On the Reciprocal Sums of Products of Two Generalized Bi-Periodic Fibonacci Numbers

Abstract

This paper concerns the properties of the generalized bi-periodic Fibonacci numbers $\left\{G_n\right\}$ generated from the recurrence relation: $G_n=a{G}_{n-1}+G_{n-2}$ (n is even) or $G_n=b{G}_{n-1}+G_{n-2}$ (n is odd). We derive general identities for the reciprocal sums of products of two generalized bi-periodic Fibonacci numbers. More precisely, we obtain formulas for the integer parts of the numbers ${\left({\sum }_{k=n}^{\infty }\frac{{(a/b)}^{\xi (k+1)}}{G_k{G}_{k+m}}\right)}^{-1},m=0,2,4,\cdots ,$ and ${\left({\sum }_{k=n}^{\infty }\frac{1}{G_k{G}_{k+m}}\right)}^{-1},m=1,3,5,\cdots .$

1. Introduction

As is well known, the Fibonacci sequence $\left\{F_n\right\}$ is generated from the recurrence relation $F_n=F_{n-1}+F_{n-2}$ ($n\ge 2)$ with the initial conditions $F_0=0$ and $F_1=1$. The Fibonacci numbers possess many interesting properties and appear in a variety of application fields [1].

Many authors tried to generalize the Fibonacci sequence. For example, Falcon and Plaza [2] considered the k-Fibonacci sequence. Edson and Yayenie [3] introduced the bi-periodic Fibonacci sequence $\left\{f_n\right\}$ defined by

$$f_0=0,f_1=1,f_n=\left\{\begin{array}{cc}a{f}_{n-1}+f_{n-2},\hfill & \mathrm{if}n\in {\mathbb{N}}_e;\hfill \\ b{f}_{n-1}+f_{n-2},\hfill & \mathrm{if}n\in {\mathbb{N}}_o,\hfill \end{array}\right.(n\ge 2),$$

(1)

where ${\mathbb{N}}_e$ (${\mathbb{N}}_o$, respectively) denotes the set of positive even (odd, respectively) integers. Filipponi [4] defined the incomplete Fibonacci sequence, and Ramírez [5] introduced the bi-periodic incomplete Fibonacci sequence.

In the remainder of this paper, we use the notation ${\left\{G_n\right\}}_{n=0}^{\infty }=S(G_0,G_1,a,b)$ to denote the generalized bi-periodic Fibonacci numbers $\left\{G_n\right\}$ generated from the recurrence relation

$$G_n=\left\{\begin{array}{cc}a{G}_{n-1}+G_{n-2},\hfill & \mathrm{i}\mathrm{f}n\in {\mathbb{N}}_e;\hfill \\ b{G}_{n-1}+G_{n-2},\hfill & \mathrm{if}n\in {\mathbb{N}}_o,\hfill \end{array}\right.(n\ge 2),$$

with initial conditions $G_0$ and $G_1$, where $G_0$ is a nonnegative integer, $G_1$, a and b are positive integers.

Recently, Ohtsuka and Nakamura [6] reported an interesting property of the Fibonacci numbers $\left\{F_n\right\}=S(0,1,1,1)$ and proved the following identities:

$$\begin{array}{ccc}\hfill ⌊{\left(\sum _{k=n}^{\infty }\frac{1}{F_k}\right)}^{-1}⌋& =& \left\{\begin{array}{cc}{F}_n-F_{n-1},\hfill & \mathrm{if}n\ge 2\mathrm{and}n\in {\mathbb{N}}_e;\hfill \\ F_n-F_{n-1}-1,\hfill & \mathrm{if}n\ge 1\mathrm{and}n\in {\mathbb{N}}_o,\hfill \end{array}\right.\hfill \end{array}$$

(2)

$$\begin{array}{ccc}\hfill ⌊{\left(\sum _{k=n}^{\infty }\frac{1}{F_k^2}\right)}^{-1}⌋& =& \left\{\begin{array}{cc}{F}_{n-1}{F}_n-1,\hfill & \mathrm{if}n\ge 2\mathrm{and}n\in {\mathbb{N}}_e;\hfill \\ F_{n-1}{F}_n,\hfill & \mathrm{if}n\ge 1\mathrm{and}n\in {\mathbb{N}}_o,\hfill \end{array}\right.\hfill \end{array}$$

(3)

where $\lfloor ·\rfloor$ is the floor function.

Following the work of Ohtsuka and Nakamura, diverse results for the numbers of the form $\left\{G_n\right\}=S(G_0,G_1,a,a)$ have been reported in the literature (see [7,8,9,10,11,12,13,14,15] and references cited therein).

On the other hand, reciprocal sums of the generalized bi-periodic numbers were considered in [16,17]. In [16], Basbuk and Yazlik proved the following identity for $\left\{G_n\right\}=S(0,1,a,b)$:

$$⌊{\left(\sum _{k=n}^{\infty }\frac{{(a/b)}^{\psi \left(k\right)}}{G_k}\right)}^{-1}⌋=\left\{\begin{array}{cc}{G}_n-G_{n-1},\hfill & \mathrm{if}n\ge 2\mathrm{and}n\in {\mathbb{N}}_e;\hfill \\ G_n-G_{n-1}-1,\hfill & \mathrm{if}n\ge 1\mathrm{and}n\in {\mathbb{N}}_o,\hfill \end{array}\right.$$

(4)

where

$$\psi \left(k\right)=\xi (k+1)-\xi (n+1)-{(-1)}^n⌊\frac{k-n}{2}⌋,$$

and $\xi \left(n\right)$ is the parity function, such that

$$\xi \left(n\right)=\left\{\begin{array}{c}0,\mathrm{i}\mathrm{f}n\in \left\{0\right\}\cup {\mathbb{N}}_e;\hfill \\ 1,\mathrm{i}\mathrm{f}n\in {\mathbb{N}}_o.\hfill \end{array}\right.$$

For $\left\{G_n\right\}=S(G_0,G_1,a,b)$, Choi and Choo [17] identified the integer parts for the numbers

$${\left(\sum _{k=n}^{\infty }\frac{{(a/b)}^{\xi (k+1)}}{G_k^2}\right)}^{-1}.$$

In this paper, we extend the results in [17] by considering the reciprocal sums of products of two generalized bi-periodic Fibonacci numbers. More precisely, we obtain general identities for the numbers

$${\left(\sum _{k=n}^{\infty }\frac{{(a/b)}^{\xi (k+1)}}{G_k{G}_{k+m}}\right)}^{-1},m\in \left\{0\right\}\cup {\mathbb{N}}_e,$$

and

$${\left(\sum _{k=n}^{\infty }\frac{1}{G_k{G}_{k+m}}\right)}^{-1},m\in {\mathbb{N}}_0.$$

2. Results

2.1. The Case where $m\in \left\{0\right\}\cup {\mathbb{N}}_e$

Lemma 1 below will be used to prove the results for the case where $m\in {\mathbb{N}}_e$.

Lemma 1.

Assume that $m\in \left\{0\right\}\cup {\mathbb{N}}_e$. Then, for $\left\{G_n\right\}=S(G_0,G_1,a,b)$, (a)–(e) below hold:

(a) 

$G_{n+2}{G}_{n+m+1}-G_n{G}_{n+m-1}=a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m}+a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1}{G}_{n+m+1}$.

(b) 

$G_{n+1}{G}_{n+m}-G_{n+2}{G}_{n+m-1}={(-1)}^n(G_m{G}_3-G_{m+1}{G}_2).$

(c) 

$a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n-1}{G}_{n+m+1}-a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m}={(-1)}^n(a{G}_{m+1}{G}_1-b{G}_{m+2}{G}_0).$

(d) 

$G_{n+1}{G}_{n+m-2}-G_n{G}_{n+m-1}={(-1)}^n(G_m{G}_3-G_{m+1}{G}_2).$

(e) 

$G_{n-1}{G}_{n+m}-G_n{G}_{n+m-1}={(-1)}^n(G_m{G}_1-G_{m+1}{G}_0).$

Proof. 

Since

$$G_n=a^{\xi (n-1)}{b}^{\xi \left(n\right)}{G}_{n-1}+G_{n-2},$$

then, (a) follows from the identity

$$\begin{array}{ccc}\hfill G_n{G}_{n+m+1}& =& (G_{n+2}-a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1})G_{n+m+1}\hfill \\ & =& G_n(a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_{n+m}+G_{n+m-1}).\hfill \end{array}$$

(b)–(e) are special cases of ([18], Theorem 2.2). □

Theorem 1.

Consider the generalized bi-periodic Fibonacci numbers $\left\{G_n\right\}=S(G_0,G_1,a,b)$ and let

$${\Phi }_m:=b(G_m{G}_3-G_{m+1}{G}_2).$$

If $m\in \left\{0\right\}\cup {\mathbb{N}}_e$, then (a) and (b) below hold:

(a) 

If

$$\frac{{\Phi }_m}{ab+2}\notin \mathbb{Z},$$

define

$$g_m:=⌊\frac{{\Phi }_m}{ab+2}⌋+\Delta ,$$

where

$$\Delta =\left\{\begin{array}{c}1,if{\Phi }_m>0;\hfill \\ 0,if{\Phi }_m<0.\hfill \end{array}\right.$$

(i) If ${\Phi }_m>0$, then there exist positive integers $n_0$ and $n_1$ such that

$$⌊{\left(\sum _{k=n}^{\infty }\frac{{\left(\frac{a}{b}\right)}^{\xi (k+1)}}{G_k{G}_{k+m}}\right)}^{-1}⌋=\left\{\begin{array}{cc}b{G}_n{G}_{n+m-1}+g_m-1,\hfill & ifn\ge n_{0}andn\in {\mathbb{N}}_e;\hfill \\ b{G}_n{G}_{n+m-1}-g_m,\hfill & ifn\ge n_{1}andn\in {\mathbb{N}}_o.\hfill \end{array}\right.$$

(5)

(ii) If ${\Phi }_m<0$, then there exist positive integers $n_2$ and $n_3$ such that

$$⌊{\left(\sum _{k=n}^{\infty }\frac{{\left(\frac{a}{b}\right)}^{\xi (k+1)}}{G_k{G}_{k+m}}\right)}^{-1}⌋=\left\{\begin{array}{cc}b{G}_n{G}_{n+m-1}+g_m,\hfill & ifn\ge n_{2}andn\in {\mathbb{N}}_e;\hfill \\ b{G}_n{G}_{n+m-1}-g_m-1,\hfill & ifn\ge n_{3}andn\in {\mathbb{N}}_o.\hfill \end{array}\right.$$

(6)

(b) 

If

$$\frac{{\Phi }_m}{ab+2}\in \mathbb{Z},$$

define

$${\widehat{g}}_m:=\frac{{\Phi }_m}{ab+2},$$

and

$${\Gamma }_m:={\widehat{g}}_{m}b(b{G}_{m+2}{G}_0-a{G}_{m+1}{G}_1)-{\widehat{g}}_m^2.$$

(i) If ${\Gamma }_m\ge 0$, then there exist positive integers $n_4$ and $n_5$, such that

$$⌊{\left(\sum _{k=n}^{\infty }\frac{{\left(\frac{a}{b}\right)}^{\xi (k+1)}}{G_k{G}_{k+m}}\right)}^{-1}⌋=\left\{\begin{array}{cc}b{G}_n{G}_{n+m-1}+{\widehat{g}}_m,\hfill & ifn\ge n_{4}andn\in {\mathbb{N}}_e;\hfill \\ b{G}_n{G}_{n+m-1}-{\widehat{g}}_m,\hfill & ifn\ge n_{5}andn\in {\mathbb{N}}_o.\hfill \end{array}\right.$$

(7)

(ii) If ${\Gamma }_m<0$, then there exist positive integers $n_4$ and $n_5$, such that

$$⌊{\left(\sum _{k=n}^{\infty }\frac{{\left(\frac{a}{b}\right)}^{\xi (k+1)}}{G_k{G}_{k+m}}\right)}^{-1}⌋=\left\{\begin{array}{cc}b{G}_n{G}_{n+m-1}+{\widehat{g}}_m-1,\hfill & ifn\ge n_{6}andn\in {\mathbb{N}}_e;\hfill \\ b{G}_n{G}_{n+m-1}-{\widehat{g}}_m-1,\hfill & ifn\ge n_{7}andn\in {\mathbb{N}}_o.\hfill \end{array}\right.$$

(8)

Proof. 

(a) To prove (5), assume that ${\Phi }_m>0$. Then

$${\Phi }_m-g_m(ab+2)<0.$$

Firstly, consider

$$\begin{array}{ccc}\hfill X_1& =& \frac{1}{b{G}_n{G}_{n+m-1}+{(-1)}^n{g}_m}-\frac{1}{b{G}_{n+2}{G}_{n+m+1}+{(-1)}^n{g}_m}-\frac{{\left(\frac{a}{b}\right)}^{\xi (n+1)}}{G_n{G}_{n+m}}-\frac{{\left(\frac{a}{b}\right)}^{\xi \left(n\right)}}{G_{n+1}{G}_{n+m+1}}\hfill \\ & =& \frac{Y_1}{(b{G}_n{G}_{n+m-1}+{(-1)}^n{g}_m)(b{G}_{n+2}{G}_{n+m+1}+{(-1)}^n{g}_m)G_n{G}_{n+1}{G}_{n+m}{G}_{n+m+1}},\hfill \end{array}$$

where, by Lemma 1 (a)

$$Y_1=\left\{{\left(\frac{a}{b}\right)}^{\xi \left(n\right)}{G}_n{G}_{n+m}+{\left(\frac{a}{b}\right)}^{\xi (n+1)}{G}_{n+1}{G}_{n+m+1}\right\}{\widehat{Y}}_1,$$

with

$$\begin{array}{c}\hfill {\widehat{Y}}_1=b^2(G_n{G}_{n+1}{G}_{n+m}{G}_{n+m+1}-G_n{G}_{n+2}{G}_{n+m-1}{G}_{n+m+1})-{(-1)}^n{g}_{m}b(G_n{G}_{n+m-1}+G_{n+2}{G}_{n+m+1})-g_m^2.\end{array}$$

By Lemma 1 (b,c), we have

$$\begin{array}{ccc}\hfill G_n{G}_{n+1}{G}_{n+m}{G}_{n+m+1}-G_n{G}_{n+2}{G}_{n+m-1}{G}_{n+m+1}& =& (G_{n+1}{G}_{n+m}-G_{n+2}{G}_{n+m-1})G_n{G}_{n+m+1}\hfill \\ & =& {(-1)}^n(G_m{G}_3-G_{m+1}{G}_2)G_n{G}_{n+m+1},\hfill \end{array}$$

and

$$\begin{array}{ccc}& & G_n{G}_{n+m-1}+G_{n+2}{G}_{n+m+1}\hfill \\ & =& G_n(G_{n+m+1}-a^{\xi (n+m)}{b}^{\xi (n+m+1)}{G}_{n+m})+(a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1}+G_n)G_{n+m+1}\hfill \\ & =& 2G_n{G}_{n+m+1}-a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m}+a^{\xi (n+1)}{b}^{\xi \left(n\right)}(a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n+G_{n-1})G_{n+m+1}\hfill \\ & =& (ab+2)G_n{G}_{n+m+1}-a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m}+a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n-1}{G}_{n+m+1}\hfill \\ & =& (ab+2)G_n{G}_{n+m+1}+{(-1)}^n(a{G}_{m+1}-b{G}_{m+2}{G}_0).\hfill \end{array}$$

Then

$$\begin{array}{ccc}\hfill {\widehat{Y}}_1& =& {(-1)}^n{b}^2(G_m{G}_3-G_{m+1}{G}_2)G_n{G}_{n+m+1}\hfill \\ & & -{(-1)}^n{g}_{m}b\left\{(ab+2)G_n{G}_{n+m+1}+{(-1)}^n(a{G}_{m+1}-b{G}_{m+2}{G}_0)\right\}-g_m^2\hfill \\ & =& {(-1)}^{n}b\left\{{\Phi }_m-g(ab+2)\right\}{G}_n{G}_{n+m+1}+g_{m}b(b{G}_{m+2}{G}_0-a{G}_{m+1}{G}_1)-g_m^2.\hfill \end{array}$$

If $n\in {\mathbb{N}}_e$, then there exists a positive integer $m_0$ such that, for $n\ge m_0$, $X_1<0$, and

$$\frac{1}{b{G}_n{G}_{n+m-1}+g_m}-\frac{1}{b{G}_{n+2}{G}_{n+m+1}+g_m}<\frac{{\left(\frac{a}{b}\right)}^{\xi (n+1)}}{G_n{G}_{n+m}}+\frac{{\left(\frac{a}{b}\right)}^{\xi \left(n\right)}}{G_{n+1}{G}_{n+m+1}}.$$

Repeatedly applying the above inequality, we have

$$\frac{1}{b{G}_n{G}_{n+m-1}+g_m}<\sum _{k=n}^{\infty }\frac{{\left(\frac{a}{b}\right)}^{\xi (k+1)}}{G_k{G}_{k+m}},\mathrm{if}n\ge m_0\mathrm{and}n\in {\mathbb{N}}_e.$$

(9)

Similarly, we obtain, for some positive integer $m_1$,

$$\sum _{k=n}^{\infty }\frac{{\left(\frac{a}{b}\right)}^{\xi (k+1)}}{G_k{G}_{k+m}}<\frac{1}{b{G}_n{G}_{n+m-1}-g_m},\mathrm{if}n\ge m_1\mathrm{and}n\in {\mathbb{N}}_o.$$

(10)

Next, consider

$$\begin{array}{ccc}\hfill X_2& =& \frac{1}{b{G}_n{G}_{n+m-1}+{(-1)}^n{g}_m-1}-\frac{1}{b{G}_{n+1}{G}_{n+m}+{(-1)}^{n+1}{g}_m-1}-\frac{{\left(\frac{a}{b}\right)}^{\xi (n+1)}}{G_n{G}_{n+m}}\hfill \\ & =& \frac{Y_2}{(b{G}_n{G}_{n+m-1}+{(-1)}^n{g}_m-1)(b{G}_{n+1}{G}_{n+m}+{(-1)}^{n+1}{g}_m-1)G_n{G}_{n+m}},\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill Y_2& =& b{G}_n{G}_{n+1}{G}_{n+m}^2-a^{\xi (n+1)}{b}^{1+\xi \left(n\right)}{G}_n{G}_{n+1}{G}_{n+m-1}{G}_{n+m}-b{G}_n^2{G}_{n+m-1}{G}_{n+m+1}\hfill \\ & & -{(-1)}^n{g}_m\left\{2G_n{G}_{n+m}+a^{\xi (n+1)}{b}^{\xi \left(n\right)}(G_{n+1}{G}_{n+m}-G_n{G}_{n+m-1})\right\}\hfill \\ & & +a^{\xi (n+1)}{b}^{\xi \left(n\right)}(G_n{G}_{n+m-1}+G_{n+1}{G}_{n+m})+a^{\xi (n+1)}{b}^{\xi \left(n\right)-1}(g_m^2-1).\hfill \end{array}$$

Using Lemma 1 (d,e), we have

$$\begin{array}{ccc}& & b{G}_n{G}_{n+1}{G}_{n+m}^2-a^{\xi (n+1)}{b}^{1+\xi \left(n\right)}{G}_n{G}_{n+1}{G}_{n+m-1}{G}_{n+m}-b{G}_n^2{G}_{n+m-1}{G}_{n+m}\hfill \\ & =& b{G}_n{G}_{n+1}{G}_{n+m}(G_{n+m}-a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+m-1})-b{G}_n^2{G}_{n+m-1}{G}_{n+m}\hfill \\ & =& b{G}_n{G}_{n+m}(G_{n+1}{G}_{n+m-2}-G_n{G}_{n+m-1})\hfill \\ & =& {(-1)}^{n}b(G_m{G}_3-G_{m+1}{G}_2)G_n{G}_{n+m},\hfill \end{array}$$

and

$$\begin{array}{ccc}& & 2G_n{G}_{n+m}+a^{\xi (n+1)}{b}^{\xi \left(n\right)}(G_{n+1}{G}_{n+m}-G_n{G}_{n+m-1})\hfill \\ & =& 2G_n{G}_{n+m}+a^{\xi (n+1)}{b}^{\xi \left(n\right)}(a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n+G_{n-1})G_{n+m}-a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_n{G}_{n+m-1})\hfill \\ & =& (ab+2)G_n{G}_{n+m}+a^{\xi (n+1)}{b}^{\xi \left(n\right)}(G_{n-1}{G}_{n+m}-G_n{G}_{n+m-1})\hfill \\ & =& (ab+2)G_n{G}_{n+m}+{(-1)}^n{a}^{\xi (n+1)}{b}^{\xi \left(n\right)}(G_m{G}_1-G_{m+1}{G}_0).\hfill \end{array}$$

Hence we obtain

$$\begin{array}{ccc}\hfill Y_2& =& {(-1)}^n\left\{{\Phi }_m-g_m(ab+2)\right\}{G}_n{G}_{n+m}+a^{\xi (n+1)}{b}^{\xi \left(n\right)}(G_n{G}_{n+m-1}+G_{n+1}{G}_{n+m})\hfill \\ & & -a^{\xi (n+1)}{b}^{\xi \left(n\right)}{g}_m(G_m{G}_1-G_{m+1}{G}_0)+a^{\xi (n+1)}{b}^{\xi \left(n\right)-1}(g_m^2-1),\hfill \end{array}$$

and there exists a positive integer $m_2$ such that, for $n\ge m_2$, $X_2>0$, and

$$\frac{{\left(\frac{a}{b}\right)}^{\xi (n+1)}}{G_n{G}_{n+m}}<\frac{1}{b{G}_n{G}_{n+m-1}+{(-1)}^n{g}_m-1}-\frac{1}{b{G}_{n+1}{G}_{n+m}+{(-1)}^{n+1}{g}_m-1}.$$

Repeatedly applying the above inequality, we have

$$\sum _{k=n}^{\infty }\frac{{\left(\frac{a}{b}\right)}^{\xi (k+1)}}{G_k{G}_{k+m}}<\frac{1}{b{G}_n{G}_{n+m-1}+{(-1)}^n{g}_m-1},\mathrm{if}n\ge m_2.$$

(11)

Similarly, consider

$$\begin{array}{ccc}\hfill X_3& =& \frac{1}{b{G}_n{G}_{n+m-1}+{(-1)}^n{g}_m+1}-\frac{1}{b{G}_{n+1}{G}_{n+m}+{(-1)}^{n+1}{g}_m+1}-\frac{{\left(\frac{a}{b}\right)}^{\xi (n+1)}}{G_n{G}_{n+m}}\hfill \\ & =& \frac{Y_3}{(b{G}_n{G}_{n+m-1}+{(-1)}^n{g}_m+1)(b{G}_{n+1}{G}_{n+m}+{(-1)}^{n+1}{g}_m+1)G_n{G}_{n+m}},\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill Y_3& =& Y_2-2a^{\xi (n+1)}{b}^{\xi \left(n\right)}(G_n{G}_{n+m-1}+G_{n+1}{G}_{n+m})\hfill \\ & =& {(-1)}^n\left\{{\Phi }_m-g_m(ab+2)\right\}{G}_n{G}_{n+m}-a^{\xi (n+1)}{b}^{\xi \left(n\right)}(G_n{G}_{n+m-1}+G_{n+1}{G}_{n+m})\hfill \\ & & -a^{\xi (n+1)}{b}^{\xi \left(n\right)}{g}_m(G_m{G}_1-G_{m+1}{G}_0)+a^{\xi (n+1)}{b}^{\xi \left(n\right)-1}(g_m^2-1).\hfill \end{array}$$

There exists a positive integer $m_3$ such that, for $n\ge m_3$, $X_3<0$, and

$$\frac{1}{b{G}_n{G}_{n+m-1}+{(-1)}^n{g}_m+1}-\frac{1}{b{G}_{n+1}{G}_{n+m}+{(-1)}^{n+1}{g}_m+1}<\frac{{\left(\frac{a}{b}\right)}^{\xi (n+1)}}{G_n{G}_{n+m}},$$

from which we have

$$\frac{1}{b{G}_n{G}_{n+m-1}+{(-1)}^n{g}_m+1}<\sum _{k=n}^{\infty }\frac{{\left(\frac{a}{b}\right)}^{\xi (k+1)}}{G_k{G}_{k+m}},\mathrm{if}n\ge m_3.$$

(12)

Then (5) follows from (9)–(12).

Now, suppose that ${\Phi }_m<0$. In this case, we have

$${\Phi }_m-g_m(ab+2)>0,$$

and (9)–(12) are respectively modified as

$$\sum _{k=n}^{\infty }\frac{{\left(\frac{a}{b}\right)}^{\xi (k+1)}}{G_k{G}_{k+m}}<\frac{1}{b{G}_n{G}_{n+m-1}+g_m},\mathrm{if}n\ge m_4\mathrm{and}n\in {\mathbb{N}}_e,$$

(13)

$$\frac{1}{b{G}_n{G}_{n+m-1}-g_m}<\sum _{k=n}^{\infty }\frac{{\left(\frac{a}{b}\right)}^{\xi (k+1)}}{G_k{G}_{k+m}},\mathrm{if}n\ge m_5\mathrm{and}n\in {\mathbb{N}}_o,$$

(14)

$$\sum _{k=n}^{\infty }\frac{{\left(\frac{a}{b}\right)}^{\xi (k+1)}}{G_k{G}_{k+m}}<\frac{1}{b{G}_n{G}_{n+m-1}+{(-1)}^n{g}_m-1},\mathrm{if}n\ge m_6,$$

(15)

and

$$\frac{1}{b{G}_n{G}_{n+m-1}+{(-1)}^n{g}_m+1}<\sum _{k=n}^{\infty }\frac{{\left(\frac{a}{b}\right)}^{\xi (k+1)}}{G_k{G}_{k+m}},\mathrm{if}n\ge m_7.$$

(16)

Then, (6) easily follows and the proof of (a) is completed.

(b) Suppose that

$$\frac{{\Phi }_m}{ab+2}\in \mathbb{Z}.$$

We recall the proof of (a). If ${\mathsf{\Gamma }}_m\ge 0$, then replacing $g_m$ by ${\widehat{g}}_m$, we have ${\widehat{Y}}_1={\mathsf{\Gamma }}_m\ge 0$, and there exists a positive integer $m_8$ such that $X_1>0$ if $n\ge m_8$. Hence, we obtain

$$\sum _{k=n}^{\infty }\frac{{\left(\frac{a}{b}\right)}^{\xi (k+1)}}{G_k{G}_{k+m}}<\frac{1}{b{G}_n{G}_{n+m-1}+{(-1)}^n{\widehat{g}}_m},\mathrm{if}n\ge m_8.$$

(17)

Similarly, there exists a positive integer $m_9$ such that $X_3<0$ if $n\ge m_9$, from which we have

$$\frac{1}{b{G}_n{G}_{n+m}+{(-1)}^n\widehat{g}+1}<\sum _{k=n}^{\infty }\frac{{\left(\frac{a}{b}\right)}^{\xi (k+1)}}{G_k{G}_{k+m}},\mathrm{if}n\ge m_9.$$

(18)

Then, (7) follows from (17) and (18). (8) can be proved similarly, and details are omitted. □

If $a=b$, then Theorem 1 reduces ([7], Theorem 2.1) with $m\in \left\{0\right\}\cup {\mathbb{N}}_e$ and $\left\{G_n\right\}=\left\{H_n\right\}=S(G_0,G_1,a,a)$.

If $m=0$, then Theorem 1 reduces ([17], Theorem 2).

2.2. The Case Where $m\in {\mathbb{N}}_o$

To deal with the case where $m\in {\mathbb{N}}_o$, we need the following lemma.

Lemma 2.

Assume that $m\in {\mathbb{N}}_o$. Then, for $\left\{G_n\right\}=S(G_0,G_1,a,b)$, (a)–(e) below hold:

(a) 

$G_{n+2}{G}_{n+m+1}-G_n{G}_{n+m-1}=a^{\xi (n+1)}{b}^{\xi \left(n\right)}(G_n{G}_{n+m}+G_{n+1}{G}_{n+m+1})$.

(b) 

$a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1}{G}_{n+m}-a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_{n+2}{G}_{n+m-1}={(-1)}^n(a{G}_m{G}_3-b{G}_{m+1}{G}_2).$

(c) 

$G_{n-1}{G}_{n+m+1}-G_n{G}_{n+m}={(-1)}^n(G_{m+2}{G}_0-G_{m+1}{G}_1).$

(d) 

$a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1}{G}_{n+m-2}-a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}={(-1)}^n(a{G}_m{G}_3-b{G}_{m+1}{G}_2).$

(e) 

$a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n-1}{G}_{n+m}-a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}={(-1)}^n(a{G}_m{G}_1-b{G}_{m+1}{G}_0).$

Proof. 

(a) follows from the identity

$$\begin{array}{ccc}\hfill G_n{G}_{n+m+1}& =& (G_{n+2}-a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1})G_{n+m+1}\hfill \\ & =& G_n(a^{\xi (n+m)}{b}^{\xi (n+m+1)}{G}_{n+m}+G_{n+m-1}).\hfill \end{array}$$

(b)–(e) are special cases of ([18], Theorem 2.2). □

Theorem 2.

Consider the generalized bi-periodic Fibonacci numbers $\left\{G_n\right\}=S(G_0,G_1,a,b)$ and let

$${\Delta }_m:=a{G}_m{G}_3-b{G}_{m+1}{G}_2.$$

If $m\in {\mathbb{N}}_0$, then (a) and (b) below hold:

(a) 

If

$$\frac{{\Delta }_m}{ab+2}\notin \mathbb{Z},$$

define

$$h_m:=⌊\frac{{\Delta }_m}{ab+2}⌋+\Delta ,$$

where

$$\Delta =\left\{\begin{array}{c}1,if{\Delta }_m>0;\hfill \\ 0,if{\Delta }_m<0.\hfill \end{array}\right.$$

(i) If ${\Delta }_m>0$, then there exist positive integers $n_0$ and $n_1$ such that

$$⌊{\left(\sum _{k=n}^{\infty }\frac{1}{G_k{G}_{k+m}}\right)}^{-1}⌋=\left\{\begin{array}{cc}b{G}_n{G}_{n+m-1}+h_m-1,\hfill & ifn\ge n_{0}andn\in {\mathbb{N}}_e;\hfill \\ a{G}_n{G}_{n+m-1}-h_m,\hfill & ifn\ge n_{1}andn\in {\mathbb{N}}_o.\hfill \end{array}\right.$$

(19)

(ii) If ${\Delta }_m<0$, then there exist positive integers $n_2$ and $n_3$ such that

$$⌊{\left(\sum _{k=n}^{\infty }\frac{1}{G_k{G}_{k+m}}\right)}^{-1}⌋=\left\{\begin{array}{cc}b{G}_n{G}_{n+m-1}+h_m,\hfill & ifn\ge n_{2}andn\in {\mathbb{N}}_e;\hfill \\ a{G}_n{G}_{n+m-1}-h_m-1,\hfill & ifn\ge n_{3}andn\in {\mathbb{N}}_o.\hfill \end{array}\right.$$

(20)

(b) 

If

$$\frac{{\Delta }_m}{ab+2}\in \mathbb{Z},$$

define

$${\widehat{h}}_m:=\frac{{\Delta }_m}{ab+2},$$

and

$${\Psi }_m:={\widehat{h}}_{m}ab(G_{m+1}{G}_1-G_{m+2}{G}_0)-{\widehat{h}}_m^2.$$

(i) If ${\Psi }_m\ge 0$, then there exist positive integers $n_4$ and $n_5$ such that

$$⌊{\left(\sum _{k=n}^{\infty }\frac{1}{G_k{G}_{k+m}}\right)}^{-1}⌋=\left\{\begin{array}{c}a{G}_n{G}_{n+m-1}+{\widehat{h}}_m,ifn\ge n_{4}andn\in {\mathbb{N}}_e;\hfill \\ b{G}_n{G}_{n+m-1}-{\widehat{h}}_m,ifn\ge n_{5}andn\in {\mathbb{N}}_o.\hfill \end{array}\right.$$

(21)

(ii) If ${\Psi }_m<0$, then there exist positive integers $n_6$ and $n_7$ such that

$$⌊{\left(\sum _{k=n}^{\infty }\frac{1}{G_k{G}_{k+m}}\right)}^{-1}⌋=\left\{\begin{array}{c}a{G}_n{G}_{n+m-1}+{\widehat{h}}_m-1,ifn\ge n_{6}andn\in {\mathbb{N}}_e;\hfill \\ b{G}_n{G}_{n+m-1}-{\widehat{h}}_m-1,ifn\ge n_{7}andn\in {\mathbb{N}}_o.\hfill \end{array}\right.$$

(22)

Proof. 

(a) To prove (19), assume that ${\Delta }_m>0$. Then

$${\Delta }_m-h_m(ab+2)<0.$$

Firstly, consider

$$\begin{array}{ccc}\hfill X_1& =& \frac{1}{a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}+{(-1)}^n{h}_m}-\frac{1}{a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_{n+2}{G}_{n+m+1}+{(-1)}^n{h}_m}-\frac{1}{G_n{G}_{n+m}}-\frac{1}{G_{n+1}{G}_{n+m+1}}\hfill \\ & =& \frac{Y_1}{(a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}+{(-1)}^n{h}_m)(a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_{n+2}{G}_{n+m+1}+{(-1)}^n{h}_m)G_n{G}_{n+1}{G}_{n+m}{G}_{n+m+1}},\hfill \end{array}$$

where, by Lemma 2 (a)

$$Y_1=(G_n{G}_{n+m}+G_{n+1}{G}_{n+m+1}){\widehat{Y}}_1,$$

with

$$\begin{array}{ccc}\hfill {\widehat{Y}}_1& =& ab{G}_n{G}_{n+1}{G}_{n+m}{G}_{n+m+1}-a^{2\xi \left(n\right)}{b}^{2\xi (n+1)}{G}_n{G}_{n+2}{G}_{n+m-1}{G}_{n+m+1}\hfill \\ & & -{(-1)}^n{h}_m{a}^{\xi \left(n\right)}{b}^{\xi (n+1)}(G_n{G}_{n+m-1}+G_{n+2}{G}_{n+m+1})-h_m^2.\hfill \end{array}$$

By Lemma 2 (b,c), we have

$$\begin{array}{ccc}& & ab{G}_n{G}_{n+1}{G}_n{G}_{n+m}-a^{2\xi \left(n\right)}{b}^{2\xi (n+1)}{G}_n{G}_{n+2}{G}_{n+m-1}{G}_{n+m+1}\hfill \\ & =& (ab{G}_{n+1}{G}_{n+m}-a^{2\xi \left(n\right)}{b}^{2\xi (n+1)}{G}_{n+2}{G}_{n+m-1})G_n{G}_{n+m+1}\hfill \\ & =& a^{\xi \left(n\right)}{b}^{\xi (n+1)}(a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1}{G}_{n+m}-a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_{n+2}{G}_{n+m-1})G_n{G}_{n+m+1}\hfill \\ & =& {(-1)}^n{a}^{\xi \left(n\right)}{b}^{\xi (n+1)}(a{G}_m{G}_3-b{G}_{m+1}{G}_2)G_n{G}_{n+m+1},\hfill \end{array}$$

and

$$\begin{array}{ccc}& & G_n{G}_{n+m-1}+G_{n+2}{G}_{n+m+1}\hfill \\ & =& G_n(G_{n+m+1}-a^{\xi (n+m)}{b}^{\xi (n+m+1)}{G}_{n+m})+(a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1}+G_n)G_{n+m+1}\hfill \\ & =& 2G_n{G}_{n+m+1}-a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_n{G}_{n+m}+a^{\xi (n+1)}{b}^{\xi \left(n\right)}(a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n+G_{n-1})G_{n+m+1}\hfill \\ & =& (ab+2)G_n{G}_{n+m+1}+a^{\xi (n+1)}{b}^{\xi \left(n\right)}(G_{n-1}{G}_{n+m+1}-G_n{G}_{n+m})\hfill \\ & =& (ab+2)G_n{G}_{n+m+1}+{(-1)}^n{a}^{\xi (n+1)}{b}^{\xi \left(n\right)}(G_{m+2}{G}_0-G_{m+1}{G}_1).\hfill \end{array}$$

Then

$$\begin{array}{ccc}\hfill {\widehat{Y}}_1& =& {(-1)}^n{a}^{\xi \left(n\right)}{b}^{\xi (n+1)}(a{G}_m{G}_3-b{G}_{m+1}{G}_2)G_n{G}_{n+m+1}\hfill \\ & & -{(-1)}^n{h}_m{a}^{\xi \left(n\right)}{b}^{\xi (n+1)}\left\{(ab+2)G_n{G}_{n+m+1}+{(-1)}^n{a}^{\xi (n+1)}{b}^{\xi \left(n\right)}(G_{m+2}{G}_0-G_{m+1}{G}_1)\right\}-h_m^2\hfill \\ & =& {(-1)}^n{a}^{\xi \left(n\right)}{b}^{\xi (n+1)}\left\{{\Delta }_m-h_m(ab+2)\right\}{G}_n{G}_{n+m+1}+h_{m}ab(G_{m+1}{G}_1-G_{m+2}{G}_0)-h_m^2.\hfill \end{array}$$

If $n\in {\mathbb{N}}_e$, then there exists a positive integer $m_0$ such that, for $n\ge m_0$, $X_1<0$, and

$$\frac{1}{a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}+h_m}-\frac{1}{a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_{n+2}{G}_{n+m+1}+h_m}<\frac{1}{G_n{G}_{n+m}}+\frac{1}{G_{n+1}{G}_{n+m+1}}.$$

Repeatedly applying the above inequality, we have

$$\frac{1}{a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}+h_m}<\sum _{k=n}^{\infty }\frac{1}{G_k{G}_{k+m}},\mathrm{if}n\ge m_0\mathrm{and}n\in {\mathbb{N}}_e.$$

(23)

Similarly, we obtain, for some positive integer $m_1$,

$$\sum _{k=n}^{\infty }\frac{1}{G_k{G}_{k+m}}<\frac{1}{a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}-h_m},\mathrm{if}n\ge m_1\mathrm{and}n\in {\mathbb{N}}_o.$$

(24)

Next, consider

$$\begin{array}{ccc}\hfill X_2& =& \frac{1}{a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}+{(-1)}^n{h}_m-1}-\frac{1}{a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1}{G}_{n+m}+{(-1)}^{n+1}{h}_m-1}-\frac{1}{G_n{G}_{n+m}}\hfill \\ & =& \frac{Y_2}{(a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}+{(-1)}^n{h}_m-1)(a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1}{G}_{n+m}+{(-1)}^{n+1}{h}_m-1)G_n{G}_{n+m}},\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill Y_2& =& a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_n{G}_{n+1}{G}_{n+m}^2-ab{G}_n{G}_{n+1}{G}_{n+m-1}{G}_{n+m}-a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n^2{G}_{n+m-1}{G}_{n+m+1}\hfill \\ & & -{(-1)}^n{h}_m(2G_n{G}_{n+m}+a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1}{G}_{n+m}-a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1})\hfill \\ & & +a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}+a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1}{G}_{n+m})+h_m^2-1.\hfill \end{array}$$

Using Lemma 2 (d,e), we have

$$\begin{array}{ccc}& & a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_n{G}_{n+1}{G}_{n+m}^2-ab{G}_n{G}_{n+1}{G}_{n+m-1}{G}_{n+m}-a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n^2{G}_{n+m-1}{G}_{n+m+1}\hfill \\ & =& a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_n{G}_{n+1}{G}_{n+m}(G_{n+m}-a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_{n+m-1})-a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n^2{G}_{n+m-1}{G}_{n+m}\hfill \\ & =& G_n{G}_{n+m}(a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1}{G}_{n+m-2}-a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1})\hfill \\ & =& {(-1)}^n(a{G}_m{G}_3-b{G}_{m+1}{G}_2)G_n{G}_{n+m},\hfill \end{array}$$

and

$$\begin{array}{ccc}& & 2G_n{G}_{n+m}+a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1}{G}_{n+m}-a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1})\hfill \\ & =& 2G_n{G}_{n+m}+a^{\xi (n+1)}{b}^{\xi \left(n\right)}(a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n+G_{n-1})G_{n+m}-a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1})\hfill \\ & =& (ab+2)2G_n{G}_{n+m}+a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n-1}{G}_{n+m}-a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1})\hfill \\ & =& (ab+2)2G_n{G}_{n+m}+{(-1)}^n(a{G}_m{G}_1-b{G}_{m+1}{G}_0).\hfill \end{array}$$

Hence we obtain

$$\begin{array}{ccc}\hfill Y_2& =& {(-1)}^n\left\{{\Delta }_m-h_m(ab+2)\right\}{G}_n{G}_{n+m}+a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}+a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1}{G}_{n+m})\hfill \\ & & -h_m(a{G}_m{G}_1-b{G}_{m+1}{G}_0)+h_m^2-1,\hfill \end{array}$$

and there exists a positive integer $m_2$ such that, for $n\ge m_2$, $X_2>0$, and

$$\frac{1}{G_n{G}_{n+m}}<\frac{1}{a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}+{(-1)}^n{h}_m-1}-\frac{1}{a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1}{G}_{n+m}+{(-1)}^{n+1}{h}_m-1}.$$

Repeatedly applying the above inequality, we have

$$\sum _{k=n}^{\infty }\frac{1}{G_k{G}_{k+m}}<\frac{1}{a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}+{(-1)}^n{h}_m-1},\mathrm{if}n\ge m_2.$$

(25)

Finally, consider

$$\begin{array}{ccc}\hfill X_3& =& \frac{1}{a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}+{(-1)}^n{h}_m+1}-\frac{1}{a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1}{G}_{n+m}+{(-1)}^{n+1}{h}_m+1}-\frac{1}{G_n{G}_{n+m}}\hfill \\ & =& \frac{Y_3}{(a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}+{(-1)}^n{h}_m+1)(a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1}{G}_{n+m}+{(-1)}^{n+1}{h}_m+1)G_n{G}_{n+m}},\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill Y_3& =& Y_2-2(a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}+a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1}{G}_{n+m})\hfill \\ & =& {(-1)}^n\left\{{\Delta }_m-h_m(ab+2)\right\}{G}_n{G}_{n+m}-a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}-a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1}{G}_{n+m})\hfill \\ & & -h_m(a{G}_m{G}_1-b{G}_{m+1}{G}_0)+h_m^2-1.\hfill \end{array}$$

There exists a positive integer $m_3$ such that, for $n\ge m_3$, $X_3<0$, and

$$\frac{1}{a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}+{(-1)}^n{h}_m+1}-\frac{1}{a^{\xi (n+1)}{b}^{\xi \left(n\right)}{G}_{n+1}{G}_{n+m}+{(-1)}^{n+1}{h}_m+1}<\frac{1}{G_n{G}_{n+m}}.$$

Repeatedly applying the above inequality, we have

$$\frac{1}{a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}+{(-1)}^n{h}_m+1}<\sum _{k=n}^{\infty }\frac{1}{G_k{G}_{k+m}},\mathrm{if}n\ge m_2.$$

(26)

Then (19) follows from (23)–(26).

Now suppose that ${\Delta }_m<0$. In this case, we have

$${\Delta }_m-h_m(ab+2)>0,$$

and (23)–(26) are respectively modified as

$$\sum _{k=n}^{\infty }\frac{1}{G_k{G}_{k+m}}<\frac{1}{a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}+h_m},\mathrm{if}n\ge m_4\mathrm{and}n\in {\mathbb{N}}_e,$$

(27)

$$\frac{1}{a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}-h_m}<\sum _{k=n}^{\infty }\frac{1}{G_k{G}_{k+m}},\mathrm{if}n\ge m_5\mathrm{and}n\in {\mathbb{N}}_o,$$

(28)

$$\sum _{k=n}^{\infty }\frac{1}{G_k{G}_{k+m}}<\frac{1}{a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}+{(-1)}^n{h}_m-1},\mathrm{if}n\ge m_6,$$

(29)

and

$$\frac{1}{a^{\xi \left(n\right)}{b}^{\xi (n+1)}{G}_n{G}_{n+m-1}+{(-1)}^n{h}_m+1}<\sum _{k=n}^{\infty }\frac{1}{G_k{G}_{k+m}},\mathrm{if}n\ge m_7.$$

(30)

Then, (20) easily follows and the proof of (a) is completed.

(b) (21) and (22) can be proved as in Theorem 1, and details are omitted. □

If $a=b$, then Theorem 2 reduces ([7], (Theorem 2.1) with $m\in {\mathbb{N}}_o$ and $\left\{G_n\right\}=\left\{H_n\right\}=S(G_0,G_1,a,a)$.

3. Discussion

This paper concerned the properties of the generalized bi-periodic Fibonacci numbers $\left\{G_n\right\}$ generated from the recurrence relation: $G_n=a{G}_{n-1}+G_{n-2}$ (n is even) or $G_n=b{G}_{n-1}+G_{n-2}$ (n is odd). We derived quite a general identities related to reciprocal sums of products of two generalized bi-periodic Fibonacci numbers. More precisely, we obtained formulas for the integer parts of the numbers

$${\left(\sum _{k=n}^{\infty }\frac{{\left(\frac{a}{b}\right)}^{\xi (k+1)}}{G_k{G}_{k+m}}\right)}^{-1},m=0,2,4,\cdots ,$$

and

$${\left(\sum _{k=n}^{\infty }\frac{1}{G_k{G}_{k+m}}\right)}^{-1},m=1,3,5,\cdots .$$

The identities obtained in this paper include many existing results as special cases. As already noted in [16], an open problem is whether we can obtain similar results for the same numbers of higher order. It seems that we can also derive similar identities for the numbers of the form

$${\left(\sum _{k=n}^{\infty }\frac{1}{G_k{H}_{k+m}}\right)}^{-1},m=0,1,2,3,\cdots ,$$

where $\left\{G_n\right\}=S(G_0,G_1,a,b)$ and $\left\{H_n\right\}=S(H_0,H_1,a,b)$, which is left as another open problem.