q-SERIES IDENTITIES AND VALUES OF CERTAIN L-FUNCTIONS

George E. Andrews, Jorge Jiménez-Urroz and Ken Ono

September 1, 2000.

1. Introduction and Statement of Results.

As usual, define Dedekind’s eta function η(z) by the infinite product

(1.1)  η(z) := q^{1/24} ∏_{n=1}^{∞}(1 − q^{n}). (q := e^{2πiz} throughout)

In a recent paper, Zagier [Z; Theorem 2] proved that (note. empty products equal 1 throughout)

(1.2)  ∑_{n=0}^{∞} (η(24z) − q(1 − q^{24})(1 − q^{48}) ⋯ (1 − q^{24n})) = η(24z)D(q) + E(q)

where the series D(q) and E(q) are defined by

D(q) = −\frac{1}{2} + ∑_{n=1}^{∞} \frac{q^{24n}}{1 − q^{24n}} = −\frac{1}{2} + ∑_{n=1}^{∞} d(n)q^{24n} = −\frac{1}{2} + q^{24} + 2q^{48} + 2q^{72} + 3q^{96} + …,

E(q) = \frac{1}{2} ∑_{n=1}^{∞} (\frac{12}{n})nq^{n^{2}} = \frac{1}{2}q − \frac{5}{2}q^{25} − \frac{7}{2}q^{49} + \frac{11}{2}q^{121} + ….

Here d(n) denotes the number of positive divisors of n. This identity plays an important role in Zagier’s work on Vassiliev invariants in knot theory [Z].

Two other similar identities were known, and they were noticed by the first author in connection with one of Ramanujan’s mock theta functions. In [A2], the first author proved that

(1.3)  ∑_{n=0}^{∞} (\frac{η(48z)}{η(24z)} − q(1 + q^{24})(1 + q^{48}) ⋯ (1 + q^{24n})) = \frac{η(48z)}{η(24z)}D(q) + \frac{M_{1}(q)}{2},

Typeset by AMS-TEX

The first and third authors thank the National Science Foundation for its generous research support. The second author thanks PB98-0067 for their support. The third author thanks the Alfred P. Sloan Foundation and the David and Lucile Packard Foundation for their support.

(1.4) ∑_{n=0}^{∞} (\frac{η(48z)}{η(24z)} − \frac{q}{(1 − q^{24})(1 − q^{72}) ⋯ (1 − q^{24(2n+1)})}) = \frac{η(48z)}{η(24z)}D(q^{2}) + \frac{M_{1}(q)}{2}

where M_{1}(q) is the mock theta function given by

(1.5) M_{1}(q) = q + ∑_{n=1}^{∞} \frac{q^{12n^{2}+12n+1}}{(1 + q^{24})(1 + q^{48}) ⋯ (1 + q^{24n})} = q + q^{25} − q^{49} + 2q^{73} − · · · .

The q-series of the function M_{1}(q) was the focus of two extensive studies [A-D-H, C]. Although M_{1}(q) is not the Fourier expansion of a modular form, these works show that the coefficients of M_{1}(q) are given by a Hecke character for the quadratic field ℚ(\sqrt{6}). In particular, M_{1}(q) enjoys nice properties that one expects for certain weight 1 cusp forms. For these reasons, we shall refer to M_{1}(q) and M_{2}(q) (defined in (1.8)) as mock theta functions although they do not exactly fit Ramanujan’s original definition [A3; p. 291].

In view of identities (1.2-4), it is natural to investigate the general behavior of q-series which are obtained by summing the iterated differences between an infinite product and its truncated products. Here we establish two general theorems which yield infinitely many such identities, and we illustrate how such identities are useful in determining the values at negative integers for certain L-functions.

We shall employ the standard notation

(1.6)  (A; q)_{n} = ∏_{j=0}^{∞} \frac{(1 − Aq^{j})}{(1 − Aq^{n+j})},

and throughout we assume that |q| < 1 and that the other parameters are restricted to domains that do not contain any singularities of the series or products under consideration.

Theorem 1.

∑_{n=0}^{∞} (\frac{(t; q)_{∞}}{(a; q)_{∞}} − \frac{(t; q)_{n}}{(a; q)_{n}} )

= ∑_{n=1}^{∞} \frac{(q/a; q)_{n}(a/t)^{n}}{(q/t; q)_{n}}

+ \frac{(t; q)_{∞}}{(a; q)_{∞}} (∑_{n=1}^{∞} \frac{q^{n}}{1 − q^{n}} + ∑_{n=1}^{∞} \frac{q^{n}t^{−1}}{1 − q^{n}t^{−1}} − ∑_{n=0}^{∞} \frac{tq^{n}}{1 − tq^{n}} − ∑_{n=0}^{∞} \frac{aq^{n}t^{−1}}{1 − aq^{n}t^{−1}} ) .

Theorem 2.

∑_{n=0}^{∞} (\frac{(a; q)_{∞}(b; q)_{∞}}{(q; q)_{∞}(c; q)_{∞}} − \frac{(a; q)_{n}(b; q)_{n}}{(q; q)_{n}(c; q)_{n}} )

= \frac{(b; q)_{∞}(a; q)_{∞}}{(c; q)_{∞}(q; q)_{∞}} (∑_{n=1}^{∞} \frac{q^{n}}{1 − q^{n}} − ∑_{n=0}^{∞} \frac{aq^{n}}{1 − aq^{n}} − ∑_{n=1}^{∞} \frac{(c/b; q)_{n}b^{n}}{(a; q)_{n}(1 − q^{n})}) .

q-SERIES IDENTITIES 3

Many interesting specializations of these two theorems yield identities for modular forms that are eta-products (including identities (1.2-4)). Here we highlight ten of these identities. First we fix notation. We let \sqrt{Θ} be the operator defined by

(1.7)  \sqrt{Θ} (∑_{n=0}^{∞} a(n)q^{n}) = ∑_{n=0}^{∞} \sqrt{n}a(n)q^{n}.

It is easy to see that the series E(q) in (1.2) is given by

E(q) = \sqrt{Θ} (η(24z)) /2.

In addition to the mock theta function M_{1}(q), we shall require the mock theta function M_{2}(q) defined by

(1.8)  M_{2}(q) = ∑_{n=1}^{∞} \frac{(−1)^{n}q^{24n^{2}−1}}{(1 − q^{24})(1 − q^{72}) ⋯ (1 − q^{24(2n−1)})} = −q^{23} − q^{47} − · · · .

See [A-D-H] for a detailed study of this function.

The ten eta-products F_{1}(z), F_{2}(z), …, F_{10}(z) we consider are of the form

F_{i}(z) = η^{a_{i}}(δ_{i}z)η^{b_{i}}(2δ_{i}z)

with a_{i} ≠ 0. Obviously, each F_{i}(z) is a modular form of weight (a_{i} + b_{i})/2. For each F_{i}(z) we define quantities c_{i} and f_{i}(j), which are not necessarily unique, for which

(1.9)  F_{i}(z) = c_{i} ∏_{j=1}^{∞} f_{i}(j).

These are listed in the table below.

Table 1.

\begin{matrix}i & F_{i}(z) & δ_{i} & c_{i} & f_{i}(j) \\ 1 & 1/η(24z) & 24 & q^{−1} & 1/(1 − q^{24j}) \\ 2 & η(2z)/η^{2}(z) & 1 & 1 & (1 + q^{j})/(1 − q^{j}) \\ 3 & η(8z)/η^{2}(16z) & 8 & q^{−1} & (1 − q^{16j−8})/(1 − q^{16j}) \\ 4 & η(48z)/η(24z) & 24 & q & 1 + q^{24j} \\ 5 & η(48z)/η(24z) & 24 & q/(1 − q^{24}) & 1/(1 − q^{24(2j+1)}) \\ 6 & η(24z)/η(48z) & 24 & q^{−1} & 1/(1 + q^{24j}) \\ 7 & η(24z)/η(48z) & 24 & q^{−1} & 1 − q^{24(2j−1)} \\ 8 & η(24z) & 24 & q & 1 − q^{24j} \\ 9 & η^{2}(z)/η(2z) & 1 & 1 & (1 − q^{j})/(1 + q^{j}) \\ 10 & η^{2}(16z)/η(8z) & 8 & q/(1 − q^{8}) & (1 − q^{16j})/(1 − q^{16j+8}) \end{matrix}

If δ ∈ {1, 8, 24}, then let d_{δ}(n) be the divisor function defined by

(1.10)  d_{δ}(n) = \{ \begin{matrix}d(n) = ∑_{d|n} 1 & if δ = 24, \\ ∑_{d|n}(−1)^{d} & if δ = 8, \\ ∑_{d|n odd} 1 & if δ = 1. \end{matrix}

Also, for each i define α_{i} by

(1.11)  α_{i} = \{ \begin{matrix}−\frac{1}{2} & if (a_{i} + 2b_{i})δ_{i} = 24, \\ 0 & otherwise. \end{matrix}

Notice that α_{i} = −1/2 if and only if the order of vanishing of F_{i}(z) at ∞ is 1. The last quantity we require is γ_{i} which is defined by

(1.12)  γ_{i} = \{ \begin{matrix}2 & if i = 5, 7, \\ 1 & otherwise. \end{matrix}

Theorem 3. If 1 ≤ i ≤ 10, then

∑_{n=0}^{∞} (F_{i}(z) − c_{i} ∏_{j=1}^{n} f_{i}(j)) = (1 + [1/δ_{i}]) F_{i}(z)D_{i}(q) + G_{i}(q)

where [•] denotes the greatest integer function,

D_{i}(q) = α_{i} + ∑_{n=1}^{∞} d_{δ_{i}}(n)q^{δ_{i}γ_{i}n}

and

G_{i}(q) = \{ \begin{matrix}0 & if i = 1, 2, 3, \\ M_{1}(q)/2 & if i = 4, 5, \\ 2M_{2}(q)/γ_{i} & if i = 6, 7, \\ (−α_{i} + [2/δ_{i}]) \sqrt{Θ} (F_{i}(z)) & if i = 8, 9, 10. \end{matrix}

The three forms F_{1}(z), F_{2}(z) and F_{3}(z) have weight -1/2 and the four forms F_{4}(z), F_{5}(z), F_{6}(z) and F_{7}(z) have weight 0. The remaining three forms have weight 1/2. The series G_{4}(z), G_{5}(z), G_{6}(z) and G_{7}(z) are mock theta functions, whereas G_{8}(q), G_{9}(q) and G_{10}(q) are the half-derivatives of F_{8}(z), F_{9}(z) and F_{10}(z). In other words, the “error series” G_{i}(q) in Theorem 3 satisfy

F_{i}(z) ⟶ G_{i}(q) ∼ \{ \begin{matrix}0 & if F_{i}(z) has weight -1/2, \\ Mock Theta function & if F_{i}(z) has weight 0, \\ \sqrt{Θ}(F_{i}(z)) & if F_{i}(z) has weight 1/2. \end{matrix}

Although these identities are elegant in their own right, they are also often useful in calculating the values of L-functions at negative integers. In particular, they lead to analogs of the classical result

\frac{t}{e^{t} − 1} = 1 + ∑_{n=1}^{∞}(−1)^{n+1}ζ(1 − n) · \frac{t^{n}}{(n − 1)!},

where ζ(s) is the Riemann zeta-function. In this direction, Zagier used (1.2) to show that

(1.13) −e^{−t/24} ∑_{n=0}^{∞}(1 − e^{−t})(1 − e^{−2t}) ⋯ (1 − e^{−nt}) = \frac{1}{2} ∑_{n=0}^{∞}(−1/24)^{n} · L(χ_{12}, −2n − 1) · \frac{t^{n}}{n!} ,

where χ_{12} is the Dirichlet character with modulus 12 defined by

χ_{12}(n) := \{ \begin{matrix}1 & if n ≡ 1, 11 (mod 12), \\ −1 & if n ≡ 5, 7 (mod 12), \\ 0 & otherwise. \end{matrix}

Here we illustrate the generality of this phenomenon by proving the following theorems.

Theorem 4. As a power series in t, we have

−\frac{1}{4} · ∑_{n=0}^{∞} (\frac{(1 − e^{−t})(1 − e^{−2t}) ⋯ (1 − e^{−nt})}{(1 + e^{−t})(1 + e^{−2t}) ⋯ (1 + e^{−nt})}) = ∑_{n=0}^{∞}(−1)^{n}(4^{n+1} − 1) · ζ(−2n − 1) · \frac{t^{n}}{n!} .

In addition to ζ(s), we shall consider the Dirichlet L-function

(1.14)  L(χ_{2}, s) := ∑_{n=1}^{∞} \frac{χ_{2}(n)}{n^{s}},

where

(1.15)  χ_{2}(n) := \{ \begin{matrix}1 & if n ≡ 1, 7 (mod 8), \\ −1 & if n ≡ 3, 5 (mod 8), \\ 0 & otherwise. \end{matrix}

Theorem 5. As a power series in t, we have

−2e^{−t/8} ∑_{n=0}^{∞} (\frac{(1 − e^{−2t})(1 − e^{−4t}) ⋯ (1 − e^{−2nt})}{(1 + e^{−t})(1 + e^{−3t}) ⋯ (1 + e^{−(2n+1)t})}) = ∑_{n=0}^{∞}(−1/8)^{n} · L(χ_{2}, −2n − 1) · \frac{t^{n}}{n!} .

We shall also consider the Hecke L-function

(1.16)  L(ρ, s) = ∑_{n=1}^{∞} \frac{a(n)}{n^{s}} := ∑_{𝔞⊆ℤ[\sqrt{6}]} χ(𝔞)N𝔞^{−s}

where χ is the order 2 character of conductor 4(3 + \sqrt{6}) on ideals in ℤ[\sqrt{6}] defined by

(1.17)  χ(𝔞) := \{ \begin{matrix}i^{yx}(\frac{12}{x}) & if y is even, \\ i^{yx+1}(\frac{12}{x}) & if y is odd, \end{matrix}

when 𝔞 = (x + y\sqrt{6}). If 1 ≤ r < 48 is an integer, then let L_{r}(ρ, s) be the partial L-function defined by

(1.18)  L_{r}(ρ, s) := ∑_{n≡r (mod 48)} \frac{a(n)}{n^{s}}.

By the orthogonality of the Dirichlet characters modulo 48 and the analytic continuation of the associated twists of L(ρ, s), each L_{r}(ρ, s) has an analytic continuation to ℂ.

Theorem 6. As a power series in t, we have

−2e^{t/24} ∑_{n=0}^{∞}(1−e^{−t})(1−e^{−3t}) ⋯ (1−e^{−(2n−1)t}) = ∑_{n=0}^{∞}(−1/24)^{n}·(L_{23}(ρ, −n) + L_{47}(ρ, −n))·\frac{t^{n}}{n!} .

Theorem 7. As a power series in t, we have

−2e^{−t/24} ∑_{n=0}^{∞}(1 − e^{−t})(1 + e^{−2t}) ⋯ (1 + (−1)^{n}e^{−nt})

= ∑_{n=0}^{∞}(−1/24)^{n} · (L_{1}(ρ, −n) − L_{25}(ρ, −n)) · \frac{t^{n}}{n!} .

In §2 we recall certain facts about q-series and basic hypergeometric series, and we prove Theorems 1 and 2. In §3 we prove Theorem 3 and in §4 we prove Theorems 4, 5, 6, and 7. In §5 we examine the partition theoretic consequences of the identities for F_{1}(z) and F_{8}(z). In §6 we give a few more identities which are related to eta-products. The most interesting of these is

∑_{n=0}^{∞} (\frac{1}{η^{2}(24z)} − \frac{1}{q^{2}(1 − q^{24})^{2}(1 − q^{48})^{2} ⋯ (1 − q^{24n})^{2}} ) = \frac{1}{η^{2}(24z)} ∑_{n=1}^{∞} (d(n) + m(n)) q^{24n},

where m(n) denotes the number of middle divisors of n. A divisor is a middle divisor if it lies in the interval [\sqrt{n/2}, \sqrt{2n}).

Acknoweledgements

The authors thank Scott Ahlgren, Andrew Granville, Don Zagier and the referee for a variety of useful comments and suggestions.

2. Preliminaries and important facts.

The ten identities in Theorem 3 rely on Theorems 1 and 2. Here we prove Theorems 1 and 2, and we begin with the following observation which follows from Abel’s Theorem.

Proposition 2.1. Suppose that f(z) = ∑_{n=0}^{∞} α(n)z^{n} is analytic for |z| < 1. If α is a complex number for which

(1) ∑_{n=0}^{∞}(α − α(n)) < +∞,

(2) lim_{n→+∞} n(α − α(n)) = 0, then

(2.1)  lim_{z→1^{−}} \frac{d}{dz} (1 − z)f(z) = ∑_{n=0}^{∞}(α − α(n)).

In all of our applications, f(z) will have a pole of order 1 at z = 1, so the lim_{z→1^{−}} can be replaced by a simple evaluation. We shall employ the standard notation of basic hypergeometric series [G-R; pp. 3-4, 125]:

(2.2)  _{r+1}ϕ_{r} ( \begin{matrix}a_{0}, a_{1}, …, a_{r}; \\ b_{1}, …, b_{r} \end{matrix} q, z) = ∑_{j=0}^{∞} \frac{(a_{0}; q)_{j}(a_{1}; q)_{j} ⋯ (a_{r}; q)_{j}z^{j}}{(q; q)_{j}(b_{1}; q)_{j} ⋯ (b_{r}; q)_{j}} ,

(2.3)  _{r}ψ_{r} ( \begin{matrix}a_{1}, a_{2}, …, a_{r}; \\ b_{1}, b_{2}, …, b_{r} \end{matrix} q, z) = ∑_{j=−∞}^{∞} \frac{(a_{1}; q)_{j}(a_{2}; q)_{j} ⋯ (a_{r}; q)_{j}z^{j}}{(b_{1}; q)_{j}(b_{2}; q)_{j} ⋯ (b_{r}; q)_{j}} .

We shall also require Heine’s transformation [G-R; p. 9]:

(2.4)  _{2}ϕ_{1} ( \begin{matrix}a, b; \\ c \end{matrix} q, z) = \frac{(b; q)_{∞}(az; q)_{∞}}{(c; q)_{∞}(z; q)_{∞}} × _{2}ϕ_{1} ( \begin{matrix}c/b, z; \\ az \end{matrix} q, b);

Ramanujan’s summation [G-R; p. 126]:

(2.5)  _{1}ψ_{1} ( \begin{matrix}a; \\ b \end{matrix} q, z) = \frac{(q; q)_{∞}(b/a, q)_{∞}(az; q)_{∞}(q/az; q)_{∞}}{(b; q)_{∞}(q/a; q)_{∞}(z; q)_{∞}(b/az; q)_{∞}} ,

and the Rogers-Fine identity [F; p. 15]:

(2.6)  (1 − t) ∑_{n=0}^{∞} \frac{(a; q)_{n}t^{n}}{(b; q)_{n}} = ∑_{n=0}^{∞} \frac{(a; q)_{n}(atq/b; q)_{n}(1 − atq^{2n})b^{n}t^{n}q^{n^{2}−n}}{(b; q)_{n}(tq; q)_{n}} .

Throughout we assume that |q| < 1 and that the other parameters are restricted to domains that do not contain any singularities of the series or products under consideration. For succinctness of notation, we define the differential operator ϵ by

(2.7)  ϵf(z) = f′(1).

Proof of Theorem 1. By Proposition 2.1, we have that

∑_{n=0}^{∞} (\frac{(t; q)_{∞}}{(a; q)_{∞}} − \frac{(t; q)_{n}}{(a; q)_{n}} ) = ϵ(1 − z) ∑_{n=0}^{∞} \frac{(t; q)_{n}z^{n}}{(a; q)_{n}} .

By (2.2-6), this equals

= ϵ(1 − z) (_{1}ψ_{1} ( \begin{matrix}t; \\ a \end{matrix} q, z) − ∑_{n=−∞}^{−1} \frac{(t; q)_{n}z^{n}}{(a; q)_{n}} )

= ϵ (\frac{(q; q)_{∞}(a/t; q)_{∞}(tz; q)_{∞}(q/tz; q)_{∞}}{(a; q)_{∞}(q/t; q)_{∞}(zq; q)_{∞}(a/tz; q)_{∞}} − (1 − z) ∑_{n=1}^{∞} \frac{(q/a; q)_{n}(a/(tz))^{n}}{(q/t; q)_{n}} ) .

Differentiating this last expression with respect to z and then setting z = 1 yields the result.

Q.E.D.

Proof of Theorem 2. By Proposition 2.1, (2.2) and (2.4), we have that

∑_{n=0}^{∞} (\frac{(a; q)_{∞}(b; q)_{∞}}{(q; q)_{∞}(c; q)_{∞}} − \frac{(a; q)_{n}(b; q)_{n}}{(q; q)_{n}(c; q)_{n}} )

= ϵ(1 − z) _{2}ϕ_{1} ( \begin{matrix}a, b; \\ c \end{matrix} q, z)

= ϵ (\frac{(b; q)_{∞}(az; q)_{∞}}{(c; q)_{∞}(zq; q)_{∞}} (1 + ∑_{n=1}^{∞} \frac{(c/b; q)_{n}(z; q)_{n}b^{n}}{(q; q)_{n}(az; q)_{n}} )) .

Noting that ϵ(z; q)_{n} = −(q; q)_{n−1} when n > 0, we differentiate this last expression with respect to z and then set z = 1. This yields the result.

Q.E.D.

3. Proof of Theorem 3.

In this section we prove each of the ten identities using the facts in §2. Case of F_{1}(z): This appears implicitly in [F; p. 14]. It is the instance of Theorem 2 where a = 0 and b = c. Case of F_{2}(z): This is the instance of a = −q and b = c in Theorem 2. Case of F_{3}(z): In Theorem 2, replace q by q^{2}, then set b = c and a = q. Case of F_{4}(z): This result was proved in [A2; eq. (1.4)]. It follows from Theorem 1 with t = −q with a → 0.

q-SERIES IDENTITIES 9

Case of F_{5}(z): This result was proved in [A2; eq. (1.5)]. In Theorem 2 replace q by q^{2}, then set a = q^{2}, b = 0 and c = q^{3}. This yields, after multiplication by (1 − q)^{−1},

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

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

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

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

By [F; p. 14, eq. (12.42)], we find that this equals

= \frac{1}{2(q; q^{2})_{∞}} lim_{τ→0} _{3}ϕ_{2} ( \begin{matrix}q/τ, q/τ, −1; \\ −q, −q \end{matrix} q, −τ^{2}) + \frac{1}{(q; q^{2})_{∞}} (−\frac{1}{2} + ∑_{n=1}^{∞} \frac{q^{n}}{1 − q^{n}} ) .

By [G-R; p. 241, eq. (III.9)] with a = b = q/τ, c = −1 and d = e = −q, this equals

= \frac{1}{2(q; q^{2})_{∞}(−q; q)_{∞}} lim_{τ→0} _{3}ϕ_{2} ( \begin{matrix}q/τ, −τ, q; \\ −q, −qτ \end{matrix} q, −τ) + \frac{1}{(q; q^{2})_{∞}} (−\frac{1}{2} + ∑_{n=1}^{∞} \frac{q^{n}}{1 − q^{n}} )

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

This is the identity. Case of F_{6}(z): In Theorem 2 let a = q, b = 0 and c = −q. This yields

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

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

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

By [F; p. 14, eq. (12.42)], this equals

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

= \frac{2q}{(−q; q)_{∞}(1 − q)^{2}} lim_{τ→0} _{3}ϕ_{2} ( \begin{matrix}q/τ, q/τ, q; \\ q^{3}, q^{3} \end{matrix} q^{2}, τ^{2}q^{3}) + \frac{1}{(−q; q)_{∞}} ∑_{n=1}^{∞} \frac{q^{n}}{1 − q^{n}}

By [G-R; p. 241, eq. (III.10)] with q replaced by q^{2} followed by letting a = c = −q/τ, b = q and d = e = q^{3}, this equals

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

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

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

This is the identity for F_{6}(z).

Case of F_{7}(z) : In Theorem 1 replace q by q^{2}, and then set a = 0 and t = q.

Case of F_{8}(z) : This is identity (1.2), and it is Theorem 2 in [Z]. We include a proof for completeness. In Proposition 2.1 set α = (q; q)_{∞} and let α(n) := (q; q)_{n}. This yields

∑_{n=0}^{∞} ((q; q)_{∞} − (q; q)_{n}) = ϵ(1 − z) ∑_{n=0}^{∞}(q; q)_{n}z^{n}

= −ϵ ∑_{n=1}^{∞}(q; q)_{n−1}q^{n}z^{n}

= − ∑_{n=1}^{∞}(q; q)_{n−1}nq^{n}. (3.1)

Now it is immediate (because the partial sums equal the partial products) that

(3.2,)  1 − ∑_{n=1}^{∞} zq^{n}(zq; q)_{n−1} = (zq; q)_{∞}

and applying ϵ to (3.2) we find that

(3.3)  − ∑_{n=1}^{∞} q^{n} (ϵ(zq; q)_{n−1}) = −1 − (q; q)_{∞} (−1 + ∑_{j=1}^{∞} \frac{q^{j}}{1 − q^{j}} ) .

Now set b = 0, a = z^{2}q and t = z^{2} in (2.6), and after simplification we obtain upon multiplication by z

(3.4) z − z^{3} ∑_{n=1}^{∞}(z^{2}q; q)_{n−1}(z^{2}q)^{n} = z + ∑_{n=1}^{∞}(−1)^{n} (q^{(3n^{2}−3n)/2}z^{6n−1} + q^{(3n^{2}+n)/2}z^{6n+1}) .

q-SERIES IDENTITIES 11

Noting that ϵ (zf(z^{2})) = f(1) + 2ϵf(z) we apply ϵ to (3.4) and find that

1+ ∑_{n=1}^{∞}(−1)^{n}(6n − 1)q^{(3n^{2}−n)/2} + ∑_{n=1}^{∞}(−1)^{n}(6n + 1)q^{(3n^{2}+n)/2}

= (q; q)_{∞} + 2ϵ (1 − z ∑_{n=1}^{∞}(zq; q)_{n−1}z^{n}q^{n})

= (q; q)_{∞} − 2 ((q; q)_{∞} − 1) − 2 ∑_{n=1}^{∞}(q; q)_{n−1}nq^{n} − 2 ∑_{n=1}^{∞} q^{n} (ϵ(zq; q)_{n−1})

(3.5) = 2 − (q; q)_{∞} + 2 ∑_{n=0}^{∞} ((q; q)_{∞} − (q; q)_{n}) + 2 (−1 − (q; q)_{∞} (−1 + ∑_{j=1}^{∞} \frac{q^{j}}{1 − q^{j}} )) ,

by (3.1) and (3.3). This is the identity. Case of F_{9}(z): In Theorem 2, let a = b = q and c = −q. This yields

(3.6) ∑_{n=0}^{∞} (\frac{(q; q)_{∞}}{(−q; q)_{∞}} − \frac{(q; q)_{n}}{(−q; q)_{n}} ) = −2\frac{(q; q)_{∞}}{(−q; q)_{∞}} ∑_{n=1}^{∞} \frac{(−q; q)_{n−1}q^{n}}{(q; q)_{n}(1 − q^{n})}.

Now in (2.6) set t = z, a = zq, b = −zq (cf. [F; p. 15, eq. (14.31)]). This yields after simplification

(3.7)  (1 − z) ∑_{n=0}^{∞} \frac{(zq; q)_{n}z^{n}}{(−zq; q)_{n}} = 1 + 2 ∑_{n=1}^{∞}(−z^{2})^{n}q^{n^{2}}.

Applying ϵ to (3.7) yields

4 ∑_{n=1}^{∞}(−1)^{n}nq^{n^{2}} =

= ϵ(1 − z)_{2}ϕ_{1} ( \begin{matrix}q, zq; \\ −zq \end{matrix} q, z) .

By (2.4) with a = q, b = zq and c = −zq, and also (3.6), we find that this equals

= ϵ\frac{(zq; q)_{∞}}{(−zq; q)_{∞}} _{2}ϕ_{1} ( \begin{matrix}−1, z; \\ zq \end{matrix} q, zq) . (3.8)

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

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

Case of F_{10}(z): In Theorem 2 replace q by q^{2}, and then set a = b = q^{2} and c = q^{3}. This yields, after multiplication by (1 − q)^{−1}

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

Now in (2.6) replace q by q^{2}, then set a = zq and b = zq^{2} (cf. [F; p. 16, eq. (14.4)]). This yields, after simplification,

(3.10)  ∑_{n=0}^{∞} \frac{(zq; q^{2})_{n}z^{n}q^{n}}{(zq^{2}; q^{2})_{n}} = ∑_{n=0}^{∞} z^{n}q^{(n^{2}+n)/2}.

Applying ϵ to (3.10), we find that

∑_{n=1}^{∞} nq^{(n^{2}+n)/2} = ϵ _{2}ϕ_{1} ( \begin{matrix}zq, q^{2}; \\ zq^{2} \end{matrix} q^{2}, zq) .

This equals, by (2.4) with q replaced by q^{2} and z replaced by zq with a = zq, b = q^{2} and c = zq^{2},

= ϵ\frac{(q^{2}; q^{2})_{∞}(z^{2}q^{2}; q^{2})_{∞}}{(zq^{2}; q^{2})_{∞}(zq; q^{2})_{∞}} _{2}ϕ_{1} ( \begin{matrix}z, zq; \\ z^{2}q^{2} \end{matrix} q^{2}, q^{2}) (3.11)

= ϵ(q^{2}; q^{2})_{∞}(−zq; q)_{∞} _{2}ϕ_{1} ( \begin{matrix}z, zq; \\ z^{2}q^{2} \end{matrix} q^{2}, q^{2})

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

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

We use (3.9) in the last step above.

4. Proof of Theorems 4, 5, 6, and 7.

In this section we prove Theorems 4, 5, 6, and 7. The proofs are similar to the proof of [Th. 3, Z], and so we give a brief proof of Theorem 4 and we give sketches of the remaining cases. In each case, it is well known that the relevant L-function has an analytic continuation to ℂ (with the exception of a simple pole at s = 1 for ζ(s)) and a functional equation via a Mellin transformation.

Proof of Theorems 4, 5, 6, and 7. Case of Theorem 4: The identity for F_{9}(z) in Theorem 3 is

(4.1) ∑_{n=0}^{∞} (F_{9}(z) − \frac{(1 − q)(1 − q^{2}) ⋯ (1 − q^{n})}{(1 + q)(1 + q^{2}) ⋯ (1 + q^{n})}) = 2F_{9}(z) (∑_{n=1}^{∞} d_{1}(n)q^{n}) + 2\sqrt{Θ} (F_{9}(z)) .

q-SERIES IDENTITIES 13

It is well known that

(4.2)  F_{9}(z) = 1 + 2 ∑_{n=1}^{∞}(−1)^{n}q^{n^{2}}.

Notice that F_{9}(z) vanishes to infinite order as q → 1. Therefore, we replace q by e^{−t} with t ↘ 0. Now define coefficients c_{9}(n) and b_{9}(n) by the asymptotic expansions

(4.3)  − ∑_{n=0}^{∞} \frac{(1 − e^{−t})(1 − e^{−2t}) ⋯ (1 − e^{−nt})}{(1 + e^{−t})(1 + e^{−2t}) ⋯ (1 + e^{−nt})} = ∑_{n=0}^{∞} c_{9}(n)t^{n}

and

H_{9} (e^{−t}) = 4 ∑_{n=1}^{∞}(−1)^{n}ne^{−n^{2}t} ∼ ∑_{n=0}^{∞} b_{9}(n)t^{n} as t ↘ 0.

Now we observe, by (4.1) and (4.2), that c_{9}(n) = b_{9}(n) for all n. On the other hand

∫_{0}^{∞} H_{9} (e^{−t}) t^{s−1}dt = 4 ∑_{n=1}^{∞}(−1)^{n}n ∫_{0}^{∞} e^{−n^{2}t}t^{s−1}dt.

By replacing T by n^{2}t, we find that

(4.4)  ∫_{0}^{∞} H_{9} (e^{−t}) t^{s−1}dt = 4Γ(s) ∑_{n=1}^{∞} \frac{(−1)^{n}}{n^{2s−1}} = 4Γ(s)(4^{1−s} − 1)ζ(2s − 1).

To compute the coefficients b_{9}(n), notice that

(4.5) ∫_{0}^{∞} H_{9} (e^{−t}) t^{s−1}dt = ∫_{0}^{∞} ( ∑_{n=0}^{N−1} b_{9}(n)t^{n} + O(t^{N})) t^{s−1}dt = ∑_{n=0}^{N−1} \frac{c_{9}(n)}{s + n} + 𝔉_{9}(s),

where 𝔉_{9}(s) is analytic for Re(s) > −N. The residue at s = −n of (4.4) is

b_{9}(n) = \frac{(−1)^{n}}{n!} · 4(4^{1+n} − 1) · ζ(−2n − 1).

This completes the proof of Theorem 4. Case of Theorem 5: It is well known that

(4.6)  F_{10}(z) = ∑_{n=0}^{∞} q^{(2n+1)^{2}}.

It is easy to see that the identity in Theorem 3 for F_{10}(z) is equivalent to

∑_{n=0}^{∞} (F_{10}(z/8) − \frac{q^{1/8}}{1 − q} × \frac{(1 − q^{2})(1 − q^{4}) ⋯ (1 − q^{2n})}{(1 − q^{3})(1 − q^{5}) ⋯ (1 − q^{2n+1})})
(4.7)  = F_{10}(z/8) (−\frac{1}{2} + ∑_{n=1}^{∞} d_{8}(n)q^{n}) + \frac{1}{2} ∑_{n=0}^{∞}(2n + 1)q^{(2n+1)^{2}/8}.

Observe that F_{10}(z/8) vanishes to infinite order as q → −1. Define the coefficients c_{10}(n) and b_{10}(n) by the asymptotic expansions

−e^{−t/8} ∑_{n=0}^{∞} \frac{(1 − e^{−2t})(1 − e^{−4t}) ⋯ (1 − e^{−2nt})}{(1 + e^{−t})(1 + e^{−3t}) ⋯ (1 + e^{−(2n+1)t})} = ∑_{n=0}^{∞} c_{10}(n)t^{n}

and

H_{10} (e^{−t}) = \frac{1}{2} ∑_{n=0}^{∞}(2n + 1)(−1)^{n(n+1)/2}e^{−t(2n+1)^{2}/8} ∼ ∑_{n=0}^{∞} b_{10}(n)t^{n} as t ↘ 0.

Therefore, by replacing q by −e^{−t} with t ↘ 0 and ζ_{8} := e^{πi/8}, (4.6) and (4.7) imply that c_{10}(n) = b_{10}(n) for all n and, on the other hand

∫_{0}^{∞} H_{10} (e^{−t}) t^{s−1}dt = \frac{1}{2} ∑_{n=0}^{∞}(2n + 1)(−1)^{n(n+1)/2} ∫_{0}^{∞} e^{−(2n+1)^{2}t/8}t^{s−1}dt

= \frac{1}{2} ∑_{n=1}^{∞} χ_{2}(n)n ∫_{0}^{∞} e^{−n^{2}t/8}t^{s−1}dt

= \frac{1}{2}8^{s}Γ(s)L(χ_{2}, 2s − 1). (4.8)

The rest of the proof is identical to the proof of Theorem 4.

Case of Theorem 6: By [C], it is known that the coefficients a(n) defining L(ρ, s) in (1.16) are defined by

(4.9)  ∑_{n=1}^{∞} a(n)q^{n} = M_{1}(q) + 2M_{2}(q).

In this case we use the identity for F_{7}(z) in Theorem 3 (4.10)

∑_{n=0}^{∞} (F_{7}(z) − q^{−1}(1 − q^{24})(1 − q^{72}) ⋯ (1 − q^{24(2n−1)})) = F_{7}(z) (∑_{n=1}^{∞} d(n)q^{24n}) + M_{2}(q).

Moreover, notice in (1.8) that the non-zero coefficients of M_{2}(q) are supported on those exponents n ≡ 23 (mod 24). Now observe that F_{7}(z/24) vanishes to infinite order as q → 1. The rest of the argument is virtually identical to those above.

Case of Theorem 7: In this case we consider the identity for F_{4}(z) in Theorem 3 (4.11)

∑_{n=0}^{∞} (F_{4}(z) − q(1 + q^{24})(1 + q^{48}) ⋯ (1 + q^{24n})) = F_{4}(z) (−\frac{1}{2} + ∑_{n=1}^{∞} d(n)q^{24n}) + \frac{1}{2}M_{1}(q).

Notice that F_{4}(z/24) vanishes to infinite order as q → −1. Arguing as before, we consider the asymptotic t-series expansion of

−ζ_{24}e^{−t/24} ∑_{n=0}^{∞}(1 − e^{−t})(1 + e^{−2t}) ⋯ (1 + (−1)^{n}e^{−nt})

where ζ_{24} = e^{πi/24}. The rest of the proof is a routine exercise using ζ_{24}, the Mellin transform, and the fact that the non-zero coefficients of M_{1}(q) are supported on those exponents n ≡ 1 (mod 24).

Q.E.D.

5. Partition theoretic consequences.

Recall that a partition of a non-negative integer N is any nonincreasing sequence of positive integers whose sum is N. If p(N) denotes the number of partitions of N, then we have

(5.1) \frac{1}{η(24z)} = ∑_{N=0}^{∞} p(N)q^{24N−1} = q^{−1} ∏_{n=1}^{∞} \frac{1}{1 − q^{24n}} = q^{−1} + q^{23} + 2q^{47} + 3q^{71} + · · · .

If p_{e}(N) (resp. p_{o}(N)) denote the number of partitions of N into an even (resp. odd) number of distinct parts, then Euler’s Pentagonal Number Theorem asserts that the Fourier expansion of η(24z) ∈ S_{1/2}(Γ_{0}(576), χ) is

(5.2)  η(24z) = ∑_{N=0}^{∞} (p_{e}(N) − p_{o}(N)) q^{24N+1} = q − q^{25} − q^{49} + ⋯ = ∑_{n=1}^{∞} χ(n)q^{n^{2}}

where

(5.3)  χ(n) := \{ \begin{matrix}1 & if n ≡ ±1 (mod 12), \\ −1 & if n ≡ ±5 (mod 12), \\ 0 & otherwise. \end{matrix}

As a consequence for our identity for F_{1}(z) = 1/η(24z), we obtain the following partition theoretic result which is equivalent to the observation made by Erdös [E] in the beginning of his study of the asymptotics of p(N) by elementary means.

Theorem 5.1. If n is a positive integer, then let a_{n}(N) denote the number of partitions of N into parts not exceeding n. For every positive integer N we have

(N + 1)p(N) = ∑_{n=1}^{N} p(N − n)d(n) + ∑_{n=1}^{N} a_{n}(N).

Proof. Here we use the identity for F_{1}(z) = 1/η(24z) in the form

(5.4)  ∑_{n=0}^{∞} (P (q) − \frac{1}{(1 − q)(1 − q^{2}) ⋯ (1 − q^{n})}) = P (q)D(q),

where

P (q) = ∑_{n=0}^{∞} p(n)q^{n} = ∏_{n=1}^{∞} \frac{1}{1 − q^{n}} ,

D(q) = ∑_{n=1}^{∞} d(n)q^{n}.

If m is a non-negative integer, then the series

(5.5)  P (q) − \frac{1}{(1 − q)(1 − q^{2}) ⋯ (1 − q^{m})} = q^{m+1} + ∑_{n=m+2}^{∞} b_{m}(n)q^{n}

for some sequence of integers b_{m}(n).

Now let N be a positive integer, then it is easy to see that the coefficient of N on the right hand side of (5.4) is

∑_{n=1}^{N} p(N − n)d(n).

Therefore, to prove the result it suffices to show that the coefficient of q^{N} on the left hand side of of (5.4) is

(5.6)  (N + 1)p(N) − ∑_{n=1}^{N} a_{n}(N).

By letting m = N as in (5.5), it is easy to see that the coefficient of q^{N} on the left hand side of (5.4) is the coefficient of q^{N} in

∑_{n=0}^{N} (P (q) − \frac{1}{(1 − q)(1 − q^{2}) ⋯ (1 − q^{n})}) .

Claim (5.6) follows from the obvious fact that

∑_{N=0}^{∞} a_{n}(N)q^{n} = \frac{1}{(1 − q)(1 − q^{2}) ⋯ (1 − q^{n})}.

This completes the proof.

Q.E.D.

As with Theorem 5.1, Zagier’s identity for F_{8}(z) = η(24z) has an interesting partition theoretic consequence which is similar in flavor to Euler’s Pentagonal Number Theorem (5.2). We require some notation. If π is a partition, then let m_{π} denote its largest part. Moreover, if N is an integer, then let S_{e}(N) (resp. S_{o}(N)) denote the set of partitions of N into an even (resp. odd) number of distinct parts. Using this notation, define the two partition functions A_{e}(N) and A_{o}(N) by

A_{e}(N) := ∑_{π∈S_{e}(N)} m_{π}, (5.7)

A_{o}(N) := ∑_{π∈S_{o}(N)} m_{π} (5.8)

Theorem 5.2. If N is a positive integer, then

A_{e}(N) − A_{o}(N) =

= ∑_{k∈ℤ}(−1)^{k}d (N − (3k^{2} + k)/2) + \{ \begin{matrix}(−1)^{k}(3k) & if N = \frac{3k^{2}+k}{2} with k ≥ 0, \\ (−1)^{k}(3k − 1) & if N = \frac{3k^{2}−k}{2} with k ≥ 1, \\ 0 & otherwise. \end{matrix}

Proof. If E(q) is the q-series defined by

E(q) = ∏_{n=1}^{∞}(1 − q^{n}) = 1 − q − q^{2} + ⋯,

then we claim that

(5.9) ∑_{n=0}^{∞} (E(q) − (1 − q)(1 − q^{2}) ⋯ (1 − q^{n})) = − ∑_{n=1}^{∞} nq^{n}(1 − q)(1 − q^{2}) ⋯ (1 − q^{n−1}).

As described above, we have that

E(q) = ∏_{n=1}^{∞}(1 − q^{n}) = ∑_{n=0}^{∞} (p_{e}(n) − p_{o}(n)) q^{n}.

Since E(q) is the generating function for the number of partitions into an even number of distinct parts minus the number into an odd number of distinct parts, it is easy to see that

E(q) − (1 − q)(1 − q^{2}) ⋯ (1 − q^{n})

is the generating function whose coefficient of q^{N} is the number of partitions of N into an even number of distinct parts where at least one part exceeds n minus the number of partitions of N into an odd number of distinct parts where at least one part exceeds n. Consequently, each partition π into distinct parts is counted with multiplicity (−1)^{n_{π}}m_{π} where n_{π} is the number of parts in π. This implies (5.9).

It is an easy exercise to deduce Theorem 5.2 from the identity in Theorem 1 for F_{8}(z) = η(24z).

Q.E.D.

6. Related Results

The uniform nature of the ten results proved in Theorem 3 makes it natural to group those identities together. There are, however, a variety of results that can be deduced from Theorems 1 and 2. Here we record a number of further identities.

Theorem 6.1. Let m(n) denote the number of divisors of n in the interval [\sqrt{n/2}, \sqrt{2n}). Moreover, let d_{e}(n) (resp. d_{o}(n)) denote the number of even (resp. odd) divisors of n. The following identities are true:

(6.1)  ∑_{n=0}^{∞} (\frac{1}{η^{2}(24z)} − \frac{1}{q^{2}(q^{24}; q^{24})_{n}^{2}} ) = \frac{1}{η^{2}(24z)} ∑_{n=1}^{∞} (d(n) + m(n)) q^{24n},

(6.2)  ∑_{n=0}^{∞} (\frac{η^{2}(2z)}{η^{4}(z)} − \frac{(−q; q)_{n}^{2}}{(q; q)_{n}^{2}} ) = \frac{4η^{2}(2z)}{η^{4}(z)} ∑_{m=0}^{∞} d_{o}(2m + 1)q^{2m+1},

(6.3)  ∑_{n=0}^{∞} (\frac{η^{2}(8z)}{η^{4}(16z)} − \frac{(q^{8}; q^{16})_{n}^{2}}{q^{2}(q^{16}; q^{16})_{n}^{2}} ) = \frac{η^{2}(8z)}{η^{4}(16z)} ∑_{n=1}^{∞} (d_{e}(n) − 2d_{o}(n)) q^{8n},

(6.4) ∑_{n=0}^{∞} (\frac{1}{η(24z)} − \frac{1}{q(q^{24}; q^{24})_{2n}} ) = \frac{1}{η(24z)} (∑_{n=1}^{∞} d(n)q^{48n} + ∑_{n=1}^{∞} \frac{(−1)^{n−1}q^{24n^{2}}}{1 − q^{48n}} ) ,

(6.5)  ∑_{n=0}^{∞} (\frac{η(2z)}{η^{2}(z)} − \frac{(−q; q)_{2n}}{(q; q)_{2n}} ) = \frac{2η(2z)}{η^{2}(z)} ∑_{n=1}^{∞} \frac{q^{2n} − (−q)^{n}}{1 − q^{4n}} ,

(6.6)

∑_{n=0}^{∞} (\frac{η(8z)}{η^{2}(16z)} − \frac{(q^{8}; q^{16})_{2n}}{q(q^{16}; q^{16})_{2n}} ) = \frac{η(8z)}{4η^{2}(16z)} (1 − ∑_{n=−∞}^{∞} q^{8n^{2}} + 2 ∑_{n=1}^{∞} (d_{e}(n) − d_{o}(n)) q^{8n}) .

Proof. Here we prove these identities.

Case of (6.1): Let a = b = 0 and c = q in Theorem 2. This yields

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

In order to complete the proof of (6.1), we must establish that

(6.8)  ∑_{n=1}^{∞} m(n)q^{n} = ∑_{n=1}^{∞} \frac{(−1)^{n−1}q^{(n^{2}+n)/2}}{1 − q^{n}} .

q-SERIES IDENTITIES 19

To see (6.8), we first note that m(n) equals d(n) minus the number of divisors in the two intervals [1, \sqrt{n/2}) and [\sqrt{2n}, n]. Hence, we have that

∑_{n=1}^{∞} m(n)q^{n} = ∑_{n=1}^{∞} \frac{q^{n}}{1 − q^{n}} − ∑_{n=1}^{∞} q^{(n)(2n)} (1 + 2q^{n} + 2q^{2n} + 2q^{3n} + ⋯)

= ∑_{n=1}^{∞} \frac{q^{n}}{1 − q^{n}} − ∑_{n=1}^{∞} \frac{(1 + q^{n})q^{2n^{2}}}{1 − q^{n}} . (6.9)

Next in [G-R, eq. (III.17), p. 242] set a = e = z and let b, c, d and f → +∞. This yields

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

Hence we get

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

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

= ϵ\frac{(q; q)_{∞}}{(zq; q)_{∞}} (1 + ∑_{n=1}^{∞} \frac{(z; q)_{n}^{2}(1 − zq^{2n})q^{2n^{2}}z^{n}}{(q; q)_{n}^{2}(1 − z)} )

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

= ∑_{n=1}^{∞} m(n)q^{n}, (6.10)

which proves (6.8) and therefore (6.1). Case of (6.2): To prove this identity let a = b = −q and c = q in Theorem 2. The result now follows easily by combining the resulting Lambert series. Case of (6.3): To prove this identity replace q by q^{2} and set a = b = q and c = q^{2} in Theorem 2. The result now follows easily by combining the resulting Lambert series. Case of (6.4): To prove this identity replace q by q^{2} and set a = b = 0 and c = q^{2} in Theorem 2. Case of (6.5): This identity follows from Theorem 2 by replacing q by q^{2} and the setting a = −q, b = −q^{2} and c = q. Case of (6.6): Surprisingly, (6.6) is more intricate. Here we use Proposition 2.1 with

α(n) = (q; q^{2})_{2n}/(q^{2}; q^{2})_{2n}.

Consequently (noting that ϵf(z) = \frac{1}{2}ϵf(z^{2})), we find that

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

= ϵ(1 − z) ∑_{n=0}^{∞} \frac{(q; q^{2})_{2n}z^{n}}{(q^{2}; q^{2})_{2n}}

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

= \frac{1}{4}ϵ \{(1 + z)\frac{(qz; q^{2})_{∞}}{(zq^{2}; q^{2})_{∞}} + (1 − z) · \frac{(−zq; q^{2})_{∞}}{(−zq^{2}; q^{2})_{∞}} \}

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

Identity (6.6) now follows by recalling [A1; p. 21, eq. (2.2.10)], with z = 1, that

\frac{(−q; q^{2})_{∞}(q^{2}; q^{2})_{∞}}{(q; q^{2})_{∞}(−q^{2}; q^{2})_{∞}} = ∑_{n=−∞}^{∞} q^{n^{2}}.

Q.E.D.

Finally, we recall the following attractive and very elementary result on partitions into consecutive integers (cf. [L, p. 85, Problem 4]).

Proposition 6.2. The number of partitions of n into consecutive integers equals the number of odd divisors of n.

Proof. This result is easily deduced from the formula for the sum of an arithmetic progression. It is also directly deduced from the generating function identity [M, p. 28]

∑_{m=1}^{∞} \frac{q^{2m−1}}{1 − q^{2m−1}} = ∑_{m=1}^{∞} \frac{q^{(m^{2}+m)/2}}{1 − q^{m}} .

Q.E.D.

From our proof of (6.1), we may easily deduce the following result for c_{e}(n) (resp. c_{o}(n)) the number of partitions of n into an even (resp. odd) number of consecutive integers.

Theorem 6.3. For every positive integer n we have

c_{o}(n) − c_{e}(n) = m(n).

Proof. This is immediate from (6.10):

∑_{n=1}^{∞} m(n)q^{n} = ∑_{n=1}^{∞} \frac{(−1)^{n−1}q^{(n^{2}+n)/2}}{1 − q^{n}} = ∑_{n=1}^{∞} (c_{o}(n) − c_{e}(n)) q^{n}.

Q.E.D.

References

[A1] G. E. Andrews, The theory of partitions, Cambridge University Press, 1984.

[A2] G. E. Andrews, Ramanujan’s “Lost” Notebook V: Euler’s partition identity, Advances in Mathematics 61 (1986), 156-164.

[A3] G. E. Andrews, Mock theta functions, Proc. Symp. Pure Math. 49 (1989), 283-298. [A-D-H] G. E. Andrews, F. Dyson and D. Hickerson, Partitions and indefinite quadratic forms, Inven-

tiones Mathematicae 91 (1988), 391-407.

[C] H. Cohen, q-identities for Maass waveforms, Inventiones Mathematicae 91 (1988), 409-422.

[E] P. Erdös, On an elementary proof of some asymptotic formulas in the theory of partitions, Annals of Mathematics 43 (1942), 437-450.

[F] N. J. Fine, Basic hypergeometric series and applications, Math. Sruveys and Monographs, Vol. 27, Amer. Math.Soc., Providence, 1988.

[G-R] G. Gasper and M. Rahman, Basic hypergeometric series, Ency. of Math. and its Appl., Vol. 2,

Addison-Wesley, Reading, 1976.

[L] W. J. Leveque, Topics in Number Theory, Vol. 1, Addison-Wesley, Reading, 1956.

[M] P. A. MacMahon, Combinatory Analysis, Vol. 2, Cambridge Univ. Press, Cambridge, 1916 (reissued by Chelsea, New York, 1960).

[Z] D. Zagier, Vassiliev invariants and a strange identity related to the Dedekind eta-function, to appear, Topology.

Department of Mathematics, Penn State University, University Park, Pa. 16802 E-mail address: andrews@math.psu.edu

Dept. Matemáticas, Facultad de Ciencias, Universidad Autónoma de Madrid, 28049, Madrid, España.

E-mail address: jorge.jimenez@uam.es

Department of Mathematics, Penn State University, University Park, Pa. 16802 E-mail address: ono@math.psu.edu

Department of Mathematics, University of Wisconsin, Madison, Wisconsin 53706 E-mail address: ono@math.wisc.edu