1 Wilf–Zeilberger (WZ) pairs
Herbert Wilf and Doron Zeilberger invented the concept of a WZ pair. Two hyper-geometric (in n and k) terms
$F(n,k)$
and
$G(n,k)$
form a WZ pair if the identity
$$ \begin{align*} F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k) \end{align*} $$
holds. A Maple code written by Zeilberger, in the package
$$ \begin{align*} \texttt{with(SumTools[Hypergeometric]);} \end{align*} $$
finds the mate of a term (that forms a WZ pair with it), whenever such exists, by means of a rational certificate
$C(n,k)$
so that
$G(n,k)=C(n,k) F(n,k)$
[Reference Wilf and Zeilberger12].
The simple concept of a WZ pair has very useful and important properties. We display two of them.
Theorem 1.1. Assume that
$F(0,k)=0$
and
$F(+\infty ,k)=\lim _{n\to +\infty }F(n,k)=0$
for any complex k. Then,
$$ \begin{align*} \sum_{n=0}^{\infty} G(n,k+1)=\sum_{n=0}^{\infty} G(n,k) \quad\text{for all}\; k \in \mathbb{C}. \end{align*} $$
In particular, the function
$$ \begin{align*} g(k)=\sum_{n=0}^{\infty} G(n,k) \end{align*} $$
is periodic of period
$k=1$
.
Proof. For k a complex number,
$$ \begin{align*} \sum_{n=0}^{\infty} (F(n+1,k)-F(n,k)) &=F(+\infty,k)-F(0,k) =0 \\ &=\sum_{n=0}^{\infty} (G(n,k+1)-G(n,k)) =g(k+1)-g(k). \\[-3.8pc] \end{align*} $$
Due to its periodicity, the function
$g(k)$
cannot have poles: those will be also (periodic) poles of the hypergeometric terms
$F(n,k)$
which is impossible. However, there could be removable singularities.
We call WZ pairs meeting the conditions
$F(0,k)=F(+\infty ,k)=0$
for all
$k\in \mathbb C$
flawless. The hypotheses of Carlson’s theorem hold very often and then, we obtain a stronger result:
$$ \begin{align*} g(k)=\sum_{n=0}^{\infty} G(n,k)=\text{constant} \quad\text{for all}\; k \in \mathbb{C}. \end{align*} $$
For justification, we use the following ‘periodic version’ of Carlson’s theorem.
Theorem 1.2 (Carlson’s theorem; see [Reference Almkvist, Kotsireas and Zima1, Appendix]).
Let
$k=x+iy$
, where x and y are the real and imaginary parts of k, respectively. If
$g(k)$
is a holomorphic function such that
$g(k)=g(k+1)$
and
$g(k)=\mathrm {O}(e^{c|y|})$
for
$|y|$
sufficiently large with some real number
$c<2\pi $
, then
$g(k)$
is a constant function.
Theorem 1.3. For every WZ pair,
$$ \begin{align*} \sum_{n=0}^{\infty} G(n,0)-\lim_{k \to \infty} \sum_{n=0}^k G(n,k) =\sum_{k=0}^{\infty} F(0,k)-\lim_{n \to \infty} \sum_{k=0}^n F(n,k), \end{align*} $$
whenever both sides converge.
The theorem is used by Au in [Reference Au3]. Very interesting pairs for applying this theorem are those such that both limits are equal to
$0$
and
$F(0,k)$
is a rational function.
2 Transformations
From a WZ pair we can get others; this mechanism forms the grounds of the WZ-seed methodology of Au explored in [Reference Au2, Reference Au3]. The following transformations will be useful for our needs.
-
(1) Transformations
$F(n,k) \to F(sn, k \pm tn)$
with
$s,t\in \mathbb Z_{>0}$
change the hypergeometric series and could produce acceleration or deceleration of its convergence rate. -
(2) Transformations
$G(n,k) \to G(n \pm tk, sk)$
with
$s,t\in \mathbb Z_{>0}$
obviously do not alter the series at
$k=0$
, but some of the new WZ pairs may have better properties than the original one. -
(3) Duality corresponds to the transformation
$G(n,k) \to G(-n,-k)$
[Reference Guillera, Kotsireas and Zima8, Section 4]. Observe that it converts divergent series into convergent ones and vice versa.
Usually, WZ pairs are not flawless (see [Reference Au3, Reference Petkovšek, Wilf and Zeilberger11]), but can sometimes be converted to such pairs via suitable transformations.
3 Ramanujan-like series and their duals
In the examples of this paper, we only consider rational Ramanujan-like series or their duals.
3.1 Rational Ramanujan-like series
These can be written in the form
$$ \begin{align*} \sum_{n=0}^{\infty} R(n) =\sum_{n=0}^{\infty} \bigg( \prod_{i=0}^{2m} \frac{(s_i)_n}{(1)_n} \bigg) z^n \sum_{k=0}^m a_k n^k = \frac{\sqrt{(-1)^m \chi}}{\pi^m}, \end{align*} $$
where
$(a)_0=1$
and
$(a)_n=a(a+1) \cdots (a+n-1)$
for
$n \in \mathbb {Z}_{>0}$
is the rising factorial or Pochhammer symbol, z is rational,
$a_0, a_1,\ldots ,a_m$
are positive rationals and
$\chi $
is the discriminant of a certain quadratic field (imaginary or real), which is an integer. Let
$z=z_0$
. If
$|z_0|>1$
, then the series are divergent, but we understand them as the analytic continuations of the series in z at
$z=z_0$
. We can evaluate them with Maple executing
$$ \begin{align*} \texttt{sum(R(n),n=0..infinity,formal);} \end{align*} $$
which converts them into hypergeometric form.
The series of degree
$m=1$
were discovered and proved using modular equations by a method highlighted by the genius Srinivasa Ramanujan. In 2002–2003, we discovered several series of higher degree [Reference Guillera5]. In addition, inspired by [Reference Ekhad and Zeilberger4], we managed to prove in [Reference Guillera6] with the WZ method the following famous Ramanujan formula for
$1/\pi $
:
$$ \begin{align*} \sum_{n=0}^{\infty} \frac{\big(\frac12\big)_n^3}{(1)_n^3} (42n+5) \bigg( \frac{1}{64} \bigg)^n=\frac{16}{\pi}. \end{align*} $$
In addition, we discovered some new Ramanujan-like series of degree
$2$
in [Reference Guillera5] and proved the following one in [Reference Guillera6]:
$$ \begin{align*} \sum_{n=0}^{\infty} \frac{\big(\frac12\big)_n^5}{(1)_n^5} (820n^2+180n+13) \bigg( - \frac{1}{1024} \bigg)^n=\frac{128}{\pi^2}, \end{align*} $$
with the same strategy using the WZ method.
3.2 The duals
The dual of a Ramanujan-like series for a reciprocal power of
$\pi $
is obtained by replacing n with
$-n$
in accordance with the following dictionary:
$$ \begin{align*} (1)_{-n} \to \frac{n (-1)^n}{(1)_n} \quad \text{and} \quad (a)_{-n} \to \frac{(-1)^n}{(1-a)_n} \quad \text{for } 0<a<1. \end{align*} $$
If the sum of a rational Ramanujan-like series equals
$\sqrt {(-1)^m \chi }/\pi ^m$
, then the sum of its dual is a rational multiple of the Dirichlet value
$L_{\chi }(m+1)$
[Reference Guillera10]. Some examples are
$L_{-4}(2)=L((\frac {-4}{\cdot }),2)$
(Catalan’s constant),
$L_1(3)=\zeta (3)$
,
$L_1(5)=\zeta (5)$
and
$L_1(4)=\zeta (4)$
(a rational multiple of
$\pi ^4$
).
To summarise, the dual of a Ramanujan-like formula has the following form:
$$ \begin{align*} \sum_{n=1}^{\infty} \bigg( \prod_{i=0}^{2m} \frac{(1)_n}{(s_i)_n} \bigg) \frac{z^{-n}}{(-n)^{m+1}} \sum_{k=0}^m a_k (-n)^{k}=r L_{\chi}(m+1). \end{align*} $$
The series converges at
$z=z_0$
if
$|z_0|>1$
and analytic continuation is required if
$|z_0|<1$
. We can determine the rational number r using supercongruences [Reference Guillera10].
Finding flawless WZ pairs for these duals seems to be a challenging task.
4 Examples inspired by Au’s pairs
4.1 Proof of a Ramanujan-like formula of degree 4 conjectured by J. Cullen
We write the WZ pair of [Reference Au3, Example I] in a different way. We start from the entry
$$ \begin{align*} U(n,k)=\frac{\big(\frac12\big)_n^5 \big(\frac12+k\big)_n^5 \big(\frac12 -k\big)_n \big(1+\frac{k}{4}\big)_n \big(\frac14+\frac{k}{4}\big)_n \big(\frac12+\frac{k}{4}\big)_n \big(\frac34+\frac{k}{4}\big)_n}{(1)_n^5 \big(1+\frac{k}{2}\big)_n^5 \big(\frac12 + \frac{k}{2}\big)_n^5} \frac{\big(\frac12\big)_k^4}{(1)_k^4}, \end{align*} $$
which involves only Pochhammer symbols of the forms
$(a+bk)_n$
and
$(c)_k$
, where a and c are rationals in the interval
$(0,1]$
, take
$$ \begin{align*} z=(\tfrac12)^{12}, \quad y=1, \end{align*} $$
and construct the rational functions
$$ \begin{align*} S(n,k)=2^{16} \frac{n^5 (4k + 6n + 1)(3 + 2k)(1 + 2k)}{(4n + k)(4k^2 + 8 k + 3)(2k + 1 - 2n)} \end{align*} $$
and
$R(n,k)$
using the recipe of [Reference Guillera9]. Then, the function
$F(n,k)=U(n,k) S(n,k) z^n y^k$
has the WZ mate
$G(n,k)=U(n,k) R(n,k) z^n y^k$
. The conditions
$F(0,k)=F(+\infty ,k)=0$
of Theorem 1.1 are satisfied and we deduce that
$$ \begin{align*} \sum_{n=0}^{\infty} G(n,k)=\sum_{n=0}^{\infty} G(n,k+1) \quad \text{for all}\; k \in \mathbb{C}. \end{align*} $$
With the help of Carlson’s theorem, we arrive at the following stronger result:
$$ \begin{align*} \sum_{n=0}^{\infty} G(n,k) = \text{constant} \quad \text{for all}\; k \in \mathbb{C}. \end{align*} $$
To determine the value of the constant, we take
$k=1/2$
and notice that then, we obtain the factor
$(0)_n$
in the numerator of
$U(n,1/2)$
, and hence in the numerator of
$G(n,1/2)$
. As
$(0)_n=0$
, if n is different from
$0$
, the only nonzero term in the sum
$\sum _{n=0}^{\infty } G(n,1/2)$
is
$G(0,1/2)$
and we obtain
$$ \begin{align*} \sum_{n=0}^{\infty} G(n,k)=\sum_{n=0}^{\infty} G\bigg(n,\frac12\bigg)=G \bigg(0,\frac12 \bigg)=\frac{2048}{\pi^4}. \end{align*} $$
Thus, for
$k=0$
,
$$ \begin{align*} \sum_{n=0}^{\infty} \frac{\big(\frac12\big)_n^7 \big(\frac14\big)_n \big(\frac34\big)_n}{n!^9} \,\frac{1}{2^{12 n}}(43680n^4 +20632n^3+4340n^2+466n+21)=\frac{2048}{\pi^4}. \end{align*} $$
The Maple code in [Reference Guillera9] generalises the pattern conjectured in [Reference Guillera7] for degree
$1$
to Ramanujan-like series of higher degrees. Finally, we notice that replacing
$F(n,k)$
with
$F(n,k-n)$
in the above formula leads to [Reference Au3, Example II].
4.2 Proof of one of our Ramanujan-like conjectured formulae of degree 2
We write the WZ pair of [Reference Au3, Example VII] starting with
$$ \begin{align*} U(n,k)=\frac{\big(\frac12\big)_n^3 \big(1+\frac{k}{3}\big)_n \big(\frac13+\frac{k}{3}\big)_n \big(\frac23+\frac{k}{3}\big)_n \big(\frac16+\frac{2k}{3}\big)_n \big(\frac12+\frac{2k}{3}\big)_n \big(\frac56+\frac{2k}{3}\big)_n }{(1)_n^3 \big(1+\frac{k}{2}\big)_n^3 \big(\frac12+\frac{k}{2}\big)_n^3} \frac{\big(\frac14\big)_k \big(\frac34\big)_k}{(1)_k^2}, \end{align*} $$
the values
$$ \begin{align*} z=-\frac{3^6}{4^6}, \quad y=1, \end{align*} $$
and constructing the rational functions
$$ \begin{align*} S(n,k)=\frac{768 n^3}{k + 3n} \end{align*} $$
and
$R(n,k)$
, as explained in [Reference Guillera9]. Then, the function
$F(n,k)=U(n,k) S(n,k) z^n y^k$
has the WZ mate
$G(n,k)=U(n,k) R(n,k) z^n y^k$
. As in our first example, we deduce that
$$ \begin{align*} \sum_{n=0}^{\infty} G(n,k) = \text{constant} \quad \text{for all}\; k \in \mathbb{C}. \end{align*} $$
To determine the constant, we let
$k \to +\infty $
, taking into account that
$(a+bk)_n \sim (bk)^n$
, and we obtain the identity that specialises to
$$ \begin{align*} \sum_{n=0}^{\infty} \frac{ \big(\frac12\big)_n \big(\frac13\big)_n \big(\frac23\big)_n \big(\frac16\big)_n \big(\frac56\big)_n }{n!^5} \bigg(-\frac{3^6}{4^6}\bigg)^n (1930n^2+549n+45) = \frac{384}{\pi^2}. \end{align*} $$
4.3 Proof of a formula for
$\zeta (4)$
Inspired by [Reference Au3, Example IV], we write
$$ \begin{align*} U(n,k)=\frac{(1)_n^3 (1+k)_n^3(1+2k)_n\big( 1+\frac{k}{2}\big)_n\big(\frac12+\frac{k}{2}\big)_n} {\big( \frac12 \big)_n^3 \big( \frac12+k \big)_n^3 \big(1+ \frac{2k}{3} \big)_n \big( \frac13+\frac{2k}{3} \big)_n \big( \frac23+\frac{2k}{3} \big)_n} \frac{(1)_k^4}{\big( \frac12\big)_k^4} \end{align*} $$
and
$$ \begin{align*} F(n,k)=U(n,k)S(n,k) z^n y^k, \end{align*} $$
where
$z=16/27$
,
$y=1$
and
$$ \begin{align*} S(n,k)=\frac{n\big( 2n+2k+\frac32\big)}{(1+2k)(2n+2k+1)^3(3n+2k+1)(3n+2k+2)}. \end{align*} $$
The function
$F(n,k)$
has a WZ companion
$G(n,k)$
such that
$$ \begin{align*} G(n,k)=U(n,k)R(n,k)z^n y^k, \end{align*} $$
where
$R(n,k)$
is a certain rational function that a Maple code written by Zeilberger determines.
However, as
$F(n,k)$
vanishes for
$n=0$
and as
$n \to +\infty $
, we deduce that
$$ \begin{align*} \sum_{n=0}^{\infty} G(n,k) = \frac{\pi^4}{32} \quad \text{for all}\; k \in \mathbb{C}, \end{align*} $$
where we have determined the constant value of the sum by letting
$k \to +\infty $
. This includes the simpler particular case corresponding to
$k=0$
[Reference Au3, Example IV].
4.4 Proof of a formula for
$\zeta (2)$
Playing with the term
$F(n,k)$
of the WZ pair used in Example 3 and modifying some of the powers, we discover a new WZ pair with
$$ \begin{align*} U(n,k)=\frac{(1)_n (1+k)_n (1+2k)_n\big( 1+\frac{k}{2}\big)_n\big(\frac12+\frac{k}{2}\big)_n} {\big( \frac12 \big)_n \big( \frac12+k \big)_n \big(1+ \frac{2k}{3} \big)_n \big( \frac13+\frac{2k}{3} \big)_n \big( \frac23+\frac{2k}{3} \big)_n} \frac{(1)_k^2}{\big( \frac12\big)_k^2} \end{align*} $$
and
$$ \begin{align*} F(n,k)=U(n,k)S(n,k) z^n y^k, \end{align*} $$
where
$z=4/27$
,
$y=1$
and
$$ \begin{align*} S(n,k)=\frac{n\big( 2n+2k+\frac32\big)}{(1+2k)(2n+2k+1)(3n+2k+1)(3n+2k+2)}. \end{align*} $$
The function
$F(n,k)$
has a WZ companion
$G(n,k)$
such that
$$ \begin{align*} G(n,k)=U(n,k)R(n,k)z^n y^k, \end{align*} $$
where
$R(n,k)$
is the following rational function:
$$ \begin{align*} R(n,k)=\frac{22n^3 + (34k + 49)n^2 + (12k^2 + 48k + 35)n + 8(k + 1)^2}{2(1 + 2n)(3 + 2k + 3n)(2n + 2k + 1)(3n + 2k + 1)(3n + 2k + 2)}, \end{align*} $$
which is discovered automatically by the Maple code. However, as
$F(n,k)$
vanishes for
$n=0$
and as
$n \to +\infty $
, we deduce that
$$ \begin{align*} \sum_{n=0}^{\infty} G(n,k) = \frac{\pi^2}{8} \quad \text{for all}\; k \in \mathbb{C}, \end{align*} $$
where we have determined the constant value of the sum by letting
$k \to +\infty $
. This includes the simpler particular case corresponding to
$k=0$
[Reference Guillera, Kotsireas and Zima8, (14)].
4.5 Proof of the famous Gourevitch conjectured formula
In [Reference Au2, Appendix B], Au finds a WZ pair suitable for proving the Ramanujan-like series for
$1/\pi ^3$
conjectured by Gourevitch. His proof is somewhat complicated. The main obstacle there is that
$$ \begin{align*} F(0,k) \ne 0 \quad\text{and}\quad \lim_{n \to \infty} \sum_{k=0}^{\infty} F(n,k) \ne 0. \end{align*} $$
The issue can be resolved by passing to
$\tilde G(n,k)=G(n-k,k)$
in place of
$G(n,k)$
. This leads to the following beautiful Ramanujan-like generalisation:
$$ \begin{align*} &\sum_{n=0}^{\infty} \frac{\big(\frac12+3k\big)_n^2 \big(\frac12-k\big)_n^2 \big(\frac12+k\big)_n^3}{(1)_n^2(1+2k)_n^2(1+k)_n^3} \frac{\big(\frac12\big)_k \big(\frac16\big)_k^2 \big(\frac56\big)_k^2}{(1)_k^5} \bigg(\frac{1}{64}\bigg)^n \bigg(\frac{729}{1024}\bigg)^k \\ &\quad\times (6n+6k+1)(28n^2+8n+1+12k^2+56nk+8k) = \frac{32}{\pi^3}, \end{align*} $$
valid for all complex values of k thanks to Carlson’s theorem (the value of the constant is obtained by taking
$k=1/2$
). This (WZ provable) generalisation proves the Gourevitch formula by specialising to
$k=0$
, namely,
$$ \begin{align*} \sum_{n=0}^{\infty} \frac{\big(\frac12\big)_n^7}{(1)_n^7} \bigg(\frac{1}{64}\bigg)^n (6n+1)(28n^2+8n+1) = \frac{32}{\pi^3}. \end{align*} $$
Full details of this derivation can be found in [Reference Guillera9].
5 Formulae with
$|z|>1$
If
$|z|>1$
, the hypergeometric series in the formulae diverge. In 1908, Barnes showed that the complex integral representation
$$ \begin{align*} \sum_{n=0}^{\infty} \frac{(a)_n(b)_n}{(1)_n(c)_n} z^n = \frac{1}{2 \pi i} \int_{-i \infty}^{i \infty} \frac{(a)_s (b)_s}{(c)_s} \Gamma(-s) ({-z})^s \,ds, \end{align*} $$
valid for the series within the domain
$|z|<1$
,
$z\notin [0,-1]$
, gives its analytic continuation to
$z\in \mathbb C\setminus [0,\infty )$
. Here, the integration path connecting
$-i\infty $
with
$i\infty $
is chosen in such a way that the poles of
$\Gamma (-s)$
are all to the left of the contour, while the poles of
$(a)_s (b)_s=\Gamma (a+s)\Gamma (b+s)/(\Gamma (a)\Gamma (b))$
are all to the right of it; the branch of
$(-z)^s=\exp (s\ln (-z))$
is chosen to be real-valued for
$z<0$
.
Inspired by the WZ pair in [Reference Au3, Example XII], we take its dual and write the new function
$F(n,k)$
as
$F(n,k)=U(n,k) S(n,k) z^n$
, where
$z=-3125/1024$
,
$$ \begin{align*} U(n,k)=\frac{\big( \frac12 \big)_n^5\big( \frac12+k\big)_n^4\big( \frac15+\frac{2k}{5} \big)_n\big( \frac25+\frac{2k}{5} \big)_n\big( \frac35+\frac{2k}{5} \big)_n\big( \frac45+\frac{2k}{5} \big)_n\big( 1 + \frac{2k}{5}\big)_n}{(1)_n^5 (1+2k)_n \big( 1+\frac{k}{2}\big)_n^4\big( \frac12+\frac{k}{2}\big)_n^4} \frac{\big(\frac12 \big)_k^4}{(1)_k^4} \end{align*} $$
and
$$ \begin{align*} S(n,k)=\frac{640n^5(6n+4k+1)}{(n+2k+1)(5n+2k)}. \end{align*} $$
From
$F(n,k)$
, we can get
$R(n,k)$
such that
$G(n,k)=U(n,k) R(n,k) z^n$
, and using the technique explained in [Reference Guillera, Kotsireas and Zima8], after replacing n with s and k with t, we deduce that
$$ \begin{align*} \frac{1}{2\pi i} \int_{-i \infty}^{i \infty} U(s,t) (1)_s \Gamma(-s) R(s,t) \bigg( \frac{3125}{1024} \bigg)^s \,ds = \frac{1280}{\pi^4} \quad \text{for all}\; t \in \mathbb{C}, \end{align*} $$
which gives the analytic continuation of
$$ \begin{align*} \sum_{n=0}^{\infty} U(n,k) R(n,k) z^n, \quad |z|<1, \end{align*} $$
at
$z=-3125/1024$
, which we can check with Maple for every complex number k writing
$$ \begin{align*} \texttt{sum(G(n,k),n=0..infinity,formal);} \end{align*} $$
specialising at
$t=0$
gives
$$ \begin{align*} &\frac{1}{2\pi i} \int_{-i \infty}^{i \infty} \frac{\big( \frac12 \big)_s^5 \big( \frac15 \big)_s \big( \frac25 \big)_s\big( \frac35 \big)_s\big( \frac45 \big)_s}{(1)_s^8} \Gamma(-s) \bigg( \frac{3125}{1024} \bigg)^s \\ &\quad\times (5532s^4+5600s^3+2275s^2+425s+30) \, ds = \frac{1280}{\pi^4}. \end{align*} $$
In a completely similar way, from [Reference Au3, Example IX],
$$ \begin{align*} \frac{1}{2\pi i} \int_{-i \infty}^{i \infty} \frac{\big( \frac12 \big)_s \big( \frac15 \big)_s \big( \frac25 \big)_s\big( \frac35 \big)_s\big( \frac45 \big)_s}{(1)_s^4} \Gamma(-s) \bigg( \frac{3125}{256} \bigg)^s (483n^2+245n+30) \, ds = \frac{80}{\pi^2}. \end{align*} $$
Ackowledgement
I am very grateful to Wadim Zudilin for his comments and suggestions on this paper.
