ON q-ANALOGUES OF SOME SERIES FOR π AND π^{2}

QING-HU HOU^{†}, CHRISTIAN KRATTENTHALER^{‡} AND ZHI-WEI SUN^{†}

Abstract. We obtain a new q-analogue of the classical Leibniz series ∑_{k=0}^{∞}(−1)^{k}/(2k + 1) = π/4, namely

∑_{k=0}^{∞} \frac{(−1)^{k}q^{k(k+3)/2}}{1 − q^{2k+1}} = \frac{(q^{2}; q^{2})_{∞}(q^{8}; q^{8})_{∞}}{(q; q^{2})_{∞}(q^{4}; q^{8})_{∞}} ,

where q is a complex number with |q| < 1. We also show that the Zeilberger-type series ∑_{k=1}^{∞}(3k − 1)16^{k}/(k\binom{2k}{k})^{3} = π^{2}/2 has two qanalogues with |q| < 1, one of which is

∑_{n=0}^{∞} q^{n(n+1)/2} \frac{1 − q^{3n+2}}{1 − q} · \frac{(q; q)_{n}^{3}(−q; q)_{n}}{(q^{3}; q^{2})_{n}^{3}} = (1 − q)^{2} \frac{(q^{2}; q^{2})_{∞}^{4}}{(q; q^{2})_{∞}^{4}} .

1. Introduction

Let q be a complex number with |q| < 1. As usual, for n ∈ ℕ = {0, 1, 2, …} and a complex number a, we define the q-shifted factorial by

(a; q)_{n} = ∏_{k=0}^{n−1}(1 − aq^{k}).

(An empty product is considered to take the value 1, and thus (a; q)_{0} = 1.) We also adopt the standard notion

(a; q)_{∞} = lim_{n→∞}(a; q)_{n} = ∏_{k=0}^{∞}(1 − aq^{k}).

By the definition of the q-gamma function [6, p. 20], we have

\frac{(q^{2}; q^{2})_{∞}}{(q; q^{2})_{∞}} = Γ_{q^{2}} (\frac{1}{2}) (1 − q^{2})^{−1/2}.

2010 Mathematics Subject Classification. Primary 05A30, 33D15; Secondary 11B65, 33F10.

Key words and phrases. q-series, identity, WZ method, series for π and π^{2}. †The first and third author are supported by the National Natural Science Foundation

of China (grants 11771330 and 11571162, respectively).

‡The second author is partially supported by the Austrian Science Foundation FWF, grant S50-N15, in the framework of the Special Research Program “Algorithmic and Enumerative Combinatorics”.

Therefore,

lim_{q→1}(1 − q)\frac{(q^{2}; q^{2})_{∞}^{2}}{(q; q^{2})_{∞}^{2}} = Γ (\frac{1}{2})^{2} lim_{q→1} \frac{1 − q}{1 − q^{2}} = \frac{π}{2}  (1.1)

and

lim_{q→1}(1 − q^{2})\frac{(q^{4}; q^{4})_{∞}^{2}}{(q^{2}; q^{4})_{∞}^{2}} = \frac{π}{2} .

In view of this, Ramanujan’s formula

∑_{k=0}^{∞} \frac{(−q)^{k}}{1 − q^{2k+1}} = \frac{(q^{4}; q^{4})_{∞}^{2}}{(q^{2}; q^{4})_{∞}^{2}}  (1.2)

(equivalent to Example (iv) in [2, p. 139]) can be viewed as a q-analogue of Leibniz’s identity

∑_{k=0}^{∞} \frac{(−1)^{k}}{2k + 1} = \frac{π}{4} .  (1.3)

Recently, Guo and Liu [8] used the WZ method to deduce the identities

∑_{n=0}^{∞} q^{n^{2}} \frac{1 − q^{6n+1}}{1 − q} · \frac{(q; q^{2})_{n}^{2}(q^{2}; q^{4})_{n}}{(q^{4}; q^{4})_{n}^{3}} = \frac{(1 + q)(q^{2}; q^{4})_{∞}(q^{6}; q^{4})_{∞}}{(q^{4}; q^{4})_{∞}^{2}}  (1.4)

and

∑_{n=0}^{∞}(−1)^{n}q^{3n^{2}} \frac{1 − q^{6n+1}}{1 − q} · \frac{(q; q^{2})_{n}^{3}}{(q^{4}; q^{4})_{n}^{3}} = \frac{(q^{3}; q^{4})_{∞}(q^{5}; q^{4})_{∞}}{(q^{4}; q^{4})_{∞}^{2}} ,  (1.5)

which are q-analogues of Ramanujan’s formulas [3, p. 352]

∑_{n=0}^{∞}(6n + 1)\frac{(1/2)_{n}^{3}}{n!^{3}4^{n}} = ∑_{n=0}^{∞}(6n + 1)\frac{\binom{2n}{n}3}{256^{n}} = \frac{4}{π}

and

∑_{n=0}^{∞}(6n + 1)(−1)^{n} \frac{(1/2)_{n}^{3}}{n!^{3}8^{n}} = ∑_{n=0}^{∞}(6n + 1)\frac{\binom{2n}{n}3}{(−512)^{n}} = \frac{2\sqrt{2}}{π},

where (a)_{n} = ∏_{k=0}^{n−1}(a + k) is the Pochhammer symbol. Note that

\frac{(1/2)_{n}}{n!} = (−1)^{n}\binom{−1/2}{n} = \frac{\binom{2n}{n}}{4^{n}} for all n ∈ ℕ.

Sun [12] provided q-analogues of Euler’s formulas ζ(2) = π^{2}/6 and ζ(4) = π^{4}/90, but no q-analogue of Euler’s formula for ζ(2m) with m ⩾ 3 is known.

In 1993, Zeilberger [13] used the WZ method to show that

∑_{n=0}^{∞}(21n + 13)\frac{n!^{6}}{8(2n + 1)!^{3}} = ∑_{k=1}^{∞} \frac{21k − 8}{k^{3}\binom{2k}{k}^{3}} = \frac{π^{2}}{6} .

A complicated q-analogue of the identity ∑_{k=1}^{∞}(21k − 8)/(k^{3}\binom{2k}{k})^{3} = ζ(2) was given by Pilehrood and Pilehrood [9] in 2011. Following Zeilberger’s work, in 2008 Guillera [7, Identity 1] employed the WZ method to obtain the Zeilberger-type series

∑_{k=1}^{∞} \frac{(3k − 1)16^{k}}{k^{3}\binom{2k}{k}^{3}} = \frac{π^{2}}{2} .  (1.6)

In this paper we study q-analogues of the identities (1.3) and (1.6).

Now we state our main results.

Theorem 1.1. For |q| < 1 we have

∑_{k=0}^{∞} \frac{(−1)^{k}q^{k(k+3)/2}}{1 − q^{2k+1}} = \frac{(q^{2}; q^{2})_{∞}(q^{8}; q^{8})_{∞}}{(q; q^{2})_{∞}(q^{4}; q^{8})_{∞}} .  (1.7)

The above identity gives a new q-analogue of Leibniz’s identity (1.3) since

lim_{q→1}(1 − q^{2})\frac{(q^{2}; q^{2})_{∞}(q^{8}; q^{8})_{∞}}{(q; q^{2})_{∞}(q^{4}; q^{8})_{∞}} = lim_{q→1} Γ_{q^{2}} (\frac{1}{2}) Γ_{q^{8}} (\frac{1}{2}) = Γ (\frac{1}{2})^{2} = π.

Theorem 1.2. For |q| < 1 we have

∑_{n=0}^{∞} q^{2n(n+1)}(1 + q^{2n+2} − 2q^{4n+3})\frac{(q^{2}; q^{2})_{n}^{3}}{(q; q^{2})_{n+1}^{3}(−1; q)_{2n+3}} = \frac{1}{2} ∑_{n=0}^{∞} \frac{q^{2n}}{(1 − q^{2n+1})^{2}}

(1.8)

and

∑_{n=0}^{∞} q^{n(n+1)/2} \frac{1 − q^{3n+2}}{1 − q} · \frac{(q; q)_{n}^{3}(−q; q)_{n}}{(q^{3}; q^{2})_{n}^{3}} = (1 − q)^{2} \frac{(q^{2}; q^{2})_{∞}^{4}}{(q; q^{2})_{∞}^{4}} .  (1.9)

Multiplying both sides of (1.8) by (1 − q)^{2} and then letting q → 1, we obtain

\frac{1}{4} ∑_{n=0}^{∞}(3n + 2)\frac{2^{4n}n!^{6}}{(2n + 1)!^{3}} = \frac{1}{2} ∑_{n=0}^{∞} \frac{1}{(2n + 1)^{2}} = \frac{1}{2} (1 − \frac{1}{4}) ζ(2) = \frac{π^{2}}{16},

which is equivalent to (1.6). In view of (1.1) and the fact that

lim_{q→1} q^{n(n+1)/2} \frac{1 − q^{3n+2}}{1 − q} · \frac{(q; q)_{n}^{3}(−q; q)_{n}}{(q^{3}; q^{2})_{n}^{3}} = \frac{(3n + 2)16^{n+1}}{2(n + 1)^{3}\binom{2n+2}{n+1}^{3}} ,

the identity (1.9) is also a q-analogue of (1.6). The expansions of both sides of (1.9) are

1 + 2q − q^{2} + 3q^{4} − 6q^{5} + 3q^{6} + 8q^{7} − 16q^{8} + 8q^{9} + 10q^{10} + · · · .

In [11, Conjecture 1.4], Sun presented several conjectural identities similar to Zeilberger-type series (one of which is ∑_{k=1}^{∞}(10k − 3)8^{k}/(k^{3}\binom{2k}{k}^{2}\binom{3k}{k}) = π^{2}/2), but we could not find q-analogues of them.

We are going to show Theorems 1.1 and 1.2 in Sections 2 and 3 respectively. Finally, in Section 4, we give alternative proofs for (1.4) and (1.5).

2. Proof of Theorem 1.1

As usual, for x ∈ ℤ we let T_{x} denote the triangular number x(x + 1)/2. Lemma 2.1. Let n ∈ ℕ, and define

t_{2}(n) := |{(x, y) ∈ ℕ^{2} : T_{x} + 4T_{y} = n}|.

Then

t_{2}(n) = _{∑_{d|8n+5} ; d<\sqrt{8n+5}}(−1)^{(d−1)/2}.  (2.1)

Proof. By Theorem 3.2.1 of [4, p. 56], for any positive integer m we have

r_{2}(m) = 4 ∑_{2∤d|m}(−1)^{(d−1)/2},

where r_{2}(m) := |{(x, y) ∈ ℤ^{2} : x^{2} + y^{2} = m}|. Observe that

t_{2}(n) =|{(x, y) ∈ ℕ^{2} : (2x + 1)^{2} + 4(2y + 1)^{2} = 8n + 5}|
=\frac{1}{4}|\{(x, y) ∈ ℤ^{2} : x^{2} + (2y)^{2} = 8n + 5\}| = \frac{1}{8}r_{2}(8n + 5)
=\frac{1}{2} ∑_{d|8n+5}(−1)^{(d−1)/2} = _{∑_{d|8n+5} ; d<\sqrt{8n+5}} \frac{(−1)^{(d−1)/2} + (−1)^{((8n+5)/d−1)/2}}{2}
=_{∑_{d|8n+5} ; d<\sqrt{8n+5}}(−1)^{(d−1)/2}.

This proves (2.1).

As usual, for |q| < 1 we define

ψ(q) := ∑_{n=0}^{∞} q^{T_{n}}.

By a known formula of Gauß (cf. (1.3.14) of [4, p. 11]),

ψ(q) = \frac{(q^{2}; q^{2})_{∞}}{(q; q^{2})_{∞}} .  (2.2)

First proof of (1.7). Let L and R denote the left-hand side and the righthand side of (1.7), respectively. In view of Gauß’ identity (2.2) and (2.1), we have

R = ψ(q)ψ(q^{4}) = ∑_{n=0}^{∞} t_{2}(n)q^{n} = ∑_{n=0}^{∞} (_{∑_{d|8n+5} ; d<\sqrt{8n+5}}(−1)^{(d−1)/2})q^{n}.

On the other hand,

L = ∑_{k=0}^{∞}(−1)^{k} ∑_{m=0}^{∞} q^{k(k+3)/2+(2k+1)m}

= ∑_{k=0}^{∞} ∑_{m=0}^{∞}(−1)^{k}q^{((2k+1)(2k+1+4(2m+1))−5)/8} = ∑_{n=0}^{∞}_{∑_{d|8n+5} ; d<\sqrt{8n+5}}(−1)^{(d−1)/2}q^{n}.

Therefore (1.7) is valid.

Second proof of (1.7). Recall the standard basic hypergeometric notation

rϕ_{s}[\begin{matrix}a_{1}, …, a_{r} \\ b_{1}, …, b_{s} \end{matrix}; q, z] = ∑_{ℓ=0}^{∞} \frac{(a_{1}; q)_{ℓ} ⋯ (a_{r}; q)_{ℓ}}{(q; q)_{ℓ}(b_{1}; q)_{ℓ} ⋯ (b_{s}; q)_{ℓ}} ((−1)^{ℓ}q^{\binom{ℓ}{2}})^{s−r+1} z^{ℓ}

(with (a_{1}; q)_{ℓ} ⋯ (a_{r}; q)_{ℓ} often abbreviated as (a_{1}, …, a_{r}; q)_{ℓ}). We start with the transformation formula (cf. [6, Eq. (3.10.4)])

10ϕ_{9}[\begin{matrix}a, \sqrt{a}q, −\sqrt{a}q, b, x, −x, y, −y, \\ \sqrt{a}, −\sqrt{a}, aq/b, aq/x, −aq/x, aq/y, −aq/y, \end{matrix}

\begin{matrix}−q^{−n}, q^{−n} \\ −aq^{n+1}, aq^{n+1} \end{matrix}; q, −\frac{a^{3}q^{2n+3}}{bx^{2}y^{2}} ]

= \frac{(a^{2}q^{2}, a^{2}q^{2}/x^{2}y^{2}; q^{2})_{n}}{(a^{2}q^{2}/x^{2}, a^{2}q^{2}/y^{2}; q^{2})_{n}} _{5}ϕ_{4}[\begin{matrix}q^{−2n}, x^{2}, y^{2}, −aq/b, −aq^{2}/b \\ x^{2}y^{2}/a^{2}q^{2n}, a^{2}q^{2}/b^{2}, −aq, −aq^{2} \end{matrix}; q^{2}, q^{2}] ,

where n is a nonnegative integer. In this identity, set a = q, x = y = \sqrt{q}, and let n → ∞. In this way, we obtain

4ϕ_{5}[\begin{matrix}q, b, \sqrt{q}, −\sqrt{q} \\ q^{2}/b, q^{3/2}, −q^{3/2}, 0, 0 \end{matrix}; q, \frac{q^{4}}{b} ]

= \frac{(q^{2}; q^{2})_{∞} (q^{4}; q^{2})_{∞}}{(q^{3}; q^{2})_{∞}^{2}}_{4}ϕ_{3}[\begin{matrix}q, −q^{2}/b, −q^{3}/b, q \\ −q^{3}, −q^{2}, q^{4}/b^{2} \end{matrix}; q^{2}, q^{2}] .

By performing the limit as b → 0, the above transformation formula is reduced to

3ϕ_{3}[\begin{matrix}q, \sqrt{q}, −\sqrt{q} \\ −q^{3/2}, q^{3/2}, 0 \end{matrix}; q, q^{2}] = \frac{(q^{2}; q^{2})_{∞} (q^{4}; q^{2})_{∞}}{(q^{3}; q^{2})_{∞}^{2}} _{2}ϕ_{2}[\begin{matrix}q, q \\ −q^{2}, −q^{3} \end{matrix}; q^{2}, q^{3}] .

If the left-hand side is written out explicitly, we see that it agrees with the left-hand side in (1.7) up to a multiplicative factor of 1 − q. On the other hand, the _{2}ϕ_{2}-series on the right-hand side can be evaluated by means of the summation formula (cf. [6, Ex. 1.19(i); Appendix (II.10)])

2ϕ_{2}[\begin{matrix}a, q/a \\ −q, b \end{matrix}; q, −b] = \begin{matrix}(ab, bq/a; q^{2}) & ∞ \\ (b; q) & ∞ \end{matrix} .

Thus, we arrive at

∑_{k=0}^{∞} \frac{(−1)^{k}q^{k(k+3)/2} (1 − q)}{1 − q^{2k+1}} = \frac{(q^{2}; q^{2})_{∞} (q^{4}; q^{2})_{∞} (−q^{4}; q^{4})_{∞}^{2}}{(−q^{3}; q^{2})_{∞} (q^{3}; q^{2})_{∞}^{2}} ,

which is indeed equivalent to (1.7).

3. Proof of Theorem 1.2

Proof of (1.8). We construct a q-analogue of the WZ pair given by Guillera [7, Identity 1].

Recall that a pair of bivariate functions (F (n, k), G(n, k)) is called a WZ pair [10, Chapter 7] if

F (n + 1, k) − F (n, k) = G(n, k + 1) − G(n, k).

It was shown (cf. [1]) that

∑_{n=0}^{∞} G(n, 0) = lim_{k→∞} ∑_{n=0}^{∞} G(n, k) + ∑_{k=0}^{∞} F (0, k).  (3.1)

We make the following construction. Let

F_{q}(n, k) = 4 · \frac{1 − q^{2n}}{1 − q} · B_{q}(n, k)

and

G_{q}(n, k) = \frac{4(1 + q^{2n+1} − 2q^{4n+2k+1})}{(1 − q)(1 + q^{2n})(1 + q^{2n+1})}B_{q}(n, k),

where

B_{q}(n, k) = \frac{(q; q^{2})_{k}^{2}(q; q^{2})_{n}^{3}}{(q^{2n+2}; q^{2})_{k}^{2}(q^{2}; q^{2})_{n}^{3}(−1; q)_{2n}} q^{2n^{2}+4nk}.

We can extend the definition of B_{q}(n, k) from nonnegative integers n, k to any real numbers n, k by defining

(a; q)_{n} = \frac{(a; q)_{∞}}{(aq^{n}; q)_{∞}} .

Let a be a positive real number. It is straightforward to check that (F_{q}(n + a, k), G_{q}(n + a, k)) is a WZ pair. Observing that B_{q}(n, k) contains the factor q^{4nk}, we get

lim_{k→∞} ∑_{n=0}^{∞} G_{q}(n + a, k) = 0.

Thus,

∑_{n=0}^{∞} G_{q}(n + a, 0) = ∑_{n=0}^{∞} F_{q}(a, k).  (3.2)

Setting a = 1/2 and noting that

(q; q^{2})_{n+1/2} = \frac{(q; q^{2})_{∞}}{(q · q^{2n+1}; q^{2})_{∞}} = \frac{(q; q^{2})_{∞}(q^{2}; q^{2})_{n}}{(q^{2}; q^{2})_{∞}}

and

(q^{2}; q^{2})_{n+1/2} = \frac{(q^{2}; q^{2})_{∞}}{(q^{2} · q^{2n+1}; q^{2})_{∞}} = \frac{(q^{2}; q^{2})_{∞}(q; q^{2})_{n+1}}{(q; q^{2})_{∞}}

for any n ∈ ℕ, we infer that

F_{q} (\frac{1}{2}, k) = \frac{2q^{1/2}}{1 − q} · \frac{q^{2k}}{(1 − q^{2k+1})^{2}} · \frac{(q; q^{2})_{∞}^{6}}{(q^{2}; q^{2})_{∞}^{6}} ,

and

G_{q} (n + \frac{1}{2}, 0) =\frac{4q^{2n^{2}+2n+1/2}(1 + q^{2n+2} − 2q^{4n+3})}{(1 − q)}

× \frac{(q; q^{2})_{∞}^{3}(q^{2n+3}; q^{2})_{∞}^{3}}{(−1; q)_{2n+3}(q^{2n+2}; q^{2})_{∞}^{3}(q^{2}; q^{2})_{∞}^{3}} .

After cancelling the common factors, we arrive at (1.8).

Remark 3.1. Guillera [7] obtained the identity (1.6) via the WZ pair

F (n, k) = 8nB(n, k), G(n, k) = (6n + 4k + 1)B(n, k),

where

B(n, k) = \frac{(2k)!^{2}(2n)!^{3}}{2^{8n+4k}(n + k)!^{2}k!^{2}n!^{4}} .

Since

lim_{q→1} F_{q}(n, k) = F (n, k) and lim_{q→1} G_{q}(n, k) = G(n, k),

the pair (F_{q}(n, k), G_{q}(n, k)) is indeed a q-analogue of the pair (F (n, k), G(n, k)).

Proof of (1.9). We start from the quadratic transformation formula (cf. [6, Eq. (3.8.13)])

∑_{n=0}^{∞} \frac{(1 − aq^{3n}) (a, d, aq/d; q^{2})_{n} (b, c, aq/bc; q)_{n}}{(1 − a) (q, aq/d, d; q)_{n} (aq^{2}/b, aq^{2}/c, bcq; q^{2})_{n}} q^{n}

= \frac{(aq^{2}, bq, cq, aq^{2}/bc; q^{2})_{∞}}{(q, aq^{2}/b, aq^{2}/c, bcq; q^{2})_{∞}}_{3}ϕ_{2}[\begin{matrix}b, c, aq/bc \\ dq, aq^{2}/d \end{matrix}; q^{2}, q^{2}] . (3.3)

We set a = q^{2} and b = c = q. This yields

∑_{n=0}^{∞} \frac{(1 − q^{3n+2}) (q^{2}, d, q^{3}/d; q^{2})_{n} (q, q, q; q)_{n}}{(1 − q^{2}) (q, q^{3}/d, d; q)_{n} (q^{3}, q^{3}, q^{3}; q^{2})_{n}} q^{n}

= \frac{(q^{4}, q^{2}, q^{2}, q^{2}; q^{2})_{∞}}{(q, q^{3}, q^{3}, q^{3}; q^{2})_{∞}}_{3}ϕ_{2}[\begin{matrix}q, q, q \\ dq, q^{4}/d \end{matrix}; q^{2}, q^{2}] . (3.4)

At this point, we observe that the limit d → 0 applied to the left-hand side of (3.4) produces exactly the left-hand side of (1.9). Thus, it remains to show the limit d → 0 applied to the right-hand side of (3.4) yields the right-hand side of (1.9).

In order to see this, we take recourse to the transformation formula (cf. [6, Eq. (3.3.1) or Appendix (III.34)])

3ϕ_{2}[\begin{matrix}A, B, C \\ D, E \end{matrix}; q, q] = \frac{(BCq/E, q/E; q)_{∞}}{(Cq/E, Bq/E; q)_{∞}}_{3}ϕ_{2}[\begin{matrix}D/A, B, C \\ D, BCq/E \end{matrix}; q, \frac{Aq}{E} ]

− \frac{(q/E, A, B, C, Dq/E; q)_{∞}}{(Cq/E, Bq/E, D, E/q, Aq/E; q)_{∞}}_{3}ϕ_{2}[\begin{matrix}Cq/E, Bq/E, Aq/E \\ Dq/E, q^{2}/E \end{matrix}; q, q] .

Here we replace q by q^{2} and set A = B = C = q, D = dq, and E = q^{4}/d, to obtain

3ϕ_{2}[\begin{matrix}q, q, q \\ dq, q^{4}/d \end{matrix}; q^{2}, q^{2}] = \frac{(d, d/q^{2}; q^{2})_{∞}}{(d/q, d/q; q^{2})_{∞}}_{2}ϕ_{1}[\begin{matrix}q, q \\ dq \end{matrix}; q^{2}, \frac{d}{q}]

− \frac{(d/q^{2}, q, q, q, d^{2}/q; q^{2})_{∞}}{(d/q, d/q, dq, q^{2}/d, d/q; q^{2})_{∞}}_{3}ϕ_{2}[\begin{matrix}d/q, d/q, d/q \\ d^{2}/q, d \end{matrix}; q^{2}, q^{2}] .

From here it is evident that

lim_{d→0}_{3}ϕ_{2}[\begin{matrix}q, q, q \\ dq, q^{4}/d \end{matrix}; q^{2}, q^{2}] = 1.  (3.5)

Indeed, the first term on the right-hand side of (3.5) converges trivially to 1, while in the second term everywhere the substitution d = 0 is fine — and thus produces a well-defined finite value —, except for the factor (q^{2}/d, q^{2})_{∞} in the denominator. However, as d → 0 (to be precise, we take d = q^{2}/D with D → −∞), this term becomes unbounded, whence the whole term tends to 0. Thus, performing this limit, we get

∑_{n=0}^{∞} \frac{(1 − q^{3n+2}) (q^{2}; q^{2})_{n} (q, q, q; q)_{n}}{(1 − q^{2}) (q; q)_{n} (q^{3}, q^{3}, q^{3}; q^{2})_{n}} q^{n+\binom{n}{2}}

= \frac{(q^{4}, q^{2}, q^{2}, q^{2}; q^{2})_{∞}}{(q, q^{3}, q^{3}, q^{3}; q^{2})_{∞}} = \frac{(1 − q)^{3} (q^{2}, q^{2}, q^{2}, q^{2}; q^{2})_{∞}}{(1 − q^{2}) (q, q, q, q; q^{2})_{∞}} ,

as desired.

4. Alternative proofs of the identities of Guo and Liu

In this last section, we provide alternative proofs of (1.4) and (1.5), showing that they can be obtained as special/limiting cases of a quadratic summation formula due to Gasper and Rahman [5].

Proof of (1.4). We start with the quadratic summation formula (cf. [6, Eq. (3.8.12)])

∑_{k=0}^{∞} \frac{1 − aq^{3k}}{1 − a} · \frac{(a, b, q/b; q)_{k}}{(q^{2}, aq^{2}/b, abq; q^{2})_{k}} · \frac{(d, f, a^{2}q/df; q^{2})_{k}}{(aq/d, aq/f, df/a; q)_{k}} · q^{k}

+ \frac{(aq, f/a, b, f/q; q)_{∞}}{(a/f, fq/a, aq/d, df/a; q)_{∞}} · \frac{(d, a^{2}q/bd, a^{2}q/df, fq^{2}/d, df^{2}q/a^{2}; q^{2})_{∞}}{(aq^{2}/b, abq, fq/ab, bf/a, aq^{2}/bf; q^{2})_{∞}}

×_{3}ϕ_{2}[\begin{matrix}f, bf/a, fq/ab \\ fq^{2}/d, df^{2}q/a^{2} \end{matrix}; q^{2}, q^{2}]

= \frac{(aq, f/a; q)_{∞}}{(aq/d, df/a; q)_{∞}} · \frac{(aq^{2}/bd, abq/d, bdf/a, dfq/ab; q^{2})_{∞}}{(aq^{2}/b, abq, bf/a, fq/ab; q^{2})_{∞}} .

Now replace f by a^{2}q^{2N+1}/d, with N a positive integer. The effect is that, because of the factor (a^{2}q/df; q^{2})_{∞}, this kills off the second term on the lefthand side. In other words, now this is indeed a genuine summation formula. Now replace q by q^{2} and choose a = b = q. Then the above identity reduces to

∑_{k=0}^{∞} \frac{1 − q^{6k+1}}{1 − q} · \frac{(d, q^{4N+4}/d, q^{−4N}; q^{4})_{k}}{(q^{4}, q^{4}, q^{4}; q^{4})_{k}} · \frac{(q, q, q; q^{2})_{k}}{(q^{3}/d, dq^{−4N−1}, q^{4N+3}; q^{2})_{k}} · q^{2k}

= \frac{(q^{3}, q^{4N+3}/d; q^{2})_{∞}}{(q^{3}/d, q^{4N+3}; q^{2})_{∞}} · \frac{(q^{4}/d, q^{4}/d, q^{4N+4}, q^{4N+4}; q^{4})_{∞}}{(q^{4}, q^{4}, q^{4N+4}/d, q^{4N+4}/d; q^{4})_{∞}} . (4.1)

Finally, we set d = q^{2} and let N → ∞. Upon little simplification, we arrive at (1.4).

Proof of (1.5). We proceed in a similar manner. In (4.1), we set d = q^{−2N} with N a positive integer. This yields the identity

∑_{k=0}^{∞} \frac{1 − q^{6k+1}}{1 − q} · \frac{(q^{−2N}, q^{6N+4}, q^{−4N}; q^{4})_{k}}{(q^{4}, q^{4}, q^{4}; q^{4})_{k}} · \frac{(q, q, q; q^{2})_{k}}{(q^{2N+3}, q^{−6N−1}, q^{4N+3}; q^{2})_{k}} · q^{2k}

= \frac{(q^{3}; q^{2})_{N}(q^{2N+4}; q^{4})_{N}^{2}}{(q^{4}; q^{4})_{N}^{2}(q^{4N+3}; q^{2})_{N}} .

Letting N → ∞, we then obtain (1.5).

Acknowledgment. The authors would like to thank the anonymous referee for helpful comments.

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(Qing-Hu Hou) School of Mathematics, Tianjin University, Tianjin 300350, People’s Republic of China

E-mail address: qh hou@tju.edu.cn
Homepage:  http://cam.tju.edu.cn/~hou

(Christian Krattenthaler) Fakultät für Mathematik, Universität Wien, Oskar-Morgenstern-Platz 1, A-1090 Vienna, Austria

E-mail address: christian.krattenthaler@univie.ac.at
Homepage:  http://www.mat.univie.ac.at/~kratt

(Zhi-Wei Sun) Department of Mathematics, Nanjing University, Nanjing 210093, People’s Republic of China

E-mail address: zwsun@nju.edu.cn
Homepage:  http://maths.nju.edu.cn/~zwsun