On the reciprocal sums of the generalized Fibonacci sequences

*Correspondence: sky.wzgfff@163.com Department of Mathematics, Northwest University, Xi’an, Shaanxi, P.R. China Abstract The Fibonacci sequence has been generalized in many ways. One of them is defined by the relation un = aun–1 + un–2 if n is even, un = bun–1 + un–2 if n is odd, with initial values u0 = 0 and u1 = 1, where a and b are positive integers. In this paper, we consider the reciprocal sum of un and then establish some identities relating to ‖(∑∞k=n 1 uk )–1‖, where ‖x‖ denotes the nearest integer to x. MSC: Primary 11B39


Introduction
For any integer n ≥ , the well-known Fibonacci sequence F n is defined by the secondorder linear recurrence sequence F n+ = F n+ + F n , where F  =  and F  = . The Fibonacci sequence has been generalized in many ways, for example, by changing the initial values, by changing the recurrence relation, and so on. Edson and Yayenie [] defined a further generalized Fibonacci sequence u n depending on two real parameters used in a non-linear recurrence relation, namely, u n = au n- + u n- if n is even and n ≥ , bu n- + u n- if n is odd and n ≥ , with initial values u  =  and u  = , where a, b are positive integers. This new sequence is actually a family of sequences where each new choice of a and b produces a distinct sequence. When a = b = , we have the classical Fibonacci sequence and when a = b = , we obtain the Pell numbers. Even further, if we set a = b = k for some positive integer k, we obtain the k-Fibonacci numbers. Various properties of the Fibonacci numbers and related sequences have been studied by many authors, see [-]. Recently, Ohtsuka and Nakamura [] studied the partial infinite sums of reciprocal Fibonacci numbers and proved that F n- if n is even and n ≥ , F n- - if n is odd and n ≥ , where x (the floor function) denotes the greatest integer less than or equal to x.
Some related works can also be found in [-]. In particular, in [], the authors studied a problem which is a little different from that of [], namely that of determining the ©2013 Zhang and Wu; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. http://www.advancesindifferenceequations.com/content/2013/1/377 nearest integer to ( ∞ k=n  v k ) - . Specifically, suppose that x = x +   (the nearest integer function), and {v n } n≥ is an integer sequence satisfying the recurrence formula Because the Fibonacci sequence has been generalized to a higher-order recursive sequence, any study on linear recursive sequences has little significance in this context, and we have to consider other non-linear recursive sequences. The main purpose of this paper is concerned with finding expressions for In fact, this problem is difficult because each item of this sequence relies on the previous relation. In order to resolve the question, we consider the reciprocal sums in two directions: on the one hand to the subsequence u pk+q and on the other to the product form u k u k+c+ , where p, q, c are non-negative integers and p ≥ . The results are as follows.
Theorem  Let {u n } be a second-order sequence defined by (). For any even p ≥  and non-negative integer q < p, there exists a positive integer n  such that

Proofs of the theorems
We need the following lemma.
Lemma (Generalized Binet's formula) The terms of the generalized Fibonacci sequence u n are given by Proof of Theorem  From the geometric series as → , we have From Lemma and the identity αβ = -ab, we have if q is even (so that pk + q is even), if q is odd (so that pk + q is odd).

Let
if q is odd. Thus, Hence, Thus, Taking the reciprocal of this expression yields Therefore, for any even p ≥  and integer  < q < p, there exists n ≥ n  sufficiently large such that the modulus of the last error term of identity () becomes less than /. This completes the proof of Theorem .
Proof of Theorem  In the first place, suppose that k ≥  is even. From Lemma we have The identities (αβ)  = a  b  + ab and αβ = -ab now yield Further, if k ≥  is odd, the same identity is similarly obtained. Thus, in both cases we have Hence, Taking the reciprocal of this expression yields ∞ k=n a k b k+c+ u k u k+c+ On the other hand, u n u n+c+ a n b n+c+ -u n- u n+c a n- b n+c = α c+ a c+ b c+ + a c b c+ · α ab n -α c+ a c+ b c+ + a c b c+ · α ab Combining () and (), finally we have ∞ k=n a k b k+c+ u k u k+c+ - -u n u n+c+ a n b n+c+ -u n- u n+c a n- b n+c = O  a n b n .
(  ) It follows that for any integer c ≥ , there exists n ≥ n  sufficiently large such that the modulus of the last error term of identity () becomes less than /. This completes the proof of Theorem . http://www.advancesindifferenceequations.com/content/2013/1/377