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An Identity of Ramanujan and The Representation of Integers as Sums of
Triangular Numbers
All the correspondence should be sent to
Prof. Zhi-Guo Liu
Department of Mathematics,
East China Normal University,
Shanghai, 200062,
P. P. China
Email address:
Running Title:
On Sums of Triangular Numbers
1
An Identity of Ramanujan and The Representation of Integers
as Sums of Triangular Numbers
Zhi-Guo Liu
Department of Mathematics, East China Normal University,
Shanghai, 200062,
People’s Republic of China
In memory of Robert Rankin
Abstract
Let k be a positive number and tk(n) denote the number of representations of n as a sum
of k triangular numbers. In this paper, we will calculate t2k(n) in the spirit of Ramanujan. We
first use the complex theory of elliptic functions to prove a theta-function identity. Then from
this identity we derive two Lambert series identities, one of them is a well-known identity of
Ramanujan. Using a variant form of Ramanujan’s identity, we study two classes of Lambert
series and derive some theta function identities related to these Lambert series . We calculate
t12(n), t16(n), t20(n), t24(n), and t28(n) using these Lambert series identities. We also rederive a
recent result of H. H. Chan and K. S. Chua [6] about t32(n). In addition, we derive some identities
among the Ramanujan function τ(n), the divisor function σ11(n), and t24(n). Our methods do
not depend upon the theory of modular forms and are somewhat more transparent.
Key words: elliptic functions, theta functions, Lambert series, triangular numbers, Ramanujan τ(n)-
function, Jacobi’s identity, modular forms.
2000 Mathematics Subject Classification:Primary– 11F11, 11F12, 11F27, 33E05.
1 Introduction
Throughout this paper we will use q to denote exp(πiτ) with Imτ > 0. If k ≥ 1 is a positive integer,
we then let rk(n) denote the number of representations of n as a sum of k squares. We also, let tk(n)
denote the number of representations of sum of k triangular numbers. Following Ramanujan (see, for
2
example [4, p. 3, Equation(1.5)]), we use the theta functions φ(q) and ψ(q) defined by
φ(q) =∞∑
n=−∞qn2
and ψ(q) =∞∑
n=0
q12 n(n+1). (1.1)
Consequently, the generating functions for rk(n) and tk(n) are
φk(q) =∞∑
n=0
rk(n)qn and ψk(q) =∞∑
n=0
tk(n)qn. (1.2)
The study of rk(n) and tk(n) has a long history and many mathematicians have made contributions
to this object. Now it still is one of the fundamental object in number theory . From (1.2) we know
that deriving explicit expansions for φk(q) and ψk(q) are the keys of calculating rk(n) and tk(n). The
calculations of φk(q) are closely related to those of ψk(q). One can obtain the formula for φk(q) using
some simple modular transformations to the formula for ψk(q) ; conversely, we can obtain the formula
for ψk(q) from φk(q). This phenomenon was perhaps first noticed by the author in [17]. In many
cases, the calculation of ψk(q) is simpler than that of φk(q). To evaluate φk(q) and ψk(q) , one usual
uses the theory of elliptic functions and the theory of modular forms.
In this paper we will provide a different approach. Our methods do not depend upon the theory of
modular forms and are also different from the method of elliptic functions. The main tool of ours is
identity (5.1) below. It is surprising that this identity is very useful and powerful in the computations
of ψ2k(q). Our method is simpler than other known methods and one key advantage of our method
is that one can follow easily.
In Section 2 we introduce some basic facts about the classical theta functions. In Section 3 we
prove a theta-function identity by using the complex theory of elliptic functions. In Section 4 we
prove one well-known theta-function identity involving Lambert series by using the theta-function
identity proved in Section 3, and calculate t2k(n) for 2k = 4, 6, 8. In Section 5 we derive a variant
form of an identity of Ramanujan. In Section 6 and 7 we study two classes of Lambert series using
the aforementioned identity. In Section 8 we derive two identities for ψ12(q). In Section 9 we obtain
three identities for ψ16(q), ψ20(q), and ψ28(q), respectively. In Section 10 we establish some identities
relating ψ24(q) and ψ32(q). In the final Section 11 we establish some identities among τ(n), t24(n),
and σ11(n).
Lastly, we introduce two lemmas that will be used in the sequel of this paper.
Lemma 1 Let σk(n) denote the sum of the kth powers of the divisors of n , namely,
σk(n) =∑
d|ndk. (1.3)
Then we have ∞∑n=1
nkqn
1− qn=
∞∑n=1
σk(n)qn (1.4)
3
and ∞∑n=0
(2n + 1)kq2n+1
1− q2(2n+1)=
∞∑n=0
σk(2n + 1)q2n+1, (1.5)
By convention, we use σ(n) to denote the number of positive divisors of the positive integer n.
Proof. Expanding 1/(1− qn) as a geometric series and reversing the order of summation, we readily
find that (1.4). The proof of (1.5) is similar and so we omit it.
Lemma 2 There holds the identity
∞∑n=1
nkqn
1− q2nsin
nπ
2=
∞∑n=0
∑
d|2n+1
dk sin(12dπ)
q2n+1. (1.6)
The proof is simple and we leave it to the reader. In this paper will also make use of the following
standard notation:
(z; q)∞ =∞∏
n=0
(1− zqn). (1.7)
2 Some basic facts about Jacobi’s theta functions
In this section we will introduce some fundamental facts about classical theta functions. Let q = eπiτ ,
where Im τ > 0. Then for a complex number z, the Jacobi theta functions [24, pp. 463-464] are
defined by
θ1(z|q) = −iq14
∞∑n=−∞
(−1)nqn(n+1)e(2n+1)iz = 2q14
∞∑n=0
(−1)nqn(n+1) sin(2n + 1)z, (2.1)
θ2(z|q) = q14
∞∑n=−∞
qn(n+1)e(2n+1)iz = 2q14
∞∑n=0
qn(n+1) cos(2n + 1)z, (2.2)
θ3(z|q) =∞∑
n=−∞qn2
e2niz = 1 + 2∞∑
n=1
qn2cos 2nz, (2.3)
θ4(z|q) =∞∑
n=−∞(−1)nqn2
e2niz = 1 + 2∞∑
n=1
(−1)nqn2cos 2nz. (2.4)
Employing the Jacobi triple product identity, one can derive the infinite product expansions for
Jacobi’s theta functions θ1(z|q), θ2(z|q), θ3(z|q), and θ4(z|q), namely,
θ1(z|q) = 2q14 (sin z)(q2; q2)∞(q2e2iz; q2)∞(q2e−2iz; q2)∞, (2.5)
θ2(z|q) = 2q14 (cos z)(q2; q2)(−q2e2iz; q2)∞(−q2e−2iz; q2), (2.6)
θ3(z|q) = (q2; q2)∞(−qe2iz; q2)∞(qe−2iz; q2)∞, (2.7)
θ4(z|q) = (q2; q2)∞(qe2iz; q2)∞(qe−2iz; q2)∞ (2.8)
4
(see, for example, [24, pp. 469-470]). Using these product expansions for the Jacobi theta functions,
by a direct computation, we readily find that
θ′1(0|q)θ1(2z|q) = 2θ1(z|q)θ2(z|q)θ3(z|q)θ4(z|q). (2.9)
From this identity we see that with the (quasi) periods π and πτ the zero points of θ1(2z|q) in the
period parallelogram are
0,π
2,
π + πτ
2,
πτ
2. (2.10)
When we let z = 0 in (2.1)-(2.8), they yield the following special cases:
θ′1(0|q) = 2q14
∞∑n=0
(−1)n(2n + 1)qn(n+1) = 2q14 (q2; q2)3∞, (2.11)
θ2(0|q) = 2q14 ψ(q2) = 2q
14
∞∑n=0
qn(n+1) = 2q14 (q2; q2)∞(−q2; q2)2∞, (2.12)
θ3(0|q) = φ(q) =∞∑
n=−∞qn2
= (q2; q2)∞(−q; q2)2∞, (2.13)
θ4(0|q) = φ(−q) =∞∑
n=−∞(−1)nqn2
= (q2; q2)∞(q; q2)2∞, (2.14)
where θ′1(z|q) denotes the partial derivative of θ1(z|q) with respect to z.
With respect to the (quasi) periods π and πτ , we have the functional equations
θ1(z + π|q) = −θ1(z|q) and θ1(z + πτ |q) = −q−1e−2πizθ1(z|q). (2.15)
The four Jacobi theta functions are obviouly related and we have the following relations:
θ1(z +12π|q) = θ2(z|q), θ1(z +
12(π + πτ)|q) = q−
14 e−πizθ3(z|q),
θ1(z +12πτ |q) = iq−
14 e−πizθ4(z|q). (2.16)
Taking the logarithmic derivative of both sides of the last identity in (2.16) with respect to z, we find
thatθ′1θ1
(z +
πτ
2
∣∣∣q)
= −i +θ′4θ4
(z|q). (2.17)
Differentiate the above equation with respect to z, k times, and we find that
(θ′1θ1
)(k) (z +
πτ
2
∣∣∣q)
=(
θ′4θ4
)(k)
(z|q), k ≥ 1. (2.18)
Employing the infinite product expansions for θ1(z|q) and θ4(z|q) , we can deduce respectively the
trigonometric series expansions for the logarithmic derivatives of θ1(z|τ) and θ4(z|τ), namely,
θ′1θ1
(z|q) = cot z + 4∞∑
n=1
q2n
1− q2nsin 2nz (2.19)
5
andθ′4θ4
(z|q) = 4∞∑
n=1
qn
1− q2nsin 2nz. (2.20)
It is well-known that the Laurent expansion formula for cot z is
cot z =1z− z
3− z3
45− 2z5
945+ · · · . (2.21)
Hence, we substitute (2.21) into (2.19) and reverse the order of summation to get
θ′1θ1
(z|q) =1z− 1
3
(1− 24
∞∑n=1
nq2n
1− q2n
)z − 1
45
(1 + 240
∞∑n=1
n3q2n
1− q2n
)z3
− 2945
(1− 504
∞∑n=1
n5q2n
1− q2n
)z5 + · · · . (2.22)
3 An Identity for Theta Function θ1(z|q)We begin this section with the following theta-functions identity. It is, perhaps, a new theta function
identity. This identity plays a fundamental role in this paper.
Theorem 1 Let θ1(z|q) be defined by (2.1). Then for any complex numbers u, v, and w, we have
θ21(v|q)θ1(w + u|q)θ1(w − u|q)− θ2
1(u|q)θ1(w + v|q)θ1(w − v|q)
= θ21(w|q)θ1(v + u|q)θ1(v − u|q). (3.1)
To prove the identity, we need the following fundamental theorem of elliptic functions (see, for example,
[7, p. 22, Theorem2]). This theorem is quite useful in proving theta-function identities . Recently, in
[12, 13, 14, 15, 16], we have used it to derive many important theta function identities.
Theorem 2 The sum of all the residues of an elliptic function at the poles inside a period-parallelogram
is zero.
Proof. Let u, v, w be three distinct parameters and v, w are different from 0, 12π, 1
2 (π+πτ), 12πτ . We
consider the function
f(z) =θ1(2z|q)θ1(z − u|q)θ1(z + u|q)
θ21(z|q)θ1(z − v|q)θ1(z + v|q)θ1(z − w|q)θ1(z + w|q) , (3.2)
where 0 < v, w < π. Using (2.15), we can readily verify that f(z) is an elliptic function with periods
π and πτ . The poles of f(z) in the period parallelogram with vertices located at 0, π, π + πτ, πτ are
at z = 0, v, π − v, w, and π − w; and all of them are simple poles. Let res(f ; α) denote the residue of
f(z) at α. From Theorem 2, we have
res(f ; 0) + res(f ; v) + res(f ;π − v) + res(f ; w) + res(f ; π − w) = 0. (3.3)
6
We now begin to compute the residues of f(z) at z = 0, v, π − v, w, and π − w, respectively. By
L’Hopital’s rule, we find that
res(f ; 0) = limz→0
zf(z)
= limz→0
θ1(z − u|q)θ1(z + u|q)θ1(z − v|q)θ1(z + v|q)θ1(z − w|q)θ1(z + w|q)
× limz→0
zθ1(2z|q)θ21(z|q)
= − 2θ21(u|q)
θ′1(0|q)θ21(w|q)θ2
1(v|q). (3.4)
Employing the same type argument as that of (3.4), we find that
res(f ; v) = res(f ; π − v) == − θ1(v − u|q)θ1(v + u|q)θ′1(0|q)θ2
1(v|q)θ1(w − v|q)θ1(w + v|q) (3.5)
and
res(f ; w) = res(f ;π − w) =θ1(w − u|q)θ1(w + u|q)
θ′1(0|q)θ21(w|q)θ1(w − v|q)θ1(w + v|q) . (3.6)
Substituting (3.4), (3.5) and (3.6) into (3.3), (3.1) follows. By analytic continuation, we know (3.1)
holds for all u, v, w. This completes the proof of Thereom 1.
Corollary 1 Let θ2(z|q), θ3(z|q), and θ4(z|q) be defined by (2.2), (2.3), and (2.4), respectively. Then
θ22(u|q)θ2(v + w|q)θ2(v − w|q) + θ2
4(w|q)θ4(u + v|q)θ4(u− v|q)= θ2
3(v|q)θ3(u + w|q)θ3(u− w|q). (3.7)
Proof. Replacing u by u+ 12π, v by v+ 1
2 (π+πτ), and w by w+ 12πτ in (3.1), respectively, simplifying
the resulting equation using (2.16), we obtain (3.7). This complete the proof of Corollary 1.
Taking u = v = w = 0 in (3.7), we obtain Jacobi’s quartic identity [24, p. 467]
θ43(0|q) = θ4
2(0|q) + θ44(0|q). (3.8)
An interesting proof of this identity is given in [9]. In [16], the author has generalized it to a theta
function identity with three parameters , namely,
θ3(x|q)θ3(y|q)θ3(z|q)θ3(x + y + z|q)= θ1(x|q)θ1(y|q)θ1(z|q)θ1(x + y + z|q)
+θ2(x|q)θ2(y|q)θ2(z|q)θ2(x + y + z|q)+θ4(x|q)θ4(y|q)θ4(z|q)θ4(x + y + z|q). (3.9)
From which we can simply derive many modular identities of degree 2 through 7.
7
4 A theta-function identity involving Lambert series and the
computations of t4(n), t6(n), and t8(n)
4.1 A theta-function identity involving Lambert series and the compuation
of t4(n)
Theorem 3 Let θ1(z|q) be defined as in (2.1). Then we have(
θ′1θ1
)′(u|q)−
(θ′1θ1
)′(v|q) = θ′1(0|q)2
θ1(u− v|q)θ1(u + v|q)θ21(u|q)θ2
1(v|q). (4.1)
This is a well-known theta-function identity (see, for example, [23, p. 325, Equation(1.7)] and [16, p.
143, Equation(7.1)]). Here we will rederive it from (3.1).
Proof. Differentiating (3.1) with respect to w, twice, by using the method of logarithmic differenti-
ation, we obtain
θ21(v|q)θ1(w + u|q)θ1(w − u|q) {
φ21(w) + φ′1(w)
}
−θ21(u|q)θ1(w + v|q)θ1(w − v|q) {
φ22(w) + φ′2(w)
}
= 2{θ′1(w|q)2 + θ1(w|q)θ′′1 (w|q)} θ1(v + u|q)θ1(v − u|q), (4.2)
where
φ1(w) =θ′1θ1
(w + u|q) +θ′1θ1
(w − u|q),
φ2(w) =θ′1θ1
(w + v|q) +θ′1θ1
(w − v|q). (4.3)
When w = 0, (4.2) reduces to (4.1). This completes the proof of Theorem 3.
Replacing u by u + 12πτ , and v by v + 1
2πτ in (4.1), and then using (2.15) , (2.16), and (2.18) in the
resulting equation, we obtain
Corollary 2 Let θ1(z|q) and θ4(z|q) be defined by (2.1) and (2.4), respectively. Then(
θ′4θ4
)′(u|q)−
(θ′4θ4
)′(v|q) = −θ′1(0|q)2
θ1(u− v|q)θ1(u + v|q)θ24(u|q)θ2
4(v|q). (4.4)
Replacing the left sides of (4.1) and (4.4) by the trigonometric series expansions of the logarithmic
derivatives of θ1(z|q) and θ4(z|q), respectively, we find that
Corollary 3 We have
cot2 v − cot2 u + 8∞∑
n=1
nq2n
1− q2n(cos 2nu− cos 2nv) = θ′1(0|q)2
θ1(u− v|q)θ1(u + v|q)θ21(u|q)θ2
1(v|q)(4.5)
and
8∞∑
n=1
nqn
1− q2n(cos 2nu− cos 2nv) = −θ′1(0|q)2
θ1(u− v|q)θ1(u + v|q)θ24(u|q)θ2
4(v|q). (4.6)
8
Setting u = 12π and v = 1
4π in (4.5), and then writing q2 as q, we find that
θ44(0|q) = 1 + 8
∞∑n=1
(−1)n nqn
1− qn− 16
∞∑n=1
(−1)n nq2n
1− q2n
= 1 + 8∞∑
n=1
(−1)n nqn
1− qn− 8
∞∑n=1
(−1)n
{nqn
1− qn− nqn
1 + qn
}
= 1 + 8∞∑
n=1
(−1)n nqn
1 + qn. (4.7)
Replacing q by −q in the above equation, after simple reduction, we obtain the following identity due
to Jacobi [4, p. 15]. A recent proof of this identity can be found in [1].
Theorem 4 Let φ(q) be defined as in (1.1). Then
φ4(q) = θ43(0|q) = 1 + 8
∞∑n=1
nqn
1 + (−q)n
= 1 + 8∞∑
n=1
nqn
1− qn− 32
∞∑n=1
nq4n
1− q4n. (4.8)
Consequently, we have
r4(n) = 8σ1(n)− 32σ1
(n
4
). (4.9)
By setting u = 0 and v = 12π in (4.6) , we obtain the following identity.
Theorem 5 Let ψ(q) be defined as in (1.2). Then
qψ4(q2) =116
θ42(0|q) =
∞∑n=0
(2n + 1)q2n+1
1− q2(2n+1), (4.10)
and thus
t4(n) = σ1(2n + 1). (4.11)
Equation (4.11) is [20, Theorem 3]. For an account of (4.12) one may consult [5].
4.2 Lambert series expansions for θ42(0|q) + θ4
3(0|q)From (4.8) and (4.10), we can obtain the following identity, which plays a pivotal role in the compu-
tation of t2k(n).
Theorem 6 There holds the identity
θ42(0|q) + θ4
3(0|q) = 1 + 24∞∑
n=1
nqn
1 + qn. (4.12)
9
Proof. By (4.8) and (4.10), we have
θ42(0|q) + θ4
3(0|q)
= 1 + 8∞∑
n=1
nqn
1 + (−q)n+ 16
∞∑n=0
(2n + 1)q2n+1
1− q2(2n+1)
= 1 + 8∞∑
n=1
2nq2n
1 + q2n+ 8
∞∑n=0
(2n + 1)q2n+1
1− q2n+1
+8∞∑
n=0
(2n + 1)q2n+1
1 + q2n+1+ 8
∞∑n=0
(2n + 1)q2n+1
1− q2n+1
= 1 + 8∞∑
n=1
nqn
1 + qn+ 16
∞∑n=0
(2n + 1)q2n+1
1− q2n+1
= 1 + 8∞∑
n=1
nqn
1 + qn+ 16
∞∑n=0
{nqn
1− qn− 2nq2n
1− q2n
}
= 1 + 24∞∑
n=1
nqn
1 + qn. (4.13)
This proves (4.12) and so we complete the proof of Theorem 6.
4.3 The computations of t6(n) and t8(n)
Dividing both sides of (4.5) and (4.6), respectively, by u− v and then letting v → u gives
Theorem 7 With θ1(z|q) and θ4(z|q) be defined by (2.1) and (2.4), respectively. Then
(θ′1θ1
)′′(z|q) = cot2 u(1 + cot2 u)− 16
∞∑n=1
n2q2n
1− q2nsin 2nu = θ′1(0|q)3
θ1(2u|q)θ41(u|q)
(4.14)
and
−(
θ′4θ4
)′′(z|q) = 16
∞∑n=1
n2qn
1− q2nsin 2nu = θ′1(0|q)3
θ1(2u|q)θ44(u|q)
. (4.15)
Now we use (4.15) to calculate t6(n) and t8(n). Setting u = 14π in (4.15), we obtain
14θ22(0|q2)θ4
4(0|q2) =∞∑
n=1
n2qn
1− q2nsin
(nπ
2
)(4.16)
Appealing to (1.6), we have
14θ22(0|q2)θ4
4(0|q2) =∞∑
n=0
( ∑
d|2n+1
d2 sin(12dπ)
)q2n+1. (4.17)
Writing q as iq(i =√−1), (4.17) becomes
14θ22(0|q2)θ4
3(0|q2) =∞∑
n=0
(−1)n( ∑
d|2n+1
d2 sin(12dπ)
)q2n+1. (4.18)
10
Subtracting (4.17) from (4.18) and using Jacobi’s identity, (3.8), we obtain
θ62(0|q2) = −8
∞∑n=0
( ∑
d|4n+3
d2 sin(12dπ)
)q4n+3. (4.19)
Hence we obtain the following theorem.
Theorem 8 Let ψ(q) be defined as in (1.2). Then
8ψ6(q) = −∞∑
n=0
( ∑
d|4n+3
d2 sin(12dπ)
)qn. (4.20)
Consequently, we have
t6(n) = −18
∑
d|4n+3
d2 sin(12dπ) =
18
∑d|4n+3
d≡3 (mod 4)
d2 − 18
∑d|4n+3
d≡1 (mod 4)
d2. (4.21)
There is an interesting history about t6(n). One may consult [18, pp. 78-79]. Equation (4.21) is
equivalent to [20, Theorem 4]. Other recent proofs have also been given by Kac and Wakimoto [10]
and B. C. Berndt [5].
Dividing both sides of (4.15) by u, and then lettig u → 0, we obtain the following identity:
∞∑n=1
n3qn
1− q2n=
116
θ42(0|q)θ4
3(0|q) = qψ8(q). (4.22)
Employing (1.4), we find that
∞∑n=1
n3qn
1− q2n=
∞∑n=1
n3qn
1− qn−
∞∑n=1
n3q2n
1− q2n
=∞∑
n=1
(σ3(n)− σ3(
n
2))qn. (4.23)
Combining (4.22) and (4.23) gives
Theorem 9 There holds the identity
qψ8(q) =∞∑
n=1
(σ3(n)− σ3(
n
2))qn. (4.24)
Consequently, we have
t8(n) = σ3(n)− σ3(n
2). (4.25)
Identity (4.22) is due to Legendre [11, p. 133]. It was stated without proof by Ramanujan [21, p.
144]. Ewell [8] proved it by using Jacobi triple product identity and another theta function identity.
Equation (4.23) is equivalent to [20, Theorem 5].
11
5 An equivalent form of an identity of Ramanujan
In this section we will prove the following identity. This identity has appeared in [16], but the proof
given here is different from that in [16].
Theorem 10 There holds the identity
{1− 24
∞∑n=1
nq2n
1− q2n+ 24
∞∑n=1
nqn
1− q2ncos nu
}2
= 1 + 240∞∑
n=1
n3q2n
1− q2n+ 48
∞∑n=1
n3qn
1− q2ncos nu. (5.1)
Proof. We multiply both sides of (3.1) by w5θ1(2w|q)/θ61(w|q). Then we obtain
θ21(v|q)f1(w)− θ2
1(u|q)f2(w) = θ1(v + u|q)θ1(v − u|q)f3(w), (5.2)
where
f1(w) = w5θ1(w + u|q)θ1(w − u|q)θ1(2w|q)θ61(w|q)
,
f2(w) = w5θ1(w + v|q)θ1(w − v|q)θ1(2w|q)θ61(w|q)
,
f3(w) = w5 θ1(2w|q)θ41(w|q)
. (5.3)
Differentiating f1(w) with respect to w, four times, by using the method of logarithmic differentiation,
we readily find that
f(4)1 (w) = f1(w)
{φ4(w) + 6φ2(w)φ′(w) + 4φ(w)φ′′(w)
+3φ′(w)2 + φ′′′(w)}
, (5.4)
where
φ(w) =5w− 6
θ′1θ1
(w|q) + 2θ′1θ1
(w|q) +θ′1θ1
(w − u|q) +θ′1θ1
(w + u|q). (5.5)
Applying (2.22) in (5.5), we find that
φ(w) =23
(1− 24
∞∑n=1
nq2n
1− q2n
)w − 2
9
(1 + 240
∞∑n=1
n3q2n
1− q2n
)w3 + O(w5)
+θ′1θ1
(w − u|q) +θ′1θ1
(w + u|q). (5.6)
From the above equation, we readily find that
φ(0) = 0, φ′′(0) = 0, φ′(0) =23A(u), φ′′′(0) = −4
3B(u), (5.7)
12
where
A(u) = 1− 24∞∑
n=1
nq2n
1− q2n+ 3
(θ′1θ1
)′(u|q)
= − 124
{112
+18
cot2 u +∞∑
n=1
nq2n
1− q2n(1− cos 2nu)
}, (5.8)
and
B(u) = 1 + 240∞∑
n=1
n3q2n
1− q2n− 3
2
(θ′1θ1
)′′′(u|q)
=1
576
{(112
+18
cot2 u
)2
+112
∞∑n=1
n3q2n
1− q2n(5 + cos 2nu)
}. (5.9)
Employing (5.7)-(5.9) in (5.4) and then setting z = 0, we find that
f(4)1 (0) = − 8θ2
1(u|q)3θ′1(0|q)5
{A2(u)−B(u)
}. (5.10)
We proceed as in (5.4)-(5.10) to obtain
f(4)2 (0) = − 8θ2
1(v|q)3θ′1(0|q)5
{A2(v)−B(v)
}. (5.11)
From (2.22) and (4.14), we can deduce that
f3(w) = 2θ′1(0|q)3w2 + O(w6). (5.12)
It follows that
f(4)3 (0) = 0. (5.13)
Differentiating (5.2) with respect to w, five times, and then taking w = 0, we have
θ21(v|q)f (4)
1 (0)− θ21(u|q)f (4)
2 (0) = θ1(v + u|q)θ1(v − u|q)f (4)3 (0). (5.14)
Substituting (5.10), (5.11), and (5.13) into (5.14), we obtain, for all u, v not equal integral multiples
of π,
A2(u)−B(u) = A2(v)−B(v). (5.15)
Using (5.8) and (5.9), we find that
576 limv→0
(A2(v)−B(v))
= limv→0
{ ∞∑n=1
nq2n
1− q2n
(13
sin2 nv +12
cot2 v sin2 nv
)}
+ limv→0
{ ∞∑n=1
nq2n
1− q2n(1− cos 2nz)
}2
13
− 112
limv→0
∞∑n=1
n3q2n
1− q2n(5 + cos 2nz)
=12
∞∑n=1
n3q2n
1− q2n− 1
2
∞∑n=1
n3q2n
1− q2n
= 0. (5.16)
Thus, letting v → 0 in (5.15) and then using (5.16), we obtain
A2(u)−B(u) = 0. (5.17)
Substituting (5.8) and (5.9) into the above equation, we obtain
{1− 24
∞∑n=1
nq2n
1− q2n+ 3
(θ′1θ1
)′(u|q)
}2
= 1 + 240∞∑
n=1
n3q2n
1− q2n− 3
2
(θ′1θ1
)′′′(u|q). (5.18)
Replacing the logarithmic derivative of θ1(z|q) by its trigonometric series expansion in this equation,
we obtain the following identity of Ramanujan [21, p. 139].
Theorem 11 We have{
18
cot2 u +112
+∞∑
n=1
nq2n
1− q2n(1− cos 2nu)
}2
={
18
cot2 u +112
}2
+112
∞∑n=1
n3q2n
1− q2n(5 + cos 2nu). (5.19)
Writing u as u + 12πτ and then using (2.18), (5.18) becomes
{1− 24
∞∑n=1
nq2n
1− q2n+ 3
(θ′4θ4
)′(u|q)
}2
= 1 + 240∞∑
n=1
n3q2n
1− q2n− 3
2
(θ′4θ4
)′′′(u|q). (5.20)
Substituting (2.20) into the above equation, and then writing 2u as u, we obatin (5.1). This completes
the proof of Theorem 10.
6 The first class of Lambert seires
In this section we will consider the Lambert seires T2k(q) defined by
T2(q) = 1 + 24∞∑
n=1
nqn
1 + qnand T2k(q) =
∞∑n=1
n2k−1qn
1− q2n, k ≥ 2. (6.1)
14
We now begin to deduce a recurrence formula for T2k(q) using (5.20). It is well known that the
Maclaurin series for cos nu is
cos nu =∞∑
k=0
(−1)k (nu)2k
(2k)!. (6.2)
Substituting (6.2) into (5.1), inverting the order of summation, we obtain the identity(
T2(q) + 24∞∑
k=1
(−1)k u2k
(2k)!T2k+2(q)
)2
= 1 + 240∞∑
n=1
n3q2n
1− q2n+ 48T4(q) + 48
∞∑
k=1
(−1)k u2k
(2k)!T2k+4(q). (6.3)
Comparing the coefficients of u2m on either side, we obtain the following recurence formula among
T2k(q), k = 1, 2, 3, · · ·. This formula allows us to express T2k(q) in terms of T2(q) and T4(q).
Theorem 12 If T2k(q) are defined by (6.1). Then we have
T 22 (q) = 1 + 240
∞∑n=1
n3q2n
1− q2n+ 48T4(q), (6.4)
T2m+4(q) = T2(0|q)T2m+2(q)
+12m−1∑
k=1
(2m)!(2k)!(2m− 2k)!
T2k+2(q)T2m+2−2k(q). (6.5)
In particular,
T6(q) = T2(q)T4(q), (6.6)
T8(q) = T2(q)T6(q) + 72T 24 (q), (6.7)
T10(q) = T2(q)T8(q) + 360T4(q)T6(q), (6.8)
T12(q) = T2(q)T10(q) + 672T4(q)T8(q) + 840T 26 (q). (6.9)
Utilizing (4.10), (4.22), and (6.5) we can write T2k(q) in terms of θ2(0|q) and θ3(0|q). Here we shall
use (6.6)-(6.9) to express T2k(q) in terms of theta functions θ2(0|q) and θ3(0|q) for 6 ≤ 2k ≤ 12. For
simplicity, we define
x(q) :=θ42(0|q)
θ43(0|q)
and z(q) := θ43(0|q) = φ4(q). (6.10)
Then (4.10) and (4.22) can be written in the following forms:
T2(q) = z(q)(1 + x(q)), (6.11)
T4(q) = qψ8(q) =116
z2(q)x(q). (6.12)
Substituting the above two equations into (6.4), we immediately obtain the following identity [4, p.
49, Equation(12. 21)]
1 + 240∞∑
n=1
n3q2n
1− q2n= z2(q)
(1− x(q) + x2(q)
). (6.13)
15
Substituting (6.11) and (6.12) into (6.6), we find that
T6(q) =116
z3(q)(x(q) + x2(q)
). (6.14)
Similarly, from (6.7), (6.8), and (6.9), we find respectively that
T8(q) =132
z4(q)(2x(q) + 13x2(q) + 2x3(q)
), (6.15)
T10(q) =116
z5(q)(x(q) + 30x2(q) + 30x3(q) + x4(q)
), (6.16)
T12(q) =132
z6(q)(2x(q) + 251x2(q) + 876x3(q) + 251x4(q) + 2x5(q)
). (6.17)
7 The second class of Lambert series
In this section we will investigate the Lambert series defined as
S2k(q) =∞∑
n=0
(2n + 1)2k−1qn+1/2
1− q2n+1. (7.1)
And the main result of this section is the following recurrence relation. Using this recurrence , we can
express S2k(q) in terms of z(q) and x(q).
Theorem 13 Let T2k(q) be defined by (6.1) and S2k(q) by (7.1). Then we have
S2m+4(q) = T2(q)S2m+2(q) + 48m∑
k=1
(2m)!4k
(2k)!(2m− 2k)!T2k+2(q)S2m+2−2k(q). (7.2)
Proof. We recall the identity in (5.1), namely,{
1− 24∞∑
n=1
nq2n
1− q2n+ 24
∞∑n=1
nqn
1− q2ncos nu
}2
= 1 + 240∞∑
n=1
n3q2n
1− q2n+ 48
∞∑n=1
n3qn
1− q2ncos nu. (7.3)
Replacing q by −q, this identity then becomes{
1− 24∞∑
n=1
nq2n
1− q2n+ 24
∞∑n=1
(−1)n nqn
1− q2ncosnu
}2
= 1 + 240∞∑
n=1
n3q2n
1− q2n+ 48
∞∑n=1
(−1)n n3qn
1− q2ncosnu. (7.4)
We subtract (7.4) from (7.3) and then replace q2 by q to get∞∑
n=0
(2n + 1)3qn+1/2
1− q2n+1cos(2n + 1)u
=
( ∞∑n=0
(2n + 1)qn+1/2
1− q2n+1cos(2n + 1)u
)
×(
1− 24∞∑
n=1
nqn
1− qn+ 48
∞∑n=1
nqn
1− q2ncos 2nu
). (7.5)
16
Substituting
cos(2n + 1)u = 1 +∞∑
k=1
(−1)k (2n + 1)2k
(2k)!u2k and cos 2nu = 1 +
∞∑
k=1
(−1)k (2n)2k
(2k)!u2k (7.6)
into (7.5), inverting the order of summation, we obtain the identity
∞∑
k=0
(−1)k u2k
(2k)!S2k+4(q)
=
( ∞∑
k=0
(−1)k u2k
(2k)!S2k+2(q)
)×
(T2(q) + 48
∞∑
k=1
(−1)k4k u2k
(2k)!T2k+2(q)
). (7.7)
Equating the coefficients of u2m on both sides of (7.7), we obtain (7.2). This complete the proof of
Theorem 13.
The first few special cases of (7.2) are
S4(q) = T2(q)S2(q), (7.8)
S6(q) = S4(q)T2(q) + 192S2(q)T4(q), (7.9)
S8(q) = T2(q)S6(q) + 1152T4(q)S4(q) + 768S2(q)T6(q), (7.10)
S10(q) = T2(q)S8(q) + 2880T4(q)S6(q) + 11520T6(q)S4(q) + 3072T8(q)S2(q), (7.11)
S12(q) = T2(q)S10(q) + 5376T4(q)S8(q) + 53760T6(q)S6(q)
+86016T8(q)S4(q) + 12288T10(q)S2(q). (7.12)
Using the product expansions for θ2(0|q) in (2.12) and θ3(0|q) in (2.13), we readily find that
4θ2(0|q)θ3(0|q) = θ22(0|q1/2). (7.13)
Replacing q2 by q in (4.10) and then using (7.13), we obtain
S2(q) =14θ22(0|q)θ2
3(0|q). (7.14)
Applying (6.10) to the right side of this equation, we find that
S2(q) =14z(q)
√x(q). (7.15)
Substituting (6.11) and (7.15) into (7.8), we find that
S4(q) =14z2(q)
√x(q)(1 + x(q)). (7.16)
In the same way, we also find that
S6(q) =14z2(q)
√x(q)(1 + 14x(q) + x2(q)), (7.17)
17
S8(q) =14z4(q)
√x(q)
(1 + 135x(q) + 135x2(q) + x3(q)
), (7.18)
S10(q) =14z4(q)
√x(q)
(1 + 1228x(q) + 5478x2(q) + 1228x3(q) + x4(q)
), (7.19)
S12(q) =14z4(q)
√x(q)
(1 + 11069x(q) + 165826x2(q) + 165826x3(q)
+11069x4(q) + x5(q)). (7.20)
Using Jacobi’s quartic identity, (3.8), and the definitions of x(q) and z(q), we find that
z(−q) = z(q)(1− x(q)) and x(−q) = − x(q)1− x(q)
. (7.21)
By direct computaions, we find that
S2(q2) =148
(T2(q)− T2(−q)) and S2k(q2) =12
(T2k(q)− T2k(−q)) , k ≥ 2. (7.22)
Combining (6.11),(7.21), and the first identity in (7.22), we find that
S2(q2) =116
z(q)x(q). (7.23)
Combining (6.12),(7.21), and the second identity in (7.22) with k = 2, we find that
S4(q2) =132
z2(q)x(q)(2− x(q)). (7.24)
Similarly, we also have
S6(q2) =116
z3(q)x(q)(1− x(q) + x2(q)), (7.25)
S8(q2) =164
z4(q)x(q)(2− x(q))(2− 2x(q) + 17x2(q)), (7.26)
S10(q2) =116
z5(q)x(q)(1− 2x(q) + 78x2(q)− 77x3(q) + 31x4(q)), (7.27)
S12(q2) =132
z6(q)x(q)(2− x(q))(1− x(q) + x2(q))(1− x(q) + 691x2(q)). (7.28)
8 Two identities for ψ12(q)
Theorem 14 Let t2k(n) and σ2k(n) be defined by (1.2) and (1.3), respectively. Then
∞∑n=0
σ5(2n + 1)q2n+1 = 256∞∑
n=0
t12(n)q2n+3 + q(q2; q2)12∞. (8.1)
Furthermore, if we define a(n) as the coefficient of qn in
q(q2; q2)12∞ =∞∑
n=0
a(n)q2n+1, (8.2)
then we have
256t12(n) = σ5(2n + 3)− a(n + 1). (8.3)
18
Equation (8.1) is the same as [20, Theorem 7]. For an account one may consult [18].
Proof. Using Jacobi’s quartic identity, (3.8), and the product expansion formulas for θ2(0|q), θ3(0|q),and θ4(0|q), we readily find that
z3(q)x(q)(1− x(q)) = 16q(q2; q2)12∞ and z(q)x(q) = θ42(0|q) = 16qψ4(q2). (8.4)
Hence the left side of (7.25) is equal to
116
z3(q)x(q)(1− x(q)) +116
z3(q)x3(q) = q(q2; q2)12∞ + 256qψ12(q2). (8.5)
It follows that
S6(q2) =∞∑
n=0
(2n + 1)5q2n+1
1− q2(2n+1)= q(q2; q2)12∞ + 256q3ψ12(q2). (8.6)
Substituting∞∑
n=0
(2n + 1)5q2n+1
1− q2(2n+1)=
∞∑n=0
σ5(2n + 1)q2n+1 (8.7)
and
q3ψ12(q2) =∞∑
n=0
t12(n)q2n+3 (8.8)
into (8.6), equating the coefficients of q2n+3, we obtain (8.1), thereby we complete the proof of Theorem
14.
Now we state the other formula for ψ12(q), which is different from the identity in (8.1).
Theorem 15 Let T2k(q) and S2k(q) be defined by (6.1) and (7.1), respectively, then
T4(q2)S2(q2) = q3ψ12(q2). (8.9)
Proof. Using the product expansion formulas for θ2(0|q) and θ3(0|q), we find that
z(q)√
x(q) = 4q1/2ψ4(q). (8.10)
Multiplying (6.12) and (7.15), we immdediately have
T4(q)S2(q) =164
(z(q)
√x(q)
)3
= q3/2ψ12(q). (8.11)
Replacing q by q2 in the above equation, we complete the proof of Theorem 15.
Comparing (8.6) and (8.9), we find the following identity, which may be a new identity.
Theorem 16 With T2k(q) and S2k(q) defined by (6.1) and (7.1), respectively,
q(q2; q2)12∞ = S6(q2)− 256S2(q2)T4(q2). (8.12)
19
9 Identities relating ψ16(q), ψ20(q), and ψ28(q)
9.1 An identity relating ψ16(q)
Theorem 17 Let S2k(q) be the Lambert series defined by (7.1). Then
192∞∑
n=0
t16(n)q2n+4 = 192q4ψ(q2)16 = S2(q2)S6(q2)− S24(q2). (9.1)
Consequently, we have
192t16(n− 1) =n∑
k=0
(σ1(2k + 1)σ5(2n− 2k + 1)− σ3(2k + 1)σ3(2n− 2k + 1)
). (9.2)
Proof. By (7.16), (7.17), and (7.18) and some elementary evaluations, we find that
S2(q)S6(q)− S24(q)
=116
z4(q)x(q)(1 + 14x(q) + x2(q))− 116
z4(q)x(q)(1 + x(q))2
=34(z(q)
√x(q))4. (9.3)
Substituting (8.10) into the left side of the above equation and then changing q as q2, we obtain (9.1).
We compare the coefficients of qn on both sides of (9.1) to obtain (9.2). This complete the proof of
Theorem 17.
9.2 An identity relating ψ20(q)
Theorem 18 Let T2k(q) and S2k(q) be the Lambert series defined by (6.1) and (7.1), respectively.
Then
72∞∑
n=0
t20(n)q2n+5 = 120q5ψ(q2)20 = T8(q2)S2(q2)− S4(q2)T6(q2). (9.4)
Proof. We recall the two identities in (6.7) and (7.8), namely,
T8(q) = T2(q)T6(q) + 72T 24 (q), (9.5)
S4(q) = T2(q)S2(q). (9.6)
Eliminating T2(q) between the above two equations, we arrive at
72S2(q)T 24 (q) = T8(q)S2(q)− S4(q)T6(q). (9.7)
Substituting (6.12) and (7.15) into the left side of the above equation, we find that
9128
(z(q)√
x(q))5 = T8(q)S2(q)− S4(q)T6(q). (9.8)
Using (8.10) in the left side of the above equation and then changing q as q2, we obtain (9.4). Thus
we complete the proof of Theorem 18 .
20
9.3 An identity for t28(n)
Theorem 19 Let T2k(q) and S2k(q) be defined by (6.1) and (7.1), respectively. Then we have
21T8(q2)S6(q2) + 2T4(q2)S10(q2)− 23T12(q2)S2(q2) = 725760q7ψ28(q2). (9.9)
Proof. From (6.12), (6.15), (6.17), (7.14), (7.17), and (7.19), by direct computations, we find that
21T8(q)S6(q) + 2T4(q)S10(q)− 23T12(q)S2(q) =283564
(z(q)√
x(q))7. (9.10)
Substituting (8.10) into the left side of the above equation and then replacing q by q2, we obtain (9.9).
This completes the proof of Theorem 19.
10 Identities relating ψ24(q) and ψ32(q)
10.1 Two formulas for ψ24(q)
Theorem 20 We have
T24(q) =∞∑
n=1
n11qn
1− q2n= 2072q2(q2; q2)24∞ + q(q; q)24∞ + 176896q3ψ24(q). (10.1)
Proof. By elementary computation, we find for any a that
2 + 251a + 876a2 + 251a3 + 2a4 = 1382a2 + 259a(1− a)2 + 2(1− a)4. (10.2)
Thus from (6.17), we have
T12(q)
=132
z6(q)x(q)(2 + 251x(q) + 876x2(q) + 251x3(q) + 2x4(q)
)
=132
z6(q)x(q)(1382x(q)2 + 259x(q)(1− x(q))2 + 2(1− x(q))4
)
=69116
(z(q)
√x(q)
)2
+25932
(z3(q)x(q)(1− x(q))
)2+
116
z6(q)x(q)(1− x(q))4. (10.3)
Using Jacobi’s quartic identity, (3.8), and the product expansion formulas for θ2(0|q), θ3(0|q), and
θ4(0|q), we readily find that
z6(q)x(q)(1− x(q))4 = 16q(q; q)24∞. (10.4)
Substituting (8.4) and (10.4) into the left side of (10.1), we find that identity in (10.1) holds. This
completes the proof of Theorem 20.
We recall the Ramanujan τ(n)-function [22, p. 197, Equation(6.1.13)] defined as the coefficient of qn
in
q(q; q)24∞ =∞∑
n=1
τ(n)qn. (10.5)
21
Then
q2(q2; q2)24∞ =∞∑
n=1
τ(n
2)qn, (10.6)
if we agree that τ(x) means 0 when x ia not an integer. Substituting
q3ψ24(q) =∞∑
n=3
t24(n− 3)qn, (10.7)
(10.5), and (10.6) into (10.1), we obtain the following theorem [20, Theorem 8].
Theorem 21 We have∞∑
n=1
n11qn
1− q2n=
∞∑n=1
(σ11(n)− σ11
(n
2
))qn
= 176896∞∑
n=1
t24(n− 3)qn +∞∑
n=1
τ(n)qn + 2072∞∑
n=1
τ(n
2)qn. (10.8)
Consequently,
176896t24(n− 3) = σ9(n)− σ9(n
2)− τ(n)− 2072τ(
n
2). (10.9)
Here, for convenience, we assume t24(n) = 0 for n is not a nonnegative integer.
Theorem 22 Let T2k(q) be defined by (6.1). Then we have
72q3ψ24(q) = T4(q)T8(q)− T 26 (q). (10.10)
Proof. Multiplying both sides of (6.7) by T4(q) yields
T4(q)T8(q) = T2(q)T4(q)T6(q) + 72T 34 (q). (10.11)
Eliminating T2(q) between this equation and (6.6), we immediately have
72T 34 (q) = T4(q)T8(q)− T 2
6 (q). (10.12)
Substituting (6.2) into the left hand side of the above equation yields (10.10). We complete the proof
of Theorem 22
This identity can also be found in [6].
10.2 An identity for ψ32(q)
Theorem 23 Let T2k(q) be defined by (6.1). Then we have
75600T 44 (q) = 75600q4ψ32(q)
= T4(q)T12(q) +214
T 28 (q)− 25
4T6(q)T10(q). (10.13)
22
This beautiful identity is due to Chan and Chua [6], in which the authors use modular forms to prove
the identity. Here we will give a simple proof.
Proof. Eliminating T2(q) between (6.7) and (6.8), we arrive at
T 24 (q)T8(q)− 5T4(q)T 2
6 (q) =172
(T 2
8 (q)− T6(q)T10(q)). (10.14)
Eliminating T2(q) between (6.6) and (6.9), we arrive at
T4(q)T12(q) = T6(q)T10(q) + 672T 24 (q)T8(q) + 840T4(q)T 2
6 (q). (10.15)
Multiplying both sides of (10.12) by T4(q) gives
72T 44 (q) = T 2
4 (q)T8(q)− T4(q)T 26 (q). (10.16)
Eliminating T 24 (q)T8(q) and T4(q)T 2
6 (q) among (10.14), (10.15), and (10.16), we obtain (10.13). This
completes the proof of Theorem 23.
11 Some identities among τ(n), t24(n), and σ11(n)
Throughout this section we assume that p is prime. The dissection operator Up [2, p. 161] operating
on f(q) =∑∞
n=0 a(n)qn is define by
Up{f(q)} =∞∑
n=0
a(np)qn
=1p
p−1∑
j=0
f(ωjp q
1p ), (11.1)
where ωp=exp( 2πip ). It is obvious that for an integer k ≥ 1,
Ukp {f(q)} =
∞∑n=0
a(npk)qn. (11.2)
Theorem 24 We have
176896t24(2kn− 3) = 211k(σ11(n)− σ11(
n
2))− τ(2kn)− 2072τ(2k−1n). (11.3)
When k = 0, the above equation reduces to the following formula for t24(n) [20, Theorem 8]:
176896t24(n− 3) = σ11(n)− σ11(n
2)− τ(n)− 2072τ
(n
2
). (11.4)
Replacing n by 2n + 1, the above equation reduces to
176896t24(2n− 2) = σ11(2n + 1)− τ(2n + 1). (11.5)
It follows that [19, p. 103, Corollary 6.4.7]
τ(2n + 1) ≡ σ11(2n + 1) (mod 176896). (11.6)
23
Proof. Using∞∑
n=1
n11qn
1− q2n+
∞∑n=1
(−1)n n11qn
1− q2n= 212
∞∑n=1
n11q2n
1− q4n, (11.7)
we find that
Uk2
{ ∞∑n=1
n11qn
1− q2n
}= 211k
∞∑n=1
n11qn
1− q2n. (11.8)
We recall the identity in (10.8), namely,
∞∑n=1
n11qn
1− q11n
= 176896∞∑
n=1
t24(n− 3)qn +∞∑
n=1
τ(n)qn + 2072∞∑
n=1
τ(n
2)qn. (11.9)
We operate both sides of this identity by Uk2 , and we find that
211k∞∑
n=1
n11qn
1− q2n
= 176896∞∑
n=1
t24(2kn− 3)qn +∞∑
n=1
τ(2kn)qn + 2072∞∑
n=1
τ(2k−1n)qn. (11.10)
Substituting∞∑
n=1
n11qn
1− q2n=
∞∑n=1
(σ11(n)− σ11(
n
2))
qn (11.11)
into (11.10), and then comparing the coefficients of qn, we obtain (11.3). We complete the proof of
Theorem 24.
Theorem 25 We have
σ11(pk)(σ11(n)− σ11(
n
2))− p11σ11(pk−1)
(σ11(
n
p)− σ11(
n
2p))
= 176896t24(pkn− 3) + τ(pkn) + 2072τ(12pkn), (11.12)
where p is an odd prime. When n = 1, the above equation reduces to the following identity [20, p.
89, Equation (9)]:
176896t24(pk − 3) = σ11(pk)− τ(pk), (11.13)
where p is an odd prime.
Proof. If p doesn’t divide n, it is well known that
(1− xy)(1− xyωnp )(1− xyω2n
p ) · · · (1− xyω(p−1)np ) = 1− xpyp, (11.14)
24
where ωp=exp( 2πip ). Taking logarithmic derivative of the above equation with respect to x, and then
setting x = 1, we obtain
y
1− y+
yωnp
1− yωnp
+yω2n
p
1− yω2np
+ · · ·+ yω(p−1)np
1− yω(p−1)np
=pyp
1− yp. (11.15)
Hence, if we take
f(q) =∞∑
n=1
nrqn
1− qn, (11.16)
then using (11.15), we find thatp−1∑
k=0
f(ωkpq)
= pr+1∞∑
n=1
nrqnp
1− qnp+
∞∑n 6≡0 (mod p)
n=1
nrωrnp qn
1− qnωrnp
= pr+1∞∑
n=1
nrqnp
1− qnp+ p
∞∑n 6≡0 (mod p)
n=1
nrqnp
1− qnp
= p(1 + pr)∞∑
n=1
nrqnp
1− qnp− pr+1
∞∑n=1
nrqnp2
1− qnp2 . (11.17)
Thus, we have
Up
{ ∞∑n=1
nrqn
1− qn
}= (1 + pr)
∞∑n=1
nrqn
1− qn− pr
∞∑n=1
nrqnp
1− qnp. (11.18)
It follows that
Ukp
{ ∞∑n=1
nrqn
1− qn
}
={
p(k+1)r − 1pr − 1
} { ∞∑n=1
nrqn
1− qn
}− pr
{pkr − 1pr − 1
} { ∞∑n=1
nrqnp
1− qnp
}
= σr(pk)∞∑
n=1
nrqn
1− qn− prσr(pk−1)
∞∑n=1
nrqnp
1− qnp. (11.19)
If p is an odd prime, writing q as q2, the above equation becomes
Ukp
{ ∞∑n=1
nrq2n
1− q2n
}= σr(pk)
∞∑n=1
nrq2n
1− q2n− prσr(pk−1)
∞∑n=1
nrq2np
1− q2np. (11.20)
Subtracting (11.20) from (11.19), we immediately have
Ukp
{ ∞∑n=1
nrqn
1− q2n
}= σr(pk)
∞∑n=1
nrqn
1− q2n− prσr(pk−1)
∞∑n=1
nrqnp
1− q2np. (11.21)
Acting on both sides of (11.9) by Ukp , and using (11.2) and (11.21), we find that
σ11(pk)∞∑
n=1
n11qn
1− q2n− p11σ11(pk−1)
∞∑n=1
n11qnp
1− q2np
= 176896∞∑
n=1
t24(pkn− 3)qn +∞∑
n=1
τ(pkn)qn + 2072∞∑
n=1
τ(12pkn)qn. (11.22)
25
Equating the coefficients of qn on both sides, we obtain (11.12). This completes the proof of Theorem
25.
Acknowledgements This work is supported, in part, by Academic Research Fund R146000027112
from the National University of Singapore. The author is grateful for the referee for his many helpful
suggestions. He also would like thanks Li-Chien Shen and Yi-Fan Yang for their comments.
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