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volume 4, issue 5, article 83, 2003.

Received 21 May, 2003;

accepted 21 August, 2003 Communicated by:J. Sándor

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Journal of Inequalities in Pure and Applied Mathematics

RATIONAL IDENTITIES AND INEQUALITIES INVOLVING FIBONACCI AND LUCAS NUMBERS

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

c

2000Victoria University ISSN (electronic): 1443-5756 067-03

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Rational Identities and Inequalities involving Fibonacci

and Lucas Numbers José Luis Díaz-Barrero

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J. Ineq. Pure and Appl. Math. 4(5) Art. 83, 2003

http://jipam.vu.edu.au

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.

The author would like to thank the anonymous referee for the suggestions and com- ments that have improved the final version of the paper.

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 combi- natorial 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.

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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. LetF_ndenote then^thFibonacci number. That is,F₀ = 0, F₁ = 1 and forn ≥2, F_n =Fn−1+Fn−2.Then, for all positive integersr,

(2.1)

n

X

k=1

1 +F_r+k^` F_r+k







n

Y

j=1 j6=k

1 F_r+k−F_r+j







= (−1)ⁿ⁺¹ F_r+1F_r+2· · ·F_r+n holds, with0≤` ≤n−1.

Proof. To prove the preceding identity we consider the integral

(2.2) I = 1

2πi I

γ

1 +z^` A_n(z)

dz z , whereA_n(z) = Qn

j=1(z−F_r+j).

Letγ be the curve defined by γ = {z ∈ C : |z| < F_r+1}.Evaluating the preceding integral in the exterior of theγ contour, we obtain

I₁ = 1 2πi

I

γ

(1 +z^` z

n

Y

j=1

1 (z−Fr+j)

) dz =

n

X

k=1

R_k, where

R_k = lim

z→F_r+k





 1 +z^`

z

n

Y

j=1 j6=k

1 (z−F_r+j)







= 1 +F_r+k^` F_r+k

n

Y

j=1 j6=k

1

(F_r+k−F_r+j).

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Then,I₁ becomes

I₁ =

n

X

k=1







1 +F_r+k^` F_r+k

n

Y

j=1 j6=k

1 (F_r+k−F_r+j)





 .

Evaluating (2.2) in the interior of theγ contour, we get I2 = 1

2πi I

γ

(1 +z^` z

n

Y

j=1

1 (z−F_r+j)

) dz

= lim

z→0

1 +z^` A_n(z)

= 1

A_n(0) = (−1)ⁿ F_r+1F_r+2· · ·F_r+n.

By Cauchy’s theorem on contour integrals we have thatI₁+I₂ = 0and the proof is complete.

A similar identity also holds for Lucas numbers. It can be stated as:

Corollary 2.2. Let L_n denote then^th Lucas number. That is, L₀ = 2, L₁ = 1 and forn ≥2, L_n=L_n−1+L_n−2.Then, for all positive integersr,

(2.3)

n

X

k=1

1 +L^`_r+k L_r+k





 Y

j=1 j6=k

1 L_r+k−L_r+j







= (−1)ⁿ⁺¹ L_r+1L_r+2· · ·+L_r+n holds, with0≤` ≤n−1.

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In particular (2.1) and (2.3) can be used (see [3]) to obtain Corollary 2.3. Forn ≥2,

(F_n²+ 1)F_n+1F_n+2

(F_n+1−F_n)(F_n+2−F_n) + F_n(F_n+1² + 1)F_n+2 (F_n−F_n+1)(F_n+2−F_n+1)

+ F_nF_n+1(F_n+2² + 1)

(F_n−F_n+2)(F_n+1−F_n+2) = 1.

Corollary 2.4. Forn ≥2, L_n+1L_n+2

(L_n+1−L_n)(L_n+2−L_n)+ L_n+2L_n

(L_n−L_n+1)(L_n+2−L_n+1) + L_nL_n+1

(L_n−L_n+2)(L_n+1−L_n+2) = 1.

In the sequelF_n and L_n denote the n^th Fibonacci and Lucas numbers, re- spectively.

Theorem 2.5. Ifn ≥3,then we have

n

X

i=1

1 Lⁿ⁻²_i







n

Y

j=1 j6=i

1−L_j

L_i −1

+Lⁿ⁻¹_i





=L_n+2−3.

Proof. First, we observe that the given statement can be written as

n

X

i=1





 1 Lⁿ⁻²_i

n

Y

j=1 j6=i

1− L_j

L_i −1





+

n

X

i=1

L_i =L_n+2−3.

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SincePn

i=1L_i =L_n+2−3,as can be easily established by mathematical induc- tion, then it will suffice to prove

(2.4)

n

X

i=1





 1 Lⁿ⁻²_i

n

Y

j=1 j6=i

1− L_j

L_i −1





= 0.

We will argue by using residue techniques. We consider the monic complex polynomialA(z) =Qn

k=1(z−L_k)and we evaluate the integral I = 1

2πi I

γ

z A(z)dz

over the interior and exterior domains limited byγ, a circle centered at the origin and radiusL_n+1,i.e.,γ ={z ∈C:|z|< L_n+1}.

Integrating in the region inside theγ contour we have I₁ = 1

2πi I

γ

z A(z)dz

=

n

X

i=1

Res z

A(z), z =L_i

=

n

X

i=1







n

Y

j=1 j6=i

Li

L_i−L_j







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=

n

X

i=1





 1 Lⁿ⁻²_i

n

Y

j=1 j6=i

1− L_j

L_i ⁻¹





.

Integrating in the region outside of theγ contour we get I2 = 1

2πi I

γ

z

A(z)dz = 0.

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.

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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. Letnbe a positive integer, then the following inequalities hold L^L_nⁿ⁺¹+L^L_n+1ⁿ < L^L_nⁿ+L^L_n+1ⁿ⁺¹,

(a)

L^L_n+1ⁿ⁺²−L^L_n+1ⁿ < L^L_n+2ⁿ⁺²−L^L_n+2ⁿ . (b)

Proof. To prove part (a) we consider the integral

I₁ =

Z Ln+1

Ln

L^x_n+1logL_n+1−L^x_nlogL_n dx.

SinceLn< Ln+1 ifn ≥1,then forLn≤x≤Ln+1 we have L^x_nlogL_n< L^x_n+1logL_n< L^x_n+1logL_n+1 andI₁ >0.On the other hand, evaluating the integral, we obtain

I₁ =

Z Ln+1

Ln

L^x_n+1logL_n+1−L^x_nlogL_n dx

=

L^x_n+1−L^x_nLn+1

Ln

=

L^L_nⁿ+L^L_n+1ⁿ⁺¹

−

L^L_nⁿ⁺¹ +L^L_n+1ⁿ

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and (a) is proved. To prove (b), we consider the integral I2 =

Z Ln+2

Ln

L^x_n+2logLn+2−L^x_n+1logLn+1

dx.

SinceL_n+1 < L_n+2,then forL_n≤x≤L_n+2we have L^x_n+1logLn+1 < L^x_n+2logLn+2

andI2 >0.On the other hand, evaluatingI2,we get I₂ =

Z Ln+2

Ln

L^x_n+2logL_n+2−L^x_n+1logL_n+1 dx

=

L^x_n+2−L^x_n+1Ln+2

Ln

=

L^L_n+2ⁿ⁺² −L^L_n+2ⁿ

−

L^L_n+1ⁿ⁺²−L^L_n+1ⁿ . This completes the proof.

Corollary 3.2. Forn ≥1,we have

L^L_nⁿ⁺¹ +L^L_n+1ⁿ⁺²+L^L_n+2ⁿ < L^L_nⁿ+L^L_n+1ⁿ⁺¹ +L^L_n+2ⁿ⁺². Proof. The statement immediately follows from the fact that

L^L_nⁿ+L^L_n+1ⁿ⁺¹+L^L_n+2ⁿ⁺²

−

L^L_nⁿ⁺¹+L^L_n+1ⁿ⁺² +L^L_n+2ⁿ

=h

L^L_nⁿ +L^L_n+1ⁿ⁺¹

−

L^L_nⁿ⁺¹+L^L_n+1ⁿ i +

h

L^L_n+2ⁿ⁺²−L^L_n+2ⁿ

−

L^L_n+1ⁿ⁺² −L^L_n+1ⁿ i

and Theorem3.1.

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Theorem 3.3. Letnbe a positive integer, then the following inequality L^L_nⁿ⁺¹L^L_n+1ⁿ⁺²L^L_n+2ⁿ < L^L_nⁿL^L_n+1ⁿ⁺¹L^L_n+2ⁿ⁺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_nⁿ⁺¹L^L_n+1ⁿ⁺²L^L_n+2ⁿ <

L_n+L_n+1+L_n+2 3

Ln+Ln+1+Ln+2

. In fact, settingx₁ =L_n, x₂ =L_n+1, x₃ =L_n+2and

w1 = L_n+1

L_n+L_n+1+L_n+2, w₂ = L_n+2

L_n+L_n+1+L_n+2, w₃ = Ln

L_n+L_n+1+L_n+2 and applying the AM-GM inequality, we have

L^L_nⁿ⁺¹^/(Lⁿ^+Lⁿ⁺¹^+Lⁿ⁺²⁾L^L_n+1ⁿ⁺²^/(Lⁿ^+Lⁿ⁺¹^+Lⁿ⁺²⁾L^L_n+2ⁿ^/(Lⁿ^+Lⁿ⁺¹^+Lⁿ⁺²⁾

< L_nL_n+1

L_n+L_n+1+L_n+2 + L_n+1L_n+2

L_n+L_n+1+L_n+2 + L_n+2L_n L_n+L_n+1+L_n+2

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or

L^L_nⁿ⁺¹L^L_n+1ⁿ⁺²L^L_n+2ⁿ <

L_nL_n+1+L_n+1L_n+2+L_n+2L_n L_n+L_n+1+L_n+2

Ln+Ln+1+Ln+2

. Inequality (3.1) will be established if we prove that

L_nL_n+1+L_n+1L_n+2+L_n+2L_n L_n+L_n+1+L_n+2

Ln+Ln+1+Ln+2

<

L_n+L_n+1+L_n+2 3

Ln+Ln+1+Ln+2

or, equivalently,

LnLn+1+Ln+1Ln+2+Ln+2Ln

L_n+L_n+1+L_n+2 < Ln+Ln+1+Ln+2

3 .

That is,

L²_n+L²_n+1+L²_n+2 > L_nL_n+1+L_n+1L_n+2+L_n+2L_n. The last inequality immediately follows by adding up the inequalities

L²_n+L²_n+1 ≥2L_nL_n+1, L²_n+1+L²_n+2 >2Ln+1Ln+2,

L²_n+2+L²_n>2L_n+2L_n and the result is proved.

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Finally, we will prove (3.2)

Ln+Ln+1+Ln+2

3

Ln+Ln+1+Ln+2

< L^L_nⁿL^L_n+1ⁿ⁺¹L^L_n+2ⁿ⁺². In fact, setting

x₁ =L_n, x₂ =L_n+1, x₃ =L_n+2, w₁ =L_n/(L_n+L_n+1+L_n+2),

w₂ =L_n+1/(L_n+L_n+1+L_n+2), and w₃ =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

=

3

L_n+L_n+1+L_n+2 ⁻¹

= 1

1

Ln+Ln+1+Ln+2 +_L ¹

n+Ln+1+Ln+2 + _L ¹

n+Ln+1+Ln+2

< L^L_nⁿ^/(Lⁿ^+Lⁿ⁺¹^+Lⁿ⁺²⁾L^L_n+1ⁿ⁺¹^/(Lⁿ^+Lⁿ⁺¹^+Lⁿ⁺²⁾L^L_n+2ⁿ⁺²^/(Lⁿ^+Lⁿ⁺¹^+Lⁿ⁺²⁾. Hence,

L_n+L_n+1+L_n+2 3

Ln+Ln+1+Ln+2

< L^L_nⁿL^L_n+1ⁿ⁺¹L^L_n+2ⁿ⁺² and (3.2) is proved. This completes the proof of the theorem.

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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. Letnbe a positive integer, then the following inequality

n

X

k=1

F_k+2

F_2k+2 ≥ nⁿ⁺¹ (n+ 1)ⁿ

n

Y

k=1





 F⁻

n+1 n

k+1 −L⁻

n+1 n

k+1

F_k+1⁻¹ −L⁻¹_k+1







holds.

Proof. From the AM-GM inequality, namely, 1

n

X

k=1

x_k ≥

n

Y

k=1

x

1 n

k, where x_k >0, k = 1,2, . . . , n, and taking into account that for allj ≥2,0< L⁻¹_j < F_j⁻¹,we get

(3.3)

Z F₂⁻¹ L⁻¹₂

Z F₃⁻¹ L⁻¹₃

. . . Z F_n+1⁻¹

L⁻¹_n+1

1 n

n+1

X

`=2

x_`

!

dx₂dx₃. . . dx_n+1

≥ Z F₂⁻¹

L⁻¹₂

Z F₃⁻¹ L⁻¹₃

. . . Z F_n+1⁻¹

L⁻¹_n+1 n+1

Y

`=1

x

1 n

`

!

dx₂dx₃. . . dx_n+1.

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Evaluating the preceding integrals (3.3) becomes

(3.4)

n+1

X

`=2

(F₂⁻¹−L⁻¹₂ )· · ·(F_`−1⁻¹ −L⁻¹_`−1)

·(F_`⁻²−L⁻²_` )(F_`+1⁻¹ −L⁻¹_`+1)· · ·(F_n+1⁻¹ −L⁻¹_n+1)

≥ 2nⁿ⁺¹ (n+ 1)ⁿ

n+1

Y

`=2

F⁻

n+1 n

` −L⁻

n+1 n

`

or, equivalently,

n+1

Y

`=2

(F_`⁻¹−L⁻¹_` )

n+1

X

`=2

(F_`⁻¹+L⁻¹_` )≥ 2nⁿ⁺¹ (n+ 1)ⁿ

n+1

Y

`=2

F⁻

n+1 n

` −L⁻

n+1 n

`

. Settingk =`−1in the preceding inequality, taking into account thatF_k+L_k = 2F_k+1, F_kL_k=F_2kand after simplification, we obtain

n

X

k=1

F_k+2

F_2k+2 ≥ nⁿ⁺¹ (n+ 1)ⁿ

n

Y

k=1





 F⁻

n+1 n

k+1 −L⁻

n+1 n

k+1

F_k+1⁻¹ −L⁻¹_k+1





 and the proof is completed.

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References

[1] S. VAJDA, Fibonacci and Lucas Numbers and the Golden Section, New York, Wiley, 1989.

[2] A.T. BENJAMINANDJ. J. QUINN, Recounting Fibonacci and Lucas iden- tities, 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. LITTLEWOODANDG. 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]