Volume 10 (2009), Issue 2, Article 39, 5 pp.
A BOUND FOR CERTAIN BIBASIC SUMS AND APPLICATIONS
MINGJIN WANG AND HONGSHUN RUAN
D EPARTMENT OF A PPLIED M ATHEMATICS
J IANGSU P OLYTECHNIC U NIVERSITY
C HANGZHOU C ITY 213164
J IANGSU P ROVINCE , P.R. C HINA
wmj@jpu.edu.cn
rhs@em.jpu.edu.cn
Received 28 June, 2007; accepted 15 May, 2009
Communicated by S.S. Dragomir
A BSTRACT. In this paper, we use the terminating case of the Euler formula, the limiting case of
the q-Gauss sum and the Grüss inequality to derive a bound for certain bibasic sums. Applications of the inequality are also given.
Key words and phrases: Basic hypergeometric function; q-binomial coefficient; Euler formula; q-Gauss sum; Grüss inequality.
2000 Mathematics Subject Classification. Primary 26D15; Secondary 33D15; 33D65.
1. I NTRODUCTION AND S OME K NOWN R ESULTS
q-Series, which are also called basic hypergeometric series, play a very important role in
many fields, such as affine root systems, Lie algebras and groups, number theory, orthogonal
polynomials and physics, etc. Inequality techniques are useful tools in the study of q-series, see
[1, 7, 8]. In [1], the authors gave some inequalities for hypergeometric functions. In this paper,
we give a new inequality about q-series. First, we recall some definitions, notations and known
results which will be used in this paper. Throughout this paper, it is supposed that 0 < q < 1.
The q-shifted factorials are defined as
(1.1)
(a; q)0 = 1,
(a; q)n =
n−1
Y
k=0
k
(1 − aq ),
(a; q)∞ =
∞
Y
(1 − aq k ).
k=0
We also adopt the following compact notation for multiple q-shifted factorials:
(1.2)
(a1 , a2 , . . . , am ; q)n = (a1 ; q)n (a2 ; q)n · · · (am ; q)n ,
Supported by STF of Jiangsu Polytechnic University.
The author would like to express his deep appreciation to the referee for helpful suggestions. In particular, the author thanks the referee for
helping to improve the presentation of this paper.
214-07
2
M INGJIN WANG AND H ONGSHUN RUAN
where n is an integer or ∞. The q-binomial coefficient is defined by
hni
(q; q)n
=
(1.3)
.
k q (q; q)k (q; q)n−k
Although it is not obvious from (1.3), it is a well-known fact [2] that the q-binomial coefficient
is a polynomial in q of degree k(n − k) with nonnegative integer coefficients. The q-Gauss sum
[2, 3, 4]
∞
X
(a, b; q)k c k (c/a, c/b; q)∞
(1.4)
=
,
(q, c; q)k ab
(c, c/ab; q)∞
k=0
which has the limiting case [4]
n
(1.5)
X hni
1
xk
=
q k(k−1)
.
(x; q)n
k
(x;
q)
q
k
k=0
We also need the following Euler formula [2]
(1.6)
(x; q)∞ =
∞
X
1
(−1)k q 2 k(k−1)
k=0
xk
,
(q; q)k
which has the terminating form [2, 4]
(1.7)
(x; q)n =
n
X
k=0
(−1)k
hni
k
1
q 2 k(k−1) xk .
q
The following is well known in the literature as the Grüss inequality [6]:
Z b
Z b
Z b
1
1
1
(1.8)
f (x)g(x)dx −
f (x)dx ·
g(x)dx b − a
b−a a
b−a a
a
(M − m)(N − n)
,
≤
4
provided that f, g : [a, b] → R are integrable on [a, b] and m ≤ f (x) ≤ M, n ≤ g(x) ≤ N for
all x ∈ [a, b], where m, M, n, N are given constants.
The discrete version of the Grüss inequality can be stated as: If a ≤ ai ≤ A and b ≤ bi ≤ B
(i = 1, 2, . . . , n), then we have
!
!
n
n
n
1 X
(A − a)(B − b)
1X
1X
(1.9)
ai b i −
ai ·
bi ≤
,
n
n i=1
n i=1
4
i=1
where a, A, b, B are given real constants.
2. A B OUND FOR B IBASIC S UMS
In this section, by means of the terminating form of the Euler formula, the limiting case of
the q-Gauss sum and the Grüss inequality, we derive a bound of the following bibasic sums
n h i h i
X
1
xk y k
n
n
(2.1)
pk(k−1) q 2 k(k−1)
.
k p k q
(x; p)k
k=0
For any real number x, let [x] denote the greatest integer less than or equal x. The main result
of this paper is the following theorem, which gives an upper bound of (2.1). It is obvious that,
under the conditions of the following theorem, the lower bound of (2.1) is zero.
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 39, 5 pp.
http://jipam.vu.edu.au/
A B OUND
FOR
C ERTAIN B IBASIC S UMS
3
Theorem 2.1. Let 0 < p < 1, 0 < q < 1, 0 < x < 1 and 0 < y < 1, then for any positive
integer n we have
(2.2)
n h i h i
k k
X
n
n
(−y; q)n
n+1
n
n
k(k−1) 12 k(k−1) x y
p
q
,
≤
+
k p k q
(x; p)k
(n + 1)(x; p)n 4(x; p)n k0 p k0 q
k=0
where k0 = [ n−1
].
2
Proof. Let k0 = [ n−1
]. Since,
2
n
k+1 q
n
k q
1 − q n−k
(q; q)n /(q; q)k+1 (q; q)n−k−1
=
,
(q; q)n /(q; q)k (q; q)n−k
1 − q k+1
=
we get
 n
[ k+1 ]q


 [ nk ] ≥ 1, when k ≤ k0 ,
q
n

[
]q

 k+1
< 1, when k > k0 .
n
[ k ]q
So we have
1≤
(2.3)
hni
k
q
n
≤
k0
,
k = 0, 1, . . . , n.
,
k = 0, 1, . . . , n.
q
Similarly,
1≤
(2.4)
hni
k
p
n
≤
k0
p
Under the conditions of the theorem, we also have
(2.5)
(2.6)
(2.7)
1
0 < pk(k−1) < 1,
0 < q 2 k(k−1) < 1,
1
1
1
1
1
0<
=
·
···
≤
,
k−1
(x; p)k
1 − x 1 − xp
1 − xp
(x; p)n
0 < xk < 1,
0 < y k < 1,
where k = 0, 1, . . . , n.
Let
(2.8)

xk
 ak = nk p pk(k−1) (x;p)
k
 b = n q 12 k(k−1) y k
k
k q
in the discrete version of the Grüss inequality (1.9). Combining (2.3), (2.4), (2.5), (2.6) and
(2.7) one gets
1
n
n
0 < ak <
and 0 < bk <
.
(x; p)n k0 p
k0 q
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 39, 5 pp.
http://jipam.vu.edu.au/
4
M INGJIN WANG AND H ONGSHUN RUAN
Substituting ak and bk into the discrete version of the Grüss inequality (1.9), gives
n h i h i
1 X
k k
n
n
k(k−1) 12 k(k−1) x y
q
p
(2.9) n + 1
k p k q
(x; p)k
k=0
(
)(
)
n
n
xk
1 X h n i 1 k(k−1) k 1 X k(k−1) h n i
q2
−
p
y n + 1 k=0
k p (x; p)k
n + 1 k=0 k q
n
n
1
.
≤
4(x; p)n k0 p k0 q
Using (1.5) and (1.7) one gets
n h i
X
n
(2.10)
k=0
n h
X
(2.11)
k=0
k
pk(k−1)
p
xk
1
=
,
(x; p)k
(x; p)n
n i 1 k(k−1) k
y = (−y; q)n .
q2
k q
Substituting (2.10) and (2.11) into (2.9), we have
n h i h i
1 X
k k
1
n
n
x
y
(−y;
q)
n
(2.12) pk(k−1) q 2 k(k−1)
−
2 (x; p) n + 1
k
(x;
p)
(n
+
1)
p k q
k
n
k=0
1
n
n
,
≤
4(x; p)n k0 p k0 q
which is equivalent to
(2.13)
n+1
n
n
(−y; q)n
−
(n + 1)(x; p)n 4(x; p)n k0 p k0 q
n h i h i
X
1
n
n
xk y k
≤
pk(k−1) q 2 k(k−1)
k p k q
(x; p)k
k=0
(−y; q)n
n+1
n
n
≤
+
.
(n + 1)(x; p)n 4(x; p)n k0 p k0 q
The proof is thus completed.
In the proof, we used the Euler formula and the limiting case of the q-Gauss sum. We wish
to point out, that there may be other pairs of summation theorems which also lead to interesting
results. As an application of the inequality, we can easily obtain the following one.
Corollary 2.2. Under the conditions of Theorem 2.1, we have
2
n 2
X
n
xk y k
(1 + y)n
n+1
n
(2.14)
≤
+
,
k
n
n
k (1 − x)
(n + 1)(1 − x)
4(1 − x) k0
k=0
n!
where nk = k!(n−k)!
.
Proof. From [5], we know
n
lim
=
.
q→1 i q
i
Letting p → 1 and q → 1 on both sides of inequality (2.2), we get (2.14).
hni
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 39, 5 pp.
http://jipam.vu.edu.au/
A B OUND
FOR
C ERTAIN B IBASIC S UMS
5
3. A PPLICATION OF THE I NEQUALITY
Convergence is an important problem in the study of q-series. In this section, we use the
inequality obtained in this paper to give a sufficient condition for the convergence of a bibasic
series.
Theorem 3.1. Suppose 0 < p < 1, 0 < q < 1, 0 < x < 1 and 0 < y < 1. Let {cn } be any
sequence of numbers. If
cn+1 < 1,
lim n→∞ cn then the bibasic series
∞ X
n
hni hni
k k
X
k(k−1) 12 k(k−1) x y
(3.1)
cn
p
q
k p k q
(x; p)k
n=0 k=0
converges absolutely.
Proof. Multiplying both sides of (2.2) by |cn |, one gets
n h i h i
X
1
n
n
xk y k
(3.2)
|cn |
pk(k−1) q 2 k(k−1)
k p k q
(x; p)k
k=0
(−y; q)n |cn |
(n + 1)|cn | n
n
≤
+
(n + 1)(x; p)n
4(x; p)n
k0 p k0 q
≤
(−y; q)n |cn |
(n + 1)|cn |
+
.
(n + 1)(x; p)n 4(x; p)n (p; p)∞ (q; q)∞
The ratio test shows that both
∞
X
(−y; q)n cn
(n + 1)(x; p)n
n=0
and
∞
X
n=0
(n + 1)cn
4(x; p)n (p; p)∞ (q; q)∞
are absolutely convergent. Together with (3.2), this immediately yields that the series in (3.1) is
absolutely convergent.
R EFERENCES
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1723.
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1
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1
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J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 39, 5 pp.
http://jipam.vu.edu.au/