Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 5, Issue 3, Article 75, 2004
RATIONAL IDENTITIES AND INEQUALITIES
TOUFIK MANSOUR
D EPARTMENT OF M ATHEMATICS
U NIVERSITY OF H AIFA
31905 H AIFA , I SRAEL .
toufik@math.haifa.ac.il
Received 25 December, 2003; accepted 27 June, 2004
Communicated by L. Tóth
A BSTRACT. Recently, in [4] the author studied some rational identities and inequalities involving Fibonacci and Lucas numbers. In this paper we generalize these rational identities and inequalities to involve a wide class of sequences.
Key words and phrases: Rational Identities and Inequalities, Fibonacci numbers, Lucas numbers, Pell numbers.
2000 Mathematics Subject Classification. 05A19, 11B39.
1. I NTRODUCTION
The Fibonacci and Lucas sequences are a source of many interesting identities and inequalities. For example, Benjamin and Quinn [1], and Vajda [5] gave combinatorial proofs for many
such identities and inequalities. Recently, Díaz-Barrero [4] (see also [2, 3]) introduced some rational identities and inequalities involving Fibonacci and Lucas numbers. A sequence (an )n≥0
is said to be positive increasing if 0 < an < an+1 for all n ≥ 1, and complex increasing if
0 < |an | ≤ |an+1 | for all n ≥ 1. In this paper, we generalize the identities and inequalities
which are given in [4] to obtain several rational identities and inequalities involving positive
increasing sequences or complex sequences.
2. I DENTITIES
In this section we present several rational identities and inequalities by using results on contour integrals.
Theorem 2.1. Let (an )n≥0 be any complex increasing sequence such that ap 6= aq for all p 6= q.
For all positive integers r,
!
n
n
X
1 + a`r+k Y
(−1)n+1
(ar+k − arj )−1 = Qn
ar+k j=1, j6=k
j=1 ar+j
k=1
ISSN (electronic): 1443-5756
c 2004 Victoria University. All rights reserved.
The author is grateful to Díaz-Barrero for his careful reading of the manuscript.
001-04
2
T OUFIK M ANSOUR
holds, with 0 ≤ ` ≤ n − 1.
Proof. Let us consider the integral
1 + z`
dz,
γ zAn (z)
Q
where γ = {z ∈ C : |z| < |ar+1 |} and An (z) = nj=1 (z − ar+j ). Evaluating the integral I in
P
the exterior of the γ contour, we get I1 = nk=1 Rk where
!
n
n
1 + a`r+k Y
1 + z` Y
−1
Rk = lim
(z − arj )
=
(ar+k − arj )−1 .
z→ar+k
z j=1, j6=k
ar+k j=1, j6=k
1
I=
2πi
I
On the other hand, evaluating I in the interior of the γ contour, we obtain
1+z
1
(−1)n
=
= Qn
.
z→0 An (z)
An (0)
j=1 ar+j
I2 = lim
Using Cauchy’s theorem on contour integrals we get that I1 + I2 = 0, as claimed.
Theorem 2.1 for an = Fn the n Fibonacci number (F0 = 0, F1 = 1, and Fn+2 = Fn+1 + Fn
for all n ≥ 0) gives [4, Theorem 2.1], and for an = Ln the n Lucas number (L0 = 2, L1 = 1,
and Ln+2 = Ln+1 +Ln for all n ≥ 0) gives [4, Theorem 2.2]. As another example, Theorem 2.1
for an = Pn the nth Pell number (P0 = 0, P1 = 1, and Pn+2 = Pn+1 + Pn for all n ≥ 0) we get
that
!
n
n
`
X
Y
1 + Pr+k
(−1)n+1
(Pr+k − Prj )−1 = Qn
Pr+k j=1, j6=k
j=1 Pr+j
k=1
holds, with 0 ≤ ` ≤ n − 1. In particular, we obtain
Corollary 2.2. For all n ≥ 2,
2
2
Pn (Pn+1
+ 1)Pn+2
Pn Pn+1 (Pn+2
+ 1)
(Pn2 + 1)Pn+1 Pn+2
+
+
= 1.
(Pn+1 − Pn )(Pn+2 − Pn ) (Pn − Pn+1 )(Pn+2 − Pn+1 ) (Pn − Pn+2 )(Pn+1 − Pn+2 )
Theorem 2.3. Let (an )n≥0 be any complex increasing sequence such that ap 6= aq for all p 6= q.
For all n ≥ 2,
n
n
X
Y
1
aj
1−
= 0.
n−2
a
a
k
k
k=1
j=1, j6=k
Proof. Let us consider the integral
1
I=
2πi
I
z
dz,
γ An (z)
Q
where γ = {z ∈ C : |z| < |an+1 |} and An (z) = nj=1 (z − ar+j ). Evaluating the integral I in
the exterior of the γ contour, we get I1 = 0. Evaluating I in the interior of the γ contour, we
obtain
n
n
n
n
n
X
X
Y
X
Y
ak
1
aj
I2 =
Res(z/An (z); z = ak ) =
=
1−
.
n−2
a
a
a
k − aj
k
k
k=1
k=1 j=1, j6=k
j=1, j6=k
k=1
Using Cauchy’s theorem on contour integrals we get that I1 + I2 = 0, as claimed.
J. Inequal. Pure and Appl. Math., 5(3) Art. 75, 2004
http://jipam.vu.edu.au/
R ATIONAL I DENTITIES AND I NEQUALITIES
3
For example, Theorem 2.3 for an = Ln the nth Lucas number gives [4, Theorem 2.5]. As
another example, Theorem 2.3 for an = Pn the nth Pell number obtains, for all n ≥ 2,
n
n
X
Y
1
Pj
1−
= 0.
Pk
P n−2 j=1, j6=k
k=1 k
3. I NEQUALITIES
In this section we suggest some inequalities on positive increasing sequences.
Theorem 3.1. Let (an )n≥0 be any positive increasing sequence such that a1 ≥ 1. For all n ≥ 1,
a
n+1
n
aann+1 + aan+1
< aann + an+1
.
(3.1)
and
a
a
n+2
n+2
n
n
− aan+2
.
− aan+1
< an+2
an+1
(3.2)
Proof. To prove (3.1) we consider the integral
Z an+1
I=
(axn+1 log an+1 − axn log an )dx.
an
Since an satisfies 1 ≤ an < an+1 for all n ≥ 1, so for all x, an ≤ x ≤ an+1 we have that
axn log an < axn+1 log an < axn+1 logan+1 ,
hence I > 0. On the other hand, evaluating the integral I directly, we get that
a
n+1
n
n
I = (an+1
− aann+1 ) − (aan+1
− aw
n ),
hence
a
n+1
n
aann+1 + aan+1
< aann + an+1
as claimed in (3.1). To prove (3.2) we consider the integral
Z an+2
(axn+2 log an+2 − axn+1 log an+1 )dx.
J=
an
Since an satisfies 1 ≤ an+1 < an+2 for all n ≥ 0, so for all x, an+1 ≤ x ≤ an+2 we have that
axn+1 log an+1 < axn+2 log an+2 ,
hence J > 0. On the other hand, evaluating the integral J directly, we get that
a
a
n+2
n+2
n
n
I = (an+2
− aan+2
) − (an+1
− aan+1
),
hence
a
a
n+2
n+2
n
n
an+1
− aan+1
< an+2
− aan+2
as claimed in (3.2).
For example, Theorem 3.1 for an = Ln the nth Lucas number gives [4, Theorem 3.1]. As
another example, Theorem 3.1 for an = Pn the nth Pell number obtains, for all n ≥ 1,
P
Pn
n+1
< PnPn + Pn+1
,
PnPn+1 + Pn+1
where Pn is the nth Pell number.
Theorem 3.2. Let (an )n≥0 be any positive increasing sequence such that a1 ≥ 1. For all
n, m ≥ 1,
m−1
m
Y a
Y
an+j
n+j+1
an
an+m
an+j <
an+j
.
j=0
J. Inequal. Pure and Appl. Math., 5(3) Art. 75, 2004
j=0
http://jipam.vu.edu.au/
4
T OUFIK M ANSOUR
Proof. Let us prove this theorem by induction on m. Since 1 ≤ an < an+1 for all n ≥ 1 then
an+1 −an
an+1
n
aann+1 −an < an+1
, equivalently, aann+1 aan+1
< aann an+1
, so the theorem holds for m = 1.
Now, assume for all n ≥ 1
n
aan+m−1
m−2
Y
an+j+1
an+j
<
m−1
Y
a
n+j
.
an+j
j=0
j=0
On the other hand, similarly as in the case m = 1, for all n ≥ 1,
−a
a
a
n
n+m
n+m
an+m−1
< an+m
−an
.
Hence,
−a
a
n an
n+m
an+m−1
an+m−1
m−2
Y
a
a
n+j+1
n+m
an+j
< an+m
−an
j=0
m−1
Y
a
n+j
an+j
,
j=0
equivalently,
n
aan+m
m−1
Y
an+j+1
an+j
<
j=0
m
Y
a
n+j
,
an+j
j=0
as claimed.
Theorem 3.2 for an = Ln the nth Lucas number and m = 3 gives [4, Theorem 3.3].
Theorem 3.3. Let (an )n≥0 and (bn )n≥0 be any two sequences such that 0 < an < bn for all
n ≥ 1. Then for all n ≥ 1,
1+1/n
1+1/n
n
− aj
2nn+1 Y bj
.
(bj + aj ) ≥
(n + 1)n i=1
bj − aj
i=1
n
X
Proof. Using the AM-GM inequality, namely
n
n
Y 1/n
1X
xi ≥
xi ,
n i=1
i=1
where xi > 0 for all i = 1, 2, . . . , n, we get that
Z
a1
Z
an
···
b1
bn
n
1X
xi dx1 · · · dxn ≥
n i=1
Z
a1
Z
an
···
b1
bn
n
Y
1/n
xi dx1 · · · dxn ,
i=1
equivalently,
n
n
n Y
Y
1 X 2
n
1+1/n
1+1/n
2
(b − ai )
(bj − aj ) ≥
(bi
− ai
) ,
2n i=1 i
n
+
1
i=1
j=1, j6=i
hence, on simplifying the above inequality we get the desired result.
−1
Theorem 3.3 for an = L−1
n where Ln is the nth Lucas number and bn = Fn where Fn is the
nth Fibonacci number gives [4, Theorem 3.4].
J. Inequal. Pure and Appl. Math., 5(3) Art. 75, 2004
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R ATIONAL I DENTITIES AND I NEQUALITIES
5
R EFERENCES
[1] A.T. BENJAMIN AND J.J. QUINN, Recounting Fibonacci and Lucas identities, College Math. J.,
30(5) (1999), 359–366.
[2] J.L. DíAZ-BARRERO, Problem B-905, The Fibonacci Quarterly, 38(4) (2000), 373.
[3] J.L. DíAZ-BARRERO, Advanced problem H-581, The Fibonacci Quarterly, 40(1) (2002), 91.
[4] J.L. DíAZ-BARRERO, Rational identities and inequalities involving Fibonacci and Lucas numbers,
J. Inequal. in Pure and Appl. Math., 4(5) (2003), Art. 83. [ONLINE http://jipam.vu.edu.
au/article.php?sid=324]
[5] S. VAJDA, Fibonacci and Lucas numbers and the Golden Section, New York, Wiley, 1989.
J. Inequal. Pure and Appl. Math., 5(3) Art. 75, 2004
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