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volume 5, issue 3, article 75, 2004.

Received 25 December, 2003;

accepted 27 June, 2004.

Communicated by:L. Tóth

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Journal of Inequalities in Pure and Applied Mathematics

RATIONAL IDENTITIES AND INEQUALITIES

TOUFIK MANSOUR

Department of Mathematics University of Haifa

31905 Haifa, Israel.

EMail:toufik@math.haifa.ac.il

c

2000Victoria University ISSN (electronic): 1443-5756 001-04

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Rational Identities and Inequalities Toufik Mansour

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J. Ineq. Pure and Appl. Math. 5(3) Art. 75, 2004

http://jipam.vu.edu.au

Abstract

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.

2000 Mathematics Subject Classification:05A19, 11B39.

Key words: Rational Identities and Inequalities, Fibonacci numbers, Lucas numbers, Pell numbers.

The author is grateful to Díaz-Barrero for his careful reading of the manuscript.

1. Introduction

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(a_n)n≥0is said to be pos- itive increasing if 0 < a_n < a_n+1 for all n ≥ 1, and complex increasing if 0 < |a_n| ≤ |a_n+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.

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2. Identities

In this section we present several rational identities and inequalities by using results on contour integrals.

Theorem 2.1. Let(a_n)n≥0 be any complex increasing sequence such thata_p 6=

a_qfor allp6=q. For all positive integersr,

n

X

k=1

1 +a^`_r+k a_r+k

n

Y

j=1, j6=k

(a_r+k−a_r_j)⁻¹

!

= (−1)ⁿ⁺¹ Qn

j=1a_r+j holds, with0≤` ≤n−1.

Proof. Let us consider the integral

I = 1 2πi

I

γ

1 +z^` zA_n(z)dz, whereγ = {z ∈ C : |z| <|a_r+1|}andA_n(z) = Qn

j=1(z−a_r+j). Evaluating the integralIin the exterior of theγ contour, we getI₁ =Pn

k=1R_kwhere Rk = lim

z→a_r+k

1 +z^` z

n

Y

j=1, j6=k

(z−arj)⁻¹

!

= 1 +a^`_r+k a_r+k

n

Y

j=1, j6=k

(ar+k−arj)⁻¹.

On the other hand, evaluatingIin the interior of theγ contour, we obtain I₂ = lim

z→0

1 +z

An(z) = 1

An(0) = (−1)ⁿ Qn

j=1ar+j

.

Using Cauchy’s theorem on contour integrals we get that I₁ + I₂ = 0, as claimed.

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Theorem2.1 for a_n = F_n then Fibonacci number (F₀ = 0, F₁ = 1, and Fn+2 =Fn+1 +Fnfor alln ≥0) gives [4, Theorem 2.1], and foran =Lnthe n Lucas number (L₀ = 2,L₁ = 1, andL_n+2 = L_n+1+L_nfor alln≥ 0) gives [4, Theorem 2.2]. As another example, Theorem2.1 fora_n = P_n thenth Pell number (P0 = 0,P1 = 1, andPn+2 =Pn+1+Pnfor alln≥0) we get that

n

X

k=1

1 +P_r+k^` Pr+k

n

Y

j=1, j6=k

(P_r+k−P_r_j)⁻¹

!

= (−1)ⁿ⁺¹ Qn

j=1Pr+j

holds, with0≤` ≤n−1. In particular, we obtain Corollary 2.2. For alln≥2,

(P_n²+ 1)P_n+1P_n+2

(P_n+1−P_n)(P_n+2−P_n) + Pn(P_n+1² + 1)Pn+2

(P_n−P_n+1)(P_n+2−P_n+1)

+ P_nP_n+1(P_n+2² + 1)

(P_n−P_n+2)(P_n+1−P_n+2) = 1.

Theorem 2.3. Let(a_n)_n≥0 be any complex increasing sequence such thata_p 6=

a_qfor allp6=q. For alln≥2,

n

X

k=1

1 aⁿ⁻²_k

n

Y

j=1, j6=k

1− a_j

a_k

= 0.

Proof. Let us consider the integral

I = 1 2πi

I

γ

z A_n(z)dz,

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whereγ = {z ∈C : |z| <|a_n+1|}andA_n(z) = Qn

j=1(z −a_r+j). Evaluating the integralIin the exterior of theγ contour, we getI1 = 0. EvaluatingIin the interior of theγ contour, we obtain

I₂ =

n

X

k=1

Res(z/A_n(z);z =a_k)

=

n

X

k=1 n

Y

j=1, j6=k

ak

a_k−a_j

=

n

X

k=1

1 aⁿ⁻²_k

n

Y

j=1, j6=k

1− a_j

a_k

.

Using Cauchy’s theorem on contour integrals we get that I₁ + I₂ = 0, as claimed.

For example, Theorem 2.3 for a_n = L_n the nth Lucas number gives [4, Theorem 2.5]. As another example, Theorem 2.3 for a_n = P_n the nth Pell number obtains, for alln ≥2,

n

X

k=1

1 P_kⁿ⁻²

n

Y

j=1, j6=k

1− P_j

P_k

= 0.

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3. Inequalities

In this section we suggest some inequalities on positive increasing sequences.

Theorem 3.1. Let(an)n≥0 be any positive increasing sequence such thata1 ≥ 1. For alln ≥1,

(3.1) aâ_nⁿ⁺¹+aâ_n+1ⁿ < aâ_nⁿ+aâ_n+1ⁿ⁺¹. and

(3.2) aâ_n+1ⁿ⁺² −aâ_n+1ⁿ < aâ_n+2ⁿ⁺²−aâ_n+2ⁿ . Proof. To prove (3.1) we consider the integral

I = Z an+1

an

(a^x_n+1loga_n+1−a^x_nloga_n)dx.

Sincea_nsatisfies1≤a_n < a_n+1for alln ≥1, so for allx,a_n≤x ≤a_n+1we have that

a^x_nloga_n< a^x_n+1loga_n< a^x_n+1loga_n+1,

henceI >0. On the other hand, evaluating the integralI directly, we get that I = (aâ_n+1ⁿ⁺¹ −aâ_nⁿ⁺¹)−(aâ_n+1ⁿ −a^w_nⁿ),

hence

aâ_nⁿ⁺¹+aâ_n+1ⁿ < aâ_nⁿ +aâ_n+1ⁿ⁺¹

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as claimed in (3.1). To prove (3.2) we consider the integral J =

Z an+2

an

(a^x_n+2loga_n+2−a^x_n+1loga_n+1)dx.

Sincea_nsatisfies1≤a_n+1 < a_n+2for alln ≥0, so for allx,a_n+1 ≤x≤a_n+2 we have that

a^x_n+1loga_n+1 < a^x_n+2loga_n+2,

henceJ >0. On the other hand, evaluating the integralJ directly, we get that I = (aâ_n+2ⁿ⁺²−aâ_n+2ⁿ )−(aâ_n+1ⁿ⁺²−aâ_n+1ⁿ ),

hence

aâ_n+1ⁿ⁺²−aâ_n+1ⁿ < aâ_n+2ⁿ⁺²−aâ_n+2ⁿ 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 a_n = P_n the nth Pell number obtains, for alln ≥1,

P_n^Pⁿ⁺¹+P_n+1^Pⁿ < P_n^Pⁿ+P_n+1^Pⁿ⁺¹, wherePnis thenth Pell number.

Theorem 3.2. Let(a_n)n≥0 be any positive increasing sequence such thata₁ ≥ 1. For alln, m≥1,

a^a_n+mⁿ

m−1

Y

j=0

a^a_n+j^n+j+1 <

m

Y

j=0

a^a_n+j^n+j.

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Proof. Let us prove this theorem by induction on m. Since1≤ a_n < a_n+1for alln≥1thenaâ_nⁿ⁺¹^−aⁿ < aâ_n+1ⁿ⁺¹^−aⁿ, equivalently,aâ_nⁿ⁺¹aâ_n+1ⁿ < aâ_nⁿaâ_n+1ⁿ⁺¹, so the theorem holds form = 1. Now, assume for alln ≥1

a^a_n+m−1ⁿ

m−2

Y

j=0

a^a_n+j^n+j+1 <

m−1

Y

j=0

a^a_n+j^n+j.

On the other hand, similarly as in the casem = 1, for alln≥1, a^a_n+m−1^n+m^−aⁿ < a^a_n+m^n+m^−aⁿ.

Hence,

a^a_n+m−1^n+m^−aⁿa^a_n+m−1ⁿ

m−2

Y

j=0

a^a_n+j^n+j+1 < a^a_n+m^n+m^−aⁿ

m−1

Y

j=0

a^a_n+j^n+j,

equivalently,

a^a_n+mⁿ

m−1

Y

j=0

a^a_n+j^n+j+1 <

m

Y

j=0

a^a_n+j^n+j,

as claimed.

Theorem3.2foran =Lnthenth Lucas number andm = 3gives [4, Theo- rem 3.3].

Theorem 3.3. Let (a_n)n≥0 and (b_n)n≥0 be any two sequences such that 0 <

a_n< b_nfor alln≥1. Then for alln ≥1,

n

X

i=1

(b_j +a_j)≥ 2nⁿ⁺¹ (n+ 1)ⁿ

n

Y

i=1

b^1+1/n_j −a^1+1/n_j b_j −a_j .

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Proof. Using the AM-GM inequality, namely

1 n

n

X

i=1

xi ≥

n

Y

i=1

x^1/n_i ,

wherex_i >0for alli= 1,2, . . . , n, we get that Z a1

b1

· · · Z an

bn

1 n

n

X

i=1

x_idx₁· · ·dx_n≥ Z a1

b1

· · · Z an

bn

n

Y

i=1

x^1/n_i dx₁· · ·dx_n,

equivalently, 1 2n

n

X

i=1

(b²_i −a²_i)

n

Y

j=1, j6=i

(b_j −a_j)≥

n

Y

i=1

n

n+ 1(b^1+1/n_i −a^1+1/n_i )

,

hence, on simplifying the above inequality we get the desired result.

Theorem3.3fora_n=L⁻¹_n whereL_nis thenth Lucas number andb_n=F_n⁻¹ whereF_nis thenth Fibonacci number gives [4, Theorem 3.4].

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References

[1] A.T. BENJAMINANDJ.J. QUINN, Recounting Fibonacci and Lucas iden- tities, 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 Quar- terly, 40(1) (2002), 91.

[4] J.L. DíAZ-BARRERO, Rational identities and inequalities involving Fi- bonacci 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.