Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 4, Article 136, 2006
AN INEQUALITY ABOUT q-SERIES
MINGJIN WANG
D EPARTMENT OF M ATHEMATICS
E AST C HINA N ORMAL U NIVERSITY
S HANGHAI , 200062
P EOPLE ’ S R EPUBLIC OF C HINA
wmj@jpu.edu.cn
Received 10 February, 2006; accepted 16 March, 2006
Communicated by H. Bor
A BSTRACT. In this paper, we first generalize the traditional notation (a; q)n to [g(x); q]n and
then obtain an inequality about q-series and some infinite products by means of the new conception. Because many q-series are not summable, our results are useful to study q-series and its
application.
Key words and phrases: Inequality, q-series.
2000 Mathematics Subject Classification. Primary 60E15; Secondary 33D15.
1. I NTRODUCTION
q-series, which are also called basic hypergeometric series, play an very important role in
many fields. Such as affine root systems, Lie algebras and groups, number theory, orthogonal
polynomials, physics (such as representations of quantum groups and Baxter’s work on the hard
hexagon model). Most of the research work on q-series is to set up identity. But there are also
great many q-series whose sums cannot be obtained easily. On these occasions, we must use
other methods to study q-series. Using inequalities is one of the choices. In this paper, we
obtain an inequality about q-series and some infinite products. Our results are useful for the
study of q-series.
2. A N I NEQUALITY A BOUT q-S ERIES
In this section we will introduce a new concept and obtain an inequality about q-series. First
we give some lemmas.
Lemma 2.1. If 0 < q < 1, 0 < a < 1, then for any natural number n, we have
1 − aq n
1 − aq
1<
≤
.
n
1−q
1−q
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
037-06
2
M INGJIN WANG
0
1−a
Proof. Let g(x) = 1−ax
, 0 ≤ x < 1, then g (x) = (1−x)
2 > 0. So g(x) is a strictly increasing
1−x
function on [0, 1). For any natural number n, we have
0 < q n ≤ q < 1.
So,
g(0) < g(q n ) ≤ g(q).
That is
1<
1 − aq n
1 − aq
≤
.
n
1−q
1−q
Lemma 2.2. If 0 < a <
1−q
,
1+q
0 < q < 1, then for any real number 0 < x ≤ 1, we have
2(1 − aqx)
1 + ax ax −
> 0.
1−q
Proof. Let
2(1 − aqx)
g(x) = 1 + ax ax −
1−q
Under the condition 0 < a <
0
g (x) =
1−q
,
1+q
=1+
a
[a(1 + q)x2 − 2x].
1−q
we know 0 < a(1 + q) < 1 − q, so
2a
2a
[a(1 + q)x − 1] <
[(1 − q)x − 1].
1−q
1−q
0
Since 0 < 1 − q < 1, 0 < x ≤ 1, we know 0 < (1 − q)x < 1. Therefore g (x) < 0 and g(x) is
a strictly decreasing function on (0, 1]. We have
2(1 − aq)
1
=
[(1 + q)a2 − 2a + (1 − q)].
g(x) > g(1) = 1 + a a −
1−q
1−q
Letting
(1 + q)a2 − 2a + (1 − q) = 0,
we have
1−q
a1 =
,
a2 = 1.
1+q
1−q
So, when 0 < a < 1+q
,
(1 + q)a2 − 2a + (1 − q) > 0,
that is,
2(1 − aqx)
1 + ax ax −
> 0.
1−q
√
Lemma 2.3. If 0 < q < 2 − 1, we have
1−q
q<
.
1+q
Proof. Let
√ √ g(q) = q(1 + q) − (1 − q) = q + 1 + 2 q + 1 − 2 .
√
When 0 < q < 2 − 1, g(q) < 0, so we have
1−q
q<
.
1+q
J. Inequal. Pure and Appl. Math., 7(4) Art. 136, 2006
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q-S ERIES
3
Definition 2.1. Suppose g(x) is a function on [0, 1], we denote [g(x); q]n by
[g(x); q]n = (1 − g(q 0 ))(1 − g(q 1 )) · · · (1 − g(q n−1 )).
We also use the notation [g(x); q]∞ to express infinite product. That is
[g(x); q]∞ = (1 − g(q 0 ))(1 − g(q 1 )) · · · (1 − g(q n )) · · · .
If g(x) = ax, then
[g(x); q]n = (1 − a)(1 − aq) · · · (1 − aq n−1 ) = (a; q)n .
So [g(x); q]n is the expansion of (a; q)n , where (a; q)n is the q-shifted factorial. Please note that
our notation [g(x); q]n in this paper is different from traditional notation (a; q)n .
√
Theorem 2.4. Suppose 0 < a < 1−q
,
0
<
q
<
2 − 1, 0 < z < 1, then the following inequality
1+q
holds
∞
X (a; q)2
[g2 (x, a); q]∞
[g1 (x, a); q]∞
n n
≤
z ≤
,
2
(1 − z)[g1 (x, q); q]∞
(q; q)n
(1 − z)[g2 (x, q); q]∞
n=0
(2.1)
where
g1 (x, a) = −ax(ax − 2)z,
2(1 − aqx)
z.
g2 (x, a) = −ax ax −
1−q
When a = q, the equality holds.
Proof. Let f (a, z) =
∞
P
n=0
f (a, z) = 1 +
∞
X
(a; q)2
=1+
=1+
z
∞
X
(aq; q)2n−1
n=1
∞
X
n=1
∞
X
n=1
− 2a
(q; q)2n
[(1 − aq n ) − a(1 − q n )]2 z n
(aq; q)2n−1
2
[(1 − aq n )2 − 2a(1 − q n )(1 − aq n ) + a2 (1 − q n ) ]z n
2
(q; q)n
∞
X
(aq; q)2n−1 n
(aq; q)2n n
2
z
+
a
z
2
(q; q)2n
(q;
q)
n−1
n=1
∞
X
(aq; q)2n−1
n=1
=1+
since 1 − a = (1 − aq n ) − a(1 − q n )
n n
(q; q)2n
n=1
=1+
(a;q)2n n
z ,
(q;q)2n
(q; q)2n
∞
X
(aq; q)2
n n
z
2
(q;
q)
n
n=1
(1 − q n )(1 − aq n )z n
+a
J. Inequal. Pure and Appl. Math., 7(4) Art. 136, 2006
2
∞
X
(aq; q)2n−1
n=1
(q; q)2n−1
n
z − 2a
∞
X
(aq; q)2n−1 1 − aq n
n=1
(q; q)2n−1 1 − q n
zn.
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4
M INGJIN WANG
From Lemma 2.1, we know
f (a, z) = 1 +
1−aq n
1−q n
∞
X
(aq; q)2
> 1. So
n n
z
2
(q;
q)
n
n=1
+ a2
≤ f (aq, z) + a2 zf (aq, z)
∞
X
(aq; q)2n−1
z n − 2a
(q; q)2n−1
n=1
∞
X
(aq; q)2n−1 n
− 2a
z
2
(q;
q)
n−1
n=1
∞
X
(aq; q)2n−1 1 − aq n
(q; q)2n−1
n=1
1−
qn
zn
= f (aq, z) + a2 zf (aq, z) − 2azf (aq, z)
= (1 + a(a − 2)z)f (aq, z).
By iterating this functional inequality n − 1 times we get that
f (a, z) ≤ [g1 (x, a); q]n f (aq n , z),
n = 1, 2, . . . ,
where g1 (x, a) = −ax(ax − 2)z. Which on letting n → ∞ and using q n → 0 gives
f (a, z) ≤ [g1 (x, a); q]∞ f (0, z).
(2.2)
Again from Lemma 2.1, we know
f (a, z) = 1 +
1−aq n
1−q n
≤
∞
X
(aq; q)2
1−aq
.
1−q
n n
z
2
(q;
q)
n
n=1
∞
X
− 2a
+a
So
2
∞
X
(aq; q)2n−1
n=1
(aq; q)2n−1
(q; q)2n−1
n=1
(q; q)2n−1
zn
1 − aq n n
z
1 − qn
∞
1 − aq X (aq; q)2n−1 n
≥ f (aq, z) + a zf (aq, z) − 2a
z
1 − q n=1 (q; q)2n−1
1 − aq
2
= 1 + a z − 2az
f (aq, z)
1−q
2(1 − aq)
= 1+a a−
z f (aq, z).
1−q
2
Using Lemma 2.2, we know that, for any natural number n
2(1 − aq n+1 )
n
n
1 + aq aq −
z
1−q
1
2(1 − aq n+1 )
n
n
=z
+ aq aq −
z
1−q
2(1 − aq n+1 )
n
n
≥ z 1 + aq (aq −
> 0.
1−q
2(1−aqx)
z, and by iterating this functional inequality n − 1 times
Let g2 (x, a) = −ax ax − 1−q
we get that
f (a, z) ≥ [g2 (x, a); q]n f (aq n , z),
n = 1, 2, . . . .
Which on letting n → ∞ and using q n → 0 gives
(2.3)
f (a, z) ≥ [g2 (x, a); q]∞ f (0, z).
Combined with (2.2) and (2.3) gives
(2.4)
[g2 (x, a); q]∞ f (0, z) ≤ f (a, z) ≤ [g1 (x, a); q]∞ f (0, z).
J. Inequal. Pure and Appl. Math., 7(4) Art. 136, 2006
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q-S ERIES
√
Using Lemma 2.3, when 0 < q < 2 − 1, we have q <
(2.2) and (2.3) gives the following inequality
5
1−q
.
1+q
So, let a = q, also combining
f (q, z)
f (q, z)
≤ f (0, z) ≤
.
[g1 (x, q); q]∞
[g2 (x, q); q]∞
Because of f (q, z) =
1
,
1−z
we have
1
1
≤ f (0, z) ≤
.
(1 − z)[g1 (x, q); q]∞
(1 − z)[g2 (x, q); q]∞
(2.5)
(2.4) and (2.5) yield the following inequality
[g2 (x, a); q]∞
[g1 (x, a); q]∞
≤ f (a, z) ≤
.
(1 − z)[g1 (x, q); q]∞
(1 − z)[g2 (x, q); q]∞
That is
∞
X (a; q)2
[g2 (x, a); q]∞
[g1 (x, a); q]∞
n n
≤
z
≤
.
2
(1 − z)[g1 (x, q); q]∞
(q;
q)
(1
−
z)[g
(x,
q);
q]
2
∞
n
n=0
If a = q, we have
∞
X (a; q)2
[g2 (x, a); q]∞
[g1 (x, a); q]∞
1
n n
=
z =
=
.
2
(1 − z)[g1 (x, q); q]∞
(q; q)n
(1 − z)[g2 (x, q); q]∞
1−z
n=0
So the equality holds. We complete our proof.
Corollary 2.5. Under the conditions of the theorem, we have
[g (x, a); q]
∞ [g2 (x, q); q]∞
1
2
∞
2
(az; q)∞
X
(qz; q)2∞
(a; q)n (az; q)∞
n
(2.6)
−
z ≤
(q;
q)
(qz;
q)
[g2 (x, q); q]∞ (z; q)∞ (qz; q)∞
n
∞
n=0
.
Proof. Since
∞ X
(a; q)n
n=0
2
(az; q)∞
−
zn
(q; q)n
(qz; q)∞
∞ X
(a; q)2n
(a; q)n (az; q)∞ (az; q)2∞ n
=
−2
+
z
(q; q)2n
(q; q)n (qz; q)∞
(qz; q)2∞
n=0
=
=
=
∞
X
(a; q)2
n n
z
2
(q;
q)
n
n=0
∞
X
(a; q)2n n
z
2
(q;
q)
n
n=0
∞
X
(a; q)2n n
z
(q; q)2n
n=0
−2
∞
∞
(az; q)∞ X (a; q)n n (az; q)2∞ X n
z +
z
(qz; q)∞ n=0 (q; q)n
(qz; q)2∞ n=0
−2
(az; q)∞ (az; q)∞ (az; q)2∞ 1
+
(qz; q)∞ (z; q)∞
(qz; q)2∞ 1 − z
−2
1 (az; q)2∞
1 (az; q)2∞
+
1 − z (qz; q)2∞ 1 − z (qz; q)2∞
J. Inequal. Pure and Appl. Math., 7(4) Art. 136, 2006
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6
M INGJIN WANG
=
≤
=
∞
X
(a; q)2
n n
z
2
(q; q)n
n=0
1 (az; q)2∞
−
1 − z (qz; q)2∞
1 (az; q)2∞
[g1 (x, a); q]∞
−
(1 − z)[g2 (x, q); q]∞ 1 − z (qz; q)2∞
[g (x, a); q]
∞ [g2 (x, q); q]∞
1
2
(az; q)∞
(qz; q)2∞
[g2 (x, q); q]∞ (z; q)∞ (qz; q)∞
we gain the inequality we seek.
Theorem 2.6. Under the conditions of the theorem, the following inequality holds
(az; q)2∞
[g1 (x, a); q]∞
≤
.
(qz; q)2∞
[g2 (x, q); q]∞
(2.7)
Proof. From the proof of (2.6), we have
2
∞ [g1 (x, a); q]∞
1 (az; q)2∞ X (a; q)n (az; q)∞
−
≥
−
zn ≥ 0
2
(1 − z)[g2 (x, q); q]∞ 1 − z (qz; q)∞
(q; q)n
(qz; q)∞
n=0
so the inequality (2.7) holds.
Corollary 2.7. Suppose 0 < a <
following inequality holds
(2.8)
∞
X
(a; q)n (b; q)n
(q; q)2n
n=0
+
zn ≤
1−q
,
1+q
0<b<
1−q
0
1+q
<q <
√
2 − 1, 0 < z < 1, then the
(az, bz; q)∞
(z; q)2∞
[g (x, a); q]
∞ [g2 (x, q); q]∞
1
2
(az; q)∞
(qz; q)2∞
1
2
[g (x, b); q]
∞ [g2 (x, q); q]∞
1
·
2
(bz; q)∞
(qz; q)2∞
[g2 (x, q); q]∞ (z; q)∞ (qz; q)∞
1
2
.
Proof. Noting that
∞
X
(a; q)n (b; q)n
n=0
(q; q)2n
zn > 0
and
(az, bz; q)∞
> 0,
(z; q)2∞
we have
(2.9)
∞
X
(a; q)n (b; q)n
n=0
(q; q)2n
(az, bz; q)∞
(z; q)2∞
∞
X
(a; q)n (b; q)n n (az, bz; q)∞ ≤
z −
n=0
(q; q)2n
(z; q)2∞ ∞ X
(a;
q)
(az;
q)
(b;
q)
(bz;
q)
n
∞
n
∞
=
−
−
zn
n=0 (q; q)n
(qz; q)∞ (q; q)n (qz; q)∞
∞ X
(a; q)n (az; q)∞ (b; q)n
(bz; q)∞ n
≤
(q; q)n − (qz; q)∞ · (q; q)n − (qz; q)∞ z
n=0
zn −
J. Inequal. Pure and Appl. Math., 7(4) Art. 136, 2006
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q-S ERIES
7
∞ X
(a; q)n (az; q)∞ n
2
=
(q; q)n − (qz; q)∞ z
n=0
(b; q)n
n
(bz;
q)
∞
z2.
−
· (q; q)n (qz; q)∞ Using the Cauchy inequality and (2.6), we have
∞ X
(a; q)n (az; q)∞ n (b; q)n
n
(bz;
q)
∞
z2 · 2
(2.10)
−
−
(q; q)n
(q; q)n (qz; q)∞ z
(qz;
q)
∞
n=0
(∞ ) 12
X (a; q)n (az; q)∞ 2
zn
≤
−
(q;
q)
(qz;
q)
n
∞
n=0
(∞ 2 ) 12
X (b; q)n
(bz; q)∞
zn
·
−
(q;
q)
(qz;
q)
n
∞
n=0
1
 2
[g
(x,
a);
q]
[g
(x,
q);
q]

1
∞
2
∞






2
 (az; q)2∞
(qz; q)∞ 
≤

[g2 (x, q); q]∞ (z; q)∞ (qz; q)∞ 








·
=
 
[g1 (x, b); q]∞ [g2 (x, q); q]∞



 (bz; q)2∞
(qz; q)2∞
1
2





[g2 (x, q); q]∞ (z; q)∞ (qz; q)∞









[g (x, a); q]
∞ [g2 (x, q); q]∞
1
2
(az; q)∞
(qz; q)2∞
1
2
[g (x, b); q]
∞ [g2 (x, q); q]∞
1
·
2
(bz; q)∞
(qz; q)2∞
[g2 (x, q); q]∞ (z; q)∞ (qz; q)∞
1
2
.
Combining (2.9) and (2.10) gives
∞
X
(a; q)n (b; q)n
n=0
(q; q)2n
zn ≤
+
(az, bz; q)∞
(z; q)2∞
[g (x, a); q]
∞ [g2 (x, q); q]∞
1
2
(az; q)∞
(qz; q)2∞
1
2
[g (x, b); q]
∞ [g2 (x, q); q]∞
1
·
2
(bz; q)∞
(qz; q)2∞
[g2 (x, q); q]∞ (z; q)∞ (qz; q)∞
1
2
.
This is the inequality we seek.
R EFERENCES
[1] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Edition 1 and 2. Cambridge University Press, Cambridge, 1934, 1952.
[2] G. GASPER, Lecture Notes for an Introductory Minicourse on q-Series, September 19, 1995 version.
[3] G. GASPER AND M. RAHMAN, Basic Hypergeometric Series, Encyclopedia of Mathematics and
Its Applications, 35. Cambridge University Press, Cambridge and New York, 1990.
J. Inequal. Pure and Appl. Math., 7(4) Art. 136, 2006
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