Recently, in [4] the author studied some rational identities and inequalities involv- ing Fibonacci and Lucas numbers

http://jipam.vu.edu.au/

Volume 5, Issue 3, Article 75, 2004

RATIONAL IDENTITIES AND INEQUALITIES

TOUFIK MANSOUR DEPARTMENT OFMATHEMATICS

UNIVERSITY OFHAIFA

31905 HAIFA, ISRAEL. toufik@math.haifa.ac.il

Received 25 December, 2003; accepted 27 June, 2004 Communicated by L. Tóth

ABSTRACT. Recently, in [4] the author studied some rational identities and inequalities involv- ing Fibonacci and Lucas numbers. In this paper we generalize these rational identities and in- equalities to involve a wide class of sequences.

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

2000 Mathematics Subject Classification. 05A19, 11B39.

1. INTRODUCTION

The Fibonacci and Lucas sequences are a source of many interesting identities and inequali- ties. 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 ra- tional identities and inequalities involving Fibonacci and Lucas numbers. A sequence(a_n)n≥0

is said to be positive 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.

2. IDENTITIES

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

Theorem 2.1. Let(an)n≥0 be any complex increasing sequence such thatap 6=aqfor 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_rj)⁻¹

!

= (−1)ⁿ⁺¹ Qn

j=1a_r+j

ISSN (electronic): 1443-5756

c 2004 Victoria University. All rights reserved.

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

001-04

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 integralI in the exterior of theγ contour, we getI₁ =Pn

k=1R_kwhere R_k= lim

z→a_r+k

1 +z^` z

n

Y

j=1, j6=k

(z−a_rj)⁻¹

!

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

n

Y

j=1, j6=k

(a_r+k−a_rj)⁻¹.

On the other hand, evaluatingI in the interior of theγcontour, we obtain I₂ = lim

z→0

1 +z

A_n(z) = 1

A_n(0) = (−1)ⁿ Qn

j=1a_r+j.

Using Cauchy’s theorem on contour integrals we get thatI₁+I₂ = 0, as claimed.

Theorem 2.1 foran =Fnthen Fibonacci number (F0 = 0,F1 = 1, andFn+2 =Fn+1+Fn

for alln ≥ 0) gives [4, Theorem 2.1], and fora_n = L_nthenLucas number (L₀ = 2, L₁ = 1, andL_n+2 =L_n+1+L_nfor alln≥0) gives [4, Theorem 2.2]. As another example, Theorem 2.1 fora_n =P_nthenth Pell number (P₀ = 0,P₁ = 1, andP_n+2 =P_n+1+P_nfor 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_rj)⁻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) + P_n(P_n+1² + 1)P_n+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, whereγ ={z ∈C : |z| <|a_n+1|}andA_n(z) = Qn

j=1(z−a_r+j). Evaluating the integralI in the exterior of theγ contour, we getI₁ = 0. EvaluatingI in 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

a_k 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 thatI₁+I₂ = 0, as claimed.

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

n

X

k=1

1 P_kⁿ⁻²

n

Y

j=1, j6=k

1− P_j

Pk

= 0.

3. INEQUALITIES

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

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

(3.1) a^a_nⁿ⁺¹+a^a_n+1ⁿ < a^a_nⁿ +a^a_n+1ⁿ⁺¹. and

(3.2) a^a_n+1ⁿ⁺²−a^a_n+1ⁿ < a^a_n+2ⁿ⁺² −a^a_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 integralIdirectly, we get that I = (a^a_n+1ⁿ⁺¹−a^a_nⁿ⁺¹)−(a^a_n+1ⁿ −a^w_nⁿ),

hence

a^a_nⁿ⁺¹ +a^a_n+1ⁿ < a^a_nⁿ+a^a_n+1ⁿ⁺¹ 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+1logan+1 < a^x_n+2logan+2,

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

hence

a^a_n+1ⁿ⁺² −a^a_n+1ⁿ < a^a_n+2ⁿ⁺²−a^a_n+2ⁿ

as claimed in (3.2).

For example, Theorem 3.1 for a_n = L_n thenth Lucas number gives [4, Theorem 3.1]. As another example, Theorem 3.1 fora_n =P_nthenth Pell number obtains, for alln ≥1,

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

Theorem 3.2. Let (a_n)n≥0 be any positive increasing sequence such that a₁ ≥ 1. For all n, 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.

Proof. Let us prove this theorem by induction onm. Since1 ≤ a_n < a_n+1 for alln ≥ 1then a^a_n^n+1−an < a^a_n+1^n+1−an, equivalently, a^a_nⁿ⁺¹a^a_n+1ⁿ < a^a_nⁿa^a_n+1ⁿ⁺¹, so the theorem holds for m = 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.

Theorem 3.2 fora_n=L_nthenth Lucas number andm= 3gives [4, Theorem 3.3].

Theorem 3.3. Let (an)n≥0 and(bn)n≥0 be any two sequences such that 0 < an < bn for all n≥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 .

Proof. Using the AM-GM inequality, namely 1 n

n

X

i=1

x_i ≥

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.

Theorem 3.3 fora_n =L⁻¹_n whereL_nis thenth Lucas number andb_n =F_n⁻¹ whereF_nis the nth Fibonacci number gives [4, Theorem 3.4].

REFERENCES

[1] A.T. BENJAMINAND J.J. QUINN, Recounting Fibonacci and Lucas identities, 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 Quarterly, 40(1) (2002), 91.

[4] J.L. DíAZ-BARRERO, Rational identities and inequalities involving Fibonacci and Lucas numbers, J. Inequal. in Pure and Appl. Math., 4(5) (2003), Art. 83. [ONLINEhttp://jipam.vu.edu.

au/article.php?sid=324]

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