Journal of Inequalities in Pure and
Applied Mathematics
AN INEQUALITY ABOUT q-SERIES
MINGJIN WANG
Department of Mathematics
East China Normal University
Shanghai, 200062
People’s Republic of China
volume 7, issue 4, article 136,
2006.
Received 10 February, 2006;
accepted 16 March, 2006.
Communicated by: H. Bor
EMail: wmj@jpu.edu.cn
Abstract
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c
2000
Victoria University
ISSN (electronic): 1443-5756
037-06
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Abstract
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.
An Inequality About q-Series
2000 Mathematics Subject Classification: Primary 60E15; Secondary 33D15.
Key words: Inequality, q-series.
Mingjin Wang
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
An Inequality About q-Series . . . . . . . . . . . . . . . . . . . . . . . . . .
References
3
4
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1.
Introduction
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.
An Inequality About q-Series
Mingjin Wang
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2.
An Inequality About q-Series
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<
1 − aq
1 − aq n
≤
.
n
1−q
1−q
0
1−a
Proof. Let g(x) = 1−ax
, 0 ≤ x < 1, then g (x) = (1−x)
2 > 0. So g(x) is a
1−x
strictly increasing function on [0, 1). For any natural number n, we have
0 < q n ≤ q < 1.
An Inequality About q-Series
Mingjin Wang
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So,
n
g(0) < g(q ) ≤ g(q).
That is
1<
1 − aq
1 − aq n
≤
.
n
1−q
1−q
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1−q
,
1+q
0 < q < 1, then for any real number 0 < x ≤ 1,
2(1 − aqx)
1 + ax ax −
> 0.
1−q
Lemma 2.2. If 0 < a <
we have
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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
Since 0 < 1 − q < 1, 0 < x ≤ 1, we know 0 < (1 − q)x < 1. Therefore
0
g (x) < 0 and g(x) is a strictly decreasing function on (0, 1]. We have
1
2(1 − aq)
g(x) > g(1) = 1 + a a −
=
[(1 + q)a2 − 2a + (1 − q)].
1−q
1−q
Letting
An Inequality About q-Series
Mingjin Wang
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2
(1 + q)a − 2a + (1 − q) = 0,
we have
a1 =
So, when 0 < a <
1−q
,
1+q
a2 = 1.
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(1 + q)a2 − 2a + (1 − q) > 0,
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that is,
1 + ax ax −
2(1 − aqx)
1−q
> 0.
J. Ineq. Pure and Appl. Math. 7(4) Art. 136, 2006
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Lemma 2.3. If 0 < q <
√
2 − 1, we have
q<
1−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
q<
An Inequality About q-Series
1−q
.
1+q
Mingjin Wang
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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 )) · · · .
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If g(x) = ax, then
[g(x); q]n = (1 − a)(1 − aq) · · · (1 − aq
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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 .
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Theorem 2.4. Suppose 0 < a <
following inequality holds
1−q
,
1+q
0<q <
√
2 − 1, 0 < z < 1, then the
∞
(2.1)
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
where
g1 (x, a) = −ax(ax − 2)z,
2(1 − aqx)
g2 (x, a) = −ax ax −
z.
1−q
When a = q, the equality holds.
An Inequality About q-Series
Mingjin Wang
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Proof. Let
f (a, z) =
∞
X
(a; q)2n n
z ,
2
(q;
q)
n
n=0
since
1 − a = (1 − aq n ) − a(1 − q n ),
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f (a, z) = 1 +
=1+
∞
X
n=1
∞
X
n=1
(a; q)2n n
z
(q; q)2n
(aq; q)2n−1
[(1 − aq n ) − a(1 − q n )]2 z n
2
(q; q)n
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=1+
∞
X
(aq; q)2n−1
(q; q)2n
n=1
=1
=1
2
[(1 − aq n )2 − 2a(1 − q n )(1 − aq n ) + a2 (1 − q n ) ]z n
∞
X
(aq; q)2
∞
X
(aq; q)2n−1 n
n n
2
+
z
+
a
z
2
2
(q;
q)
(q;
q)
n−1
n
n=1
n=1
∞
2
X
(aq; q)n−1
− 2a
(1 − q n )(1 − aq n )z n
2
(q; q)n
n=1
∞
∞
X
X
(aq; q)2n−1 n
(aq; q)2n n
2
+
z
+
a
z −
2
2
(q;
q)
(q;
q)
n−1
n
n=1
n=1
From Lemma 2.1, we know
f (a, z) = 1 +
∞
X
(aq; q)2
1−aq n
1−q n
n n
z
2
(q;
q)
n
n=1
∞
X
(aq; q)2n−1 1 − aq n
(q; q)2n−1
n=1
1 − qn
An Inequality About q-Series
zn.
Mingjin Wang
> 1. So
+ a2
≤ f (aq, z) + a2 zf (aq, z)
2a
∞
X
(aq; q)2n−1
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z n − 2a
(q; q)2n−1
n=1
∞
X
(aq; q)2n−1 n
z
− 2a
(q; q)2n−1
n=1
∞
X
(aq; q)2n−1 1 − aq n
(q; q)2n−1
n=1
1 − qn
= f (aq, z) + a2 zf (aq, z) − 2azf (aq, z)
= (1 + a(a − 2)z)f (aq, z).
zn
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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
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gives
f (a, z) ≤ [g1 (x, a); q]∞ f (0, z).
(2.2)
Again from Lemma 2.1, we know
f (a, z) = 1 +
∞
X
(aq; q)2
1−aq n
1−q n
n n
z
2
(q; q)n
n=1
∞
X
− 2a
n=1
≤
+a
2
1−aq
.
1−q
So
∞
X
(aq; q)2n−1
(q; q)2n−1
n=1
zn
(aq; q)2n−1 1 − aq n n
z
(q; q)2n−1 1 − q n
∞
1 − aq X (aq; q)2n−1 n
2
≥ 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
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
2(1 − aq n+1 )
1
n
n
=z
+ aq aq −
z
1−q
2(1 − aq n+1 )
n
n
≥ z 1 + aq (aq −
> 0.
1−q
An Inequality About q-Series
Mingjin Wang
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Let g2 (x, a) = −ax ax −
ity n − 1 times we get that
2(1−aqx)
1−q
z, and by iterating this functional inequal-
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
f (a, z) ≥ [g2 (x, a); q]∞ f (0, z).
(2.3)
Combined with (2.2) and (2.3) gives
An Inequality About q-Series
Mingjin Wang
[g2 (x, a); q]∞ f (0, z) ≤ f (a, z) ≤ [g1 (x, a); q]∞ f (0, z).
√
Using Lemma 2.3, when 0 < q < 2 − 1, we have q < 1−q
. So, let a = q, also
1+q
combining (2.2) and (2.3) gives the following inequality
(2.4)
f (q, z)
f (q, z)
≤ f (0, z) ≤
.
[g1 (x, q); q]∞
[g2 (x, q); q]∞
Because of f (q, z) =
(2.5)
1
,
1−z
we have
1
1
≤ f (0, z) ≤
.
(1 − z)[g1 (x, q); q]∞
(1 − z)[g2 (x, q); q]∞
(2.4) and (2.5) yield the following inequality
[g1 (x, a); q]∞
[g2 (x, a); q]∞
≤ f (a, z) ≤
.
(1 − z)[g1 (x, q); q]∞
(1 − z)[g2 (x, q); q]∞
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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)n
(1 − z)[g2 (x, q); q]∞
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
(az; q)2∞
X
(qz; q)2∞
(a; q)n (az; q)∞
(2.6)
−
zn ≤
(q; q)n
(qz; q)∞
[g2 (x, q); q]∞ (z; q)∞ (qz; q)∞
n=0
Proof. Since
2
∞ X
(a; q)n (az; q)∞
−
zn
(q;
q)
(qz;
q)
n
∞
n=0
∞
X (a; q)2
(a; q)n (az; q)∞ (az; q)2∞ n
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
∞
∞
(az; q)∞ X (a; q)n n (az; q)2∞ X n
−2
z +
z
(qz; q)∞ n=0 (q; q)n
(qz; q)2∞ n=0
An Inequality About q-Series
Mingjin Wang
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=
=
=
≤
=
∞
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)∞ (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∞
−
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]∞
≤
.
(2.7)
(qz; q)2∞
[g2 (x, q); q]∞
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
(1 − z)[g2 (x, q); q]∞ 1 − z (qz; q)2∞
(q;
q)
(qz;
q)
n
∞
n=0
so the inequality (2.7) holds.
An Inequality About q-Series
Mingjin Wang
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Corollary 2.7. Suppose 0 < a < 1−q
,0<b<
1+q
then the following inequality holds
(2.8)
∞
X
(a; q)n (b; q)n
(q; q)2n
n=0
+
[g (x, a); q]
∞
1
2
(az; q)∞
1−q
0
1+q
<q<
√
2 − 1, 0 < z < 1,
(az, bz; q)∞
(z; q)2∞
1 [g2 (x, q); q]∞ 2 [g1 (x, b); q]∞ [g2 (x, q); q]∞
·
(qz; q)2∞
(qz; q)2∞ (bz; q)2∞
zn ≤
[g2 (x, q); q]∞ (z; q)∞ (qz; q)∞
∞
X
(a; q)n (b; q)n
(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
.
An Inequality About q-Series
Mingjin Wang
Proof. Noting that
n=0
1
2
(az, bz; q)∞
z −
(z; q)2∞
n
(q; q)2n
∞
X
(az,
bz;
q)
(a;
q)
(b;
q)
∞
n
n n
≤
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
·
≤
−
−
z
(q; q)n
(qz; q)∞ (q; q)n (qz; q)∞ n=0
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∞ 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
(bz; q)∞ n
2
2
(2.10)
(q; q)n − (qz; q)∞ z · (q; q)n − (qz; q)∞ z
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
2
 (az; q)
(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)∞









An Inequality About q-Series
Mingjin Wang
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=
[g (x, a); q]
∞ [g2 (x, q); q]∞
1
(az; q)2∞
(qz; q)2∞
1
2
[g (x, b); q]
∞ [g2 (x, q); q]∞
1
·
(bz; q)2∞
(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)∞
This is the inequality we seek.
1
2
An Inequality About q-Series
Mingjin Wang
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J. Ineq. Pure and Appl. Math. 7(4) Art. 136, 2006
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References
[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.
An Inequality About q-Series
Mingjin Wang
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J. Ineq. Pure and Appl. Math. 7(4) Art. 136, 2006
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