Journal of Inequalities in Pure and
Applied Mathematics
RATIONAL IDENTITIES AND INEQUALITIES INVOLVING
FIBONACCI AND LUCAS NUMBERS
volume 4, issue 5, article 83,
2003.
JOSÉ LUIS DíAZ-BARRERO
Applied Mathematics III
Universitat Politècnica de Catalunya
Jordi Girona 1-3, C2,
08034 Barcelona, Spain.
EMail: jose.luis.diaz@upc.es
Received 21 May, 2003;
accepted 21 August, 2003
Communicated by: J. Sándor
Abstract
Contents
JJ
J
II
I
Home Page
Go Back
Close
c
2000
Victoria University
ISSN (electronic): 1443-5756
067-03
Quit
Abstract
In this paper we use integral calculus, complex variable techniques and some
classical inequalities to establish rational identities and inequalities involving
Fibonacci and Lucas numbers.
2000 Mathematics Subject Classification: 05A19, 11B39
Key words: Rational Identities and Inequalities, Sequences of Integers, Fibonacci
and Lucas numbers.
Rational Identities and
Inequalities involving Fibonacci
and Lucas Numbers
The author would like to thank the anonymous referee for the suggestions and comments that have improved the final version of the paper.
José Luis Díaz-Barrero
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Rational Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
Title Page
3
4
9
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 2 of 16
J. Ineq. Pure and Appl. Math. 4(5) Art. 83, 2003
http://jipam.vu.edu.au
1.
Introduction
The Fibonacci sequence is a source of many nice and interesting identities and
inequalities. A similar interpretation exists for Lucas numbers. Many of these
identities have been documented in an extensive list that appears in the work
of Vajda [1], where they are proved by algebraic means, even though combinatorial proofs of many of these interesting algebraic identities are also given
(see [2]). However, rational identities and inequalities involving Fibonacci and
Lucas numbers seldom have appeared (see [3]). In this paper, integral calculus,
complex variable techniques and some classical inequalities are used to obtain
several rational Fibonacci and Lucas identities and inequalities.
Rational Identities and
Inequalities involving Fibonacci
and Lucas Numbers
José Luis Díaz-Barrero
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 3 of 16
J. Ineq. Pure and Appl. Math. 4(5) Art. 83, 2003
http://jipam.vu.edu.au
2.
Rational Identities
In what follows several rational identities are considered and proved by using
results on contour integrals. We begin with:
Theorem 2.1. Let Fn denote the nth Fibonacci number. That is, F0 = 0, F1 = 1
and for n ≥ 2, Fn = Fn−1 + Fn−2 . Then, for all positive integers r,




n
n


`
X 1 + Fr+k
Y
1
(−1)n+1
(2.1)
=
Fr+k 
 j=1 Fr+k − Fr+j 
 Fr+1 Fr+2 · · · Fr+n
k=1
j6=k
holds, with 0 ≤ ` ≤ n − 1.
Rational Identities and
Inequalities involving Fibonacci
and Lucas Numbers
José Luis Díaz-Barrero
Proof. To prove the preceding identity we consider the integral
I
1 + z ` dz
1
(2.2)
I=
,
2πi γ An (z) z
Q
where An (z) = nj=1 (z − Fr+j ).
Let γ be the curve defined by γ = {z ∈ C : |z| < Fr+1 }. Evaluating the
preceding integral in the exterior of the γ contour, we obtain
)
I (
n
n
X
1
1 + z` Y
1
I1 =
dz =
Rk ,
2πi γ
z j=1 (z − Fr+j )
k=1
where
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Rk = lim
z→Fr+k


n
 1 + z` Y


z
j=1
j6=k



n
`
Y
1 + Fr+k
1
1
=
.
(z − Fr+j ) 
Fr+k j=1 (Fr+k − Fr+j )

j6=k
Page 4 of 16
J. Ineq. Pure and Appl. Math. 4(5) Art. 83, 2003
http://jipam.vu.edu.au
Then, I1 becomes

I1 =
n 

X
`
1 + Fr+k
 Fr+k
k=1 
n
Y
j=1
j6=k



1
.
(Fr+k − Fr+j ) 

Evaluating (2.2) in the interior of the γ contour, we get
)
I (
n
1 + z` Y
1
1
I2 =
dz
2πi γ
z j=1 (z − Fr+j )
1 + z`
= lim
z→0
An (z)
1
(−1)n
=
=
.
An (0)
Fr+1 Fr+2 · · · Fr+n
By Cauchy’s theorem on contour integrals we have that I1 + I2 = 0 and the
proof is complete.
A similar identity also holds for Lucas numbers. It can be stated as:
Corollary 2.2. Let Ln denote the nth Lucas number. That is, L0 = 2, L1 = 1
and for n ≥ 2, Ln = Ln−1 + Ln−2 . Then, for all positive integers r,




n

X 1 + L`r+k Y
1
(−1)n+1
(2.3)
=
Lr+k 
 j=1 Lr+k − Lr+j 
 Lr+1 Lr+2 · · · + Lr+n
k=1
Rational Identities and
Inequalities involving Fibonacci
and Lucas Numbers
José Luis Díaz-Barrero
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 5 of 16
j6=k
holds, with 0 ≤ ` ≤ n − 1.
J. Ineq. Pure and Appl. Math. 4(5) Art. 83, 2003
http://jipam.vu.edu.au
In particular (2.1) and (2.3) can be used (see [3]) to obtain
Corollary 2.3. For n ≥ 2,
2
Fn (Fn+1
+ 1)Fn+2
(Fn2 + 1)Fn+1 Fn+2
+
(Fn+1 − Fn )(Fn+2 − Fn ) (Fn − Fn+1 )(Fn+2 − Fn+1 )
2
Fn Fn+1 (Fn+2
+ 1)
+
= 1.
(Fn − Fn+2 )(Fn+1 − Fn+2 )
Corollary 2.4. For n ≥ 2,
Ln+1 Ln+2
Ln+2 Ln
+
(Ln+1 − Ln )(Ln+2 − Ln ) (Ln − Ln+1 )(Ln+2 − Ln+1 )
Ln Ln+1
+
= 1.
(Ln − Ln+2 )(Ln+1 − Ln+2 )
In the sequel Fn and Ln denote the nth Fibonacci and Lucas numbers, respectively.
Theorem 2.5. If n ≥ 3, then we have


n
n
−1
X 1 Y
Lj

1−
+ Ln−1
 = Ln+2 − 3.
i
n−2 
Li
Li
j=1
i=1
j6=i
Proof. First, we observe that the given statement can be written as


n
n
n
−1
X 1 Y
Lj
 X
Li = Ln+2 − 3.
1−
+
 n−2
Li
Li
j=1
i=1
i=1
j6=i
Rational Identities and
Inequalities involving Fibonacci
and Lucas Numbers
José Luis Díaz-Barrero
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 6 of 16
J. Ineq. Pure and Appl. Math. 4(5) Art. 83, 2003
http://jipam.vu.edu.au
P
Since ni=1 Li = Ln+2 − 3, as can be easily established by mathematical induction, then it will suffice to prove


n
n
−1
X 1 Y
Lj

(2.4)
1−
 = 0.
 n−2
Li
Li
j=1
i=1
j6=i
We will argue by using
Qn residue techniques. We consider the monic complex
polynomial A(z) = k=1 (z − Lk ) and we evaluate the integral
I
1
z
I=
dz
2πi γ A(z)
over the interior and exterior domains limited by γ, a circle centered at the origin
and radius Ln+1 , i.e., γ = {z ∈ C : |z| < Ln+1 } .
Integrating in the region inside the γ contour we have
I
z
1
I1 =
dz
2πi γ A(z)
n
X
z
Res
, z = Li
=
A(z)
i=1


n
n
X Y L i 
=


L i − Lj
j=1
i=1
Rational Identities and
Inequalities involving Fibonacci
and Lucas Numbers
José Luis Díaz-Barrero
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 7 of 16
j6=i
J. Ineq. Pure and Appl. Math. 4(5) Art. 83, 2003
http://jipam.vu.edu.au
=
n
X
i=1

 1
 n−2
Li
n Y
j=1
j6=i

1−
Lj
Li
−1

.
Integrating in the region outside of the γ contour we get
I
z
1
I2 =
dz = 0.
2πi γ A(z)
Again, by Cauchy’s theorem on contour integrals we have I1 + I2 = 0. This
completes the proof of (2.4).
Note that (2.4) can also be established by using routine algebra.
Rational Identities and
Inequalities involving Fibonacci
and Lucas Numbers
José Luis Díaz-Barrero
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 8 of 16
J. Ineq. Pure and Appl. Math. 4(5) Art. 83, 2003
http://jipam.vu.edu.au
3.
Inequalities
Next, several inequalities are considered and proved with the aid of integral
calculus and the use of classical inequalities. First, we state and prove some
nice inequalities involving circular powers of Lucas numbers similar to those
obtained for Fibonacci numbers in [4].
Theorem 3.1. Let n be a positive integer, then the following inequalities hold
L
(a)
n+1
n
LLn n+1 + LLn+1
< LLn n + Ln+1
,
(b)
Ln+2
Ln+1
−
n
LLn+1
<
Ln+2
Ln+2
−
Ln
Ln+2
.
Proof. To prove part (a) we consider the integral
Z Ln+1 x
x
I1 =
Ln+1 log Ln+1 − Ln log Ln dx.
Ln
log Ln <
José Luis Díaz-Barrero
Title Page
Contents
Since Ln < Ln+1 if n ≥ 1, then for Ln ≤ x ≤ Ln+1 we have
Lxn
Rational Identities and
Inequalities involving Fibonacci
and Lucas Numbers
Lxn+1
log Ln <
Lxn+1
log Ln+1
and I1 > 0. On the other hand, evaluating the integral, we obtain
Z Ln+1 I1 =
Lxn+1 log Ln+1 − Lxn log Ln dx
Ln
L
= Lxn+1 − Lxn Ln+1
n
Ln+1
Ln
n
= Ln + Ln+1 − LLn n+1 + LLn+1
JJ
J
II
I
Go Back
Close
Quit
Page 9 of 16
J. Ineq. Pure and Appl. Math. 4(5) Art. 83, 2003
http://jipam.vu.edu.au
and (a) is proved. To prove (b), we consider the integral
Z Ln+2 I2 =
Lxn+2 log Ln+2 − Lxn+1 log Ln+1 dx.
Ln
Since Ln+1 < Ln+2 , then for Ln ≤ x ≤ Ln+2 we have
Lxn+1 log Ln+1 < Lxn+2 log Ln+2
and I2 > 0. On the other hand, evaluating I2 , we get
Z Ln+2 I2 =
Lxn+2 log Ln+2 − Lxn+1 log Ln+1 dx
Ln
L
= Lxn+2 − Lxn+1 Ln+2
n Ln+2
Ln+2
Ln
n
− LLn+1
.
= Ln+2 − Ln+2 − Ln+1
José Luis Díaz-Barrero
Title Page
This completes the proof.
Contents
Corollary 3.2. For n ≥ 1, we have
L
L
L
n+2
n+1
n+2
n
LLn n+1 + Ln+1
+ LLn+2
< LLn n + Ln+1
+ Ln+2
.
Proof. The statement immediately follows from the fact that
Ln+1
Ln+2
Ln+2
Ln
Ln
Ln+1
Ln + Ln+1 + Ln+2 − Ln
+ Ln+1 + Ln+2
h
i
Ln+1
n
= LLn n + Ln+1
− LLn n+1 + LLn+1
h
i
Ln+2
Ln+2
n
n
+ Ln+2
− LLn+2
− Ln+1
− LLn+1
and Theorem 3.1.
Rational Identities and
Inequalities involving Fibonacci
and Lucas Numbers
JJ
J
II
I
Go Back
Close
Quit
Page 10 of 16
J. Ineq. Pure and Appl. Math. 4(5) Art. 83, 2003
http://jipam.vu.edu.au
Theorem 3.3. Let n be a positive integer, then the following inequality
L
L
L
n+2 Ln
n+1
n+2
LLn n+1 Ln+1
Ln+2 < LLn n Ln+1
Ln+2
holds.
Proof. We will argue by using the weighted AM-GM-HM inequality (see [5]).
The proof will be done in two steps. First, we will prove
(3.1)
Ln+2 Ln
LLn n+1 Ln+1
Ln+2
<
Ln + Ln+1 + Ln+2
3
Ln +Ln+1 +Ln+2
.
Rational Identities and
Inequalities involving Fibonacci
and Lucas Numbers
José Luis Díaz-Barrero
In fact, setting x1 = Ln , x2 = Ln+1 , x3 = Ln+2 and
Ln+1
,
Ln + Ln+1 + Ln+2
Ln+2
w2 =
,
Ln + Ln+1 + Ln+2
Ln
w3 =
Ln + Ln+1 + Ln+2
Title Page
w1 =
Contents
JJ
J
II
I
Go Back
and applying the AM-GM inequality, we have
L
/(L +L
+L
)
Close
L /(L +L
+L
)
n
n
n
n+2
n+1
n+2
n+1
n+2
LnLn+1 /(Ln +Ln+1 +Ln+2 ) Ln+1
Ln+2
Ln+1 Ln+2
Ln+2 Ln
Ln Ln+1
<
+
+
Ln + Ln+1 + Ln+2 Ln + Ln+1 + Ln+2 Ln + Ln+1 + Ln+2
Quit
Page 11 of 16
J. Ineq. Pure and Appl. Math. 4(5) Art. 83, 2003
http://jipam.vu.edu.au
or
Ln+2 Ln
LLn n+1 Ln+1
Ln+2
<
Ln Ln+1 + Ln+1 Ln+2 + Ln+2 Ln
Ln + Ln+1 + Ln+2
Ln +Ln+1 +Ln+2
.
Inequality (3.1) will be established if we prove that
Ln Ln+1 + Ln+1 Ln+2 + Ln+2 Ln
Ln + Ln+1 + Ln+2
Ln +Ln+1 +Ln+2
<
Ln + Ln+1 + Ln+2
3
Ln +Ln+1 +Ln+2
Rational Identities and
Inequalities involving Fibonacci
and Lucas Numbers
José Luis Díaz-Barrero
or, equivalently,
Ln Ln+1 + Ln+1 Ln+2 + Ln+2 Ln
Ln + Ln+1 + Ln+2
<
.
Ln + Ln+1 + Ln+2
3
That is,
L2n + L2n+1 + L2n+2 > Ln Ln+1 + Ln+1 Ln+2 + Ln+2 Ln .
The last inequality immediately follows by adding up the inequalities
L2n + L2n+1 ≥ 2Ln Ln+1 ,
L2n+1 + L2n+2 > 2Ln+1 Ln+2 ,
L2n+2 + L2n > 2Ln+2 Ln
and the result is proved.
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 12 of 16
J. Ineq. Pure and Appl. Math. 4(5) Art. 83, 2003
http://jipam.vu.edu.au
Finally, we will prove
L +L +L
Ln + Ln+1 + Ln+2 n n+1 n+2
Ln+1 Ln+2
(3.2)
< LLn n Ln+1
Ln+2 .
3
In fact, setting
x1
w1
w2
w3
= Ln , x2 = Ln+1 , x3 = Ln+2 ,
= Ln /(Ln + Ln+1 + Ln+2 ),
= Ln+1 /(Ln + Ln+1 + Ln+2 ),
and
= Ln+2 /(Ln + Ln+1 + Ln+2 )
Rational Identities and
Inequalities involving Fibonacci
and Lucas Numbers
José Luis Díaz-Barrero
and using the GM-HM inequality, we have
Ln + Ln+1 + Ln+2
3
−1
3
=
Ln + Ln+1 + Ln+2
1
=
1
1
+ Ln +Ln+1
+
Ln +Ln+1 +Ln+2
+Ln+2
<
Title Page
Contents
1
Ln +Ln+1 +Ln+2
Ln+1 /(Ln +Ln+1 +Ln+2 ) Ln+2 /(Ln +Ln+1 +Ln+2 )
LLn n /(Ln +Ln+1 +Ln+2 ) Ln+1
Ln+2
.
Hence,
JJ
J
II
I
Go Back
Close
Quit
Ln + Ln+1 + Ln+2
3
Ln +Ln+1 +Ln+2
L
L
n+1
n+2
< LLn n Ln+1
Ln+2
and (3.2) is proved. This completes the proof of the theorem.
Page 13 of 16
J. Ineq. Pure and Appl. Math. 4(5) Art. 83, 2003
http://jipam.vu.edu.au
Stronger inequalities for second order recurrence sequences, generalizing the
ones given in [4] have been obtained by Stanica in [6].
Finally, we state and prove an inequality involving Fibonacci and Lucas
numbers.
Theorem 3.4. Let n be a positive integer, then the following inequality


n+1
− n+1
n  − n
n
n 
n+1
X
Y
Fk+1 − Lk+1
Fk+2
n
≥
−1
n
 Fk+1

F2k+2
(n + 1)
− L−1
k+1
k=1
k=1
Rational Identities and
Inequalities involving Fibonacci
and Lucas Numbers
holds.
José Luis Díaz-Barrero
Proof. From the AM-GM inequality, namely,
n
Title Page
n
Y 1
1X
xk ≥
xkn ,
n k=1
k=1
where
xk > 0, k = 1, 2, . . . , n,
−1
and taking into account that for all j ≥ 2, 0 < L−1
j < Fj , we get
Z
F2−1
Z
F3−1
(3.3)
L−1
2
L−1
3
−1
Fn+1
!
n+1
1X
x` dx2 dx3 . . . dxn+1
...
n
L−1
n+1
`=2
!
−1
Z F2−1 Z F3−1
Z Fn+1
n+1
Y 1
≥
...
x`n dx2 dx3 . . . dxn+1 .
Z
L−1
2
L−1
3
L−1
n+1
`=1
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 14 of 16
J. Ineq. Pure and Appl. Math. 4(5) Art. 83, 2003
http://jipam.vu.edu.au
Evaluating the preceding integrals (3.3) becomes
(3.4)
n+1
X
−1
−1
(F2−1 − L−1
2 ) · · · (F`−1 − L`−1 )
`=2
−1
−1
−1
−1
· (F`−2 − L−2
` )(F`+1 − L`+1 ) · · · (Fn+1 − Ln+1 )
n+1
2nn+1 Y − n+1
− n+1
n
n
≥
F
− L`
(n + 1)n `=2 `
or, equivalently,
n+1
Y
(F`−1
−
L−1
` )
`=2
n+1
X
(F`−1
`=2
+
L−1
` )
n+1
2nn+1 Y − n+1
− n+1
n
n
F
−
L
.
≥
`
(n + 1)n `=2 `
Setting k = `−1 in the preceding inequality, taking into account that Fk +Lk =
2Fk+1 , Fk Lk = F2k and after simplification, we obtain


n+1
− n+1
n
n  − n
n 
n+1
X
Y
Fk+1 − Lk+1
Fk+2
n
≥
−1
n
 Fk+1

F2k+2
(n + 1)
− L−1
k+1
k=1
k=1
Rational Identities and
Inequalities involving Fibonacci
and Lucas Numbers
José Luis Díaz-Barrero
Title Page
Contents
JJ
J
II
I
Go Back
Close
and the proof is completed.
Quit
Page 15 of 16
J. Ineq. Pure and Appl. Math. 4(5) Art. 83, 2003
http://jipam.vu.edu.au
References
[1] S. VAJDA, Fibonacci and Lucas Numbers and the Golden Section, New
York, Wiley, 1989.
[2] A.T. BENJAMIN AND J. J. QUINN, Recounting Fibonacci and Lucas identities, College Math. J., 30(5) (1999), 359–366.
[3] J.L. DĺAZ-BARRERO, Problem B–905, The Fibonaci Quarterly, 38(4)
(2000), 373.
[4] J.L. DĺAZ-BARRERO, Advanced Problem H–581, The Fibonaci Quarterly,
40(1) (2002), 91.
[5] G. HARDY, J.E. LITTLEWOOD AND G. PÓLYA, Inequalities, Cambridge,
1997.
[6] P. STANICA, Inequalities on linear functions and circular powers, J.
Ineq. Pure and Appl. Math., 3(3) (2002), Article 43. [ONLINE: http:
//jipam.vu.edu.au/v3n3/015_02.html]
Rational Identities and
Inequalities involving Fibonacci
and Lucas Numbers
José Luis Díaz-Barrero
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 16 of 16
J. Ineq. Pure and Appl. Math. 4(5) Art. 83, 2003
http://jipam.vu.edu.au