On the reciprocal sums of the generalized Fibonacci sequences

Abstract

The Fibonacci sequence has been generalized in many ways. One of them is defined by the relation $u_n=a{u}_{n-1}+u_{n-2}$ if n is even, $u_n=b{u}_{n-1}+u_{n-2}$ if n is odd, with initial values $u_0=0$ and $u_1=1$, where a and b are positive integers. In this paper, we consider the reciprocal sum of $u_n$ and then establish some identities relating to $\parallel {({\sum }_{k=n}^{\mathrm{\infty }}\frac{1}{u_k})}^{-1}\parallel$, where $\parallel x\parallel$ denotes the nearest integer to x.

MSC:11B39.

1 Introduction

For any integer $n\ge 0$, the well-known Fibonacci sequence $F_n$ is defined by the second-order linear recurrence sequence $F_{n+2}=F_{n+1}+F_n$, where $F_0=0$ and $F_1=1$. 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 [1] defined a further generalized Fibonacci sequence $u_n$ depending on two real parameters used in a non-linear recurrence relation, namely,

$$u_n=\{\begin{array}{ll}a{u}_{n-1}+u_{n-2}& \text{if }n\text{ is even and }n\ge 2,\\ b{u}_{n-1}+u_{n-2}& \text{if }n\text{ is odd and }n\ge 1,\end{array}$$

(1)

with initial values $u_0=0$ and $u_1=1$, 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 $a=b=1$, we have the classical Fibonacci sequence and when $a=b=2$, 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 [2–4]. Recently, Ohtsuka and Nakamura [5] studied the partial infinite sums of reciprocal Fibonacci numbers and proved that

$$\lfloor {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{F_k}\right)}^{-1}\rfloor =\{\begin{array}{ll}{F}_{n-2}& \text{if }n\text{ is even and }n\ge 2,\\ F_{n-2}-1& \text{if }n\text{ is odd and }n\ge 1,\end{array}$$

where $\lfloor x\rfloor$ (the floor function) denotes the greatest integer less than or equal to x.

Some related works can also be found in [6–13]. In particular, in [13], the authors studied a problem which is a little different from that of [5], namely that of determining the nearest integer to ${({\sum }_{k=n}^{\mathrm{\infty }}\frac{1}{v_k})}^{-1}$. Specifically, suppose that $\parallel x\parallel =\lfloor x+\frac{1}{2}\rfloor$ (the nearest integer function), and ${\{v_n\}}_{n\ge 0}$ is an integer sequence satisfying the recurrence formula

$$v_n=a_1{v}_{n-1}+a_2{v}_{n-2}+\cdots +a_s{v}_{n-s}(s\ge 2)$$

for any positive integer $a_1,a_2,\dots ,a_m$, with the initial conditions $v_0\ge 0$, $v_k\in \mathbf{N}$, $1\le k\le s-1$. Then, provided $a_1\ge a_2\ge \cdots \ge a_m\ge 1$, we can conclude that there exists a positive integer $n_0$ such that

$$\parallel {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{v_k}\right)}^{-1}\parallel =v_n-v_{n-1}$$

for all $n>n_0$.

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

$$\parallel {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{u_k}\right)}^{-1}\parallel .$$

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 $u_k{u}_{k+2c+1}$, where p, q, c are non-negative integers and $p\ge 2$. The results are as follows.

Theorem 1 Let $\{u_n\}$ be a second-order sequence defined by (1). For any even $p\ge 2$ and non-negative integer $q<p$, there exists a positive integer $n_1$ such that

$$\parallel {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{u_{pk+q}}\right)}^{-1}\parallel =u_{pn+q}-u_{pn-p+q}$$

for all $n\ge n_1$.

Theorem 2 Let $\{u_n\}$ be a second-order sequence defined by (1). For any integer $c\ge 0$, there exists a positive integer $n_2$ such that

$$\parallel {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{a^k{b}^{k+2c+1}}{u_k{u}_{k+2c+1}}\right)}^{-1}-(\frac{u_n{u}_{n+2c+1}}{a^n{b}^{n+2c+1}}-\frac{u_{n-1}{u}_{n+2c}}{a^{n-1}{b}^{n+2c}})\parallel =0$$

for all $n\ge n_2$.

Open problem In the light of our investigation, for any positive integer $s\ge 2$ and l, whether there exist identities for

$${\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{u_k^s}\right)}^{-1}\text{and}{\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{u_k{u}_{k+l}}\right)}^{-1}$$

represent two interesting, albeit challenging, open problems.

2 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

$$u_n=\frac{a^{2\lfloor \frac{n}{2}\rfloor -n+1}}{{(ab)}^{\lfloor \frac{n}{2}\rfloor }}\cdot \frac{{\alpha }^n-{\beta }^n}{\alpha -\beta },$$

where $\alpha =\frac{ab+\sqrt{a^2{b}^2+4ab}}{2}$, $\beta =\frac{ab-\sqrt{a^2{b}^2+4ab}}{2}$.

Proof See Theorem 2 of [4]. □

Proof of Theorem 1 From the geometric series as $ϵ\to 0$, we have

$$\frac{1}{1\pm ϵ}=1\mp ϵ+O\left({ϵ}^2\right)=1+O(ϵ).$$

From Lemma and the identity $\alpha \beta =-ab$, we have

$$u_{pk+q}=\{\begin{array}{ll}\frac{{\alpha }^{pk+q}-{\beta }^{pk+q}}{a^{\frac{pk+q-2}{2}}{b}^{\frac{pk+q}{2}}(\alpha -\beta )}& \text{if }q\text{ is even }(\text{so that }pk+q\text{ is even}),\\ \frac{{\alpha }^{pk+q}-{\beta }^{pk+q}}{a{b}^{\frac{pk+q-1}{2}}(\alpha -\beta )}& \text{if }q\text{ is odd }(\text{so that }pk+q\text{ is odd}).\end{array}$$

Let

$$A=\{\begin{array}{ll}\frac{1}{a^{\frac{pk+q-2}{2}}{b}^{\frac{pk+q}{2}}(\alpha -\beta )}& \text{if }q\text{ is even},\\ \frac{1}{a{b}^{\frac{pk+q-1}{2}}(\alpha -\beta )}& \text{if }q\text{ is odd}.\end{array}$$

Thus,

$$u_{pk+q}=A{\alpha }^{pk+q}+O\left(\frac{{|\beta |}^{\frac{pk}{2}}}{{\alpha }^{\frac{pk}{2}}}\right).$$

Hence,

$$\begin{array}{rl}\frac{1}{u_{pk+q}}& =\frac{1}{A{\alpha }^{pk+q}(1+O(\frac{{|\beta |}^{\frac{pk}{2}}}{{\alpha }^{\frac{3pk}{2}}}))}=\frac{1}{A{\alpha }^{pk+q}}(1+O\left(\frac{{|\beta |}^{\frac{pk}{2}}}{{\alpha }^{\frac{3pk}{2}}}\right))\\ =\frac{1}{A{\alpha }^{pk+q}}+O\left(\frac{{|\beta |}^{\frac{pk}{2}}}{{\alpha }^{\frac{5pk}{2}}}\right).\end{array}$$

Thus,

$$\sum _{k=n}^{\mathrm{\infty }}\frac{1}{u_{pk+q}}=\frac{\frac{1}{{\alpha }^{pn+q}}}{A(1-\frac{1}{{\alpha }^p})}+O\left(\frac{{|\beta |}^{\frac{pn}{2}}}{{\alpha }^{\frac{5pn}{2}}}\right)=\frac{{\alpha }^p}{A{\alpha }^{pn+q}({\alpha }^p-1)}(1+O\left(\frac{{|\beta |}^{\frac{pn}{2}}}{{\alpha }^{\frac{3pn}{2}}}\right)).$$

Taking the reciprocal of this expression yields

$$\begin{array}{rl}{\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{u_{pk+q}}\right)}^{-1}& =\frac{A{\alpha }^{pn+q}({\alpha }^p-1)}{{\alpha }^p}+O\left(\frac{{|\beta |}^{\frac{pn}{2}}}{{\alpha }^{\frac{pn}{2}}}\right)\\ =A{\alpha }^{pn+q}-A{\alpha }^{pn-p+q}+O\left(\frac{{|\beta |}^{\frac{pn}{2}}}{{\alpha }^{\frac{pn}{2}}}\right)\\ =u_{pn+q}-u_{pn-p+q}+O\left(\frac{{|\beta |}^{\frac{pn}{2}}}{{\alpha }^{\frac{pn}{2}}}\right).\end{array}$$

(2)

Therefore, for any even $p\ge 2$ and integer $0<q<p$, there exists $n\ge n_1$ sufficiently large such that the modulus of the last error term of identity (2) becomes less than $1/2$. This completes the proof of Theorem 1. □

Proof of Theorem 2 In the first place, suppose that $k\ge 2$ is even. From Lemma we have

$$u_k=\frac{1}{a^{\frac{k}{2}-1}{b}^{\frac{k}{2}}}\cdot \frac{{\alpha }^k-{\beta }^k}{\alpha -\beta },$$

and

$$u_{k+2c+1}=\frac{1}{a^{\frac{k}{2}+c}{b}^{\frac{k}{2}+c}}\cdot \frac{{\alpha }^{k+2c+1}-{\beta }^{k+2c+1}}{\alpha -\beta }.$$

The identities ${(\alpha -\beta )}^2=a^2{b}^2+4ab$ and $\alpha \beta =-ab$ now yield

$$\begin{array}{rl}{u}_k{u}_{k+2c+1}& =\frac{{\alpha }^{2k+2c+1}+{\beta }^{2k+2c+1}-{(\alpha \beta )}^k({\alpha }^{2c+1}+{\beta }^{2c+1})}{a^{k+c-1}{b}^{k+c}{(\alpha -\beta )}^2}\\ =\frac{{\alpha }^{2k+2c+1}}{(a{b}^2+4b){(ab)}^{k+c}}+O\left({\left(\frac{\alpha \beta }{ab}\right)}^k\right)\\ =\frac{{\alpha }^{2k+2c+1}}{(a{b}^2+4b){(ab)}^{k+c}}+O(1).\end{array}$$

Further, if $k\ge 1$ is odd, the same identity is similarly obtained. Thus, in both cases we have

$$\begin{array}{rl}\frac{1}{u_k{u}_{k+2c+1}}& =\frac{1}{\frac{{\alpha }^{2k+2c+1}}{(a{b}^2+4b){(ab)}^{k+c}}+O(1)}=\frac{1}{\frac{{\alpha }^{2k+2c+1}}{(a{b}^2+4b){(ab)}^{k+c}}(1+O(\frac{{(ab)}^k}{{\alpha }^{2k}}))}\\ =\frac{(a{b}^2+4b){(ab)}^{k+c}}{{\alpha }^{2k+2c+1}}+O\left(\frac{{(ab)}^{2k}}{{\alpha }^{4k}}\right).\end{array}$$

Hence,

$$\begin{array}{rl}\frac{a^k{b}^{k+2c+1}}{u_k{u}_{k+2c+1}}& =\frac{(a{b}^3+4b^2)a^{2k+c}{b}^{2k+3c}}{{\alpha }^{2k+2c+1}}+O\left(\frac{{(ab)}^{3k}}{{\alpha }^{4k}}\right)\\ =\frac{a^{c+1}{b}^{3c+3}+4a^c{b}^{3c+2}}{{\alpha }^{2c+1}}\cdot \frac{{(ab)}^{2k}}{{\alpha }^{2k}}+O\left(\frac{{(ab)}^{3k}}{{\alpha }^{4k}}\right).\end{array}$$

Let $B=\frac{a^{c+1}{b}^{3c+3}+4a^c{b}^{3c+2}}{{\alpha }^{2c+1}}$, then

$$\frac{a^k{b}^{k+2c+1}}{u_k{u}_{k+2c+1}}=\frac{B{(ab)}^{2k}}{{\alpha }^{2k}}+O\left(\frac{{(ab)}^{3k}}{{\alpha }^{4k}}\right).$$

Consequently,

$$\begin{array}{rl}\sum _{k=n}^{\mathrm{\infty }}\frac{a^k{b}^{k+2c+1}}{u_k{u}_{k+2c+1}}& =B\sum _{k=n}^{\mathrm{\infty }}\frac{{(ab)}^{2k}}{{\alpha }^{2k}}+O\left(\sum _{k=n}^{\mathrm{\infty }}\frac{{(ab)}^{3k}}{{\alpha }^{4k}}\right)\\ =B\cdot \frac{{(\frac{ab}{\alpha })}^{2n}}{1-{(\frac{ab}{\alpha })}^2}+O\left(\frac{{(ab)}^{3n}}{{\alpha }^{4n}}\right).\end{array}$$

Taking the reciprocal of this expression yields

$$\begin{array}{rl}{\left(\sum _{k=n}^{\mathrm{\infty }}\frac{a^k{b}^{k+2c+1}}{u_k{u}_{k+2c+1}}\right)}^{-1}& =\frac{1}{B\cdot \frac{{(\frac{ab}{\alpha })}^{2n}}{1-{(\frac{ab}{\alpha })}^2}(1+O(\frac{{|\beta |}^n}{{\alpha }^n}))}\\ =\frac{1-{(\frac{ab}{\alpha })}^2}{B\cdot {(\frac{ab}{\alpha })}^{2n}}(1+O\left(\frac{{|\beta |}^n}{{\alpha }^n}\right))\\ =\frac{1-{(\frac{ab}{\alpha })}^2}{B\cdot {(\frac{ab}{\alpha })}^{2n}}+O\left(\frac{1}{{\alpha }^n{|\beta |}^n}\right)\\ =\frac{1}{B}\cdot {\left(\frac{\alpha }{ab}\right)}^{2n}-\frac{1}{B}\cdot {\left(\frac{\alpha }{ab}\right)}^{2n-2}+O\left(\frac{1}{{\alpha }^n{|\beta |}^n}\right).\end{array}$$

(3)

On the other hand,

$$\begin{array}{r}\frac{u_n{u}_{n+2c+1}}{a^n{b}^{n+2c+1}}-\frac{u_{n-1}{u}_{n+2c}}{a^{n-1}{b}^{n+2c}}\\ =\frac{{\alpha }^{2c+1}}{a^{c+1}{b}^{3c+3}+4a^c{b}^{3c+2}}\cdot {\left(\frac{\alpha }{ab}\right)}^{2n}-\frac{{\alpha }^{2c+1}}{a^{c+1}{b}^{3c+3}+4a^c{b}^{3c+2}}\cdot {\left(\frac{\alpha }{ab}\right)}^{2n-2}+O\left(\frac{1}{a^n{b}^n}\right)\\ =\frac{1}{B}\cdot {\left(\frac{\alpha }{ab}\right)}^{2n}-\frac{1}{B}\cdot {\left(\frac{\alpha }{ab}\right)}^{2n-2}+O\left(\frac{1}{a^n{b}^n}\right).\end{array}$$

(4)

Combining (3) and (4), finally we have

$${\left(\sum _{k=n}^{\mathrm{\infty }}\frac{a^k{b}^{k+2c+1}}{u_k{u}_{k+2c+1}}\right)}^{-1}-(\frac{u_n{u}_{n+2c+1}}{a^n{b}^{n+2c+1}}-\frac{u_{n-1}{u}_{n+2c}}{a^{n-1}{b}^{n+2c}})=O\left(\frac{1}{a^n{b}^n}\right).$$

(5)

It follows that for any integer $c\ge 0$, there exists $n\ge n_2$ sufficiently large such that the modulus of the last error term of identity (5) becomes less than $1/2$. This completes the proof of Theorem 2. □