http://jipam.vu.edu.au/
Volume 4, Issue 5, Article 83, 2003
RATIONAL IDENTITIES AND INEQUALITIES INVOLVING FIBONACCI AND
LUCAS NUMBERS
JOSÉ LUIS DÍAZ-BARRERO
A PPLIED M ATHEMATICS III
U NIVERSITAT P OLITÈCNICA DE C ATALUNYA
J ORDI G IRONA 1-3, C2,
08034 B ARCELONA , S PAIN .
jose.luis.diaz@upc.es
Received 21 May, 2003; accepted 21 August, 2003
Communicated by J. Sándor
A BSTRACT . 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.
Key words and phrases: Rational Identities and Inequalities, Sequences of Integers, Fibonacci and Lucas numbers.
2000 Mathematics Subject Classification. 05A19, 11B39.
1. I
NTRODUCTION
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.
2. R ATIONAL I DENTITIES
In what follows several rational identities are considered and proved by using results on contour integrals. We begin with:
ISSN (electronic): 1443-5756 c 2003 Victoria University. All rights reserved.
The author would like to thank the anonymous referee for the suggestions and comments that have improved the final version of the paper.
067-03

2 J OSÉ L UIS D ÍAZ -B ARRERO
Theorem 2.1. Let F n
F n
= F n − 1
+ F n − 2
.
denote the n th Fibonacci number. That is,
Then, for all positive integers r,
F
0
= 0 , F
1
= 1 and for n ≥ 2 ,
(2.1) n
X k =1
1 + F
` r + k
F r + k


 n
Y

 j =1 j = k

F r + k
1


− F r + j


=
( − 1) n +1
F r +1
F r +2
· · · F r + n holds, with 0 ≤ ` ≤ n − 1 .
Proof. To prove the preceding identity we consider the integral
(2.2) I =
1
2 πi
I 1 + z
γ
A n
( z )
` dz
, z where A n
( z ) = Q n j =1
( z − F r + j
) .
Let γ be the curve defined by γ = { z ∈ in the exterior of the γ contour, we obtain
C
: | z | < F r +1
} .
Evaluating the preceding integral
I
1
=
1
2 πi
I
γ
(
1 + z ` z n
Y j =1
1
( z − F r + j
)
) dz = n
X
R k
, k =1 where

R k


1 + z
`
= lim z → F r + k

 z
 n
Y j =1 j = k
1


( z − F r + j
)


=
1 + F
` r + k
F r + k n
Y j =1 j = k
( F r + k
1
− F r + j
)
.
Then, I
1 becomes
I
1
=
 n
X


1 + F ` r + k
F r + k k =1


 n
Y j =1 j = k
( F r + k
1 

− F r + j
)


.
Evaluating (2.2) in the interior of the
γ contour, we get
I
2
=
1
2 πi
I
γ
(
1 + z ` z n
Y
1 j =1
( z − F r + j
)
) dz
1 + z `
= lim
= z → 0
1
A n
(0)
A n
( z )
=
F r +1
F
( − 1) n r +2
· · · F r + n
.
By Cauchy’s theorem on contour integrals we have that I
1
+ I
2
= 0 and the proof is complete.
A similar identity also holds for Lucas numbers. It can be stated as:
Corollary 2.2. Let L n
L n
= L n − 1
+ L n − 2
.
denote the n th
Lucas number. That is,
Then, for all positive integers r ,
L
0
= 2 , L
1
= 1 and for n ≥ 2 ,
(2.3)
  n
X
1 + L ` r + k
L r + k k =1


Y

 j =1 j = k
1


L r + k
− L r + j


=
( − 1) n +1
L r +1
L r +2
· · · + L r + n holds, with 0 ≤ ` ≤ n − 1 .
J. Inequal. Pure and Appl. Math., 4(5) Art. 83, 2003 http://jipam.vu.edu.au/

R ATIONAL I DENTITIES AND I NEQUALITIES
In particular (2.1) and (2.3) can be used (see [3]) to obtain
Corollary 2.3. For n ≥ 2 ,
( F
2 n
( F n +1
+ 1) F n +1
F
− F n
)( F n +2 n +2
− F n
)
+
( F n
F n
( F 2 n +1
+ 1)
− F n +1
)( F n +2
F n +2
− F n +1
)
+
F n
F n +1
( F 2 n +2
( F n
− F n +2
)( F n +1
+ 1)
− F n +2
)
= 1 .
Corollary 2.4. For n ≥ 2 ,
L n +1
L n +2
( L n +1
− L n
)( L n +2
− L n
)
+
( L n
L n +2
L n
− L n +1
)( L n +2
− L n +1
)
L n
L n +1
+
( L n
− L n +2
)( L n +1
− L n +2
)
= 1 .
In the sequel F n and L n denote the n th
Fibonacci and Lucas numbers, respectively.
Theorem 2.5. If n ≥ 3 , then we have n
X i =1
1
L n − 2 i
 n

Y
 j =1 j = i
1 −
L j
L i

− 1
+ L n − 1 i


= L n +2
− 3 .
Proof. First, we observe that the given statement can be written as
 n
X i =1
1

 L n − 2 i n
Y j =1 j = i
1 −
L j
L i

− 1


+ n
X
L i i =1
= L n +2
− 3 .
Since P n i =1
L i suffice to prove
= L n +2
− 3 , as can be easily established by mathematical induction, then it will
(2.4)
 n
X i =1
1

 L n − 2 i n
Y j =1 j = i
1 −
L j
L i
− 1



= 0 .
We will argue by using residue techniques. We consider the monic complex polynomial A ( z ) =
Q n
) and we evaluate the integral k =1
( z − L k
1
I =
2 πi
I
γ z
A ( z ) dz over the interior and exterior domains limited by γ , a circle centered at the origin and radius
L n +1
, i.e., γ = { z ∈
C
: | z | < L n +1
} .
J. Inequal. Pure and Appl. Math., 4(5) Art. 83, 2003 http://jipam.vu.edu.au/
3

4 J OSÉ L UIS D ÍAZ -B ARRERO
Integrating in the region inside the γ contour we have
I
1
=
1 I
2 πi
γ
= n
X
Res z
A ( z ) dz z
A ( z )
, z = L i i =1
= n
X
 n

Y
 i =1 j =1 j = i

L i
L i
− L j


=
 n
X i =1
1

 L n − 2 i n
Y j =1 j = i
1 −
L
L j i
− 1



.
Integrating in the region outside of the γ contour we get
I
2
=
1
2 πi
I
γ z
A ( z ) dz = 0 .
Again, by Cauchy’s theorem on contour integrals we have I
1
of (2.4).
+ I
2
= 0 .
This completes the proof
Note that (2.4) can also be established by using routine algebra.
3. I NEQUALITIES
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
(a)
(b)
L
L n +1 n
+ L
L n n +1
< L
L n n
+ L
L n +1 n +1
,
L
L n +2 n +1
− L
L n n +1
< L
L n +2 n +2
− L
L n n +2
.
Proof. To prove part (a) we consider the integral
Z
L n +1
I
1
=
Since L n
L n
< L n +1 if n ≥ 1 , then for L n
L x n +1 log L n +1
− L x n log L n
≤ x ≤ L n +1 we have dx.
L x n log L n
< L x n +1 log L n
< L x n +1 log L n +1 and I
1
> 0 .
On the other hand, evaluating the integral, we obtain
Z
L n +1
I
1
= L x n +1 log L n +1
− L x n log L n
L n
= L x n +1
− L x n
L n +1
L n
= L
L n n
+ L
L n +1 n +1
− L
L n +1 n
+ L
L n n +1 dx
and (a) is proved. To prove (b), we consider the integral
I
2
=
Z
L n +2
L n
L x n +2 log L n +2
− L x n +1 log L n +1 dx.
J. Inequal. Pure and Appl. Math., 4(5) Art. 83, 2003 http://jipam.vu.edu.au/

R ATIONAL I DENTITIES AND I NEQUALITIES
Since L n +1
< L n +2
, then for L n
≤ x ≤ L n +2 we have
L x n +1 log L n +1
< L x n +2 log L n +2 and I
2
> 0 .
On the other hand, evaluating I
2
, we get
I
2
=
Z
L n +2
L x n +2 log L n +2
− L x n +1 log L n +1
L n
= L x n +2
− L x n +1
L n +2
L n
= L
L n +2 n +2
− L
L n n +2
− L
L n +2 n +1
− L
L n n +1
.
This completes the proof.
dx
Corollary 3.2. For n ≥ 1 , we have
L
L n +1 n
+ L
L n +2 n +1
+ L
L n n +2
< L
L n n
+ L
L n +1 n +1
+ L
L n +2 n +2
.
Proof. The statement immediately follows from the fact that
L
L n n
+ L
L n +1 n +1
+ L
L n +2 n +2
− L
L n +1 n
+ L
L n +2 n +1
+ L
L n n +2
= h
L
L n n
+ L
L n +1 n +1
− L
L n +1 n
+ L
L n n +1 i
+ h
L
L n +2 n +2
− L
L n n +2
and Theorem 3.1.
− L
L n +2 n +1
− L
L n n +1 i
Theorem 3.3. Let n be a positive integer, then the following inequality
L
L n +1 n
L
L n +2 n +1
L
L n n +2
< L
L n n
L
L n +1 n +1
L
L n +2 n +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) L
L n +1 n
L
L n +2 n +1
L
L n n +2
<
L n
+ L n +1
3
+ L n +2
L n
+ L n +1
+ L n +2
.
In fact, setting x
1
= L n
, x
2
= L n +1
, x
3
= L n +2 and w
1
= w
2
= w
3
=
L n
+ L
L n +1 n +1
+ L n +2
,
L n
+ L
L n +2 n +1
+ L n +2
,
L n
L n
+ L n +1
+ L n +2 and applying the AM-GM inequality, we have or
L
L n +1 n
/ ( L n
+ L n +1
+ L n +2
)
L
L n +2
/ ( L n
+ L n +1
+ L n +2
) n +1
L
L n
/ ( L n
+ L n +1
+ L n +2
) n +2
L n
L n +1
<
L n
+ L n +1
+ L n +2
+
L n +1
L n +2
L n
+ L n +1
+ L n +2
+
L n +2
L n
L n
+ L n +1
+ L n +2
L n
+ L n +1
+ L n +2
L
L n +1 n
L
L n +2 n +1
L
L n n +2
<
L n
L n +1
+ L n +1
L n +2
+ L n +2
L n
L n
+ L n +1
+ L n +2
.
J. Inequal. Pure and Appl. Math., 4(5) Art. 83, 2003 http://jipam.vu.edu.au/
5

6 J OSÉ L UIS D ÍAZ -B ARRERO
Inequality (3.1) will be established if we prove that
L n
+ L n +1
+ L n +2 L n
L n +1
+ L n +1
L n +2
+ L n +2
L n
L n
+ L n +1
+ L n +2
<
L n
+ L n +1
3
+ L n +2 or, equivalently,
L n
L n +1
+ L n +1
L n +2
+ L n +2
L n
L n
+ L n +1
+ L n +2
<
L n
+ L n +1
+ L n +2
.
3
That is,
L
2 n
+ L
2 n +1
+ L
2 n +2
> L n
L n +1
+ L n +1
L n +2
+ L n +2
The last inequality immediately follows by adding up the inequalities
L n
.
L
2 n
+ L
2 n +1
≥ 2 L n
L n +1
,
L
2 n +1
+ L
2 n +2
> 2 L n +1
L n +2
,
L
2 n +2
+ L
2 n
> 2 L n +2
L n and the result is proved.
Finally, we will prove
(3.2)
L n
+ L n +1
3
+ L n +2
L n
+ L n +1
+ L n +2
< L
L n n
L
L n +1 n +1
L
L n +2 n +2
.
In fact, setting
L n
+ L n +1
+ L n +2 x
1
= L n
, x
2
= L n +1
, x
3
= L n +2
, w
1
= L n
/ ( L n
+ L n +1
+ L n +2
) , w
2
= L n +1
/ ( L n
+ L n +1
+ L n +2
) , and w
3
= L n +2
/ ( L n
+ L n +1
+ L n +2
) and using the GM-HM inequality, we have
L n
+ L n +1
+ L n +2
3
− 1
3
=
=
L n
+ L n +1
+ L n +2
1
L n
+ L n +1
+ L n +2
+
1
1
L n
+ L n +1
+ L n +2
+
1
L n
+ L n +1
+ L n +2
< L
L n
/ ( L n
+ L n +1
+ L n +2
) n
L
L n +1
/ ( L n
+ L n +1 n +1
+ L n +2
)
L
L n +2 n +2
/ ( L n
+ L n +1
+ L n +2
)
.
Hence,
L n
+ L n +1
+ L n +2
L n
+ L n +1
+ L n +2
< L
L n n
L
L n +1 n +1
L
L n +2 n +2
3
and (3.2) is proved. This completes the proof of the theorem.
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
X k =1
F k +2
F
2 k +2
≥ n n +1
( n + 1) n n
Y k =1



F
− n +1 k +1 n
F
− 1 k +1 n +1
− L
− k +1 n
− L
− 1 k +1


 holds.
J. Inequal. Pure and Appl. Math., 4(5) Art. 83, 2003 http://jipam.vu.edu.au/

R ATIONAL I DENTITIES AND I NEQUALITIES
Proof. From the AM-GM inequality, namely,
1 n n
X x k
≥ k =1 n
Y x k n
1
, where x k k =1
> 0 , k = 1 , 2 , . . . , n, and taking into account that for all j ≥ 2 , 0 < L
− 1 j
< F j
− 1
, we get
(3.3)
Z
F
2
− 1
Z
F
3
− 1
L
− 1
2
L
− 1
3
. . .
Z
F
− 1 n +1
L
− 1 n +1
1 n n +1
X x
`
!
` =2 dx
2 dx
3
. . . dx n +1
≥
Z
F
2
− 1
Z
F
3
− 1
L
− 1
2
L
− 1
3
. . .
− 1
Z
F n +1 n +1
Y x
` n
1
!
L
− 1 n +1 ` =1 dx
2 dx
3
. . . dx n +1
.
Evaluating the preceding integrals (3.3) becomes
(3.4) n +1
X
( F
2
− 1 − L
− 1
2
) · · · ( F
− 1
` − 1
− L
− 1
` − 1
)( F
`
− 2
` =2
− L
− 2
`
)( F
− 1
` +1
− L
− 1
` +1
) · · · ( F
− 1 n +1
− L
− 1 n +1
)
2 n n +1
≥
( n + 1) n n +1
Y
` =2
F
`
− n +1 n − L
−
` n +1 n or, equivalently, n +1
Y
( F
`
− 1
` =2
− L
− 1
`
) n +1
X
( F
`
− 1
` =2
+ L
− 1
`
) ≥
(
2 n n n +1
+ 1) n n +1
Y
` =2
F
`
− n +1 n − L
−
` n +1 n .
Setting k = ` − 1 in the preceding inequality, taking into account that F k
F k
L k
= F
2 k and after simplification, we obtain
+ L k n
X k =1
F k +2
F
2 k +2
≥ n n +1
( n + 1) n n
Y k =1



F
− n +1 k +1 n
F
− 1 k +1 n +1
− L
− k +1 n
− L
− 1 k +1



= 2 F k +1
, and the proof is completed.
R EFERENCES
[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. D´lAZ-BARRERO, Problem B–905, The Fibonaci Quarterly, 38(4) (2000), 373.
[4] J.L. D´lAZ-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
]
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J. Inequal. Pure and Appl. Math., 4(5) Art. 83, 2003 http://jipam.vu.edu.au/