Skip to main content

On the reciprocal sums of higher-order sequences

Abstract

Let { u n } be a higher-order recursive sequence. Several identities are obtained for the infinite sums and finite sums of the reciprocals of higher-order recursive sequences.

MSC:11B39.

1 Introduction

The so-called Fibonacci zeta function and Lucas zeta function defined by

ζ F (s)= n = 1 1 F n s and ζ L (s)= n = 1 1 L n s ,

where the F n and L n denote the Fibonacci numbers and Lucas numbers, have been considered in several different ways. Navas [1] discussed the analytic continuation of these series. Elsner et al. [2] obtained that for any positive distinct integer s 1 , s 2 , s 3 , the numbers ζ F (2 s 1 ), ζ F (2 s 2 ), and ζ F (2 s 3 ) are algebraically independent if and only if at least one of s 1 , s 2 , s 3 is even.

Ohtsuka and Nakamura [3] studied the partial infinite sums of reciprocal Fibonacci numbers and proved the following conclusions:

( k = n 1 F k ) 1 = { F n 2 if  n  is even and  n 2 ; F n 2 1 if  n  is odd and  n 1 . ( k = n 1 F k 2 ) 1 = { F n 1 F n 1 if  n  is even and  n 2 ; F n 1 F n if  n  is odd and  n 1 .

Where denotes the floor function.

Further, Wu and Zhang [4, 5] generalized these identities to the Fibonacci polynomials and Lucas polynomials. Similar properties were also investigated in [68]. Related properties of the Fibonacci polynomials and Lucas polynomials can be found in [912].

Recently, some authors considered the nearest integer of the sums of reciprocal Fibonacci numbers and other famous sequences and obtained several new interesting identities, see [13] and [14]. Kilic and Arikan [15] defined a k th-order linear recursive sequence { u n } for any positive integer pq and n>k as follows:

u n =p u n 1 +q u n 2 + u n 3 ++ u n k ,

and they proved that there exists a positive integer n 0 such that

( k = n 1 u k ) 1 = u n u n 1 (n n 0 ),

where denotes the nearest integer. (Clearly, x=x+ 1 2 .)

In this paper, we unify the above results by proving some theorems that include all the results, [38] and [1315], as special cases. We consider the following type of higher-order recurrence sequences. For any positive integer a 1 , a 2 ,, a m , we define m th-order linear recursive sequences { u n } for n>m as follows:

u n = a 1 u n 1 + a 2 u n 2 ++ a m 1 u n m + 1 + a m u n m ,
(1)

with initial values u i N for 0i<m and at least one of them is not zero. If m=2, a 1 = a 2 =1, then u n = F n are the Fibonacci numbers. If m=2, a 1 =2, a 2 =1, then u n = P n are the Pell numbers. Our main results are the following.

Theorem 1 Let { u n } be an mth-order sequence defined by (1) with the restriction a 1 a 2 a m 1. For any positive real number β>2, there exists a positive integer n 1 such that

( k = n β n 1 u k ) 1 = u n u n 1 (n n 1 ).

Taking β+, from Theorem 1 we may immediately deduce the following.

Corollary 1 Let { u n } be an mth-order sequence defined by (1) with the restriction a 1 a 2 a m 1. Then there exists a positive integer n 2 such that

( k = n 1 u k ) 1 = u n u n 1 (n n 2 ).

For a positive real number 1<β2, whether there exits an identity for

( k = n β n 1 u k ) 1

is an interesting open problem.

2 Several lemmas

To complete the proof of our theorem, we need the following.

Lemma 1 Let a 1 , a 2 ,, a m be positive integers with a 1 a 2 a m 1 and mN with m2. Then, for the polynomial

f(x)= x m a 1 x m 1 a 2 x m 2 a m 1 x a m ,

we have

  1. (I)

    Polynomial f(x) has exactly one positive real zero α with a 1 <α< a 1 +1.

  2. (II)

    Other m1 zeros of f(x) lie within the unit circle in the complex plane.

Proof For any positive integer a 1 a 2 a m 1 and m2, we have

f ( a 1 ) = a 1 m a 1 m a 2 a 1 m 2 a m 1 a 1 a m = a 2 a 1 m 2 a m 1 a 1 a m < 0 ,

and

f ( a 1 + 1 ) = ( a 1 + 1 ) m a 1 ( a 1 + 1 ) m 1 a 2 ( a 1 + 1 ) m 2 a m > ( a 1 + 1 ) m a 1 ( ( a 1 + 1 ) m 1 + ( a 1 + 1 ) m 2 + + 1 ) = ( a 1 + 1 ) m a 1 ( a 1 + 1 ) m 1 a 1 = 1 > 0 .

Thus there exits a positive real zero α of f(x) with a 1 <α< a 1 +1. According to Descarte’s rule of signs, f(x)=0 has at most one positive real root. So, f(x) has exactly one positive real zero α with a 1 <α< a 1 +1. This completes the proof of (I) in Lemma 1.

Observe that from (I) in Lemma 1 we have

if xR such that x>α, then f(x)>0,
(2)
if xR such that 0<x<α, then f(x)<0.
(3)

Let

g ( x ) = ( x 1 ) f ( x ) = x m + 1 ( a 1 + 1 ) x m + ( a 1 a 2 ) x m 1 + ( a 2 a 3 ) x m 2 + + ( a m 1 a m ) x + a m .

Since f(x) has exactly one positive real zero α, g(x) has two positive real zeros α and 1. Observe that

if xR such that x>α, then g(x)>0,
(4)
if xR such that 1<x<α, then g(x)<0.
(5)

To complete the proof of (II) in Lemma 1, it is sufficient to show that there is no zero on and outside of the unit circle. □

Claim 1 f(x) has no complex zero z 1 with | z 1 |>α.

Proof Assume that there exits such a zero. So, we have

f( z 1 )= z 1 m a 1 z 1 m 1 a 2 z 1 m 2 a m 1 z 1 a m =0,

then we obtain

| z 1 m | a 1 | z 1 m 1 | + a 2 | z 1 m 2 | + + a m 1 | z 1 | + a m , f ( | z 1 | ) = | z 1 m | a 1 | z 1 m 1 | a 2 | z 1 m 2 | a m 1 | z 1 | a m 0 .

This contradicts with (2). □

Claim 2 f(x) has no complex zero z 2 with 1<| z 2 |<α.

Proof Assume that there exits such a zero. Since f( z 2 )=0,

g( z 2 )= z 2 m + 1 ( a 1 +1) z 2 m +( a 1 a 2 ) z 2 m 1 ++( a m 1 a m ) z 2 + a m =0,

then we obtain

( a 1 +1)| z 2 | m | z 2 | m + 1 +( a 1 a 2 )| z 2 | m 1 ++( a m 1 a m )| z 2 |+ a m .

So, we have g(| z 2 |)0, which contradicts with (5). □

Claim 3 On the circle | z 3 |=α and | z 3 |=1, f(x) has the unique zero α.

Proof If f( z 3 )=0, then

g( z 3 )= z 3 m + 1 ( a 1 +1) z 3 m +( a 1 a 2 ) z 3 m 1 ++( a m 1 a m ) z 3 + a m =0,

then we obtain

( a 1 +1)| z 3 | m | z 3 | m + 1 +( a 1 a 2 )| z 3 | m 1 ++( a m 1 a m )| z 3 |+ a m .
(6)

If z 3 =α or z 3 =1, then g( z 3 )=0, so (6) must be an equality. Therefore, z 3 m + 1 ,( a 1 a 2 ) z 3 m 1 ,( a 2 a 3 ) z 3 m 2 ,,( a m 1 a m ) z 3 and a m all lie on the same ray issuing from the origin. Since ( a 1 a 2 ),( a 2 a 3 ),, a m , are all the elements of R + , z 3 m + 1 , z 3 m 1 , z 3 m 2 ,, z 3 must be the elements of R + . Therefore we obtain f( z 3 ) R + . On the circle | z 3 |=α and | z 3 |=1, there are two conditions z 3 =1 or z 3 =α. Since f(1)0, α is the unique zero of f(x), Claim 3 holds.

From the three claims, (II) in Lemma 1 is proven. □

Lemma 2 Let m2 and let { u n } n 0 be an integer sequence satisfying the recurrence formula (1). Then the closed formula of u n is given by

u n =c α n +O ( d n ) (n),

where c>0, d>1, and a 1 <α< a 1 +1 is the positive real zero of f(x).

Proof Let α, α 1 ,, α t be the distinct roots of f(x)=0, where f(x)=0 is the characteristic equation of the recurrence formula (1). From Lemma 1 we know that α is the simple root of f(x)=0 , then let r j , for j=1,2,,t, denote the multiplicity of the root α j . From the properties of m th-order linear recursive sequences, u n can be expressed as follows:

u n =c α n + i = 1 t P i (n) α i n ,
(7)

where

P i (n)R[n],deg P i (n)= r i 1, r 1 + r 2 ++ r t =m1,andcR.

For example, for positive integers 1u,v,wt, if α u is the simple root of f(x), then P u (n)= g 1 , where g 1 R, and deg P u (n)=0; if α v is the double root of f(x), then P v (n)= g 2 n+ g 3 , where g 2 , g 3 R, and deg P v (n)=1; if α w is the multiple root of f(x) with the multiplicity r w , then P w (n)= b 1 n r w 1 + b 2 n r w 2 ++ b r w 1 n+ b r w , where b 1 , b 2 ,, b r w R, and deg P w (n)= r w 1.

From Lemma 1 we have | α i |<1 for 1it. Since each term of tail in (7) goes to 0 as n, we can find the constant MR and dR with d>1 for n> n 0 such that

| i = 1 t P i (n) α i n | i = 1 t | P i (n) α i n |M d n ,

which completes the proof (note that if all the roots of f(x) are distinct, we can choose d 1 =max{| α 1 |,| α 2 |,,| α m 1 |} and M=m1). □

3 Proof of Theorem 1

In this section, we shall complete the proof of Theorem 1. From the geometric series as ϵ0, we have

1 1 ± ϵ =1ϵ+O ( ϵ 2 ) =1+O(ϵ).

Using Lemma 2, we have

1 u k = 1 c α k + O ( d k ) = 1 c α k ( 1 + O ( ( α d ) k ) ) = 1 c α k ( 1 + O ( ( α d ) k ) ) = 1 c α k + O ( ( α 2 d ) k ) .
(8)

Thus

k = n β n 1 u k = 1 c k = n β n 1 α k + O ( k = n β n ( α 2 d ) k ) = α c ( α 1 ) α n 1 c ( α 1 ) α β n + O ( α 2 n d n ) = α c ( α 1 ) α n + O ( α 2 n α β n + 2 n ) + O ( α 2 n d n ) = α c ( α 1 ) α n + O ( α 2 n h ) ,

where h=max{ α β n + 2 n , d n }.

Taking reciprocal, we get

( k = n β n 1 u k ) 1 = 1 α c ( α 1 ) α n ( 1 + O ( α n h ) ) = α 1 α c α n ( 1 + O ( α n h ) ) = α 1 α c α n + O ( h ) = u n u n 1 + O ( h ) .

Since h=max{ α β n + 2 n , d n }<1, there exists n n 1 sufficient large so that the modulus of the last error term becomes less than 1/2, which completes the proof.

Proof of Corollary 1 From identity (8), we have

1 u k = 1 c α k +O ( ( α 2 d ) k ) .

Thus

k = n 1 u k = 1 c k = n 1 α k +O ( k = n ( α 2 d ) k ) = α c ( α 1 ) α n +O ( ( α 2 d ) n ) .

Taking reciprocal, we get

( k = n 1 u k ) 1 = 1 α c ( α 1 ) α n ( 1 + O ( ( α d ) n ) ) = α 1 α c α n ( 1 + O ( ( α d ) n ) ) = α 1 α c α n + O ( d n ) = u n u n 1 + O ( d n ) .

So, there exists n n 2 sufficiently large so that the modulus of the last error term becomes less than 1/2, which completes the proof. □

4 Related results

The following results are obtained similarly.

Theorem 2 Let { u n } be an mth-order sequence defined by (1) with the restriction a 1 a 2 a m 1. Let p and q be positive integers with 0q<p. For any real number β>2, there exist positive integers n 3 , n 4 and n 5 depending on a 1 , a 2 , , and a m such that

(a) ( k = n β n ( 1 ) k u k ) 1 = ( 1 ) n ( u n + u n 1 ) ( n n 3 ) , (b) ( k = n β n 1 u p k + q ) 1 = u p n + q u p n p + q ( n n 4 ) , (c) ( k = n β n ( 1 ) k u p k + q ) 1 = ( 1 ) n ( u p n + q + u p n p + q ) ( n n 5 ) .

For β+, we deduce the following identity of infinite sum as a special case of Theorem  2.

Corollary 2 Let { u n } be an mth-order sequence defined by (1) with the restriction a 1 a 2 a m 1. Let p and q be positive integers with 0q<p. Then there exist positive integers n 6 , n 7 and n 8 depending on a 1 , a 2 , , and a m such that

(e) ( k = n ( 1 ) k u k ) 1 = ( 1 ) n ( u n + u n 1 ) ( n n 6 ) , (f) ( k = n 1 u p k + q ) 1 = u p n + q u p n p + q ( n n 7 ) , (g) ( k = n ( 1 ) k u p k + q ) 1 = ( 1 ) n ( u p n + q + u p n p + q ) ( n n 8 ) .

Proof We shall prove only (c) in Theorem 2 and other identities are proved similarly. From Lemma 2 we have

( 1 ) k u p k + q = ( 1 ) k c α p k + q + O ( d p k q ) = ( 1 ) k c α p k + q ( 1 + O ( ( α d ) p k q ) ) .

Thus

k = n β n ( 1 ) k u p k + q = ( 1 ) n α p c α p n + q ( α p + 1 ) + ( 1 ) n α p c α p β n + q ( α p + 1 ) + O ( ( α 2 d ) p n q ) = ( 1 ) n α p c α p n + q ( α p + 1 ) + O ( α p β n q ) + O ( α 2 p n 2 q d p n q ) = ( 1 ) n α p c α p n + q ( α p + 1 ) + O ( α 2 p n α p β n + 2 p n ) + O ( α 2 p n d p n ) = ( 1 ) n α p c α p n + q ( α p + 1 ) + O ( α 2 p n h p ) ,

where h=max{ α β n + 2 n , d n }.

Taking reciprocal, we get

( k = n β n ( 1 ) k u p k + q ) 1 = ( 1 ) n ( c α p n + q + c α p n p + q ) ( 1 + O ( α p n h p ) ) = ( 1 ) n ( u p n + q + u p n p + q ) + O ( h p ) .

Since h=max{ α β n + 2 n , d n }<1, there exists n n 5 sufficiently large so that the modulus of the last error term becomes less than 1/2, which completes the proof. □

References

  1. 1.

    Navas L: Analytic continuation of the Fibonacci Dirichlet series. Fibonacci Q. 2001, 39: 409-418.

    MathSciNet  Google Scholar 

  2. 2.

    Elsner C, Shimomura S, Shiokawa I: Algebraic relations for reciprocal sums of odd sums of Fibonacci numbers. Acta Arith. 2011, 148(3):205-223. 10.4064/aa148-3-1

    MathSciNet  Article  Google Scholar 

  3. 3.

    Ohtsuka H, Nakamura S: On the sum of reciprocal Fibonacci numbers. Fibonacci Q. 2008/2009, 46/47: 153-159.

    Google Scholar 

  4. 4.

    Wu Z, Zhang W: The sums of the reciprocal of Fibonacci polynomials and Lucas polynomials. J. Inequal. Appl. 2012, 134: 1-8.

    Google Scholar 

  5. 5.

    Wu Z, Zhang W: Several identities involving Fibonacci polynomials and Lucas polynomials. J. Inequal. Appl. 2013, 205: 1-14.

    Google Scholar 

  6. 6.

    Holliday S, Komatsu T: On the sum of reciprocal generalized Fibonacci numbers. Integers 2011, 11: 441-455.

    MathSciNet  Article  Google Scholar 

  7. 7.

    Zhang G: The infinite sum of reciprocal of the Fibonacci numbers. J. Math. Res. Expo. 2011, 31: 1030-1034.

    Google Scholar 

  8. 8.

    Zhang W, Wang T: The infinite sum of reciprocal Pell numbers. Appl. Math. Comput. 2012, 218: 6164-6167. 10.1016/j.amc.2011.11.090

    MathSciNet  Article  Google Scholar 

  9. 9.

    Falcón S, Plaza Á: On the Fibonacci k-numbers. Chaos Solitons Fractals 2007, 32: 1615-1624. 10.1016/j.chaos.2006.09.022

    MathSciNet  Article  Google Scholar 

  10. 10.

    Ma R, Zhang W: Several identities involving the Fibonacci numbers and Lucas numbers. Fibonacci Q. 2007, 45: 164-170.

    Google Scholar 

  11. 11.

    Wang T, Zhang W: Some identities involving Fibonacci, Lucas polynomials and their applications. Bull. Math. Soc. Sci. Math. Roum. 2012, 55: 95-103.

    Google Scholar 

  12. 12.

    Yi Y, Zhang W: Some identities involving the Fibonacci polynomials. Fibonacci Q. 2002, 40: 314-318.

    MathSciNet  Google Scholar 

  13. 13.

    Komatsu T: On the nearest integer of the sum of reciprocal Fibonacci numbers. Aportaciones Matematicas Investigacion 2011, 20: 171-184.

    MathSciNet  Google Scholar 

  14. 14.

    Komatsu T, Laohakosol V: On the sum of reciprocals of numbers satisfying a recurrence relation of order s. J. Integer Seq. 2010., 13: Article ID 10.5.8

    Google Scholar 

  15. 15.

    Kilic E, Arikan T: More on the infinite sum of reciprocal usual Fibonacci, Pell and higher order recurrences. Appl. Math. Comput. 2013, 219: 7783-7788. 10.1016/j.amc.2013.02.003

    MathSciNet  Article  Google Scholar 

Download references

Acknowledgements

The authors express their gratitude to the referee for very helpful and detailed comments. This work is supported by the N.S.F. (11071194, 11001218) of P.R. China and G.I.C.F. (YZZ12062) of NWU.

Author information

Affiliations

Authors

Corresponding author

Correspondence to Zhengang Wu.

Additional information

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

ZW obtained the theorems and completed the proof. HZ corrected and improved the final version. Both authors read and approved the final manuscript.

Rights and permissions

Open Access This article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Reprints and Permissions

About this article

Cite this article

Wu, Z., Zhang, H. On the reciprocal sums of higher-order sequences. Adv Differ Equ 2013, 189 (2013). https://doi.org/10.1186/1687-1847-2013-189

Download citation

Keywords

  • infinite sums
  • finite sums
  • reciprocal
  • higher-order recurrences