On the reciprocal products of generalized Fibonacci sequences

Abstract

In this paper, we use the properties of error estimation and the analytic method to study the reciprocal products of the bi-periodic Fibonacci sequence, the bi-periodic Lucas sequence, and the mth-order linear recursive sequence.

1 Introduction

The so-called Fibonacci sequence $\{ F_{n} \}$ and Lucas sequence $\{ L_{n} \}$ are defined by

$$ F_{0}=0, \qquad F_{1}=1, \qquad F_{n}=F_{n-1} +F_{n-2}, \quad n\ge 2, $$

and

$$ L_{0}=2, \qquad L_{1}=1, \qquad L_{n}=L_{n-1} +L_{n-2} , \quad n\ge 2. $$

The Fibonacci and Lucas sequences have many interesting properties and applications [1]. In addition, in [2], Ohtsuka and Nakamura considered the partial infinite sums of reciprocal Fibonacci sequence and proved that:

$$ { \Biggl\lfloor \Biggl( { \sum_{k=n}^{\infty } \frac{1}{F_{k} } } \Biggr)^{-1} \Biggr\rfloor =} \textstyle\begin{cases} F_{n-2}, & \text{if } n \text{ is even} ; \\ F_{n-2}-1, & \text{if } n \text{ is odd}, \end{cases}\displaystyle \quad n\ge 2, $$

and

$$ { \Biggl\lfloor \Biggl( { \sum_{k=n}^{\infty } \frac{1}{F_{k}^{2} } } \Biggr)^{-1} \Biggr\rfloor =} \textstyle\begin{cases} F_{n-1} F_{n}-1, & \text{if } n \text{ is even} ; \\ F_{n-1} F_{n}, & \text{if } n \text{ is odd} , \end{cases}\displaystyle \quad n\ge 2, $$

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

Many authors have studied the Fibonacci and Lucas sequences by changing initial conditions or recursive relations. For instance, for any two nonzero real numbers a and b, Edson and Yayenie [3] introduced the bi-periodic Fibonacci sequence $\{ f_{n} \}$ as:

$$ f_{0} =0, \qquad f_{1} =1, \qquad {f_{n}=} \textstyle\begin{cases} af_{n-1}+f_{n-2}, & \text{if } n \text{ is even} ; \\ bf_{n-1}+f_{n-2}, & \text{if } n \text{ is odd}, \end{cases}\displaystyle \quad n\ge 2. $$

(1)

For $a=b=1$, $\{ f_{n} \}$ reduces to the Fibonacci sequence $\{ F_{n} \}$. If $a=b=k$, then $\{ f_{n} \}$ becomes the k-Fibonacci sequence $\{ q_{n} \}$ defined in [4], etc. Similarly, for any two nonzero real numbers a and b, Bilgici [5] introduced the bi-periodic Lucas sequence $\{ l_{n} \}$ as:

$$ l_{0} =2, \qquad l_{1} =a, \qquad {l_{n}=} \textstyle\begin{cases} bl_{n-1}+l_{n-2}, & \text{if } n \text{ is even} ; \\ al_{n-1}+l_{n-2}, & \text{if } n \text{ is odd}, \end{cases}\displaystyle \quad n\ge 2. $$

(2)

For $a=b=1$, $\{ l_{n} \}$ reduces to the Lucas sequence $\{ L_{n} \}$. If $a=b=k$, then $\{ l_{n} \}$ becomes the k-Lucas sequence $\{ p_{n} \}$ defined in [6]. In [7], Tan and Leung considered a generalization of Horadam sequence $\{ w_{n} \}$, which is defined by the recurrence relation

$$w_{0}=w_{0}, \qquad w_{1}=w_{1} , \qquad w_{n}= \textstyle\begin{cases} aw_{n-1}+cw_{n-2}, & \text{if } n \text{ is even} ; \\ bw_{n-1}+cw_{n-2}, & \text{if } n \text{ is odd}, \end{cases}\displaystyle \quad n\ge 2, $$

with arbitrary initial conditions $w_{0}$, $w_{1}$ and nonzero real numbers a, b, and c. In [8], Tan considered the sequence $\{ w_{n} \}$ when $c=1$. In [9], Ramírez and Sirvent introduced a q-bi-periodic Fibonacci sequence by

$$F_{n} ^{(a,b)} (q,s)= \textstyle\begin{cases} aF_{n-1} ^{(a,b)} (q,s)+q^{n-2}sF_{n-2} ^{(a,b)} (q,s), & \text{if } n \text{ is even} ; \\ bF_{n-1} ^{(a,b)} (q,s)+q^{n-2}sF_{n-2} ^{(a,b)} (q,s),& \text{if } n \text{ is odd}, \end{cases}\displaystyle \quad n \ge 2, $$

with initial conditions $F_{0} ^{(a,b)} (q,s)=0$ and $F_{1} ^{(a,b)} (q,s) =1$ and nonzero real numbers a, b, q and s. Motivated by [9], in [10] Tan introduced a q-bi-periodic Lucas sequence by

$$L_{n} ^{(a,b)} (q,s)= \textstyle\begin{cases} bL_{n-1} ^{(a,b)} (q,s)+sL_{n-2} ^{(a,b)} (q,qs), & \text{if } n \text{ is even} ; \\ aL_{n-1} ^{(a,b)} (q,s)+sL_{n-2} ^{(a,b)} (q,qs),& \text{if } n \text{ is odd}, \end{cases}\displaystyle \quad n \ge 2, $$

with initial conditions $L_{0} ^{(a,b)} (q,s)=2$ and $L_{1} ^{(a,b)} (q,s) =q$, and nonzero real numbers a, b,q and s.

In [11], Holliday and Komatsu obtained the infinite sums of the reciprocal of k-Fibonacci sequence $\{ q_{n} \}$. In [12], Basbük and Yazlik obtained the infinite sums of the reciprocal of the bi-periodic Fibonacci sequence $\{ f_{n} \}$. Various authors studied the infinite sums of the reciprocal of the other famous sequences [13–15].

Recently, some authors studied the nearest integer of the sums of reciprocal of some linear recurrence sequences. In [16], Komatsu proved that there exists a positive integer $n_{1}$ such that:

$$ \Biggl\Vert \Biggl( \sum_{k=n}^{\infty } \frac{1}{q_{k} } \Biggr)^{-1} \Biggr\Vert =q_{n}-q_{n-1}, \quad n\ge n_{1} , $$

where $\{ q_{n} \}$ is the k-Fibonacci sequence. $\Vert \cdot \Vert$ denotes the nearest integer. Specifically, suppose that $\Vert x \Vert = \lfloor x+\frac{1}{2} \rfloor$.

On the other hand, Wu and Zhang [17] considered an mth-order linear recursive sequence $\{ u_{n} \}$ defined by

$$ u_{n}=x_{1}u_{n-1}+x_{2}u_{n-2}+ \cdots +x_{m}u_{n-m}, \quad n>m, $$

(3)

where initial values $u_{i}\in N$ for $0\le i < m$, at least one of them is different from zero, and $x_{1}, x_{2}, \dots , x_{m}$ are positive integers. The characteristic polynomial of the sequence $\{ u_{n} \}$ is given by

$$ \psi ( y )=y^{m} -x_{1}y^{m-1} -\cdots -x_{m-1}y-x_{m}. $$

For $m=2$, $x_{1}=x_{2}=1$ and initial values $u_{0}=0$, $u_{1}=1$, $\{ u_{n} \}$ reduces to the Fibonacci sequence. If $m=2$, $x_{1}=x_{2}=1$ and initial values $u_{0}=2$, $u_{1}=1$, then $\{ u_{n} \}$ becomes the Lucas sequence.

In addition, they proved that there exists a positive integer $n_{2}$ such that:

$$ \Biggl\Vert \Biggl(\sum_{k=n}^{\infty } \frac{1}{u_{k} } \Biggr)^{-1} \Biggr\Vert =u_{n}-u_{n-1}, \quad n\ge n_{2} , $$

for any positive integers $x_{1}\ge x_{2}\ge \cdots \ge x_{m}\ge 1$. For more the nearest integer of the sums of reciprocal of the recurrence sequence studies, see [18–21]. Specifically, in [19], Trojorský considered finding a sequence that is “asymptotically equivalent” to partial infinite sums and proved that

$$ \Biggl\{ \Biggl( \sum_{k=n}^{\infty } \frac{1}{P ( u_{k} ) } \Biggr) ^{-1} \Biggr\} _{n} \quad \text{and} \quad \bigl\{ P ( u_{n} )- P ( u_{n-1} ) \bigr\} _{n} $$

are asymptotically equivalent, where $P ( z )\in C [ z ]$ is a non-constant polynomial. Specifically, we say that two sequences $\{ G_{n} \}$ and $\{ H_{n} \}$ are called “asymptotically equivalent” if $\{ G_{n} \} / \{ H_{n} \}$ tends to 1 as $n \rightarrow \infty$.

In addition to the study of the infinite reciprocal sums of recursive sequence, we can also consider the infinite reciprocal products of recursive sequence. In 2006, Wu [22] studied the partial infinite products of $\frac{q_{k}^{i} -1 }{q_{k}^{i} }$. He used the element method and the properties of the floor function and proved that

$$ \Biggl\lfloor \Biggl( 1-\prod_{k=n}^{\infty } \biggl( 1- \frac{1}{q_{k} } \biggr) \Biggr)^{-1} \Biggr\rfloor =q_{n}- q_{n-1}, \quad n\ge 2 , $$

and

$$ { \Biggl\lfloor \Biggl( 1-\prod_{k=n}^{\infty } \biggl( 1- \frac{1}{q_{k}^{2} } \biggr) \Biggr)^{-1} \Biggr\rfloor =} \textstyle\begin{cases} q_{n}^{2}-q_{n-1}^{2} , & \text{if } n \text{ is even} ; \\ q_{n}^{2}-q_{n-1}^{2}-1, & \text{if } n \text{ is odd}, \end{cases}\displaystyle \quad n\ge 2, $$

where $\{ q_{n} \}$ is the k-Fibonacci sequence. For more the partial infinite products of the other sequences, see [23, 24].

Inspired by [19], in this paper, we apply a different research method from the previous one and use the properties of error estimation and the analytic method to study the reciprocal products of $\{ f_{n} \}$, $\{ l_{n} \}$ and $\{ u_{n} \}$. We derive some sequences that are asymptotically equivalent to reciprocal products including $\{ f_{n} \}$, $\{ l_{n} \}$ and $\{ u_{n} \}$. Our main results are the following:

Theorem 1

Let $\{ f_{n} \}$ be the bi-periodic Fibonacci sequence, and $\{ l_{n} \}$ be the bi-periodic Lucas sequence. For positive integers a and b with $a\ge 1$, $b\ge 1$, the sequences

$$ \Biggl\{ \Biggl( 1-\prod_{k=n}^{\infty } \biggl( 1-\frac{1}{f_{k} } \biggr) \Biggr)^{-1} \Biggr\} _{n} \quad \textit{and} \quad \{ f_{n}-f_{n-1} \} _{n} $$

(4)

are asymptotically equivalent, and the sequences

$$ \Biggl\{ \Biggl( 1-\prod_{k=n}^{\infty } \biggl( 1-\frac{1}{l_{k} } \biggr) \Biggr)^{-1} \Biggr\} _{n} \quad \textit{and} \quad \{ l_{n}-l_{n-1} \} _{n} $$

(5)

are asymptotically equivalent.

Corollary 1

We obtain the infinite products of the reciprocal of the k-Fibonacci sequence $q_{n}$ and k-Lucas sequence $p_{n}$, when $a=b=k$. Then, the sequences

$$ \Biggl\{ \Biggl( 1-\prod_{k=n}^{\infty } \biggl( 1-\frac{1}{q_{k} } \biggr) \Biggr)^{-1} \Biggr\} _{n} \quad \textit{and} \quad \{ q_{n}-q_{n-1} \} _{n} $$

(6)

are asymptotically equivalent, and the sequences

$$ \Biggl\{ \Biggl( 1-\prod_{k=n}^{\infty } \biggl( 1-\frac{1}{p_{k} } \biggr) \Biggr)^{-1} \Biggr\} _{n} \quad \textit{and} \quad \{ p_{n}-p_{n-1} \} _{n} $$

(7)

are asymptotically equivalent.

Theorem 2

Let $\{ u_{n} \}$ be an mth-order linear recursive sequence with any positive integers $x_{1}\ge x_{2}\ge \cdots \ge x_{m}\ge 1$. Then, the sequences

$$ \Biggl\{ \Biggl( 1-\prod_{k=n}^{\infty } \biggl( 1-\frac{1}{u_{k} } \biggr) \Biggr)^{-1} \Biggr\} _{n} \quad \textit{and} \quad \{ u_{n}-u_{n-1} \} _{n} $$

(8)

are asymptotically equivalent.

2 Proof of the theorems

To complete the proof of our theorems, we need the following:

Lemma 1

([3, 5], Generalized Binet’s formula)

The terms of the bi-periodic Fibonacci sequence $\{ f_{n} \}$, and bi-periodic Lucas sequence $\{ l_{n} \}$ are given by

$$ f_{n}= \frac{a^{\zeta (n+1 ) }}{ ( ab )^{ \lfloor \frac{n}{2} \rfloor } } \biggl(\frac{\alpha ^{n} -\beta ^{n} }{\alpha -\beta } \biggr), $$

and

$$ l_{n}= \frac{a^{\zeta (n ) }}{ ( ab )^{ \lfloor \frac{n+1}{2} \rfloor } } \bigl(\alpha ^{n} +\beta ^{n} \bigr), $$

where $\alpha =\frac{ab+\sqrt{a^{2}b^{2}+4ab } }{2}$ and $\beta =\frac{ab-\sqrt{a^{2}b^{2}+4ab } }{2}$, i.e. α and β are roots of the equation $x^{2} -abx-ab=0$. It is obvious that $\alpha > 1$ and $-1<\beta < 0$ with $a\ge 1$, $b\ge 1$. In addition, $\zeta ( n )$ is the parity function, such that $\zeta ( n ) =0$ if n is even and $\zeta ( n ) =1$ if n is odd.

Lemma 2

Let $\{ f_{n} \}$ be the bi-periodic Fibonacci sequence defined by (1), and $\{ l_{n} \}$ be the bi-periodic Lucas sequence defined by (2). Then, we have

$$ {f_{n}=} \textstyle\begin{cases} \frac{c\alpha ^{n} }{ ( ab )^{\frac{n}{2} } }- \frac{c\beta ^{n } }{ ( ab )^{\frac{n}{2} } }, & \textit{if } n \textit{ is even}; \\ \frac{d\alpha ^{n} }{ ( ab )^{\frac{n-1}{2} } }- \frac{d\beta ^{n} }{ ( ab )^{\frac{n-1}{2} } }, & \textit{if } n \textit{ is odd}, \end{cases} $$

where $c=\frac{a}{\alpha -\beta }$, $d=\frac{1}{\alpha -\beta }$, and

$$ {l_{n}=} \textstyle\begin{cases} \frac{\alpha ^{n} }{ ( ab )^{\frac{n}{2} } }+ \frac{\beta ^{n } }{ ( ab )^{\frac{n}{2} } }, & \textit{if } n \textit{ is even}; \\ \frac{a\alpha ^{n} }{ ( ab )^{\frac{n+1}{2} } }+ \frac{a\beta ^{n} }{ ( ab )^{\frac{n+1}{2} } }, & \textit{if } n \textit{ is odd}. \end{cases} $$

Proof

By Lemma 1, we can easily prove it. □

Lemma 3

([17])

Let $\{u_{n} \}$ be an mth-order linear recursive sequence defined by (3). The coefficients of the characteristic polynomial $\psi ( y )$ are satisfied that $x_{1}\ge x_{2}\ge \cdots \ge x_{m}\ge 1$. Then, the closed formula of $\{u_{n} \}$ is given by

$$ u_{n}=s\gamma ^{n} +\mathcal{O} \bigl( t^{-n} \bigr), \quad ( n\rightarrow \infty ), $$

where $s>0$, $t>1$, γ is the positive real zero of $\psi ( y )$ for $x_{1}<\gamma <x_{1}+1$, and “$\mathcal{O}$” (the Landau symbol) denotes if $g ( x ) > 0$ for all $x\ge a$, we write $f ( x ) =\mathcal{O} ( g ( x ) )$ to mean that the quotient $f ( x ) /g ( x )$ is bounded for $x\ge a$.

Lemma 4

Let a, b, c, d, α, and β be defined by Lemma 1or Lemma 2and s, γ, and t be defined by Lemma 3. Then, we have

$$\begin{aligned}& \begin{aligned} &\prod_{k=n}^{\infty } \biggl( 1-\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } \beta }{\alpha ^{2} } \biggr)^{k} \biggr) \biggr) \\ &\quad =1-\sum_{k=n}^{\infty }\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr)^{2n} \biggr), \end{aligned} \end{aligned}$$

(9)

$$\begin{aligned}& \begin{aligned} &\prod_{k=n}^{\infty } \biggl( 1- \frac{1}{ ( ab ) ^{\frac{1}{2} } d} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } \beta }{\alpha ^{2} } \biggr)^{k} \biggr) \biggr) \\ &\quad =1-\sum_{k=n}^{\infty } \frac{1}{ ( ab ) ^{\frac{1}{2} } d} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr)^{2n} \biggr), \end{aligned} \\ \end{aligned}$$

(10)

$$\begin{aligned}& \prod_{k=n}^{\infty } \biggl( 1- \frac{1}{s\gamma ^{k} } +\mathcal{O} \bigl( \gamma ^{-2k}t^{-k} \bigr) \biggr) =1-\sum_{k=n}^{\infty } \frac{1}{s\gamma ^{k} } +\mathcal{O} \bigl( \gamma ^{-2n} \bigr). \end{aligned}$$

(11)

Proof

We shall prove only (6) in Lemma 4, and other identities are proved similarly. The identity $ab=-\alpha \beta$ now yield $\vert \beta \vert < ( ab )^{\frac{1}{2} }= ( - \alpha \beta ) ^{\frac{1}{2} }< \alpha$, where $\alpha > 1$ and $-1<\beta < 0$. First, we prove the following equation

$$ \begin{aligned} &\prod_{k=n}^{n+m } \biggl( 1-\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } \beta }{\alpha ^{2} } \biggr)^{k} \biggr) \biggr) \\ &\quad =1-\sum_{k=n}^{n+m}\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr)^{2n} \biggr), \end{aligned} $$

(12)

We prove (9) by mathematical induction. When $m=1$,

$$ \begin{aligned} &\prod_{k=n}^{n+1} \biggl( 1-\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } \beta }{\alpha ^{2} } \biggr)^{k} \biggr) \biggr) \\ &\quad = \biggl( 1-\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{n} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } \beta }{\alpha ^{2} } \biggr)^{n} \biggr) \biggr) \\ &\qquad {}\times \biggl( 1-\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{n+1} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } \beta }{\alpha ^{2} } \biggr)^{n+1} \biggr) \biggr) \\ &\quad =1-\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{n} - \frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{n+1} + \frac{1}{c^{2}} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{2n+1}+ \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} }\beta }{\alpha ^{2} } \biggr)^{n} \biggr) \\ &\quad =1-\sum_{k=n}^{n+1}\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr)^{2n} \biggr)+\mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} }\beta }{\alpha ^{2} } \biggr)^{n} \biggr) \\ &\quad =1-\sum_{k=n}^{n+1}\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr)^{2n} \biggr). \end{aligned} $$

When $m=2$,

$$\begin{aligned}& \begin{aligned} &\prod_{k=n}^{n+2} \biggl( 1-\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } \beta }{\alpha ^{2} } \biggr)^{k} \biggr) \biggr) \\ &\quad = \Biggl( 1-\sum_{k=n}^{n+1} \frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr)^{2n} \biggr) \Biggr) \\ &\qquad {}\times \biggl( 1-\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{n+2} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } \beta }{\alpha ^{2} } \biggr)^{n+2} \biggr) \biggr) \\ &\quad =1-\sum_{k=n}^{n+1}\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} - \frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{n+2} + \frac{1}{c^{2} } \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{n+2} \Biggl( \sum _{k=n}^{n+1} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr)^{k} \Biggr) \\ &\qquad {}+\mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr)^{2n} \biggr)+ \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} }\beta }{\alpha ^{2} } \biggr)^{n} \biggr) \\ &\quad =1-\sum_{k=n}^{n+2} \frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr)^{2n} \biggr)+ \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} }\beta }{\alpha ^{2} } \biggr)^{n} \biggr) \\ &\quad =1-\sum_{k=n}^{n+1}\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr)^{2n} \biggr). \end{aligned} \end{aligned}$$

That is, (9) is true for $m=1$ or $m=2$. Suppose that for any integer m, we have

$$ \begin{aligned} &\prod_{k=n}^{n+m } \biggl( 1-\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } \beta }{\alpha ^{2} } \biggr)^{k} \biggr) \biggr) \\ &\quad =1-\sum_{k=n}^{n+m}\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr)^{2n} \biggr). \end{aligned} $$

(13)

Then, for $m+1$, we have

$$ \begin{aligned} &\prod_{k=n}^{n+m+1} \biggl( 1-\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } \beta }{\alpha ^{2} } \biggr)^{k} \biggr) \biggr) \\ &\quad = \Biggl( 1-\sum_{k=n}^{n+m} \frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr)^{2n} \biggr) \Biggr) \\ &\qquad {}\times \biggl( 1-\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{n+m+1} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } \beta }{\alpha ^{2} } \biggr)^{n+m+1} \biggr) \biggr) \\ &\quad =1-\sum_{k=n}^{n+m}\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} - \frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{n+m+1} + \frac{1}{c^{2} } \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{n+m+1} \Biggl( \sum _{k=n}^{n+m} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr)^{k} \Biggr) \\ &\qquad {}+\mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr)^{2n} \biggr)+ \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} }\beta }{\alpha ^{2} } \biggr)^{n} \biggr) \\ &\quad =1-\sum_{k=n}^{n+m+1}\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr)^{2n} \biggr)+\mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} }\beta }{\alpha ^{2} } \biggr)^{n} \biggr) \\ &\quad =1-\sum_{k=n}^{n+m+1}\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr)^{2n} \biggr). \end{aligned} $$

Taking $m \rightarrow \infty$, we have

$$ \begin{aligned} &\prod_{k=n}^{\infty } \biggl( 1-\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } \beta }{\alpha ^{2} } \biggr)^{k} \biggr) \biggr) \\ &\quad =1-\sum_{k=n}^{\infty }\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr)^{2n} \biggr), \end{aligned} $$

which completes the proof. □

Proof of Theorem 1

We shall prove only (4) in Theorem 1, and the identity (5) is proved similarly. From the geometric series as $\epsilon \rightarrow 0$, we find

$$ \frac{1}{1\pm \epsilon }=1+ \mathcal{O} ( \epsilon ). $$

If n is even, with $n\ge 2$. Using Lemma 2, we have

$$ \frac{1}{f_{k}}= \frac{1}{ \frac{c\alpha ^{k} }{ ( ab )^{\frac{k}{2} } }- \frac{c\beta ^{k } }{ ( ab )^{\frac{k}{2} } }} = \frac{1}{ \frac{c\alpha ^{k} }{ ( ab )^{\frac{k}{2} } } ( 1- ( \frac{\beta }{\alpha } )^{k} ) } = \frac{ ( ab )^{\frac{k}{2}} }{ c\alpha ^{k} } \biggl( 1+ \mathcal{O} \biggl( \frac{\beta }{\alpha } \biggr)^{k} \biggr) . $$

By Lemma 4, we obtain

$$ \begin{aligned} \prod_{k=n}^{\infty } \biggl( 1-\frac{1}{f_{k} } \biggr) &= \prod_{k=n}^{\infty } \biggl( 1-\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} }\beta }{\alpha ^{2} } \biggr)^{k} \biggr) \biggr) \\ &=1-\sum_{k=n}^{\infty }\frac{1}{c} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha} \biggr)^{2n} \biggr) \\ &=1-\frac{ ( ab ) ^{\frac{n}{2} } }{c\alpha ^{n} } \biggl( \frac{\alpha }{\alpha - ( ab )^{\frac{1}{2} } } \biggr) + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr)^{2n} \biggr). \end{aligned} $$

Taking the reciprocal of this expression yields

$$ \begin{aligned} \Biggl( 1- \prod_{k=n}^{\infty } \biggl( 1- \frac{1}{f_{k} } \biggr) \Biggr)^{-1} &= \frac{1}{\frac{ ( ab ) ^{\frac{n}{2} } }{c\alpha ^{n} } ( \frac{\alpha }{\alpha - ( ab )^{\frac{1}{2} } } ) +\mathcal{O} ( ( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } )^{2n} )} \\ &= \frac{1}{\frac{ ( ab ) ^{\frac{n}{2} } }{c\alpha ^{n} } ( \frac{\alpha }{\alpha - ( ab )^{\frac{1}{2} } } ) ( 1+\mathcal{O} ( \frac{ ( ab ) ^{\frac{1}{2} } }{\alpha } )^{n} )} \\ &=\frac{c\alpha ^{n} }{ ( ab ) ^{\frac{n}{2} } } \biggl( \frac{ \alpha - ( ab )^{\frac{1}{2} } }{\alpha } \biggr) \biggl( 1+\mathcal{O} \biggl( \frac{ ( ab ) ^{\frac{1}{2} } }{\alpha } \biggr)^{n} \biggr) \\ &= \biggl(f_{n}-f_{n-1} + \frac{c\beta ^{n} }{ ( ab )^{\frac{n}{2} } }- \frac{c\beta ^{n-1} }{ ( ab )^{\frac{n-1}{2} } } \biggr) \biggl( 1+\mathcal{O} \biggl( \frac{ ( ab ) ^{\frac{1}{2} } }{\alpha } \biggr)^{n} \biggr), \end{aligned} $$

where $\vert \beta \vert < ( ab )^{\frac{1}{2} }$ yields

$$ \biggl(f_{n}-f_{n-1} + \frac{c\beta ^{n} }{ ( ab )^{\frac{n}{2} } }- \frac{c\beta ^{n-1} }{ ( ab )^{\frac{n-1}{2} } } \biggr) \quad \text{tends to } (f_{n}-f_{n-1} ), $$

as $n \rightarrow \infty$. In addition, as $( ab )^{\frac{1}{2} }< \alpha$, we obtain

$$ \frac{ ( 1- \prod_{k=n}^{\infty } ( 1-\frac{1}{f_{k} } ) )^{-1}}{ ( f_{n}-f_{n-1} ) } \quad \text{tends to } 1, $$

as $n \rightarrow \infty$.

If n is odd, with $n\ge 1$. Using Lemma 2, we have

$$ \frac{1}{f_{k}}= \frac{1}{ \frac{d\alpha ^{k} }{ ( ab )^{\frac{k-1}{2} } }- \frac{d\beta ^{k } }{ ( ab )^{\frac{k-1}{2} } }} = \frac{1}{ \frac{d\alpha ^{k} }{ ( ab )^{\frac{k-1}{2} } } ( 1- ( \frac{\beta }{\alpha } )^{k} ) } = \frac{ ( ab )^{\frac{k-1}{2}} }{ d\alpha ^{k} } \biggl( 1+ \mathcal{O} \biggl( \frac{\beta }{\alpha } \biggr)^{k} \biggr) . $$

By Lemma 4, we obtain

$$\begin{aligned} \prod_{k=n}^{\infty } \biggl( 1-\frac{1}{f_{k} } \biggr) &= \prod_{k=n}^{\infty } \biggl( 1- \frac{1}{ ( ab ) ^{\frac{1}{2} } d} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } \beta }{\alpha ^{2} } \biggr)^{k} \biggr) \biggr) \\ &=1-\sum_{k=n}^{\infty } \frac{1}{ ( ab ) ^{\frac{1}{2} } d} \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr) ^{k} + \mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr)^{2n} \biggr) \\ &=1-\frac{ ( ab ) ^{\frac{n-1}{2} } }{d\alpha ^{n} } \biggl( \frac{\alpha }{\alpha - ( ab )^{\frac{1}{2} } } \biggr) +\mathcal{O} \biggl( \biggl( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } \biggr)^{2n} \biggr). \end{aligned}$$

Taking the reciprocal of this expression yields

$$ \begin{aligned} \Biggl( 1- \prod_{k=n}^{\infty } \biggl( 1- \frac{1}{f_{k} } \biggr) \Biggr)^{-1} &= \frac{1}{\frac{ ( ab ) ^{\frac{n-1}{2} } }{d\alpha ^{n} } ( \frac{\alpha }{\alpha - ( ab )^{\frac{1}{2} } } ) +\mathcal{O} ( ( \frac{ ( ab )^{\frac{1}{2} } }{\alpha } )^{2n} )} \\ &= \frac{1}{\frac{ ( ab ) ^{\frac{n-1}{2} } }{ d\alpha ^{n} } ( \frac{\alpha }{\alpha - ( ab )^{\frac{1}{2} } } ) ( 1+\mathcal{O} ( \frac{ ( ab ) ^{\frac{1}{2} } }{\alpha } )^{n} )} \\ &=\frac{ d\alpha ^{n} }{ ( ab ) ^{\frac{n-1}{2} } } \biggl( \frac{ \alpha - ( ab )^{\frac{1}{2} } }{\alpha } \biggr) \biggl( 1+\mathcal{O} \biggl( \frac{ ( ab ) ^{\frac{1}{2} } }{\alpha } \biggr)^{n} \biggr) \\ &= \biggl( f_{n}-f_{n-1} + \frac{d\beta ^{n} }{ ( ab )^{\frac{n}{2} } }- \frac{d\beta ^{n-1} }{ ( ab )^{\frac{n-1}{2} } } \biggr) \biggl( 1+\mathcal{O} \biggl( \frac{ ( ab ) ^{\frac{1}{2} } }{\alpha } \biggr)^{n} \biggr) , \end{aligned} $$

where $\vert \beta \vert < ( ab )^{\frac{1}{2} }$ yields

$$ \biggl(f_{n}-f_{n-1} + \frac{c\beta ^{n} }{ ( ab )^{\frac{n}{2} } }- \frac{c\beta ^{n-1} }{ ( ab )^{\frac{n-1}{2} } } \biggr) \quad \text{tends to } (f_{n}-f_{n-1} ), $$

as $n \rightarrow \infty$. In addition, as $( ab )^{\frac{1}{2} }< \alpha$, we obtain

$$ \frac{ ( 1- \prod_{k=n}^{\infty } ( 1-\frac{1}{f_{k} } ) )^{-1}}{ ( f_{n}-f_{n-1} ) } \quad \text{tends to } 1, $$

as $n \rightarrow \infty$, which completes the proof. □

Proof of Theorem 2

Using Lemma 3, we have

$$ \frac{1}{u_{k}}= \frac{1}{s\gamma ^{k}+\mathcal{O} ( t^{-k} ) }= \frac{1}{s\gamma ^{k} ( 1+\mathcal{O} ( \gamma ^{-k}t ^{-k} ) ) } =\frac{1}{s\gamma ^{k}} \bigl( 1+\mathcal{O} \bigl(\gamma ^{-k}t ^{-k} \bigr) \bigr) . $$

By Lemma 4, we obtain

$$ \begin{aligned} \prod_{k=n}^{\infty } \biggl( 1-\frac{1}{u_{k} } \biggr) &=\prod_{k=n}^{\infty } \biggl( 1-\frac{1}{s\gamma ^{k} } + \mathcal{O} \bigl( \gamma ^{-2k}t^{-k} \bigr) \biggr) \\ &=1-\sum_{k=n}^{\infty }\frac{1}{s\gamma ^{k} } + \mathcal{O} \bigl( \gamma ^{-2n} \bigr) \\ &=1-\frac{\gamma }{s\gamma ^{n} ( \gamma -1 ) }+ \mathcal{O} \bigl( \gamma ^{-2n} \bigr). \end{aligned} $$

Taking the reciprocal of this expression yields

$$\begin{aligned} \Biggl( 1- \prod_{k=n}^{\infty } \biggl( 1- \frac{1}{u_{k} } \biggr) \Biggr)^{-1} &= \frac{1}{\frac{\gamma }{s\gamma ^{n} ( \gamma -1 ) }+ \mathcal{O} ( \gamma ^{-2n} )} \\ &= \frac{1}{\frac{\gamma }{s\gamma ^{n} ( \gamma -1 ) } ( 1+\mathcal{O} ( \gamma ^{-n} ) ) } \\ &=\frac{s\gamma ^{n} ( \gamma -1 ) }{\gamma } \bigl( 1+ \mathcal{O} \bigl( \gamma ^{-n} \bigr) \bigr) \\ &= ( u_{n}-u_{n-1} ) \bigl( 1+\mathcal{O} \bigl( \gamma ^{-n} \bigr) \bigr) , \end{aligned}$$

which yields

$$ \frac{ ( 1- \prod_{k=n}^{\infty } ( 1-\frac{1}{u_{k} } ) )^{-1}}{ ( u_{n}-u_{n-1} ) } \quad \text{tends to } 1, $$

as $n \rightarrow \infty$, which completes the proof. □

3 Discussion

In this paper, we obtain the sequences that are asymptotically equivalent to reciprocal products of $\frac{f_{k}-1 }{f_{k}}$, $\frac{l_{k}-1 }{l_{k}}$ and $\frac{u_{k}-1 }{u_{k}}$, where $\{ f_{n} \}$ denotes the bi-periodic Fibonacci sequence, $\{ l_{n} \}$ denotes the bi-periodic Lucas sequence, and $\{ u_{n} \}$ denotes an mth-order linear recursive sequence. For any positive integers j, an open problem is whether there exists the similar identities for the infinity products of $\frac{f_{k}^{j} -1 }{f_{k}^{j} }$, $\frac{l_{k}^{j} -1 }{l_{k}^{j} }$ and $\frac{u_{k}^{j} -1 }{u_{k}^{j} }$.