace{.3in}

\bf 2. A scheme for proving the irrationality of $Ln_q(2)$: \rm \bigskip

The claims made in subsections 2.1-2.5 below were found using the Maple Package {\tt qEKHAD} accompanying [PWZ]. The relevant script substantiating our claims can be found in this paper's Web Pages.\smallskip

2.1. The qWZ 1-form $\omega$ is:
$$\omega = \frac {(-1)^k}{(1-q^{k+1})}\frac {(q)_n}{\binom {n+k+1}{k+1}_{q}(q^2)_{n}}
\{\delta k + \frac {q^{n+1}}{(1+q^{n+1})}\delta n\}.$$

2.2. The choice of the potential $c(n,k)$ is:
$$c(n,k)=\sum_{m=1}^{n}\frac {q^m(q)_m}{(1-q^m)(q^2)_{m}} + \sum_{m=1}^{k}\frac {(-1)^{m-1}}{(1-q^m)}\frac {(q)_n}{\binom {n+m}m_{q}(q^2)_{n}}.$$

\smallskip

2.3. The choice of the mollifier $b(n,k)$ is:
$$b(n,k)=q^{k(k+1)/2}\binom {n+k}k_{q}\binom nk_{q}.$$

2.4. We define two sequences:
$$
a(n)=\sum_{k=0}^{n}c(n,k)b(n,k), \qquad\text{and} \qquad b(n)=\sum_{k=0}^{n}b(n,k).$$

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
$$
 A(n,k)=c(n,k)B(n,k)+\frac{(-1)^kq^{2n+3}}{1-q^{n+1}}\binom {n+1}k_q\frac {q^{\binom k2}(q)_{n+1}}{(q^2)_{n+1}}P_q^2(n,k)\qquad\text{and}$$

$$\multline
P_q^1(n,k)=q\alpha_n^2\bigl[q^3\alpha_n^5+q^2(1+q)\alpha_n^4+2q(1+q^2)\alpha_n^3-(1-q+q^2)\alpha_n-3(1+q)\big]\\
+q\alpha_n^2\bigl[q\beta_k^{-1}(q^2\alpha_n^3+q(1+q)\alpha_n^2+(2-q)\alpha_n-2)+(q\alpha_n^3+(q-1)\alpha_n^2+(2q-1)\alpha_n)\alpha_k-2\bigr] \endmultline$$
$$\multline
P_q^2(n,k)=q^2\alpha_n^3+q(1+q)\alpha_n^2+(2+q)\alpha_n+2-\alpha_n\alpha_k^2\bigl[\alpha_n^3+(1-q^{-1})\alpha_n^2+(2-q)\alpha_n-2q^{-1}\bigr]\\
-\alpha_k\bigl[q^2\alpha_n^6+q(1+q)\alpha_n^5+(2+q+2q^2)\alpha_n^4+(1+q)\alpha_n^3+2\alpha_n^2-(2+q+q^{-1})\alpha_n+(q^{-1}-1)\big],\endmultline$$

$y_0(n)= -q(\alpha_n-1)(\alpha_n+1)(q^2\alpha_n^2+q\alpha_n+2)$, ${ }$ $y_2(n)=-(q\alpha_n-1)(q\alpha_n+1)(\alpha_n^2+\alpha_n+2)$, $ {}$ 
$$
y_1(n)=q^4\alpha_n^7+q^2(1+q)(q\alpha_n^6+\alpha_n^4)+q(1+q+q^2)(2q\alpha_n^5+\alpha_n^3)-(1+3q+3q^2+q^3)\alpha_n^2-(1+q^2)(2+\alpha_n),$$
and $\alpha_n=q^{n+1}$, $ { }$ $\beta_k=q^{k+1}.$ 

\bigskip
Then $$\alignat3
L(b(n,k))&=B(n,k)-B(n,k-1) &\qquad\text{and} &\qquad L(b(n,k)c(n,k))&=A(n,k)-A(n,k-1).\tag{$**$}\endalignat$$
Now, summing over k in $(**)$ shows that both sequences $a(n)$ and $b(n)$ are solutions of $Lu(n) = 0$.

\pagebreak
\midspace{.4in}

2.6. Similar arguments and estimates as in (1.6) above lead to
$$Ln_q(2)-\frac {a_n}{b_n}=O\(b_n^{-2}\).\tag2.6.1$$
In particular, the sequence of rational numbers $\frac {a_n}{b_n}$ converges moderately faster to $Ln_q(2)$.

2.7.
\bf Claim: \rm The sequences 
$$
u_n=a_n\prod_{t=1}^n(1+q^t)\prod_{s=[n/2]}^n(1-q^s) \qquad\text{and} \qquad z_n= b_n\prod_{t=1}^n(1+q^t)\prod_{s=[n/2]}^n(1-q^s)$$
are polynomials in q with integer coefficients, and moreover 
$$z_n = O\(q^{19n^2/8}\).\tag2.7.1$$
  
\bf Proof: \rm Applying (1.7.2) and (2.7.1), we estimate the denominator of $u_n$: 
$$\align
ord_p\(\frac {(1-q^m)(q^2)_{n}\binom{n+m}m_q}{\binom{n+k}k_q(q)_n}\)
&\leq ord_p\(\frac {(q^m-1)(q^2)_{n}\binom km_q}{\binom{n+k}{k-m}_q(q)_n}\)\\
&\leq ord_p\(\frac{(q^2)_n}{(q)_{n}}\) + ord_p(q^m-1) + ord_p(q)_k - ord_p(q)_m\\
&\leq ord_p\(\frac{(q^2)_n}{(q)_{n}}\) + ord_p\prod_{s=[n/2]}^n(1-q^s) +  ord_p(q)_k - ord_p(q)_m\\
&\leq ord_p\(\prod_{t=1}^n(1+q^t)\prod_{s=[n/2]}^n(1-q^s)\),\endalign$$
since $m \leq k \leq n$. This proves the claim on $u_n$. And (2.7.1) follows from (1.6.1). The rest is trivial. \bigskip
 
\bf Claim: \rm $Ln_q(2)$ - $\frac {u_n}{z_n}$ = $O\(\frac 1{z_n^{1+\delta}}\)$; where $\delta =0.26316\dots$$>$ 0.\bigskip

\bf proof: \rm 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\(z_n^{-1-(5/19)}\).$$
\smallskip
Thus, we have proved:\smallskip
\bf Theorem 2: \rm If $|q|\neq 0, 1$ is an integer, $Ln_q(2)$ is irrational with irrationality measure 4.80.\smallskip
\bf Note: \rm By invoking Theorem 7 ([Z], p. 596) with $\omega$ as in 2.1, we obtain the series acceleration:
$$
Ln_q(2)=\sum_{n=1}^{\infty}\frac {q^n(q)_n}{(1-q^n)(q^2)_n} \qquad\text{and}\qquad
Ln_q(2)=\sum_{n=1}^{\infty}\frac {(-1)^{n-1}(q)_n(1-q^{3n})}{(1-q^n)^2\binom {2n}n_q (q^2)_n}.$$

\pagebreak
\midspace{.3in}

\Refs 
\widestnumber\key{PWZ}

\ref\key A \by R. Ap\'ery \paper \it  Irrationalit\`e de $\zeta(2)$ et \rm$\zeta(3)$ \jour Asterisque \vol 61\yr1979 \pages11-13
\endref
\smallskip

\ref\key B1 \by P. Borwein \paper \it On the irrationality of $\sum 1/(q^n+r)$ \rm \jour J. Number Theory \vol 37 \yr1991 \pages253-259
\endref
\smallskip

\ref\key B2 \by P. Borwein \paper \it On the irrationality of certain series \rm \jour Proc. Camb. Phil. Soc. \vol 112 \yr1992 \pages141-146
\endref
\smallskip

\ref\key E1 \by P. Erd\"os \paper \it On arithmetical properties of Lambert Series \rm \jour J. Indian Math. Soc. (N.S.) \vol 12 \yr1948 \pages63-66
\endref
\smallskip

\ref\key E2 \by P. Erd\"os \book \it On the irrationality of certain series: problems and results \rm \publ In New Advances in Transcendence Theory (Cambridge University Press) \yr1988 \pages102-109
\endref
\smallskip

\ref\key P \by A. van der Poorten \paper \it A proof that Euler missed ..., Ap\'ery's proof of the irrationality of \rm $\zeta(3)$ \jour Math. Intel. \vol     1 \yr1979 \pages195-203
\endref
\smallskip

\ref\key PWZ \by M. Petkov\v sek, H.S. Wilf, D. Zeilberger\book \it ``A=B'' \rm \publ A.K. Peters Ltd. \yr1996 \endref
{\tt The package qEKHAD is available by the www at http://www.math.temple.edu/\~{}zeilberg/programs.html}
\smallskip

\ref\key Z \by D. Zeilberger \paper \it Closed Form (pun intended!) \rm \jour Contemporary Mathematics \vol 143 \yr1993 \pages579-607
\endref
\endRefs


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