Pfaff's Method Revisited

Abstract

In 1797, Pfaff gave a simple proof of a 3F2{}_3F_2 hypergeometric series which was much later reproved by Andrews in 1996. In the same paper, Andrews also proved other well-known hypergeometric identities using Pfaff's method. In this paper, we prove a number of terminating q-hypergeometric series-product identities using Pfaff's method thereby providing a detailed account of its wide applicability.

Full text

arXiv:2405.08705v1 [math.NT] 14 May 2024
PFAFF’S METHOD REVISITED
ARITRAM DHAR
Dedicated to George E. Andrews and Bruce C. Berndt in celebration of their 85th birthdays
ABS TR ACT. In 1797, Pfaff gave a simple proof of a 3F2hypergeometric series which was much later reproved by
Andrews in 1996. In the same paper, Andrews also proved other well-known hypergeometric identities using Pfaff’s
method. In this paper, we prove a number of terminating q-hypergeometric series-product identities using Pfaff’s
method thereby providing a detailed account of its wide applicability.
1. IN TRO DU CT IO N
Let Nbe the set of natural numbers. For complex variables aand q, define the conventional q-shifted facto-
rial by
(a;q)n:=
n−1
Y
k=0
(1 −aqk)
for any n∈Nand (a;q)0= 1. For |q|<1, define
(a;q)∞:= lim
n→∞(a;q)n.
We can also define (a;q)nfor all real numbers nby
(a;q)n:= (a;q)∞
(aqn;q)∞
.
The q-shifted factorial for negative integers can be defined by
(a;q)−n:= 1
(aq−n;q)n
=(−q/a;q)n
(q/a;q)n
q(n
2).(1.1)
Following Gasper and Rahman’s text [3], the r+1Frhypergeometric series and the r+1 φrunilateral basic
hypergeometric series are defined, respectively, by
(1.2) r+1Fra0,...,ar
b1,...,br;z:=
∞
X
n=0
(a0,...,ar)n
(1, b1,...,br)n
zn,
Date: May 15, 2024.
2020 Mathematics Subject Classification. 33D15, 33D65.
Key words and phrases. Pfaff’s method, basic hypergeometric series, terminating q-hypergeometric series sum-product identities,
recurrence relations.
1

2 ARITRAM DHAR
where
(a1, a2,...,ak)n:=
k
Y
i=1
n−1
Y
j=0
(ai+j),
and
(1.3) r+1φra0,...,ar
b1,...,br;q, z:=
∞
X
n=0
(a0,...,ar;q)n
(q, b1,...,br;q)n
zn,|z|<1,
where
(a1, a2,...,ak;q)n:=
k
Y
i=1
(ai;q)n.
The series (1.3) is called a terminating series if one of its numerator parameters is of the form q−mwhere
m∈N∪ {0}. The series (1.3) is called balanced or Saalsch ¨
utzian if
b1b2...br=qa0a1. . . ar.
The series (1.3) is said to be well-poised if the parameters satisfy the relations
qa0=a1b1=a2b2=... =arbr,
very-well-poised if, in addition,
a1=q√a0, a2=−q√a0.
In 1797, Pfaff [6] gave “the simplest proof” (as Andrews writes in [1]) of the following 3F2hypergeometric
identity,
3F2−n, a, b
c, 1 + a+b−n−c; 1=(c−a, c −b)n
(c, c −a−b)n
.
Pfaff’s result was ignored until Saalsch ¨utz rediscovered it in 1890. In the second part of a series of three
papers (which he refers to as Pfaff’s Trilogy), Andrews [1] then recreated the proof of Pfaff’s 3F2sum and
also proved some other hypergeometric identities using Pfaff’s method in the same paper, for instance, the
terminating Kummer theorem, Bailey’s theorem, the theorems of Dougall and Lakin, to name a few.
Towards the end of his paper, Andrews [1] proved a q-analog of Pfaff’s original theorem which was due to
Jackson [4]. He then concludes with a brief description saying Pfaff’s method is mostly effective for balanced
and well-poised hypergeometric series (and q-series). In this paper, we give proofs of a plethora of well-known
terminating q-hypergeometric series-product identities using Pfaff’s method.
We now present the statements of the main terminating q-hypergeometric identites which we prove in this
paper.

3
Theorem 1.1. (q-binomial theorem [3, II.4]) For any non-negative integer n, we have
(1.4) 1φ0q−n
−;q, z= (zq−n;q)n.
Theorem 1.2. (q-Chu-Vandermonde [3, II.6]) For any non-negative integer n, we have
(1.5) 2φ1a, q−n
c;q, q=(c/a;q)n
(c;q)n
an.
Theorem 1.3. (q-Pfaff-Saalsch¨
utz [3, II.12]) For any non-negative integer n, we have
(1.6) 3φ2a, b, q−n
c, abq−n+1 /c ;q, q=(c/a, c/b;q)n
(c, c/ab;q)n
.
Theorem 1.4. (q-Dixon [3, II.14]) For any non-negative integer n, we have
(1.7) 4φ3a, −q√a, b, q−n
−√a, aq/b, aqn+1 ;q, qn+1√a
b=(aq, q√a/b;q)n
(q√a, aq/b;q)n
.
Theorem 1.5. ([3, p. 110, Ex. 3.34]) For any non-negative integer n, we have
(1.8) 4φ3q−2n, c2, a, aq
a2q2, cq−n, cq−n+1 ;q2, q 2=(−q, aq/c;q)n
(−aq, q/c;q)n
.
Theorem 1.6. (Andrews [1, p. 22, eq.(7.7) and eq.(7.6)], Li-Chu [5, p.1006, eq.(2) and eq.(3)]) For any non-
negative integer n, we have
(1.9) 4φ3q−n, b, b√q, d2qn
dq, d√q, b2;q , q=bn(1 −d)(−√q, d√q/b;√q)n
(1 −qnd)(−b, d;√q)n
and
(1.10) 4φ3q−n, b, b√q, d2qn+1
dq, d√q, b2q;q, q =bn(−√q, d√q/b;√q)n
(−b√q, d√q;√q)n
.
Theorem 1.7. ([3, II.21]) For any non-negative integer n, we have
(1.11) 6φ5a, q√a, −q√a, b, c, q−n
√a, −√a, aq/b, aq/c, aqn+1 ;q, aqn+1
bc =(aq, aq/bc;q)n
(aq/b, aq/c;q)n
.

4 ARITRAM DHAR
Theorem 1.8. ([3, II.22]) For any non-negative integer n, we have
(1.12) 8φ7a, q√a, −q√a, b, c, d, e, q−n
√a, −√a, aq/b, aq/c, aq/d, aq/e, aqn+1 ;q, q =(aq, aq/bc, aq/bd, aq /cd;q)n
(aq/b, aq/c, aq/d, aq/bcd;q)n
,
where a2q=bcdeq−n.
Theorem 1.9. (Andrews-Berkovich [2, p. 535, eq.(3.1) and eq.(3.2)]) For any non-negative integer n, we have
(1.13)
10φ9a, q√a, −q√a, apq/k, −apq/k, aq/√k, −aq/√k, k/aq, kq n, q−n
√a, −√a, √kq, −√kq, √k, −√k, a2q2/k, aq−n+1 /k, aqn+1 ;q, q=(aq , k2/a2q;q)n
(k, k/a;q)n
and
(1.14)
10φ9a, q√a, −q√a, apq /k, −apq/k, a/√k, −aq/√k, k/a, k qn, q−n
√a, −√a, √kq, −√kq, q√k, −√k, a2q/k, aq−n+1 /k, aqn+1 ;q, q=(aq, √k , k2/a2;q)n
(k, k/a, q √k;q)n
.
In Section 2, we prove Theorems 1.1, 1.2, 1.3, 1.4, 1.6, 1.7, 1.8, and 1.9 using Pfaff’s original method. We
also show that upon applying a quadratic transformation, Theorem 1.5 gives rise to Theorem 1.3.
2. PRO OF S
In this section, we present the proofs of Theorems 1.1 - 1.9.
2.1. Proof of Theorem 1.1. Consider Sn(z;q)to be the the left-hand side of (1.4). Then, we have
Sn(z;q)−Sn−1(z;q) =
∞
X
j=0
(q−n;q)j
(q;q)j
zj−
∞
X
j=0
(q−n+1;q)j
(q;q)j
zj
=
∞
X
j=1
zj
(q;q)j(q−n;q)j−(q−n+1;q)j
=−q−n
∞
X
j=1
(q−n+1;q)j−1
(q;q)j−1
zj
=−zq−n
∞
X
j=0
(q−n+1;q)j
(q;q)j
zj
=−zq−nSn−1(z;q).
Thus, we have
Sn(z;q) = (1 −zq−n)Sn−1(z;q).(2.1)
Now, let S′
n(z;q)denote the right-hand side of (1.4). Then, we have
S′
n(z;q)−S′
n−1(z;q) = (zq−n;q)n−(zq−n+1 ;q)n−1

5
=−zq−n(zq−n+1;q)n−1
=−zq−nS′
n−1(z;q).
Thus, we have
S′
n(z;q) = (1 −zq−n)S′
n−1(z;q).(2.2)
Now, observe that both Sn(z;q)and S′
n(z;q)have the same initial conditions S0(z;q) = S′
0(z;q) = 1 and
both obey the same recurrences (2.1) and (2.2) respectively. Hence,
Sn(z;q) = S′
n(z;q)
and we have a proof of q-binomial theorem (Theorem 1.1) using Pfaff’s method. 
2.2. Proof of Theorem 1.2. Consider Sn(a, c;q)to be the the left-hand side of (1.5). Then, we have
Sn(a, c;q)−Sn−1(a, c;q) =
∞
X
j=0
(a, q−n;q)j
(q, c;q)j
qj−
∞
X
j=0
(a, q−n+1 ;q)j
(q, c;q)j
qj
=
∞
X
j=1
(a;q)jqj
(q, c;q)j(q−n;q)j−(q−n+1 ;q)j
=−q−n
∞
X
j=1
(a;q)j(q−n+1;q)j−1
(q;q)j−1(c;q)j
qj
=−q−n+1
∞
X
j=0
(a;q)j+1(q−n+1 ;q)j
(q;q)j(c;q)j+1
qj
=−q−n+1(1 −a)
(1 −c)
∞
X
j=0
(aq, q−n+1 ;q)j
(q, cq;q)j
qj
=−q−n+1(1 −a)
(1 −c)Sn−1(aq, cq;q).
Thus, we have
Sn(a, c;q)−Sn−1(a, c;q) = −q−n+1(1 −a)
(1 −c)Sn−1(aq, cq;q).(2.3)
Now, let S′
n(a, c;q)denote the right-hand side of (1.5). Then, we have
S′
n(a, c;q)−S′
n−1(a, c;q) = (c/a;q)n
(c;q)n
an−(c/a;q)n−1
(c;q)n−1
an−1
=−(1 −a)(c/a;q)n−1
(c;q)n
an−1
=−q−n+1(1 −a)(cq/aq;q)n−1
(1 −c)(cq;q)n−1
(aq)n−1
=−qn+1(1 −a)
(1 −c)S′
n−1(aq, cq;q).

6 ARITRAM DHAR
Thus, we have
S′
n(a, c;q)−S′
n−1(a, c;q) = −q−n+1(1 −a)
(1 −c)S′
n−1(aq, cq;q).(2.4)
Now, observe that both Sn(a, c;q)and S′
n(a, c;q)have the same initial conditions S0(a, c;q) = S′
0(a, c;q) = 1
and both obey the same recurrences (2.3) and (2.4) respectively. Hence,
Sn(a, c;q) = S′
n(a, c;q)
and we have a proof of q-Chu-Vandermonde sum (Theorem 1.2) using Pfaff’s method. 
2.3. Proof of Theorem 1.3. Consider Sn(a, b, c;q)to be the the left-hand side of (1.6). Then, we have
Sn(a, b, c;q)−Sn−1(a, b, c;q) =
∞
X
j=0
(a, b, q−n;q)j
(q, c, abq−n+1/c;q)j
qj−(a, b, q−n+1 ;q)j
(q, c, abq−n+2/c;q)j
qj
=
∞
X
j=1
(a, b;q)jqj
(q, c;q)j(q−n;q)j
(abq−n+1 /c;q)j−(q−n+1;q)j
(abq−n+2 /c;q)j
=−q−n(1 −abq/c)
(1 −abq−n+1 /c)
∞
X
j=1
(a, b;q)j(q−n+1;q)j−1
(q;q)j−1(c, abq−n+2 /c;q)j
qj
=−q−n(1 −abq/c)
(1 −abq−n+1 /c)
∞
X
j=0
(a, b;q)j+1(q−n+1 ;q)j
(q;q)j(c, abq−n+2 /c;q)j+1
qj+1
=−q−n+1(1 −a)(1 −b)(1 −abq/c)
(1 −c)(1 −abq−n+1/c)(1 −abq−n+2/c)
×
∞
X
j=0
(aq, bq, q−n+1;q)j
(q, cq, abq−n+3/c;q)j
qj
=−q−n+1(1 −a)(1 −b)(1 −abq/c)
(1 −c)(1 −abq−n+1/c)(1 −abq−n+2/c)Sn−1(aq , bq, cq;q).
Thus, we have
Sn(a, b, c;q)−Sn−1(a, b, c;q) = −q−n+1 (1 −a)(1 −b)(1 −abq/c)
(1 −c)(1 −abq−n+1 /c)(1 −abq−n+2 /c)Sn−1(aq, bq, cq ;q).(2.5)
Remark. (2.5) was also proved by Andrews [1, p. 22, eq.(7.3)].
Now, let S′
n(a, b, c;q)denote the right-hand side of (1.6). Then, we have
S′
n(a, b, c;q)−S′
n−1(a, b, c;q) = (c/a, c/b;q)n
(c, c/ab;q)n−(c/a, c/b;q)n−1
(c, c/ab;q)n−1
=cqn−1(1 −1/a)(1 −1/b)(c/a, c/b;q)n−1
(c, c/ab;q)n

7
=cqn−1(1 −a)(1 −b)(c/abq;q)n−1(cq /aq, cq/bq;q)n−1
ab(1 −c)(1 −c/ab)(cq/ab;q)n−1(cq, c/abq;q)n−1
=cqn−1(1 −a)(1 −b)(1 −c/abq)
ab(1 −c)(1 −cqn−2/ab)(1 −cqn−1/ab)S′
n−1(aq, bq, cq;q)
=−q−n+1(1 −a)(1 −b)(1 −abq/c)
(1 −c)(1 −abq−n+1/c)(1 −abq−n+2/c)S′
n−1(aq, bq, cq;q).
Thus, we have
S′
n(a, b, c;q)−S′
n−1(a, b, c;q) = −q−n+1(1 −a)(1 −b)(1 −abq/c)
(1 −c)(1 −abq−n+1 /c)(1 −abq−n+2 /c)S′
n−1(aq, bq, cq;q).(2.6)
Now, observe that both Sn(a, b, c;q)and S′
n(a, b, c;q)have the same initial conditions S0(a, b, c;q) = S′
0(a, b, c;q) =
1and both obey the same recurrences (2.5) and (2.6) respectively. Hence,
Sn(a, b, c;q) = S′
n(a, b, c;q)
and we have a proof of q-Pfaff-Saalsch¨utz sum (Theorem 1.3) using Pfaff’s method. 
2.4. Proof of Theorem 1.4. Consider Sn(a, b;q)to be the the left-hand side of (1.7). Then, we have
Sn(a, b;q)−Sn−1(a, b;q) =
∞
X
j=0
(a, −q√a, b, q−n;q)j
(q, −√a, aq/b, aqn+1;q)jqn+1 √a
bj
−
∞
X
j=0
(a, −q√a, b, q−n+1 ;q)j
(q, −√a, aq/b, aq n;q)jqn√a
bj
=
∞
X
j=1
(a, −q√a, b;q)j(qn√a/b)j
(q, −√a, aq/b;q)j(q−n;q)jqj
(aqn+1 ;q)j−(q−n+1;q)j
(aqn;q)j
=−1
(1 −aqn)
∞
X
j=1
(a, −q√a, b;q)j(q−n+1;q)j−1(1 −aqj)
(−√a, aq/b;q)j(q, aq n+1;q)j−1(1 −aqn+j)qn√a
bj
=−(1 −a)
(1 −aqn)
∞
X
j=1
(aq, −q√a, b;q)j(q−n+1 ;q)j−1
(q;q)j−1(−√a, aq/b, aqn+1;q)jqn√a
bj
=−(1 −a)
(1 −aqn)
∞
X
j=0
(aq, −q√a, b;q)j+1 (q−n+1;q)j
(q;q)j(−√a, aq/b, aqn+1;q)j+1 qn√a
bj+1
=−(1 −a)(1 −aq)(1 + q√a)(1 −b)(qn√a/b)
(1 −aqn)(1 + √a)(1 −aq/b)(1 −aqn+1)
×
∞
X
j=0
(aq2,−q2√a, bq, q−n+1;q)j
(q, −q√a, aq2/b, aqn+2;q)jqn√a
bj
=−(1 −√a)(1 + q√a)(1 −aq)(1 −b)(qn√a/b)
(1 −aq/b)(1 −aqn)(1 −aqn+1)Sn−1(aq2, bq;q).
Thus, we have

8 ARITRAM DHAR
Sn(a, b;q)−Sn−1(a, b;q) = −(1 −√a)(1 + q√a)(1 −aq)(1 −b)(qn√a/b)
(1 −aq/b)(1 −aqn)(1 −aqn+1)Sn−1(aq2, bq;q).(2.7)
Now, let S′
n(a, b;q)denote the right-hand side of (1.7). Then, we have
S′
n(a, b;q)−S′
n−1(a, b;q) = (aq, q√a/b;q)n
(q√a, aq/b;q)n−(aq, q√a/b;q)n−1
(q√a, aq/b;q)n−1
=qn√a(1 −√a)(1 −1/b)(aq, q√a/b;q)n−1
(q√a, aq/b;q)n
=−(1 −√a)(1 −b)(qn√a/b)(aq, q√a/b, q 2√a, aq2/b;q)n−1
(q√a, aq/b, aq3, q√a/b;q)n−1
×(aq3, q√a/b;q)n−1
(q2√a, aq2/b;q)n−1
=−(1 −√a)(1 −b)(qn√a/b)(aq;q)n−1
(1 −q√a)(1 −aq/b)(aq3;q)n−1
S′
n−1(aq2, bq;q)
=−(1 −√a)(1 + q√a)(1 −aq)(1 −b)(qn√a/b)
(1 −aq/b)(1 −aqn)(1 −aqn+1)S′
n−1(aq2, bq;q).
Thus, we have
S′
n(a, b;q)−S′
n−1(a, b;q) = −(1 −√a)(1 + q√a)(1 −aq)(1 −b)(qn√a/b)
(1 −aq/b)(1 −aqn)(1 −aqn+1)S′
n−1(aq2, bq;q).(2.8)
Now, observe that both Sn(a, b;q)and S′
n(a, b;q)have the same initial conditions S0(a, b;q) = S′
0(a, b;q) = 1
and both obey the same recurrences (2.7) and (2.8) respectively. Hence,
Sn(a, b;q) = S′
n(a, b;q)
and we have a proof of q-Dixon sum (Theorem 1.4) using Pfaff’s method. 
2.5. Proof of Theorem 1.5. Consider Sn(a, c;q)to be the the left-hand side of (1.8). Then, using the substi-
tution (a, b, c, d)7−→ (√a, √aq, −c, q−n)in Singh’s quadratic transformation [3, p. 361, III.21] (change of
base from q2→q) on Sn(a, c;q), we get
Sn(a, c;q) = 4φ3a, aq, c2, q−2n
a2q2, cq−n, cq−n+1 ;q2, q 2
=4φ3a, aq, −c, q−n
aq, −aq, cq−n;q, q
=3φ2a, −c, q−n
−aq, cq−n;q , q
=(−q, aq/c;q)n
(−aq, q/c;q)n
where the last line follows by substituting (a, b, c)7−→ (a, −c, −aq)in (1.6). Thus, we obtain a direct proof of
Theorem 1.5 without resorting to Pfaff’s method. 

9
2.6. Proof of Theorem 1.6. Consider An(b, d;q)and Bn(b, d;q)to be the left-hand sides of (1.9) and (1.10)
respectively. Then, we have
Bn(b, d;q)−Bn−1(b, d;q) =
∞
X
j=0
(q−n, b, b√q, d2qn+1 ;q)j
(q, dq, d√q , b2q;q)j
qj−
∞
X
j=0
(q−n+1, b, b√q, d2qn;q)j
(q, dq, d√q , b2q;q)j
qj
=
∞
X
j=1
(b, b√q;q)jqj
(q, dq, d√q , b2q;q)j(q−n, d2qn+1;q)j−(q−n+1, d2qn;q)j
= (d2qn−q−n)
∞
X
j=1
(b, b√q;q)j(q−n+1, q2qn+1 ;q)j−1
(q;q)j−1(dq, d√q, b2q;q)j
qj
= (d2qn−q−n)
∞
X
j=0
(b, b√q;q)j+1(q−n+1 , q2qn+1 ;q)j
(q;q)j(dq, d√q, b2q;q)j+1
qj+1
=q(d2qn−q−n)(1 −b)(1 −b√q)
(1 −dq)(1 −d√q)(1 −b2q)
×
∞
X
j=0
(q−n+1, bq, bq√q, d2qn+1 ;q)j
(q, dq2, dq √q, b2q2;q)j
qj
=q(d2qn−q−n)(1 −b)
(1 + b√q)(1 −dq)(1 −d√q)An−1(bq, dq;q).
Thus, we have
Bn(b, d;q)−Bn−1(b, d;q) = q(d2qn−q−n)(1 −b)
(1 + b√q)(1 −dq)(1 −d√q)An−1(bq, dq;q).(2.9)
Now, let A′
n(b, d;q)and B′
n(b, d;q)denote the right-hand sides of (1.9) and (1.10) respectively. Then, we have
B′
n(b, d;q)−B′
n−1(b, d;q) = bn(−√q, d√q/b;√q)n
(−b√q, d√q;√q)n−bn−1(−√q, d√q/b;√q)n−1
(−b√q, d√q;√q)n−1
=bn−1(b−1)(1 + dqn)(−√q, d√q/b;√q)n−1
(−b√q, d√q;√q)n
=q−n+1(b−1)(1 −d2q2n)(−bq, dq;√q)n−1
(1 −dq)(−b√q, d√q;√q)n
×(bq)n−1(1 −dq)(−√q , d√q/b;√q)n−1
(1 −dqn)(−bq, dq ;√q)n−1
=q−n+1(b−1)(1 −d2q2n)(−bq, dq;√q)n−1
(1 −dq)(−b√q, d√q;√q)n
A′
n−1(bq, dq;q)
=q(d2qn−q−n)(1 −b)
(1 + b√q)(1 −dq)(1 −d√q)A′
n−1(bq, dq;q).
Thus, we have
B′
n(b, d;q)−B′
n−1(b, d;q) = q(d2qn−q−n)(1 −b)
(1 + b√q)(1 −dq)(1 −d√q)A′
n−1(bq, dq;q).(2.10)
Now, observe that An(b, d;q)and A′
n(b, d;q)have the same initial conditions A0(b, d;q) = A′
0(b, d;q) = 1

10 ARITRAM DHAR
and that Bn(b, d;q)and B′
n(b, d;q)have the same initial conditions B0(b, d;q) = B′
0(b, d;q) = 1. Also, note
that both An(b, d;q),Bn(b, d;q)and A′
n(b, d;q),B′
n(b, d;q)obey the same recurrences (2.9) and (2.10) re-
spectively. Hence,
An(b, d;q) = A′
n(b, d;q)
and
Bn(b, d;q) = B′
n(b, d;q).
Thus, we have a proof of the two 4φ3identities due to Andrews and Li-Chu (Theorem 1.6) using Pfaff’s
method. 
2.7. Proof of Theorem 1.7. Consider Sn(a, b, c;q)to be the left-hand side of (1.11). Then, we have
Sn(a, b, c;q)−Sn−1(a, b, c;q) =
∞
X
j=0
(a, q√a, −q√a, b, c, q−n;q)j
(q, √a, −√a, aq/b, aq /c, aqn+1;q)jaqn+1
bc j
−
∞
X
j=0
(a, q√a, −q√a, b, c, q−n+1;q)j
(q, √a, −√a, aq/b, aq /c, aqn;q)jaqn
bc j
=
∞
X
j=1
(a, q√a, −q√a, b, c;q)j(aqn/bc)j
(q, √a, −√a, aq/b, aq/c;q)j(q−n;q)jqj
(aqn+1 ;q)j−(q−n+1;q)j
(aqn;q)j
=−(1 −a)
(1 −aqn)
∞
X
j=1
(aq, q√a, −q√a, b, c;q)j(q−n+1;q)j−1
(q;q)j−1(√a, −√a, aq/b, aq/c, aqn+1;q)jaqn
bc j
=−(1 −a)
(1 −aqn)
∞
X
j=0
(aq, q√a, −q√a, b, c;q)j+1(q−n+1 ;q)j
(q;q)j(√a, −√a, aq/b, aq/c, aqn+1;q)j+1 aqn
bc j+1
=−(1 −a)(1 −aq)(1 −q√a)(1 + q√a)(1 −b)(1 −c)(aqn/bc)
(1 −√a)(1 + √a)(1 −aq/b)(1 −aq/c)(1 −aqn)(1 −aqn+1 )
×
∞
X
j=0
(aq2, q2√a, −q2√a, bq, cq, q−n+1;q)j
(q, q√a, −q√a, aq 2/b, aq2/c, aqn+2;q)jaqn
bc j
=−(1 −aq)(1 −aq2)(1 −b)(1 −c)(aqn/bc)
(1 −aqn)(1 −aqn+1 )(1 −aq/b)(1 −aq/c)Sn−1(aq2, bq, cq;q).
Thus, we have
Sn(a, b, c;q)−Sn−1(a, b, c;q) = −(1 −aq)(1 −aq2)(1 −b)(1 −c)(aqn/bc)
(1 −aqn)(1 −aqn+1 )(1 −aq/b)(1 −aq/c)Sn−1(aq2, bq, cq;q).
(2.11)
Now, let S′
n(a, b, c;q)denote the right-hand side of (1.11). Then, we have
S′
n(a, b, c;q)−S′
n−1(a, b, c;q) = (aq, aq/bc;q)n
(aq/b, aq/c;q)n−(aq, aq/bc;q)n−1
(aq/b, aq/c;q)n−1
=−aqn(1 −1/b)(1 −1/c)(aq, aq/bc;q)n−1
(aq/b, aq/c;q)n

11
=−(1 −b)(1 −c)(aqn/bc)(aq, aq /bc, aq2/b, aq2/c;q)n−1
(aq/b, aq/c;q)n(aq3, aq/bc;q)n−1
×(aq3, aq/bc;q)n−1
(aq2/b, aq2/c;q)n−1
=−(1 −b)(1 −c)(aqn/bc)(aq, aq /bc, aq2/b, aq2/c;q)n−1
(aq/b, aq/c;q)n(aq3, aq/bc;q)n−1
×S′
n−1(aq2, bq, cq;q)
=−(1 −aq)(1 −aq2)(1 −b)(1 −c)(aqn/bc)
(1 −aqn)(1 −aqn+1 )(1 −aq/b)(1 −aq/c)S′
n−1(aq2, bq, cq;q).
Thus, we have
S′
n(a, b, c;q)−S′
n−1(a, b, c;q) = −(1 −aq)(1 −aq2)(1 −b)(1 −c)(aqn/bc)
(1 −aqn)(1 −aqn+1 )(1 −aq/b)(1 −aq/c)S′
n−1(aq2, bq, cq;q).
(2.12)
Now, observe that both Sn(a, b, c;q)and S′
n(a, b, c;q)have the same initial conditions S0(a, b, c;q) =
S′
0(a, b, c;q) = 1 and both obey the same recurrences (2.11) and (2.12) respectively. Hence,
Sn(a, b, c;q) = S′
n(a, b, c;q)
and we have a proof of Rogers’ 6φ5sum (Theorem 1.7) using Pfaff’s method. 
2.8. Proof of Theorem 1.8. Consider Sn(a, b, c, d;q)to be the left-hand side of (1.12). Then, we have
Sn(a, b, c, d;q)−Sn−1(a, b, c, d;q) =
∞
X
j=0
(a, q√a, −q√a, b, c, d, a2qn+1 /bcd, q−n;q)j
(q, √a, −√a, aq/b, aq /c, aq/d, bcdq−n/a, aqn+1;q)j
qj
−
∞
X
j=0
(a, q√a, −q√a, b, c, d, a2qn/bcd, q−n+1;q)j
(q, √a, −√a, aq/b, aq /c, aq/d, bcdq−n+1/a, aq n;q)j
qj
=
∞
X
j=1
(a, q√a, −q√a, b, c, d;q)jqj
(q, √a, −√a, aq/b, aq /c, aq/d;q)j
×(a2qn+1/bcd, q −n;q)j
(bcdq−n/a, aqn+1;q)j−(a2qn/bcd, q−n+1 ;q)j
(bcdq−n+1 /a, aqn;q)j
=−(q−n−a2qn/bcd +aqn−bcdq−n/a)
(1 −bcdq−n/a)(1 −aqn)
×
∞
X
j=1
(a, q√a, −q√a, b, c, d;q)j(a2qn+1 /bcd, q−n+1 ;q)j−1(1 −aqj)
(q;q)j−1(√a, −√a, aq/b, aq/c, aq/d, bcdq−n+1 /a, aqn+1;q)j
qj
=−(1 −a)(q−n−a2qn/bcd +aqn−bcdq−n/a)
(1 −bcdq−n/a)(1 −aqn)
×
∞
X
j=1
(aq, q√a, −q√a, b, c, d;q)j(a2qn+1 /bcd, q−n+1 ;q)j−1
(q;q)j−1(√a, −√a, aq/b, aq/c, aq/d, bcdq−n+1 /a, aqn+1 ;q)j
qj
=−(1 −a)(q−n−a2qn/bcd +aqn−bcdq−n/a)
(1 −bcdq−n/a)(1 −aqn)

12 ARITRAM DHAR
×
∞
X
j=0
(aq, q√a, −q√a, b, c, d;q)j+1 (a2qn+1/bcd, q−n+1;q)j
(q;q)j(√a, −√a, aq/b, aq/c, aq/d, bcdq−n+1 /a, aqn+1 ;q)j+1
qj+1
=−q(q−n−a2qn/bcd +aqn−bcdq−n/a)
(1 −bcdq−n/a)(1 −aqn)
×(1 −aq)(1 −aq2)(1 −b)(1 −c)(1 −d)
(1 −aq/b)(1 −aq/c)(1 −aq/d)(1 −bcdq−n+1 /a)(1 −aqn+1 )
×
∞
X
j=0
(aq2, q2√a, −q2√a, bq, cq, dq, a2qn+1/bcd, q−n+1;q)j
(q, q√a, −q√a, aq 2/b, aq2/c, aq2/d, bcdq−n+2 /a, aqn+2 ;q)j
qj
=−q(q−n−a2qn/bcd +aqn−bcdq−n/a)
(1 −bcdq−n/a)(1 −aqn)
×(1 −aq)(1 −aq2)(1 −b)(1 −c)(1 −d)
(1 −aq/b)(1 −aq/c)(1 −aq/d)(1 −bcdq−n+1 /a)(1 −aqn+1 )
×Sn−1(aq2, bq, cq, dq;q).
Thus, we have
Sn(a, b, c, d;q)−Sn−1(a, b, c, d;q) = φn(a, b, c, d;q)Sn−1(aq2, bq, cq, dq;q),(2.13)
where
φn(a, b, c, d;q) = −q(1 −aq)(1 −aq2)(1 −b)(1 −c)(1 −d)(q−n−a2qn/bcd +aqn−bcdq−n/a)
(1 −aqn)(1 −aqn+1 )(1 −aq/b)(1 −aq/c)(1 −aq/d)(1 −bcdq−n/a)(1 −bcdq−n+1/a).
Now, let S′
n(a, b, c, d;q)denote the right-hand side of (1.12). Then, we have
S′
n(a, b, c, d;q)−S′
n−1(a, b, c, d;q) = (aq, aq/bc, aq /bd, aq/cd;q)n
(aq/b, aq/c, aq/d, aq/bcd;q)n−(aq, aq/bc, aq /bd, aq/cd;q)n−1
(aq/b, aq/c, aq/d, aq/bcd;q)n−1
=−aqn(1 −a2q2n/bcd)(1 −1/b)(1 −1/c)(1 −1/d)
×(aq, aq/bc, aq/bd, aq/cd;q)n−1
(aq/b, aq/c, aq/d, aq/bcd;q)n
=−aqn(1 −a2q2n/bcd)(1 −1/b)(1 −1/c)(1 −1/d)
×(aq2/b, aq2/c, aq2/d, a/bcd, aq, aq/bc, aq/bd, aq/cd;q)n−1
(aq3, aq/bc, aq/bd, aq/cd;q)n−1(aq/b, aq/c, aq /d, aq/bcd;q)n
×(aq3, aq/bc, aq/bd, aq/cd;q)n−1
(aq2/b, aq2/c, aq2/d, a/bcd;q)n−1
=−aqn(1 −a2q2n/bcd)(1 −1/b)(1 −1/c)(1 −1/d)(aq, a/bcd;q)n−1
(1 −aq/b)(1 −aq/c)(1 −aq/d)(1 −aq/bcd)(aq3, aq2/bcd;q)n−1
×S′
n−1(aq2, bq, cq, dq;q)
=−q(q−n−a2qn/bcd +aqn−bcdq−n/a)
(1 −bcdq−n/a)(1 −aqn)
×(1 −aq)(1 −aq2)(1 −b)(1 −c)(1 −d)
(1 −aq/b)(1 −aq/c)(1 −aq/d)(1 −bcdq−n+1 /a)(1 −aqn+1 )
×S′
n−1(aq2, bq, cq, dq;q).
Thus, we have

13
S′
n(a, b, c, d;q)−S′
n−1(a, b, c, d;q) = φ′
n(a, b, c, d;q)S′
n−1(aq2, bq, cq, dq;q),(2.14)
where
φ′
n(a, b, c, d;q) = −q(1 −aq)(1 −aq2)(1 −b)(1 −c)(1 −d)(q−n−a2qn/bcd +aqn−bcdq−n/a)
(1 −aqn)(1 −aqn+1 )(1 −aq/b)(1 −aq/c)(1 −aq/d)(1 −bcdq−n/a)(1 −bcdq−n+1/a).
Now, observe that both Sn(a, b, c, d;q)and S′
n(a, b, c, d;q)have the same initial conditions S0(a, b, c, d;q) =
S′
0(a, b, c, d;q) = 1 and both obey the same recurrences (2.13) and (2.14) respectively since φn(a, b, c, d;q) =
φ′
n(a, b, c, d;q). Hence,
Sn(a, b, c, d;q) = S′
n(a, b, c, d;q)
and we have a proof of Jackson’s 8φ7sum (Theorem 1.8) using Pfaff’s method. 
2.9. Proof of Theorem 1.9. Consider An(a, k;q)and Bn(a, k;q)to be the left-hand sides of (1.13) and (1.14)
respectively. Then, we have
An(a, k;q)−Bn(a, k;q) =
∞
X
j=0
(a, q√a, −q√a, apq/k, −apq/k, aq /√k, −aq/√k, k/aq , kqn, q−n;q)j
(q, √a, −√a, √kq , −√kq, √k , −√k, a2q2/k, aq−n+1 /k, aqn+1 ;q)j
qj
−
∞
X
j=0
(a, q√a, −q√a, apq/k, −apq/k, a/√k, −aq/√k, k /a, kqn, q−n;q)j
(q, √a, −√a, √kq, −√kq, q √k, −√k, a2q/k, aq −n+1/k, aqn+1 ;q)j
qj
=
∞
X
j=1
(a, q√a, −q√a, apq/k, −apq/k, −aq/√k, kq n, q−n;q)jqj
(q, √a, −√a, √kq , −√kq, −√k , aq−n+1/k, aqn+1;q)j
× (aq/√k, k /aq;q)j
(√k, a2q2/k;q)j−(a/√k , k/a;q)j
(q√k, a2q/k;q)j!
=(1 −a)(√k−a2q/k +a/√k−k/aq)
(1 −√k)(1 −a2q/k)
×
∞
X
j=1
(aq, q√a, −q√a, apq /k, −apq/k, −aq/√k , kqn, q−n;q)j(aq/√k , k/a;q)j−1
(q;q)j−1(√a, −√a, √kq, −√kq, q√k, −√k, a2q2/k, aq−n+1/k, aqn+1;q)j
qj
=(1 −a)(√k−a2q/k +a/√k−k/aq)
(1 −√k)(1 −a2q/k)
×
∞
X
j=0
(aq, q√a, −q√a, apq /k, −apq/k, −aq/√k , kqn, q−n;q)j+1 (aq/√k, k /a;q)j
(q;q)j(√a, −√a, √kq, −√kq, q√k, −√k, a2q2/k, aq−n+1/k, aqn+1;q)j+1
qj+1
=q(1 −aq)(1 −aq2)(1 −kqn)(1 −q−n)(√k−a2q/k +a/√k−k/aq)
(1 −k)(1 −kq)(1 −q√k)(1 −aq/√k)(1 −aq−n+1 /k)(1 −aqn+1 )
×
∞
X
j=0
(aq2, q2√a, −q2√a, aq pq/k, −aqpq/k , aq/√k, −aq2/√k, k/a, kqn+1 , q−n+1 ;q)j
(q, q√a, −q√a, q √kq, −q√kq, q 2√k, −q√k, a2q3/k, aq−n+2 /k, aqn+2;q)j
qj
=q(1 −aq)(1 −aq2)(1 −kqn)(1 −q−n)(√k−a2q/k +a/√k−k/aq)
(1 −k)(1 −kq)(1 −q√k)(1 −aq/√k)(1 −aq−n+1 /k)(1 −aqn+1 )
×Bn−1(aq2, kq 2;q).

14 ARITRAM DHAR
Thus, we have
An(a, k;q)−Bn(a, k;q) = ψn(a, k ;q)Bn−1(aq2, kq2;q),(2.15)
where
ψn(a, k;q) = q(1 −aq)(1 −aq2)(1 −kqn)(1 −q−n)(√k−a2q/k +a/√k−k/aq)
(1 −k)(1 −kq)(1 −q√k)(1 −aq/√k)(1 −aq−n+1 /k)(1 −aqn+1 ).
Similarly, after tedious computations, it can also be shown that
Bn(a, k;q)−An−1(a, k;q) = χn(a, k ;q)Bn−1(aq2, kq2;q),(2.16)
where
χn(a, k;q) = (1 −aq)(1 −aq2)(1 −kqn) ˜χn(a, k;q)
(1 −k)(1 −q√k)(1 −kq)(1 −aqn)(1 −aqn+1 )(1 −k2qn−2/a2)(1 −kqn−1/a),
where
˜χn(a, k;q) = −k2qn−2/a2−k2qn−1/a2+k4q2n−3/a4−aqn+k2q2n−1/a −k4q3n−3/a3−√k
+k2qn−2√k/a2−k4q2n−3√k/a4+aqn√k−k2q2n−2√k/a −k2q2n−1√k/a +kqn−1/a
+k2/a2q−k3qn−2/a3+kqn−1−k3qn−2/a2+k4q2n−3/a3+qn√k−kq2n−1√k/a
+k3q2n−2√k/a3−kq2n−1√k+k2q3n−2√k/a +k3q2n−2√k/a2.
On adding (2.15) and (2.16), we get
An(a, k;q)−An−1(a, k;q) = φn(a, k ;q)Bn−1(aq2, kq 2;q),(2.17)
where
φn(a, k;q) = ψn(a, k;q) + χn(a, k ;q).
Now, let A′
n(a, k;q)and B′
n(a, k;q)denote the right-hand sides of (1.13) and (1.14) respectively. Then, we
have
A′
n(a, k;q)−B′
n(a, k;q) = (aq, k 2/a2q;q)n
(k, k/a;q)n−(aq , √k, k2/a2;q)n
(k, k/a, q √k;q)n
=(aq, k2/a2q;q)n
(k, k/a;q)n−(1 −√k)(aq , k2/a2;q)n
(1 −qn√k)(k, k/a;q)n
=(1 −qn)(√k−k2/a2q)(aq;q)n(k2/a2;q)n−1
(1 −qn√k)(k, k/a;q)n
=(1 −qn)(√k−k2/a2q)(kq2, k /a;q)n−1(aq;q)n(aq3, q√k, k 2/a2;q)n−1
(1 −√k)(k, k/a;q)n(aq3;q)n−1(kq2, k /a, q2√k;q)n−1
=q(1 −aq)(1 −aq2)(1 −kqn)(1 −q−n)(√k−a2q/k +a/√k−k/aq)
(1 −k)(1 −kq)(1 −q√k)(1 −aq/√k)(1 −aq−n+1 /k)(1 −aqn+1 )

15
×B′
n−1(aq2, kq 2;q).
Thus, we have
A′
n(a, k;q)−B′
n(a, k;q) = ψ′
n(a, k;q)B′
n−1(aq2, kq 2;q),(2.18)
where
ψ′
n(a, k;q) = q(1 −aq)(1 −aq2)(1 −kqn)(1 −q−n)(√k−a2q/k +a/√k−k/aq)
(1 −k)(1 −kq)(1 −q√k)(1 −aq/√k)(1 −aq−n+1 /k)(1 −aqn+1 ).
Again, it can be shown that
B′
n(a, k;q)−A′
n−1(a, k;q) = χ′
n(a, k;q)B′
n−1(aq2, kq 2;q),(2.19)
where
χ′
n(a, k;q) = (1 −aq)(1 −aq2)(1 −kqn) ˜χ′
n(a, k;q)
(1 −k)(1 −q√k)(1 −kq)(1 −aqn)(1 −aqn+1 )(1 −k2qn−2/a2)(1 −kqn−1/a),
where
˜χ′
n(a, k;q) = −k2qn−2/a2−k2qn−1/a2+k4q2n−3/a4−aqn+k2q2n−1/a −k4q3n−3/a3−√k
+k2qn−2√k/a2−k4q2n−3√k/a4+aqn√k−k2q2n−2√k/a −k2q2n−1√k/a +kqn−1/a
+k2/a2q−k3qn−2/a3+kqn−1−k3qn−2/a2+k4q2n−3/a3+qn√k−kq2n−1√k/a
+k3q2n−2√k/a3−kq2n−1√k+k2q3n−2√k/a +k3q2n−2√k/a2.
On adding (2.18) and (2.19), we get
A′
n(a, k;q)−A′
n−1(a, k;q) = φ′
n(a, k;q)B′
n−1(aq2, kq 2;q),(2.20)
where
φ′
n(a, k;q) = ψ′
n(a, k;q) + χ′
n(a, k;q).
Now, observe that An(a, k;q)and A′
n(a, k;q)have the same initial conditions A0(a, k;q) = A′
0(a, k;q) = 1
and that Bn(a, k;q)and B′
n(a, k;q)have the same initial conditions B0(a, k;q) = B′
0(a, k;q) = 1. Also, note
that both An(a, k;q),Bn(a, k;q)and A′
n(a, k;q),B′
n(a, k;q)obey the same recurrences (2.17) and (2.20)
respectively since φn(a, k;q) = φ′
n(a, k;q). Hence,
An(a, k;q) = A′
n(a, k;q)
and
Bn(a, k;q) = B′
n(a, k;q).
Thus, we have a proof of the two 10W9identities due to Andrews-Berkovich (Theorem 1.9) using Pfaff’s
method. 

16 ARITRAM DHAR
ACK NOW LEDGMENTS
The author would like to thank George E. Andrews for advising him to look at Pfaff’s method as a com-
ment after his talk titled “On 5ψ5identities of Bailey” at the Number Theory Seminar in the Department of
Mathematics, University of Florida on March 21, 2023. The author would also like to thank his Ph.D. advisor,
Alexander Berkovich, for suggesting him to prove Theorem 1.9 and for his continuous support and encourage-
ment.
REF ER EN CE S
1. G. E. Andrews, Pfaff’s Method II: diverse applications, J. of Computational and Appl. Math., 68 (1996) 15–23.
2. G. E. Andrews and A. Berkovich, The WP-Bailey tree and its implications, J. Lond. Math. Soc. (2), 66 (2002) 529–549.
3. G. Gasper and M. Rahman, Basic Hypergeometric Series, vol. 96, Cambridge University Press, 2004.
4. F. H. Jackson, Transformations of q-series, Messenger Math., 39 (1910) 145–153.
5. N. N. Li and W. Chu, Terminating balanced 4φ3-series and very well-poised 8φ7-series, J. Math. Anal. Appl., 478 (2019) 1005-1019.
6. J. F. Pfaff, Observationes analyticae ad L. Euler Institutiones Calculi Integralis, Vol. IV, Supplem. II et IV, Historia de 1793, Nova
Acta Acad. Sci. Petropolitanae 11 (1797) 38–57.
DEPA RTM ENT O F MAT HEM ATI CS , UN IV ERS ITY O F FLO RI DA, G AI NE SV IL LE FL 32 61 1, U SA
Email address:aritramdhar@ufl.edu