Theory and Modern Applications

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

(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 [24]. Recently, Ohtsuka and Nakamura [5] studied the partial infinite sums of reciprocal Fibonacci numbers and proved that

where $⌊x⌋$ (the floor function) denotes the greatest integer less than or equal to x.

Some related works can also be found in [613]. 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 ${\left({\sum }_{k=n}^{\mathrm{\infty }}\frac{1}{{v}_{k}}\right)}^{-1}$. Specifically, suppose that $\parallel x\parallel =⌊x+\frac{1}{2}⌋$ (the nearest integer function), and ${\left\{{v}_{n}\right\}}_{n\ge 0}$ is an integer sequence satisfying the recurrence formula

${v}_{n}={a}_{1}{v}_{n-1}+{a}_{2}{v}_{n-2}+\cdots +{a}_{s}{v}_{n-s}\phantom{\rule{1em}{0ex}}\left(s\ge 2\right)$

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 $\left\{{u}_{n}\right\}$ be a second-order sequence defined by (1). For any even $p\ge 2$ and non-negative integer $q, 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 $\left\{{u}_{n}\right\}$ be a second-order sequence defined by (1). For any integer $c\ge 0$, there exists a positive integer ${n}_{2}$ such that

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

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

${\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{u}_{k}^{s}}\right)}^{-1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{u}_{k}{u}_{k+l}}\right)}^{-1}$

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

${u}_{n}=\frac{{a}^{2⌊\frac{n}{2}⌋-n+1}}{{\left(ab\right)}^{⌊\frac{n}{2}⌋}}\cdot \frac{{\alpha }^{n}-{\beta }^{n}}{\alpha -\beta },$

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

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

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

Let

Thus,

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

Hence,

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

Thus,

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

Taking the reciprocal of this expression yields

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

Therefore, for any even $p\ge 2$ and integer $0, 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 ${\left(\alpha -\beta \right)}^{2}={a}^{2}{b}^{2}+4ab$ and $\alpha \beta =-ab$ now yield

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

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

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

Hence,

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

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

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

Consequently,

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

Taking the reciprocal of this expression yields

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

On the other hand,

$\begin{array}{r}\frac{{u}_{n}{u}_{n+2c+1}}{{a}^{n}{b}^{n+2c+1}}-\frac{{u}_{n-1}{u}_{n+2c}}{{a}^{n-1}{b}^{n+2c}}\\ \phantom{\rule{1em}{0ex}}=\frac{{\alpha }^{2c+1}}{{a}^{c+1}{b}^{3c+3}+4{a}^{c}{b}^{3c+2}}\cdot {\left(\frac{\alpha }{ab}\right)}^{2n}-\frac{{\alpha }^{2c+1}}{{a}^{c+1}{b}^{3c+3}+4{a}^{c}{b}^{3c+2}}\cdot {\left(\frac{\alpha }{ab}\right)}^{2n-2}+O\left(\frac{1}{{a}^{n}{b}^{n}}\right)\\ \phantom{\rule{1em}{0ex}}=\frac{1}{B}\cdot {\left(\frac{\alpha }{ab}\right)}^{2n}-\frac{1}{B}\cdot {\left(\frac{\alpha }{ab}\right)}^{2n-2}+O\left(\frac{1}{{a}^{n}{b}^{n}}\right).\end{array}$
(4)

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

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

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## Acknowledgements

The authors express their gratitude to the referee for very helpful and detailed comments. This work is supported by the N.S.F. (11371291), the S.R.F.D.P. (20136101110014), the N.S.F. (2013JZ001) of Shaanxi Province, and the G.I.C.F. (YZZ12062) of NWU, P.R. China.

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Correspondence to Zhengang Wu.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

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

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Zhang, H., Wu, Z. On the reciprocal sums of the generalized Fibonacci sequences. Adv Differ Equ 2013, 377 (2013). https://doi.org/10.1186/1687-1847-2013-377