Key words and phrases: Rational Identities and Inequalities, Sequences of Integers, Fibonacci and Lucas numbers

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 APPLIEDMATHEMATICSIII

UNIVERSITATPOLITÈCNICA DECATALUNYA

JORDIGIRONA1-3, C2, 08034 BARCELONA, SPAIN. jose.luis.diaz@upc.es

Received 21 May, 2003; accepted 21 August, 2003 Communicated by J. Sándor

ABSTRACT. In this paper we use integral calculus, complex variable techniques and some clas- sical 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. 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.

2. RATIONALIDENTITIES

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

Theorem 2.1. LetF_ndenote then^thFibonacci number. That is,F₀ = 0, F₁ = 1and 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 byγ = {z ∈C : |z| < F_r+1}.Evaluating the preceding integral in the exterior of theγcontour, we obtain

I1 = 1 2πi

I

γ

(1 +z^` z

n

Y

j=1

1 (z−F_r+j)

) dz =

n

X

k=1

Rk, 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). 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 = 1and for n ≥ 2, L_n=Ln−1+Ln−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.

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

(F_n²+ 1)Fn+1Fn+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, Ln+1Ln+2

(L_n+1−L_n)(L_n+2−L_n) + Ln+2Ln

(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_nandL_ndenote then^thFibonacci and Lucas numbers, respectively.

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− Lj

L_i −1





+

n

X

i=1

L_i =L_n+2−3.

SincePn

i=1L_i =L_n+2−3,as can be easily established by mathematical induction, 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 by γ, a circle centered at the origin and radius L_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







=

n

X

i=1





 1 Lⁿ⁻²_i

n

Y

j=1 j6=i

1− Lj

L_i −1





. Integrating in the region outside of theγcontour we get

I₂ = 1 2πi

I

γ

z

A(z)dz = 0.

Again, by Cauchy’s theorem on contour integrals we haveI₁+I₂ = 0.This completes the proof

of (2.4).

Note that (2.4) can also be established by using routine algebra.

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.

SinceL_n < L_n+1ifn≥1,then forL_n ≤x≤L_n+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 and (a) is proved. To prove (b), we consider the integral

I₂ =

Z Ln+2

Ln

L^x_n+2logL_n+2−L^x_n+1logL_n+1 dx.

SinceL_n+1 < L_n+2,then forL_n ≤x≤L_n+2 we have

L^x_n+1logL_n+1 < L^x_n+2logL_n+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 Theorem 3.1.

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_n^{n+1/(Ln+Ln+1+Ln+2)}L^L_n+1^{n+2/(Ln+Ln+1+Ln+2)}L^L_n+2^{n/(Ln+Ln+1+Ln+2)}

< LnLn+1

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

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

L_n+L_n+1+L_n+2 or

L^L_nⁿ⁺¹L^L_n+1ⁿ⁺²L^L_n+2ⁿ <

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

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

Ln+Ln+1+Ln+2

Ln+Ln+1+Ln+2

<

L_n+L_n+1+L_n+2 3

Ln+Ln+1+Ln+2

or, equivalently,

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

< L_n+L_n+1+L_n+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 >2L_n+1L_n+2,

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

Finally, we will prove (3.2)

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. 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 =Ln+1/(Ln+Ln+1+Ln+2), and w₃ =L_n+2/(L_n+L_n+1+L_n+2)

and using the GM-HM inequality, we have Ln+Ln+1+Ln+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_n^{n/(Ln+Ln+1+Ln+2)}L^L_n+1^{n+1/(Ln+Ln+1+Ln+2)}L^L_n+2^{n+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.

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

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. Evaluating the preceding integrals (3.3) becomes

(3.4)

n+1

X

`=2

(F₂⁻¹−L⁻¹₂ )· · ·(F_{`−1</sub><sup>−1</sup> −L<sup>−1</sup><sub>`−1})(F_{`</sub><sup>−2</sup>−L<sup>−2</sup><sub>`} )(F_{`+1</sub><sup>−1</sup> −L<sup>−1</sup><sub>`+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_{`</sub><sup>−1</sup>−L<sup>−1</sup><sub>`} )

n+1

X

`=2

(F_{`</sub><sup>−1</sup>+L<sup>−1</sup><sub>`} )≥ 2nⁿ⁺¹ (n+ 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 +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.

REFERENCES

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

[2] A.T. BENJAMIN ANDJ. 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. 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]