The influence of synaptic strength and noise on the robustness of central pattern generator

Feibiao Zhan; Jian Song; Shenquan Liu; Feibiao Zhan; Jian Song; Shenquan Liu

doi:10.3934/era.2024033

Electronic Research Archive

2024, Volume 32, Issue 1: 686-706. doi: 10.3934/era.2024033

Previous Article Next Article

Research article Special Issues

The influence of synaptic strength and noise on the robustness of central pattern generator

1.
School of Mathematics, Nanjing Audit University, Nanjing 211815, China
2.
School of Mathematics, South China University of Technology, Guangzhou 510640, China
3.
School of Mathematical and Computational Sciences, Massey University, Auckland 4442, New Zealand

Received: 25 September 2023 Revised: 23 December 2023 Accepted: 27 December 2023 Published: 09 January 2024

In this paper, we explore the mechanisms of central pattern generators (CPGs), circuits that can generate rhythmic patterns of motor activity without external input. We study the half-center oscillator, a simple form of CPG circuit consisting of neurons connected by reciprocally inhibitory synapses. We examine the role of asymmetric coupling factors in shaping rhythm activity and how different network topologies contribute to network efficiency. We have discovered that neurons with lower synaptic strength are more susceptible to noise that affects rhythm changes. Our research highlights the importance of asymmetric coupling factors, noise, and other synaptic parameters in shaping the broad regimes of CPG rhythm. Finally, we compare three topology types' regular regimes and provide insights on how to locate the rhythm activity.

Keywords:

Citation: Feibiao Zhan, Jian Song, Shenquan Liu. The influence of synaptic strength and noise on the robustness of central pattern generator[J]. Electronic Research Archive, 2024, 32(1): 686-706. doi: 10.3934/era.2024033

Related Papers:

[1]	Tingting Du, Zhengang Wu . On the reciprocal sums of products of $ m $th-order linear recurrence sequences. Electronic Research Archive, 2023, 31(9): 5766-5779. doi: 10.3934/era.2023293
[2]	Qingjie Chai, Hanyu Wei . The binomial sums for four types of polynomials involving floor and ceiling functions. Electronic Research Archive, 2025, 33(3): 1384-1397. doi: 10.3934/era.2025064
[3]	Jiafan Zhang . On the distribution of primitive roots and Lehmer numbers. Electronic Research Archive, 2023, 31(11): 6913-6927. doi: 10.3934/era.2023350
[4]	Jin Zhang, Xiaoxue Li . The sixth power mean of one kind generalized two-term exponential sums and their asymptotic properties. Electronic Research Archive, 2023, 31(8): 4579-4591. doi: 10.3934/era.2023234
[5]	Jianghua Li, Zepeng Zhang, Tianyu Pan . On two-term exponential sums and their mean values. Electronic Research Archive, 2023, 31(9): 5559-5572. doi: 10.3934/era.2023282
[6]	Jianxing Du, Xifeng Su . On the existence of solutions for the Frenkel-Kontorova models on quasi-crystals. Electronic Research Archive, 2021, 29(6): 4177-4198. doi: 10.3934/era.2021078
[7]	Li Wang, Yuanyuan Meng . Generalized polynomial exponential sums and their fourth power mean. Electronic Research Archive, 2023, 31(7): 4313-4323. doi: 10.3934/era.2023220
[8]	Guo-Niu Han, Huan Xiong . Skew doubled shifted plane partitions: Calculus and asymptotics. Electronic Research Archive, 2021, 29(1): 1841-1857. doi: 10.3934/era.2020094
[9]	Zhefeng Xu, Xiaoying Liu, Luyao Chen . Hybrid mean value involving some two-term exponential sums and fourth Gauss sums. Electronic Research Archive, 2025, 33(3): 1510-1522. doi: 10.3934/era.2025071
[10]	Dmitry Krachun, Zhi-Wei Sun . On sums of four pentagonal numbers with coefficients. Electronic Research Archive, 2020, 28(1): 559-566. doi: 10.3934/era.2020029

Abstract

1. Introduction

In the past years, many mathematicians were interested in finding the formula for the integer part of the reciprocal tails of the convergent series. That is, find the explicit value of $\left\lfloor\left(\sum_{k = n}^\infty a_k\right)^{-1}\right\rfloor$ when $\sum_{k = 1}^\infty a_k$ converges. The motivation of such research comes from the reciprocal sum of Fibonacci numbers. Let us recall that the Fibonacci sequence $\left(F_{n}\right)_{n\geq0}$ is defined by

$F_{n+2} = F_{n+1}+F_{n},\quad F_0 = 0,\,F_1 = 1,$

where $n\geq0$ . In ^[1], Ohtsuka and Nakamura proved

$\left\lfloor\left(\sum\limits_{k = n}^\infty \frac{1}{F_k}\right)^{-1}\right\rfloor = \begin{cases} F_{n-2}, & \text{if}\ n\ge2\ \text{is even}; \\ F_{n-2}-1, & \text{if}\ n\ge1\ \text{is odd}, \end{cases}$

and

$\left\lfloor\left(\sum\limits_{k = n}^\infty \frac{1}{F_k^2}\right)^{-1}\right\rfloor = \begin{cases} F_{n-1}F_n-1, &\text{if}\ n\ge2\ \text{is even}; \\ F_{n-1}F_n, &\text{if}\ n\ge1\ \text{is odd}, \end{cases}$

where $\left\lfloor x\right\rfloor$ denotes the greatest integer $\le x$ . In ^[2], Wang proved

$\left\lfloor\left(\sum\limits_{k = n}^\infty \frac{1}{F_k^3}\right)^{-1}\right\rfloor = \begin{cases} F_nF_{n-1}^2+F_{n-2}F_{n}^2+\left\lfloor\frac{1}{11}(14F_{n-2}-5F_n)\right\rfloor, &\text{if}\ n\ge2\ \text{is even}; \\ F_nF_{n-1}^2+F_{n-2}F_{n}^2+\left\lfloor\frac{1}{11}(5F_n-14F_{n-2})\right\rfloor, &\text{if}\ n\ge1\ \text{is odd}, \end{cases}$

where $F_{-1} = F_1 = 1$ . In ^[3], Hwang et al. provided the relevant formula for the reciprocal sum of the fourth power of the Fibonacci numbers, that is,

$\left\lfloor\left(\sum\limits_{k = n}^\infty \frac{1}{F_k^4}\right)^{-1}\right\rfloor = F_n^4-F_{n-1}^4+\frac{2(-1)^n}{5}F_{2n-1}-\left\{\frac{n+2}{5}\right\},$

where $\{x\} = x-\lfloor x \rfloor$ . In addition, some mathematicians studied the reciprocal sums of Fibonacci, Lucas, and Pell sequences, such as ^[4,5,6,7], while some mathematicians studied the reciprocal sums of other types of sequences, such as ^[8,9,10].

Many mathematicians also considered the asymptotic behavior of these sequences. That is, find a suitable function $g_n$ such that

$\left(\sum\limits_{k = n}^\infty a_k\right)^{-1}\sim g_n$

when $\sum_{k = 1}^\infty a_k$ converges. Here the notation $A_n\sim B_n$ means that

$\underset{n\to\infty}{\lim}\left(A_n-B_n\right) = 0.$

In ^[11], Lee et al. proved that

$\left(\sum\limits_{k = n}^\infty \frac{1}{F_{mk-l}}\right)^{-1}\sim F_{mn-l}-F_{m(n-1)-l}$

for any positive integer $m$ and $0\le l\le m-1$ . In ^[12], Marques et al. proved that for any positive integer $m$ there exists a positive constant $C_m$ such that

$\left(\sum\limits_{k = n}^\infty \frac{1}{F_{mk}^2}\right)^{-1}\sim F_{mn}^2-F_{m(n-1)}^2+(-1)^{mn}C_{m}.$

Moreover, they gave an explicit form for $C_m$ as follows:

$C_m = \begin{cases} -\frac{2(L_{2m}-2)}{25F_{2m}}\sqrt{5}, & \mbox{if}\, \,m\, \,\mbox{is even} , \\ \frac{2(L_{2m}+2)}{5L_{2m}}, & \mbox{if}\, \,m\, \,\mbox{is odd}, \end{cases}$

where $L_n$ is the $n$ th Lucas number. In ^[3], Hwang et al. studied the asymptotic behavior of the reciprocal sum of the fourth power of the Fibonacci numbers, and they proved that

$\left( \sum\limits_{k = n}^\infty \frac{1}{F_{k}^4}\right)^{-1} \sim{F_{n}^4-F_{n-1}^4}+\frac{2(-1)^n}{5}F_{2n-1}+\frac{2\sqrt{5}}{75}.$

In ^[13], Lee and Park studied the asymptotic behavior of the reciprocal sum of $F_kF_{k+m}$ , and they proved that

$\left(\sum\limits_{k = n}^\infty \frac{1}{F_kF_{k+2l}}\right)^{-1}\sim F_{n+l-1}F_{n+l}-(F_l^2+(-1)^l)\frac{(-1)^n}{3}$

and

$\left(\sum\limits_{k = n}^\infty \frac{1}{F_kF_{k+2l-1}}\right)^{-1}\sim F_{n+l-1}^2-(F_{l-1}F_l+(-1)^l)\frac{(-1)^n}{3},$

where $l$ is a positive integer. In addition, some mathematicians studied other types of asymptotic behavior, such as ^[14,15]. Inspired by the above results, we use the method of Yuan et al. ^[16] to study the asymptotic behavior of the sequences that are more general than the Fibonacci sequence. Let $\left(u_n\right)_{n\geq0}$ be the special Lucas $u$ -sequence defined by

$\begin{equation} u_{n+2} = Au_{n+1}-Bu_n,\quad u_0 = 0,\, u_1 = 1, \end{equation}$

(1.1)

where $n\geq0$ , $B = \pm1$ , and $A$ is an integer such that $A^2-4B > 0$ . The relevant properties of the Lucas $u$ -sequence can be found in Sun's book ^[17]. We know that the Binet formula is related to the sequence $\left(u_n\right)_{n\geq0}$ has the form

$\begin{equation} u_n = \frac{\alpha^n-\beta^n}{\alpha-\beta},\quad n\geq0, \end{equation}$

(1.2)

where

$\alpha,\, \beta = \frac{A\pm\sqrt{A^2-4B}}{2}.$

Let

$a_k = \frac{1}{u_{mk}^s},\,\frac{1}{u_{mk}+u_{mk+l}},\,\frac{1}{\sum\nolimits_{i = 0}^l u_{mk+i}},\,\frac{1}{u_{mk}u_{mk+2l}},\,\frac{1}{u_{mk}u_{mk+2l-1}},\,\frac{1}{u_{mk}+C},$

where $m$ and $l$ are positive integers, $s = 1, 2, 3, 4$ , and $C$ is any constant. The aim of this paper is to find a form $g_n$ such that

$\left( \sum\limits_{k = n}^\infty a_k\right)^{-1}\sim{g_n}.$

The rest of this paper is organized as follows: in Section 2, we give our main results. In Section 3, we give the proof of our main results.

2. Main results

Let $u_n$ be defined by the second-order linear recurrence sequence (1.1). In this paper, we shall prove the following eight theorems.

Theorem 2.1. For any positive integer $m$ , we have

$\left( \sum\limits_{k = n}^\infty \frac{1}{u_{mk}}\right)^{-1}\sim{u_{mn}-u_{m(n-1)}}.$

Theorem 2.2. For any positive integer $m$ , we have

$\left( \sum\limits_{k = n}^\infty \frac{1}{u_{mk}^2}\right)^{-1}\sim{u_{mn}^2-u_{m\left(n-1\right)}^2+B^{mn}C_{m}},$

where $C_{m} = \frac{2(1-B^m)}{(\alpha-\beta)^2}-\frac{2(\alpha^{2m}-1)^2}{{\left(\alpha-\beta\right)^2}\left(\alpha^{4m}-B^m\right)}$ .

Theorem 2.3. For any positive integer $m$ , we have

$\left( \sum\limits_{k = n}^\infty \frac{1}{u_{mk}^3}\right)^{-1}\sim{u_{mn}^3-u_{m(n-1)}^3}+ 3B^{mn}Q_m\left(u_{m(n+2)}-u_{m(n-3)}\right),$

where $Q_m = \frac{u_m^2}{(1-(B\alpha)^{5m})(1-(B\beta)^{5m})}$ .

Theorem 2.4. For any positive integer $m$ , we have

$\left( \sum\limits_{k = n}^\infty \frac{1}{u_{mk}^4}\right)^{-1}\sim{u_{mn}^4-u_{m(n-1)}^4}+ 4B^{mn} U_m\left(u_{m(n+1)}^2-B^mu_{m(n-2)}^2\right)+V_m,$

where $U_m = \frac{u_m^2}{(1-B^m\alpha^{6m})(1-B^m\beta^{6m})}$ and $V_m = \frac{(\alpha^{4m}-1)^2}{(\alpha-\beta)^4} \left(\frac{16(\alpha^{4m}-1)}{(\alpha^{6m}-B^m)^2} -\frac{10}{\alpha^{8m}-1}\right)$ .

Theorem 2.5. For all positive integers $m$ and $l$ , we have

$\left( \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}+u_{mk+l}}\right)^{-1} \sim u_{mn+l}-u_{m(n-1)+l}+u_{mn}-u_{m(n-1)}.$

Theorem 2.6. For all positive integers $m$ and $l$ , we have

$\left( \sum\limits_{k = n}^{\infty}\frac{1}{\sum\nolimits_{i = 0}^l u_{mk+i}}\right)^{-1} \sim \frac{1}{\alpha-1}\left(u_{mn+l+1}-u_{m(n-1)+l+1}-u_{mn}+u_{m(n-1)}\right).$

Remark 2.1. Note that when $l = 1$ , the two main terms of Theorems 2.5 and 2.6 are different. However, there is no contradiction since they are equivalent.

Theorem 2.7. For all positive integers $m$ and $l$ , we have

(ⅰ)

$\left( \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}u_{mk+2l}}\right)^{-1} \sim u_{mn+l}^2-u_{m(n-1)+l}^2- \frac{B^{mn}(\alpha^{2m}-1)^2}{\alpha^{4m}-B^m} C_{m, l},$

where $C_{m, l} = u_l^2+\frac{2B^l}{A^2-4B}\left(1- \frac{(1-B^m)(\alpha^{4m}-B^m)}{(\alpha^{2m}-1)^2}\right)$ .

(ⅱ)

$\left( \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}u_{mk+2l-1}}\right)^{-1} \sim \alpha(1-\beta^{2m})u_{mn+l-1}^2-\frac{B^{mn}(\alpha^{2m}-1)^2}{\alpha^{4m}-B^m} C_{m,l}',$

where $C_{m, l}' = u_lu_{l-1}+ \frac{B^{l-1}}{A^2-4B}\left(A -\frac{2(\alpha-\alpha\beta^{2m})(\alpha^{4m}-B^m)} {(\alpha^{2m}-1)^2}\right)$ .

Theorem 2.8. For any positive integer $m$ and constant $C$ , we have

$\left( \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}+C}\right)^{-1} \sim u_{mn}-u_{m(n-1)}+C\frac{\alpha^m-1}{\alpha^m+1}.$

Let $A = 1$ and $B = -1$ in (1.1). Then $u_n$ becomes the $n$ th Fibonacci number. So we can obtain some known results when we take some special values for $m, \, A, \, B$ .

If we take $A = 1$ and $B = -1$ in Theorem 2.2, then

$\begin{equation} C_m = \frac{2(1-B^m)}{(\alpha-\beta)^2}- \frac{2(\alpha^{2m}-1)^2}{{\left(\alpha-\beta\right)^2} \left(\alpha^{4m}-B^m\right)} = \frac{2(1-(-1)^m)}{5}- \frac{2(\alpha^{2m}-1)^2}{5 \left(\alpha^{4m}-(-1)^m\right)} . \end{equation}$

(2.1)

If $m$ is even, then it follows from (2.1) that

$\begin{aligned} C_m& = -\frac{2(\alpha^{2m}-1)^2}{5\left(\alpha^{4m}-(-1)^m\right)} = \frac{-2(\alpha^{2m}-\alpha^m\beta^m)^2}{5(\alpha^{4m}-\alpha^{2m}\beta^{2m})} = \frac{-2(\alpha^{m}-\beta^m)^2}{5(\alpha^{2m}-\beta^{2m})}\\ & = \frac{-2(\alpha^{2m}+\beta^{2m}-2)(\alpha-\beta)}{5(\alpha^{2m}-\beta^{2m})(\alpha-\beta)} = \frac{-2(L_{2m}-2)}{5\sqrt{5}u_{2m}} = \frac{-2(L_{2m}-2)}{25u_{2m}}\sqrt{5}, \end{aligned}$

where $L_n$ is the $n$ th Lucas number with $L_0 = 2, \, L_1 = 1$ . If $m$ is odd, then it follows from (2.1) that

$\begin{aligned} C_m& = \frac{4}{5}-\frac{2(\alpha^{2m}-1)^2}{5\left(\alpha^{4m}+1\right)} = \frac{4}{5}-\frac{2(\alpha^{2m}+\alpha^m\beta^m)^2}{5(\alpha^{4m}+\alpha^{2m}\beta^{2m})} = \frac{4}{5}-\frac{2(\alpha^{m}+\beta^m)^2}{5(\alpha^{2m}+\beta^{2m})}\\ & = \frac{4}{5}-\frac{2(\alpha^{2m}+\beta^{2m}-2)}{5(\alpha^{2m}+\beta^{2m})} = \frac{4L_{2m}}{5L_{2m}}-\frac{2(L_{2m}-2)}{5L_{2m}} = \frac{2(L_{2m}+2)}{5L_{2m}}. \end{aligned}$

So we have the following corollary, which is given by ^[12].

Corollary 2.1. Let $F_n$ be the $n$ th Fibonacci number with $F_0 = 0, \, F_1 = 1$ and let $L_n$ be the $n$ th Lucas number with $L_0 = 2, \, L_1 = 1$ . Then for any positive integer $m$ , we have

$\left( \sum\limits_{k = n}^\infty \frac{1}{F_{mk}^2}\right)^{-1}\sim{F_{mn}^2-F_{m\left(n-1\right)}^2+\left(-1\right)^{mn}C_{m}},$

where

$C_m = \begin{cases} -\frac{2(L_{2m}-2)}{25F_{2m}}\sqrt{5}, & \mathit{\mbox{if}}\, \, m \, \,\mathit{\mbox{is even}} , \\ \frac{2(L_{2m}+2)}{5L_{2m}}, & \mathit{\mbox{if}}\, \, m \, \,\mathit{\mbox{is odd}}. \end{cases}$

If we take $m = A = 1$ and $B = -1$ in Theorem 2.4, then

$\begin{aligned} U_m& = \frac{u_m^2}{(1-B^m\alpha^{6m})(1-B^m\beta^{6m})} = \frac{1}{(1+\alpha^6)(1+\beta^6)} = \frac{1}{2+\alpha^6+\beta^6} = \frac{1}{20} \end{aligned}$

and

$\begin{aligned} V_m& = \frac{(\alpha^{4m}-1)^2}{(\alpha-\beta)^4} \left(\frac{16(\alpha^{4m}-1)}{(\alpha^{6m}-B^m)^2} -\frac{10}{\alpha^{8m}-1}\right) = \frac{(\alpha^{4}-1)^2}{(\alpha-\beta)^4} \left(\frac{16(\alpha^{4}-1)}{(\alpha^{6}+1)^2} -\frac{10}{\alpha^{8}-1}\right)\\ & = \frac{1}{25}\left(\frac{16(\alpha^{4}-1)^3}{(\alpha^{6}+1)^2} -\frac{10(\alpha^4-1)}{\alpha^{4}+1}\right) = \frac{1}{25}\left(4\sqrt{5}-\frac{10\sqrt{5}}{3}\right) = \frac{2\sqrt{5}}{75}, \end{aligned}$

which imply that

$\begin{aligned} &u_{mn}^4-u_{m(n-1)}^4+4B^{mn}U_m\left(u_{m(n+1)}^2-B^m u_{m(n-2)}^2\right)+V_m\\ & = u_{n}^4-u_{n-1}^4+4(-1)^{n}\cdot\frac{1}{20}\cdot\left(u_{n+1}^2+u_{n-2}^2\right)+\frac{2\sqrt{5}}{75}\\ & = u_{n}^4-u_{n-1}^4+\frac{(-1)^{n}}{5}\cdot\left(u_{n+1}^2+u_{n-2}^2\right)+\frac{2\sqrt{5}}{75}\\ & = u_{n}^4-u_{n-1}^4+\frac{(-1)^{n}}{5}\cdot\left(u_{n+1}^2-u_{n-1}^2+u_{n-1}^2+u_{n-2}^2\right)+\frac{2\sqrt{5}}{75}\\ & = u_{n}^4-u_{n-1}^4+\frac{(-1)^{n}}{5}\cdot\left(u_{2n}+u_{2n-3}\right)+\frac{2\sqrt{5}}{75}\\ & = u_{n}^4-u_{n-1}^4+\frac{2(-1)^{n}}{5}u_{2n-1}+\frac{2\sqrt{5}}{75}. \end{aligned}$

So we have the following corollary, which is given by [3, Corollary 4.3].

Corollary 2.2. Let $F_n$ be the $n$ th Fibonacci number with $F_0 = 0, \, F_1 = 1$ . Then we have

$\left( \sum\limits_{k = n}^\infty \frac{1}{F_{k}^4}\right)^{-1} \sim{F_{n}^4-F_{n-1}^4}+\frac{2(-1)^n}{5}F_{2n-1}+\frac{2\sqrt{5}}{75}.$

If we take $m = A = 1$ and $B = -1$ in Theorem 2.7, then

$\begin{equation} \begin{aligned} C_{1, l}& = u_l^2+\frac{2B^l}{A^2-4B}\left(1- \frac{(1-B^m)(\alpha^{4m}-B^m)}{(\alpha^{2m}-1)^2}\right) = u_l^2+\frac{2(-1)^l}{5}\left(1-\frac{2(\alpha^{4}+1)}{(\alpha^{2}-1)^2}\right)\\ & = u_l^2+\frac{2(-1)^l}{5}\left(1-\frac{2(\alpha^{2}+\beta^2)}{(\alpha+\beta)^2}\right) = u_l^2-2(-1)^l \end{aligned} \end{equation}$

(2.2)

and

$\begin{equation} \begin{aligned} C_{1,l}'& = u_lu_{l-1}+ \frac{B^{l-1}}{A^2-4B}\left(A -\frac{2(\alpha-\alpha\beta^{2m})(\alpha^{4m}-B^m)} {(\alpha^{2m}-1)^2}\right) = u_lu_{l-1}+\frac{(-1)^{l-1}}{5}\left(1 -\frac{2(\alpha-\alpha\beta^{2})(\alpha^{4}+1)} {(\alpha^{2}-1)^2}\right)\\ & = u_lu_{l-1}+\frac{(-1)^{l-1}}{5}\left(1 -\frac{2(\alpha^{4}+1)} {(\alpha^{2}-1)^2}\right) = u_lu_{l-1}+(-1)^{l}. \\ \end{aligned} \end{equation}$

(2.3)

Note that

$\begin{equation} \frac{B^{mn}(\alpha^{2m}-1)^2}{\alpha^{4m}-B^m} = \frac{(-1)^{n}(\alpha^{2}-1)^2}{\alpha^{4}+1} = (-1)^n\frac{(\alpha+\beta)^2}{\alpha^2+\beta^2} = \frac{(-1)^n}{3}. \end{equation}$

(2.4)

Then it follows from (2.2)–(2.4) that

$\begin{aligned} u_{mn+l}^2-u_{m(n-1)+l}^2-\frac{B^{mn}(\alpha^{2m}-1)^2}{\alpha^{4m}-B^m} C_{m,l} = &u_{n+l}^2-u_{n-1+l}^2-\frac{(-1)^n}{3}C_{1,l}\\ = &u_{n+l}u_{n+l-1}+(-1)^{n+l-1}-\frac{(-1)^n}{3}(u_l^2-2(-1)^l)\\ = &u_{n+l}u_{n+l-1}-\frac{(-1)^n}{3}(u_l^2-2(-1)^l+3(-1)^l)\\ = &u_{n+l}u_{n+l-1}-\frac{(-1)^n}{3}(u_l^2+(-1)^l) \end{aligned}$

and

$\begin{aligned} \alpha(1-\beta^{2m})u_{mn+l-1}^2-\frac{B^{mn}(\alpha^{2m}-1)^2}{\alpha^{4m}-B^m}C_{m,l}'& = \alpha(1-\beta^{2})u_{n+l-1}^2-\frac{(-1)^n}{3}C_{1,l}'\\ & = u_{n+l-1}^2-\frac{(-1)^n}{3}(u_lu_{l-1}+(-1)^{l}).\\ \end{aligned}$

So we have the following corollary, which is given by [13, Sections 3 and 4].

Corollary 2.3. Let $F_n$ be the $n$ th Fibonacci number with $F_0 = 0, \, F_1 = 1$ . Then for any positive integer $l$ , we have

$\left(\sum\limits_{k = n}^\infty \frac{1}{F_kF_{k+2l}}\right)^{-1}\sim F_{n+l-1}F_{n+l}-(F_l^2+(-1)^l)\frac{(-1)^n}{3}$

and

$\left(\sum\limits_{k = n}^\infty \frac{1}{F_kF_{k+2l-1}}\right)^{-1}\sim F_{n+l-1}^2-(F_{l-1}F_l+(-1)^l)\frac{(-1)^n}{3}.$

3. Proof of the theorem

In this section, we give the proofs of the above eight theorems. We first give some identities that will be used in the proofs of our main results. Let $m$ and $k$ be positive integers. Then it follows from (1.2) that

$\begin{equation} \frac{1}{u_{mk}} = \frac{\alpha-\beta}{\alpha^{mk}-\beta^{mk}} = \frac{\alpha-\beta}{\alpha^{mk}}\left(1-\frac{B^{mk}}{\alpha^{2mk}}\right)^{-1}. \end{equation}$

(3.1)

Moreover, we have the following identities:

$\begin{align} &(1+x)^{-1} = 1-x+x^2-\frac{x^3}{1+x}, \end{align}$

(3.2a)

$\begin{align} &(1-x)^{-1} = 1+x+x^2+\frac{x^3}{1-x}, \end{align}$

(3.2b)

$\begin{align} &(1-x)^{-2} = 1+2x+3x^2+\frac{x^3(4-3x)}{(x-1)^2}, \end{align}$

(3.2c)

$\begin{align} &(1-x)^{-3} = 1+3x+6x^2+\frac{x^3(10-15x+6x^2)}{(1-x)^3}, \end{align}$

(3.2d)

$\begin{align} &(1-x)^{-4} = 1+4x+10x^2+\frac{x^3(20-45x+36x^2-10x^3)}{(x-1)^4}. \end{align}$

(3.2e)

To prove the above eight theorems, we may split the proofs into eight subsections as follows:

3.1. The proof of Theorem 2.1

In this subsection, we will provide a proof of Theorem 2.1.

Proof. By (3.1) and (3.2b), we have

$\begin{equation} \begin{split} \frac{1}{u_{mk}} & = \frac{\alpha-\beta}{\alpha^{mk}}\left(1-\frac{B^{mk}}{\alpha^{2mk}}\right)^{-1} = \frac{\alpha-\beta}{\alpha^{mk}}\left(1+\frac{B^{mk}}{\alpha^{2mk}} +\frac{1}{\alpha^{4mk}}+\frac{B^{mk}}{\alpha^{4mk}(\alpha^{2mk}-B^{mk})}\right)\\ & = (\alpha-\beta) \left(\frac{1}{\alpha^{mk}}+\frac{B^{mk}}{\alpha^{3mk}} +\frac{1}{\alpha^{5mk}}+\frac{B^{mk}}{\alpha^{5mk}(\alpha^{2mk}-B^{mk})}\right). \end{split} \end{equation}$

(3.3)

Let $n$ be a positive integer. Then it follows from (3.3) that

$\begin{equation} \begin{aligned} \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}} = &(\alpha-\beta) \left(\sum\limits_{k = n}^{\infty}\frac{1}{\alpha^{mk}}+\sum\limits_{k = n}^{\infty}\frac{B^{mk}}{\alpha^{3mk}} +\sum\limits_{k = n}^{\infty}\frac{1}{\alpha^{5mk}}+\sum\limits_{k = n}^{\infty}\frac{B^{mk}}{\alpha^{5mk}(\alpha^{2mk}-B^{mk})}\right)\\ = &(\alpha-\beta) \left(\frac{\alpha^{m}}{\alpha^{mn}(\alpha^{m}-1)}+ \frac{B^{mn}\alpha^{3m}}{\alpha^{3mn}(\alpha^{3m}-B^m)} +\frac{\alpha^{5m}}{\alpha^{5mn}(\alpha^{5m}-1)}\right)+(\alpha-\beta) \sum\limits_{k = n}^{\infty}\frac{B^{mk}}{\alpha^{5mk}(\alpha^{2mk}-B^{mk})}\\\ = &\frac{\alpha^{m}(\alpha-\beta)}{\alpha^{mn}(\alpha^{m}-1)}\left( 1+\frac{B^{mn}\alpha^{2m}(\alpha^{m}-1)}{\alpha^{2mn}(\alpha^{3m}-B^m)}+ \frac{\alpha^{4m}(\alpha^{m}-1)}{\alpha^{4mn}(\alpha^{5m}-1)} +O\left(\frac{1}{\alpha^{6mn}}\right)\right). \end{aligned} \end{equation}$

(3.4)

Here the notation $f(x) = O(g(x))$ means that there is a constant $C$ such that $|f(x)|\leq Cg(x)$ for all large enough real numbers $x$ . By (3.2a) and (3.4), we have

$\begin{split} \left( \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}}\right)^{-1} & = \frac{\alpha^{mn}(\alpha^{m}-1)}{\alpha^{m}(\alpha-\beta)}\left( 1+\frac{B^{mn}\alpha^{2m}(\alpha^{m}-1)}{\alpha^{2mn}(\alpha^{3m}-B^m)}+ \frac{\alpha^{4m}(\alpha^{m}-1)}{\alpha^{4mn}(\alpha^{5m}-1)} +O\left(\frac{1}{\alpha^{6mn}}\right)\right)^{-1}\\ & = \frac{\alpha^{mn}(\alpha^{m}-1)}{\alpha^{m}(\alpha-\beta)} \left(1-\frac{B^{mn}\alpha^{2m}(\alpha^{m}-1)}{\alpha^{2mn}(\alpha^{3m}-B^m)} +O\left(\frac{1}{\alpha^{4mn}}\right)\right)\\ & = \frac{\alpha^{mn}(\alpha^{m}-1)}{\alpha^{m}(\alpha-\beta)}- \frac{B^{mn}\alpha^{m}(\alpha^{m}-1)^2}{\alpha^{mn}(\alpha-\beta)(\alpha^{3m}-B^m)}+ O\left(\frac{1}{\alpha^{3mn}}\right)\\ & = \frac{\alpha^{mn}}{\alpha-\beta}- \frac{\alpha^{m(n-1)}}{\alpha-\beta}+O\left(\frac{1}{\alpha^{mn}}\right)\\ & = \frac{\alpha^{mn}-\beta^{mn}+\beta^{mn}}{\alpha-\beta}- \frac{\alpha^{m(n-1)}-\beta^{m(n-1)}+\beta^{m(n-1)}}{\alpha-\beta}+O\left(\frac{1}{\alpha^{mn}}\right)\\ & = u_{mn}-u_{m(n-1)}+\frac{\beta^{mn}-\beta^{m(n-1)}}{\alpha-\beta}+O\left(\frac{1}{\alpha^{mn}}\right).\\ \end{split}$

Then

$\underset{n\to\infty}{\lim}\left( \left( \sum\limits_{k = n}^\infty \frac{1}{u_{mk}}\right)^{-1} -\left( {u_{mn}-u_{m\left(n-1\right)}} \right)\right) = \underset{n\to\infty}{\lim}\left(\frac{\beta^{mn}-\beta^{m(n-1)}}{\alpha-\beta}+O\left(\frac{1}{\alpha^{mn}}\right)\right) = 0.$

So we have

$\left( \sum\limits_{k = n}^\infty \frac{1}{u_{mk}}\right)^{-1}\sim{u_{mn}-u_{m\left(n-1\right)}}.$

□

3.2. The proof of Theorem 2.2

In this subsection, we will provide a proof of Theorem 2.2.

Proof. By (3.1) and (3.2c), we have

$\begin{equation} \begin{split} \frac{1}{u_{mk}^2} & = \frac{(\alpha-\beta)^2}{\alpha^{2mk}} \left(1-\frac{B^{mk}}{\alpha^{2mk}}\right)^{-2} = \frac{(\alpha-\beta)^2}{\alpha^{2mk}} \left(1+\frac{2B^{mk}}{\alpha^{2mk}}+\frac{3}{\alpha^{4mk}}+ \frac{4B^{mk}\alpha^{2mk}-3}{\alpha^{4mk}(B^{mk}\alpha^{2mk}-1)^2}\right)\\ & = (\alpha-\beta)^2\left(\frac{1}{\alpha^{2mk}}+\frac{2B^{mk}}{\alpha^{4mk}}+\frac{3}{\alpha^{6mk}}+ \frac{4B^{mk}\alpha^{2mk}-3}{\alpha^{6mk}(B^{mk}\alpha^{2mk}-1)^2}\right)\\ & = (\alpha-\beta)^2\left(\frac{1}{\alpha^{2mk}}+\frac{2B^{mk}}{\alpha^{4mk}}+\frac{3}{\alpha^{6mk}}+R_k\right), \end{split} \end{equation}$

(3.5)

where

$R_k = \frac{4B^{mk}\alpha^{2mk}-3}{\alpha^{6mk}(B^{mk}\alpha^{2mk}-1)^2}.$

Let $n$ be a positive integer. Then it follows from (3.5) that

$\begin{equation} \begin{split} \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}^2} = &(\alpha-\beta)^2\left(\sum\limits_{k = n}^{\infty}\frac{1}{\alpha^{2mk}}+\sum\limits_{k = n}^{\infty}\frac{2B^{mk}}{\alpha^{4mk}} +\sum\limits_{k = n}^{\infty}\frac{3}{\alpha^{6mk}}+ \sum\limits_{k = n}^{\infty}R_k\right)\\ = &(\alpha-\beta)^2 \left(\frac{\alpha^{2m}}{\alpha^{2mn}(\alpha^{2m}-1)}+ \frac{2B^{mn}\alpha^{4m}}{\alpha^{4mn}(\alpha^{4m}-B^m)}+ \frac{3\alpha^{6m}}{\alpha^{6mn}(\alpha^{6m}-1)} + \sum\limits_{k = n}^{\infty}R_k\right)\\ = &\frac{(\alpha-\beta)^2\alpha^{2m}}{\alpha^{2mn}(\alpha^{2m}-1)} \left(1+ \frac{2B^{mn}\alpha^{2m}(\alpha^{2m}-1)}{\alpha^{2mn}(\alpha^{4m}-B^m)}+ \frac{3\alpha^{4m}(\alpha^{2m}-1)}{\alpha^{4mn}(\alpha^{6m}-1)}\right)\\ &+\frac{(\alpha-\beta)^2\alpha^{2m}}{\alpha^{2mn}(\alpha^{2m}-1)} \cdot\frac{\alpha^{2mn}(\alpha^{2m}-1)}{\alpha^{2m}} \sum\limits_{k = n}^{\infty}R_k\\ = &\frac{(\alpha-\beta)^2\alpha^{2m}}{\alpha^{2mn}(\alpha^{2m}-1)}\left(1+\omega\right), \end{split} \end{equation}$

(3.6)

where

$\omega = \frac{2B^{mn}}{\alpha^{2mn}}\cdot\frac{\alpha^{2m}(\alpha^{2m}-1)}{(\alpha^{4m}-B^m)}+ \frac{3}{\alpha^{4mn}}\cdot\frac{\alpha^{4m}}{(\alpha^{4m}+\alpha^{2m}+1)}+ \frac{\alpha^{2mn}(\alpha^{2m}-1)}{\alpha^{2m}} \sum\limits_{k = n}^{\infty}R_k .$

Note that

$\omega = \frac{2B^{mn}}{\alpha^{2mn}}\cdot\frac{\alpha^{2m}(\alpha^{2m}-1)}{(\alpha^{4m}-B^m)} +O\left(\frac{1}{\alpha^{4mn}}\right).$

Then we have

$\begin{equation} \omega^2-\frac{\omega^3}{1+\omega} = O\left(\frac{1}{\alpha^{4mn}}\right). \end{equation}$

(3.7)

From (3.2a), (3.6), and (3.7), it follows that

$\begin{align*} \left( \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}^2}\right)^{-1} = &\frac{\alpha^{2mn}(\alpha^{2m}-1)}{(\alpha-\beta)^2\alpha^{2m}}\left(1+\omega\right)^{-1} = \frac{\alpha^{2mn}(\alpha^{2m}-1)}{(\alpha-\beta)^2\alpha^{2m}}\left(1-\omega+\omega^2-\frac{\omega^3}{1+\omega}\right)\\ = &\frac{\alpha^{2mn}(\alpha^{2m}-1)}{(\alpha-\beta)^2\alpha^{2m}}\left(1-\omega+O\left(\frac{1}{\alpha^{4mn}}\right)\right)\\ = &\frac{\alpha^{2mn}(\alpha^{2m}-1)}{(\alpha-\beta)^2\alpha^{2m}}\left(1-\frac{2B^{mn}}{\alpha^{2mn}} \cdot\frac{\alpha^{2m}(\alpha^{2m}-1)}{(\alpha^{4m}-B^m)} +O\left(\frac{1}{\alpha^{4mn}}\right)\right)\\ = &\frac{\alpha^{2mn}(\alpha^{2m}-1)}{(\alpha-\beta)^2\alpha^{2m}}-\frac{2B^{mn}}{(\alpha-\beta)^2}\cdot\frac{(\alpha^{2m}-1)^2}{(\alpha^{4m}-B^m)}+O\left(\frac{1}{\alpha^{2mn}}\right)\\ = &\frac{\alpha^{2mn}}{(\alpha-\beta)^2}-\frac{\alpha^{2m(n-1)}}{(\alpha-\beta)^2}-\frac{2B^{mn}}{(\alpha-\beta)^2}\cdot\frac{(\alpha^{2m}-1)^2}{(\alpha^{4m}-B^m)} +O\left(\frac{1}{\alpha^{2mn}}\right)\\ = &\left(\frac{\alpha^{mn}-\beta^{mn}+\beta^{mn}}{\alpha-\beta}\right)^2- \left(\frac{\alpha^{m(n-1)}-\beta^{m(n-1)}+\beta^{m(n-1)}}{\alpha-\beta}\right)^2-\frac{2B^{mn}}{(\alpha-\beta)^2}\cdot\frac{(\alpha^{2m}-1)^2}{(\alpha^{4m}-B^m)} +O\left(\frac{1}{\alpha^{2mn}}\right)\\ = &u_{mn}^2-u_{m(n-1)}^2+ \frac{2B^{mn}(1-B^m)}{(\alpha-\beta)^2}-\frac{2B^{mn}}{(\alpha-\beta)^2}\cdot\frac{(\alpha^{2m}-1)^2}{(\alpha^{4m}-B^m)}+\frac{\beta^{2mn}(\alpha^{2m}-1)}{(\alpha-\beta)^2}+O\left(\frac{1}{\alpha^{2mn}}\right)\\ = &u_{mn}^2-u_{m(n-1)}^2+B^{mn}C_{m}+O\left(\frac{1}{\alpha^{2mn}}\right), \end{align*}$

where

$C_{m} = \frac{2(1-B^m)}{(\alpha-\beta)^2}-\frac{2(\alpha^{2m}-1)^2}{{\left(\alpha-\beta\right)^2}\left(\alpha^{4m}-B^m\right)}.$

Then

$\underset{n\to\infty}{\lim}\left( \left( \sum\limits_{k = n}^\infty \frac{1}{u_{mk}^2}\right)^{-1} -\left( {u_{mn}^2-u_{m\left(n-1\right)}^2+B^{mn}C_{m}} \right)\right) = 0.$

So we have

$\left( \sum\limits_{k = n}^\infty \frac{1}{u_{mk}^2}\right)^{-1}\sim{u_{mn}^2-u_{m\left(n-1\right)}^2+B^{mn}C_{m}},$

where $C_{m} = \frac{2(1-B^m)}{(\alpha-\beta)^2}-\frac{2(\alpha^{2m}-1)^2}{{\left(\alpha-\beta\right)^2}\left(\alpha^{4m}-B^m\right)}.$ □

3.3. The proof of Theorem 2.3

In this subsection, we will provide a proof of Theorem 2.3.

Proof. By (3.1) and (3.2d), we have

$\begin{equation} \begin{split} \frac{1}{u_{mk}^3} & = \frac{(\alpha-\beta)^3}{\alpha^{3mk}}\left(1-\frac{B^{mk}}{\alpha^{2mk}}\right)^{-3} = \frac{(\alpha-\beta)^3}{\alpha^{3mk}}\left(1+\frac{3B^{mk}}{\alpha^{2mk}}+\frac{6}{\alpha^{4mk}}+ \frac{10\alpha^{4mk}-15B^{mk}\alpha^{2mk}+6}{\alpha^{4mk}(B^{mk}\alpha^{2mk}-1)^3}\right)\\ & = (\alpha-\beta)^3\left(\frac{1}{\alpha^{3mk}}+\frac{3B^{mk}}{\alpha^{5mk}}+\frac{6}{\alpha^{7mk}}+ \frac{10\alpha^{4mk}-15B^{mk}\alpha^{2mk}+6}{\alpha^{7mk}(B^{mk}\alpha^{2mk}-1)^3}\right)\\ & = (\alpha-\beta)^3\left(\frac{1}{\alpha^{3mk}}+\frac{3B^{mk}}{\alpha^{5mk}}+\frac{6}{\alpha^{7mk}}+R_k\right),\\ \end{split} \end{equation}$

(3.8)

where

$R_k = \frac{10\alpha^{4mk}-15B^{mk}\alpha^{2mk}+6}{\alpha^{7mk}(B^{mk}\alpha^{2mk}-1)^3}.$

Let $n$ be a positive integer. Then it follows from (3.8) that

$\begin{equation} \begin{split} \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}^3} = &(\alpha-\beta)^3\left(\sum\limits_{k = n}^{\infty}\frac{1}{\alpha^{3mk}}+\sum\limits_{k = n}^{\infty}\frac{3B^{mk}}{\alpha^{5mk}} +\sum\limits_{k = n}^{\infty}\frac{6}{\alpha^{7mk}}+\sum\limits_{k = n}^{\infty}R_k\right)\\ = &(\alpha-\beta)^3\left(\frac{\alpha^{3m}}{\alpha^{3mn}(\alpha^{3m}-1)}+\frac{3B^{mn}\alpha^{5m}}{\alpha^{5mn}(\alpha^{5m}-B^m)}+ \frac{6\alpha^{7m}}{\alpha^{7mn}(\alpha^{7m}-1)}+ \sum\limits_{k = n}^{\infty}R_k\right)\\ = &\frac{(\alpha-\beta)^3\alpha^{3m}}{\alpha^{3mn}(\alpha^{3m}-1)} \left(1+\frac{3B^{mn}\alpha^{2m}(\alpha^{3m}-1)}{\alpha^{2mn}(\alpha^{5m}-B^m)}+ \frac{6\alpha^{4m}(\alpha^{3m}-1)}{\alpha^{4mn}(\alpha^{7m}-1)}\right)\\ &+\frac{(\alpha-\beta)^3\alpha^{3m}}{\alpha^{3mn}(\alpha^{3m}-1)} \cdot\frac{\alpha^{3mn}(\alpha^{3m}-1)}{\alpha^{3m}} \sum\limits_{k = n}^{\infty}R_k\\ = &\frac{(\alpha-\beta)^3\alpha^{3m}}{\alpha^{3mn}(\alpha^{3m}-1)}\left(1+\omega\right), \end{split} \end{equation}$

(3.9)

where

$\omega = \frac{3B^{mn}}{\alpha^{2mn}}\cdot\frac{\alpha^{2m}(\alpha^{3m}-1)}{(\alpha^{5m}-B^m)}+ \frac{6}{\alpha^{4mn}}\cdot\frac{\alpha^{4m}(\alpha^{3m}-1)}{(\alpha^{7m}-1)}+\frac{\alpha^{3mn}(\alpha^{3m}-1)}{\alpha^{3m}} \sum\limits_{k = n}^{\infty}R_k .$

Note that

$\omega = \frac{3B^{mn}}{\alpha^{2mn}}\cdot\frac{\alpha^{2m}(\alpha^{3m}-1)}{(\alpha^{5m}-B^m)} +O\left(\frac{1}{\alpha^{4mn}}\right).$

Then we have

$\begin{equation} \omega^2-\frac{\omega^3}{1+\omega} = O\left(\frac{1}{\alpha^{4mn}}\right). \end{equation}$

(3.10)

From (3.2a), (3.9), and (3.10), it follows that

$\begin{split} \left( \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}^3}\right)^{-1} = &\left(\frac{(\alpha-\beta)^3\alpha^{3m}}{\alpha^{3mn}(\alpha^{3m}-1)}\right)^{-1} \left(1+\omega\right)^{-1} = \frac{\alpha^{3mn}(\alpha^{3m}-1)}{(\alpha-\beta)^3\alpha^{3m}}\left(1-\omega+\omega^2-\frac{\omega^3}{1+\omega}\right)\\ = &\frac{\alpha^{3mn}(\alpha^{3m}-1)}{(\alpha-\beta)^3\alpha^{3m}}\left(1-\omega+O\left(\frac{1}{\alpha^{4mn}}\right)\right)\\ = &\frac{\alpha^{3mn}(\alpha^{3m}-1)}{(\alpha-\beta)^3\alpha^{3m}}\left(1-\frac{3B^{mn}}{\alpha^{2mn}}\cdot\frac{\alpha^{2m}(\alpha^{3m}-1)}{(\alpha^{5m}-B^m)} +O\left(\frac{1}{\alpha^{4mn}}\right)\right)\\ = &\frac{\alpha^{3mn}(\alpha^{3m}-1)}{(\alpha-\beta)^3\alpha^{3m}}-\frac{3(B\alpha)^{mn}}{(\alpha-\beta)^3}\cdot\frac{(\alpha^{3m}-1)^2}{\alpha^m(\alpha^{5m}-B^m)}+O\left(\frac{1}{\alpha^{mn}}\right)\\ = &\frac{\alpha^{3mn}}{(\alpha-\beta)^3}-\frac{\alpha^{3m(n-1)}}{(\alpha-\beta)^3} -\frac{3(B\alpha)^{mn}}{(\alpha-\beta)^3}\cdot\frac{(\alpha^{3m}-1)^2}{\alpha^m(\alpha^{5m}-B^m)}+O\left(\frac{1}{\alpha^{mn}}\right)\\ = &\left(\frac{\alpha^{mn}-\beta^{mn}+\beta^{mn}}{\alpha-\beta}\right)^3 -\left(\frac{\alpha^{m(n-1)}-\beta^{m(n-1)}+\beta^{m(n-1)}}{\alpha-\beta}\right)^3-\frac{3(B\alpha)^{mn}}{(\alpha-\beta)^3}\cdot\frac{(\alpha^{3m}-1)^2}{\alpha^m(\alpha^{5m}-B^m)}+O\left(\frac{1}{\alpha^{mn}}\right)\\ = &u_{mn}^3-u_{m(n-1)}^3+\delta+O\left(\frac{1}{\alpha^{mn}}\right),\\ \end{split}$

where

$\begin{split} \delta& = \frac{3(B\alpha)^{mn}(1-\beta^m)+3(B\beta)^{mn}(\alpha^m-1)+ \beta^{3mn}(1-(B\alpha)^{3m})}{(\alpha-\beta)^3}- \frac{3(B\alpha)^{mn}}{(\alpha-\beta)^3}\cdot\frac{(\alpha^{3m}-1)^2}{\alpha^m(\alpha^{5m}-B^m)}\\ & = \frac{3(B\alpha)^{mn}(1-\beta^m)}{(\alpha-\beta)^3}- \frac{3(B\alpha)^{mn}}{(\alpha-\beta)^3}\cdot\frac{(\alpha^{3m}-1)^2}{\alpha^m(\alpha^{5m}-B^m)} +O\left(\frac{1}{\alpha^{mn}}\right)\\ & = \frac{3(B\alpha)^{mn}}{(\alpha-\beta)^3}\left(\frac{-(\alpha^{3m}-1)^2+(1-\beta^m)\alpha^m(\alpha^{5m}-B^m)} {\alpha^m(\alpha^{5m}-B^m)}\right) +O\left(\frac{1}{\alpha^{mn}}\right)\\ & = -\frac{3(B\alpha)^{mn}}{(\alpha-\beta)^3} \left(\frac{\alpha^{4m}-2B^m\alpha^{2m}+B^{2m}}{(B\alpha)^{5m}-1}\right) +O\left(\frac{1}{\alpha^{mn}}\right)\\ & = \frac{3(B\alpha)^{mn}}{(\alpha-\beta)^3} \cdot\frac{\alpha^{2m}\left(\alpha^{m}-\beta^m\right)^2}{1-B^m\alpha^{5m}} +O\left(\frac{1}{\alpha^{mn}}\right)\\ & = \left(\frac{\alpha^m-\beta^m}{\alpha-\beta}\right)^2\cdot \frac{3(B\alpha)^{m(n+2)}}{\alpha-\beta}\cdot\frac{1}{1-(B\alpha)^{5m}} +O\left(\frac{1}{\alpha^{mn}}\right).\\ \end{split}$

Let $Q_m = \frac{u_m^2}{(1-(B\alpha)^{5m})(1-(B\beta)^{5m})}.$ Then we can obtain

$\begin{split} \delta& = 3B^{mn} Q_m\frac{\alpha^{m(n+2)}\left(1-(B\beta)^{5m}\right)}{\alpha-\beta}+O\left(\frac{1}{\alpha^{mn}}\right)\\ & = 3B^{mn}Q_m \left(\frac{\alpha^{m(n+2)}}{\alpha-\beta}-\frac{\alpha^{m(n-3)}}{\alpha-\beta}\right) +O\left(\frac{1}{\alpha^{mn}}\right)\\ & = 3B^{mn}Q_m \left(\frac{\alpha^{m(n+2)}-\beta^{m(n+2)}+\beta^{m(n+2)}}{\alpha-\beta} -\frac{\alpha^{m(n-3)}-\beta^{m(n-3)}+\beta^{m(n-3)}}{\alpha-\beta}\right) +O\left(\frac{1}{\alpha^{mn}}\right)\\ & = 3B^{mn} Q_m\left(u_{m(n+2)}-u_{m(n-3)}+\frac{\beta^{m(n+2)}-\beta^{m(n-3)}}{\alpha-\beta}\right) +O\left(\frac{1}{\alpha^{mn}}\right).\\ \end{split}$

Then

$\begin{split} &\underset{n\to\infty}{\lim}\left( \left( \sum\limits_{k = n}^\infty \frac{1}{u_{mk}^3}\right)^{-1}\ -\left( {u_{mn}^3-u_{m(n-1)}^3}+ 3B^{mn}Q_m\left(u_{m(n+2)}-u_{m(n-3)}\right) \right)\right)\\ & = \underset{n\to\infty}{\lim}\left(3B^{mn}Q_m \frac{\beta^{m(n+2)}-\beta^{m(n-3)}}{\alpha-\beta}+O\left(\frac{1}{\alpha^{mn}}\right) \right) = 0. \end{split}$

So we have

$\left( \sum\limits_{k = n}^\infty \frac{1}{u_{mk}^3}\right)^{-1}\sim{u_{mn}^3-u_{m(n-1)}^3}+ 3B^{mn}Q_m\left(u_{m(n+2)}-u_{m(n-3)}\right),$

where $Q_m = \frac{u_m^2}{(1-(B\alpha)^{5m})(1-(B\beta)^{5m})}$ . □

3.4. The proof of Theorem 2.4

In this subsection, we will provide a proof of Theorem 2.4.

Proof. By (3.1) and (3.2e), we have

$\begin{equation} \begin{split} \frac{1}{u_{mk}^4} & = \frac{(\alpha-\beta)^4}{\alpha^{4mk}} \left(1-\frac{B^{mk}}{\alpha^{2mk}}\right)^{-4} = \frac{(\alpha-\beta)^4}{\alpha^{4mk}} \left(1+\frac{4B^{mk}}{\alpha^{2mk}}+\frac{10}{\alpha^{4mk}}+R_k\right)\\ & = (\alpha-\beta)^4 \left(\frac{1}{\alpha^{4mk}}+\frac{4B^{mk}}{\alpha^{6mk}}+\frac{10}{\alpha^{8mk}}+\frac{R_k}{\alpha^{4mk}}\right),\\ \end{split} \end{equation}$

(3.11)

where

$R_k = \frac{20B^{mk}\alpha^{6mk}-45\alpha^{4mk}+36B^{mk}\alpha^{2mk}-10} {\alpha^{4mk}(B^{mk}\alpha^{2mk}-1)^4}.$

Let $n$ be a positive integer. Then it follows from (3.11) that

$\begin{equation} \begin{split} \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}^4} = &(\alpha-\beta)^4 \left(\sum\limits_{k = n}^{\infty}\frac{1}{\alpha^{4mk}}+\sum\limits_{k = n}^{\infty} \frac{4B^{mk}}{\alpha^{6mk}}+\sum\limits_{k = n}^{\infty}\frac{10}{\alpha^{8mk}} +\sum\limits_{k = n}^{\infty}\frac{R_k}{\alpha^{4mk}}\right)\\ = &(\alpha-\beta)^4\left(\frac{\alpha^{4m}}{\alpha^{4mn}(\alpha^{4m}-1)}+ \frac{4B^{mn}\alpha^{6m}}{\alpha^{6mn}(\alpha^{6m}-B^m)}+ \frac{10\alpha^{8m}}{\alpha^{8mn}(\alpha^{8m}-1)}+ \sum\limits_{k = n}^{\infty}\frac{R_k}{\alpha^{4mk}}\right)\\ = &\frac{(\alpha-\beta)^4\alpha^{4m}}{\alpha^{4mn}(\alpha^{4m}-1)} \left(1+\frac{4B^{mn}\alpha^{2m}(\alpha^{4m}-1)}{\alpha^{2mn}(\alpha^{6m}-B^m)}+ \frac{10\alpha^{4m}(\alpha^{4m}-1)}{\alpha^{4mn}(\alpha^{8m}-1)}\right)\\ &+\frac{(\alpha-\beta)^4\alpha^{4m}}{\alpha^{4mn}(\alpha^{4m}-1)} \cdot\frac{\alpha^{4mn}(\alpha^{4m}-1)}{\alpha^{4m}} \sum\limits_{k = n}^{\infty}\frac{R_k}{\alpha^{4mk}}\\ = &\frac{(\alpha-\beta)^4\alpha^{4m}}{\alpha^{4mn}(\alpha^{4m}-1)}\left(1+\omega\right), \end{split} \end{equation}$

(3.12)

where

$\omega = \frac{4B^{mn}\alpha^{2m}(\alpha^{4m}-1)}{\alpha^{2mn}(\alpha^{6m}-B^m)}+ \frac{10\alpha^{4m}(\alpha^{4m}-1)}{\alpha^{4mn}(\alpha^{8m}-1)}+ \frac{\alpha^{4mn}(\alpha^{4m}-1)}{\alpha^{4m}} \sum\limits_{k = n}^{\infty}\frac{R_k}{\alpha^{4mk}}.$

Note that

$\omega = \frac{4B^{mn}\alpha^{2m}(\alpha^{4m}-1)}{\alpha^{2mn}(\alpha^{6m}-B^m)}+ \frac{10\alpha^{4m}(\alpha^{4m}-1)}{\alpha^{4mn}(\alpha^{8m}-1)}+ O\left(\frac{1}{\alpha^{6mn}}\right).$

Then we have

$\begin{equation} \omega^2-\frac{\omega^3}{1+\omega} = \frac{16\alpha^{4m}(\alpha^{4m}-1)^2} {\alpha^{4mn}(\alpha^{6m}-B^m)^2}+ O\left(\frac{1}{\alpha^{6mn}}\right). \end{equation}$

(3.13)

By (3.2a), (3.12), and (3.13), we have

$\begin{align*} \left( \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}^4}\right)^{-1} = &\left(\frac{(\alpha-\beta)^4\alpha^{4m}}{\alpha^{4mn}(\alpha^{4m}-1)}\right)^{-1} \left(1+\omega\right)^{-1} = \frac{\alpha^{4mn}(\alpha^{4m}-1)}{(\alpha-\beta)^4\alpha^{4m}} \left(1-\omega+\omega^2-\frac{\omega^3}{1+\omega}\right)\\ = &\frac{\alpha^{4mn}(\alpha^{4m}-1)}{(\alpha-\beta)^4\alpha^{4m}} \left(1-\omega+\frac{16\alpha^{4m}(\alpha^{4m}-1)^2} {\alpha^{4mn}(\alpha^{6m}-B^m)^2}+O\left(\frac{1}{\alpha^{6mn}}\right)\right)\\ = &\frac{\alpha^{4mn}(\alpha^{4m}-1)}{(\alpha-\beta)^4\alpha^{4m}} \left(1-\frac{4B^{mn}\alpha^{2m}(\alpha^{4m}-1)}{\alpha^{2mn}(\alpha^{6m}-B^m)} +\frac{1}{\alpha^{4mn}}C_m^{\prime}+ O\left(\frac{1}{\alpha^{6mn}}\right)\right)\\ = &\frac{\alpha^{4mn}(\alpha^{4m}-1)}{(\alpha-\beta)^4\alpha^{4m}} -\frac{4B^{mn}\alpha^{2mn}(\alpha^{4m}-1)^2}{(\alpha-\beta)^4\alpha^{2m}(\alpha^{6m}-B^m)} +\frac{(\alpha^{4m}-1)}{(\alpha-\beta)^4\alpha^{4m}}C_m^{\prime}+ O\left(\frac{1}{\alpha^{2mn}}\right)\\ = &\frac{\alpha^{4mn}}{(\alpha-\beta)^4} -\frac{\alpha^{4m(n-1)}}{(\alpha-\beta)^4} -\frac{4B^{mn}\alpha^{2mn}(\alpha^{4m}-1)^2}{(\alpha-\beta)^4\alpha^{2m}(\alpha^{6m}-B^m)}+\frac{(\alpha^{4m}-1)}{(\alpha-\beta)^4\alpha^{4m}}C_m^{\prime}+ O\left(\frac{1}{\alpha^{2mn}}\right)\\ = &\left(\frac{\alpha^{mn}-\beta^{mn}+\beta^{mn}}{\alpha-\beta}\right)^4- \left(\frac{\alpha^{m(n-1)}-\beta^{m(n-1)}+\beta^{m(n-1)}}{\alpha-\beta}\right)^4\\ &-\frac{4B^{mn}\alpha^{2mn}(\alpha^{4m}-1)^2}{(\alpha-\beta)^4\alpha^{2m}(\alpha^{6m}-B^m)} +\frac{(\alpha^{4m}-1)}{(\alpha-\beta)^4\alpha^{4m}}C_m^{\prime}+ O\left(\frac{1}{\alpha^{2mn}}\right)\\ = &u_{mn}^4-u_{m(n-1)}^4+\delta+V_m+O\left(\frac{1}{\alpha^{2mn}}\right),\\ \end{align*}$

where

$C_m^{\prime} = \frac{16\alpha^{4m}(\alpha^{4m}-1)^2}{(\alpha^{6m}-B^m)^2} -\frac{10\alpha^{4m}(\alpha^{4m}-1)}{\alpha^{8m}-1},$

$\begin{split} V_m = \frac{(\alpha^{4m}-1)}{(\alpha-\beta)^4\alpha^{4m}}C_m^{\prime} & = \frac{(\alpha^{4m}-1)}{(\alpha-\beta)^4\alpha^{4m}} \left(\frac{16\alpha^{4m}(\alpha^{4m}-1)^2}{(\alpha^{6m}-B^m)^2} -\frac{10\alpha^{4m}(\alpha^{4m}-1)}{\alpha^{8m}-1}\right)\\ & = \frac{(\alpha^{4m}-1)^2}{(\alpha-\beta)^4} \left(\frac{16(\alpha^{4m}-1)}{(\alpha^{6m}-B^m)^2} -\frac{10}{\alpha^{8m}-1}\right)\\ \end{split}$

and

$\begin{split} \delta & = \frac{4B^{mn}\alpha^{2mn}}{(\alpha-\beta)^4} \left(\frac{(1-B^{m}\beta^{2m})\alpha^{2m}(\alpha^{6m}-B^m)-(\alpha^{4m}-1)^2} {\alpha^{2m}(\alpha^{6m}-B^m)}\right) = \frac{4B^{mn}\alpha^{2mn}}{(\alpha-\beta)^4} \left(\frac{\alpha^{4m}-2B^{m}\alpha^{2m}+B^{2m}} {1-B^m\alpha^{6m}}\right)\\ & = \frac{4B^{mn}\alpha^{2mn}}{(\alpha-\beta)^4}\cdot \frac{\alpha^{2m}(\alpha^m-\beta^m)^2}{1-B^m\alpha^{6m}} = \left(\frac{\alpha^m-\beta^m}{\alpha-\beta}\right)^2\cdot \frac{4B^{mn}}{(1-B^m\alpha^{6m})(1-B^m\beta^{6m})} \cdot\frac{\alpha^{2m(n+1)}(1-B^m\beta^{6m})}{(\alpha-\beta)^2}\\ & = \frac{4B^{mn}u_m^2}{(1-B^m\alpha^{6m})(1-B^m\beta^{6m})} \left(\frac{\alpha^{2m(n+1)}}{(\alpha-\beta)^2}-\frac{B^m\alpha^{2m(n-2)}}{(\alpha-\beta)^2}\right).\\ \end{split}$

Let $U_m = \frac{u_m^2}{(1-B^m\alpha^{6m})(1-B^m\beta^{6m})}$ . Then we can obtain

$\begin{align*} \delta & = 4B^{mn}U_m \left(\left(\frac{\alpha^{m(n+1)}-\beta^{m(n+1)}+\beta^{m(n+1)}}{\alpha-\beta}\right)^2 -B^m\left(\frac{\alpha^{m(n-2)}-\beta^{m(n-2)}+\beta^{m(n-2)}}{\alpha-\beta}\right)^2\right)\\ & = 4B^{mn}U_m \left(u_{m(n+1)}^2-B^m u_{m(n-2)}^2+ \frac{B^m\beta^{2m(n-2)}-\beta^{2m(n-2)}}{(\alpha-\beta)^2}\right)\\ & = 4B^{mn}U_m \left(u_{m(n+1)}^2-B^m u_{m(n-2)}^2\right)+O\left(\frac{1}{\alpha^{2m(n-2)}}\right). \end{align*}$

Then

$\begin{split} &\underset{n\to\infty}{\lim}\left( \left( \sum\limits_{k = n}^\infty \frac{1}{u_{mk}^4}\right)^{-1} -\left( {u_{mn}^4-u_{m(n-1)}^4}+ 4B^{mn}U_m\left(u_{m(n+1)}^2 -B^m u_{m(n-2)}^2\right)+V_m \right)\right)\\ & = \underset{n\to\infty}{\lim}\left(O\left(\frac{1}{\alpha^{2m(n-2)}}\right)+O\left(\frac{1}{\alpha^{2mn}}\right)\right) = 0. \end{split}$

So we have

$\left( \sum\limits_{k = n}^\infty \frac{1}{u_{mk}^4}\right)^{-1}\sim{u_{mn}^4-u_{m(n-1)}^4}+ 4B^{mn}U_m\left(u_{m(n+1)}^2-B^m u_{m(n-2)}^2\right)+V_m,$

3.5. The proof of Theorem 2.5

In this subsection, we will provide a proof of Theorem 2.5.

Proof. By (1.2) and (3.2b), we have

$\begin{equation} \begin{aligned} \frac{1}{u_{mk}+u_{mk+l}} & = \left(\frac{\alpha^{mk}-\beta^{mk}}{\alpha-\beta} +\frac{\alpha^{mk+l}-\beta^{mk+l}}{\alpha-\beta}\right)^{-1} = \left(\frac{\alpha^{mk}(1+\alpha^l)}{\alpha-\beta}\right)^{-1}\left(1- \frac{B^{mk}(1+\beta^l)}{\alpha^{2mk}(1+\alpha^l)}\right)^{-1}\\ & = \frac{\alpha-\beta}{\alpha^{mk}(1+\alpha^l)} \left(1+\frac{B^{mk}(1+\beta^l)}{\alpha^{2mk}(1+\alpha^l)}+ \frac{(1+\beta^l)^2}{\alpha^{4mk}(1+\alpha^l)^2}+R_k\right)\\ & = \frac{\alpha-\beta}{1+\alpha^l} \left(\frac{1}{\alpha^{mk}}+\frac{B^{mk}(1+\beta^l)}{\alpha^{3mk}(1+\alpha^l)}+ \frac{(1+\beta^l)^2}{\alpha^{5mk}(1+\alpha^l)^2}+\frac{R_k}{\alpha^{mk}}\right), \end{aligned} \end{equation}$

(3.14)

where

$R_k = \frac{B^{mk}(1+\beta^l)}{\alpha^{4mk}(1+\alpha^l) \left(\alpha^{2mk}(1+\alpha^l)-B^{mk}(1+\beta^l)\right)}.$

Let $n$ be a positive integer. Then it follows from (3.14) that

$\begin{equation} \begin{aligned} \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}+u_{mk+l}} = &\frac{\alpha-\beta}{1+\alpha^l}\left(\sum\limits_{k = n}^{\infty}\frac{1}{\alpha^{mk}}+ \sum\limits_{k = n}^{\infty}\frac{B^{mk}(1+\beta^l)}{\alpha^{3mk}(1+\alpha^l)}+ \sum\limits_{k = n}^{\infty}\frac{(1+\beta^l)^2}{\alpha^{5mk}(1+\alpha^l)^2}+\sum\limits_{k = n}^{\infty}\frac{R_k}{\alpha^{mk}}\right)\\ = &\frac{\alpha-\beta}{1+\alpha^l}\left(\frac{\alpha^m}{\alpha^{mn}(\alpha^m-1)}+C_{m}\right) = \frac{\alpha^m(\alpha-\beta)}{\alpha^{mn}(\alpha^m-1)(1+\alpha^l)} \left(1+\frac{\alpha^{mn}(\alpha^m-1)}{\alpha^m}C_{m}\right), \end{aligned} \end{equation}$

(3.15)

where

$C_{m} = \frac{B^{mn}\alpha^{3m}(1+\beta^l)}{\alpha^{3mn}(\alpha^{3m}-B^m)(1+\alpha^l)} +\frac{\alpha^{5m}(1+\beta^l)^2}{\alpha^{5mn}(\alpha^{5m}-1)(1+\alpha^l)^2} + \sum\limits_{k = n}^{\infty}\frac{R_k}{\alpha^{mk}}.$

Then we have

$\begin{equation} \frac{\alpha^{mn}(\alpha^m-1)}{\alpha^m}C_{m} = O\left(\frac{1}{\alpha^{2mn}}\right). \end{equation}$

(3.16)

By (3.2a), (3.15), and (3.16), we have

$\begin{aligned} \left(\sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}+u_{mk+l}}\right)^{-1} = &\left(\frac{\alpha^m(\alpha-\beta)}{\alpha^{mn}(1+\alpha^l)(\alpha^m-1)}\right)^{-1} \left(1+O\left(\frac{1}{\alpha^{2mn}}\right)\right)^{-1}\\ = &\frac{\alpha^{mn}(1+\alpha^l)(\alpha^m-1)}{\alpha^m(\alpha-\beta)}\left(1+O\left(\frac{1}{\alpha^{2mn}}\right)\right)\\ = &\frac{\alpha^{mn}+\alpha^{mn+l}-\alpha^{m(n-1)}-\alpha^{m(n-1)+l}}{\alpha-\beta}+ O\left(\frac{1}{\alpha^{mn}}\right)\\ = &\frac{\alpha^{mn+l}-\beta^{mn+l}+\beta^{mn+l}}{\alpha-\beta} -\frac{\alpha^{m(n-1)+l}-\beta^{m(n-1)+l}+\beta^{m(n-1)+l}}{\alpha-\beta}\\ &+\frac{\alpha^{mn}-\beta^{mn}+\beta^{mn}}{\alpha-\beta}- \frac{\alpha^{m(n-1)}-\beta^{m(n-1)}+\beta^{m(n-1)}}{\alpha-\beta} +O\left(\frac{1}{\alpha^{mn}}\right)\\ = &u_{mn+l}-u_{m(n-1)+l}+u_{mn}-u_{m(n-1)}+\frac{\beta^{mn}+\beta^{mn+l}-\beta^{m(n-1)}-\beta^{m(n-1)+l}}{\alpha-\beta} +O\left(\frac{1}{\alpha^{mn}}\right). \end{aligned}$

Then

$\begin{aligned} &\underset{n\to\infty}{\lim}\left( \left( \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}+u_{mk+l}}\right)^{-1} -\left( u_{mn+l}-u_{m(n-1)+l}+u_{mn}-u_{m(n-1)} \right)\right)\\ & = \underset{n\to\infty}{\lim}\left( \frac{\beta^{mn}+\beta^{mn+l}-\beta^{m(n-1)}-\beta^{m(n-1)+l}}{\alpha-\beta}+O\left(\frac{1}{\alpha^{mn}}\right) \right) = 0. \end{aligned}$

So we have

$\left( \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}+u_{mk+l}}\right)^{-1} \sim u_{mn+l}-u_{m(n-1)+l}+u_{mn}-u_{m(n-1)}.$

□

3.6. The proof of Theorem 2.6

In this subsection, we will provide a proof of Theorem 2.6.

Proof. By (1.2), we have

$\begin{equation} \begin{aligned} \sum\limits_{i = 0}^lu_{mk+i} & = \frac{1}{\alpha-\beta}\left(\alpha^{mk}\sum\limits_{i = 0}^l\alpha^i-\beta^{mk}\sum\limits_{i = 0}^l\beta^i\right) = \frac{1}{\alpha-\beta}\left(\frac{\alpha^{mk}(1-\alpha^{l+1})}{1-\alpha}- \frac{\beta^{mk}(1-\beta^{l+1})}{1-\beta}\right)\\ & = \frac{\alpha^{mk}(1-\alpha^{l+1})}{(\alpha-\beta)(1-\alpha)}\left( 1-\frac{B^{mk}(1-\alpha)(1-\beta^{l+1})}{\alpha^{2mk}(1-\beta)(1-\alpha^{l+1})}\right). \end{aligned} \end{equation}$

(3.17)

From (3.2b) and (3.17), it follows that

$\begin{equation} \begin{aligned} \frac{1}{\sum\limits_{i = 0}^lu_{mk+i}} = &\left(\frac{\alpha^{mk}(1-\alpha^{l+1})}{(\alpha-\beta)(1-\alpha)}\right)^{-1} \left(1-\frac{B^{mk}(1-\alpha)(1-\beta^{l+1})}{\alpha^{2mk}(1-\beta)(1-\alpha^{l+1})}\right)^{-1}\\ = &\frac{(\alpha-\beta)(1-\alpha)}{\alpha^{mk}(1-\alpha^{l+1})} \left(1+\frac{B^{mk}(1-\alpha)(1-\beta^{l+1})}{\alpha^{2mk}(1-\beta)(1-\alpha^{l+1})}+ \frac{(1-\alpha)^2(1-\beta^{l+1})^2}{\alpha^{4mk}(1-\beta)^2(1-\alpha^{l+1})^2}+R_k\right)\\ = &\frac{(\alpha-\beta)(1-\alpha)}{(1-\alpha^{l+1})} \left(\frac{1}{\alpha^{mk}}+ \frac{B^{mk}(1-\alpha)(1-\beta^{l+1})}{\alpha^{3mk}(1-\beta)(1-\alpha^{l+1})}+ \frac{(1-\alpha)^2(1-\beta^{l+1})^2}{{\alpha^{5mk}}(1-\beta)^2(1-\alpha^{l+1})^2}\right)\\ &+\frac{(\alpha-\beta)(1-\alpha)}{(1-\alpha^{l+1})}\cdot \frac{R_k}{\alpha^{mk}}, \end{aligned} \end{equation}$

(3.18)

where

$R_k = \frac{(1-\beta^{l+1})^3(1-\alpha)^3B^{mk} \left(4\alpha^{mk}(1-\beta)(1-\alpha^{l+1})-3B^{mk}(1-\alpha)(1-\beta^{l+1})\right)} {\alpha^{4mk}(1-\beta)^2(1-\alpha^{l+1})^2\left(\alpha^{2mk}(1-\beta)(1-\alpha^{l+1})-B^{mk} (1-\alpha)(1-\beta^{l+1})\right)}.$

Let $n$ be a positive integer. Then it follows from (3.18) that

$\begin{equation} \begin{aligned} \sum\limits_{k = n}^{\infty}\frac{1}{\sum\nolimits_{i = 0}^lu_{mk+i}} & = \frac{(\alpha-\beta)(1-\alpha)}{(1-\alpha^{l+1})} \left(\sum\limits_{k = n}^{\infty}\frac{1}{\alpha^{mk}}+C_m\right) = \frac{(\alpha-\beta)(1-\alpha)}{(1-\alpha^{l+1})} \left(\frac{\alpha^m}{\alpha^{mn}(\alpha^m-1)}+C_m\right)\\ & = \frac{\alpha^m(\alpha-\beta)(1-\alpha)}{\alpha^{mn}(\alpha^m-1)(1-\alpha^{l+1})} \left(1+\frac{\alpha^{mn}(\alpha^m-1)}{\alpha^m}C_m\right), \end{aligned} \end{equation}$

(3.19)

where

$\begin{aligned} C_m& = \sum\limits_{k = n}^{\infty} \frac{B^{mk}(1-\alpha)(1-\beta^{l+1})}{\alpha^{3mk}(1-\beta)(1-\alpha^{l+1})}+ \sum\limits_{k = n}^{\infty}\frac{(1-\alpha)^2(1-\beta^{l+1})^2} {\alpha^{5mk}(1-\beta)^2(1-\alpha^{l+1})^2} +\sum\limits_{k = n}^{\infty} \frac{R_k}{\alpha^{mk}}\\ & = \frac{B^{mn}\alpha^{3m}(1-\alpha)(1-\beta^{l+1}))} {\alpha^{3mn}(\alpha^{3m}-B^m)(1-\beta)(1-\alpha^{l+1})} +\frac{\alpha^{5m}(1-\alpha)^2(1-\beta^{l+1})^2} {\alpha^{5mn}(\alpha^{5m}-1)(1-\beta)^2(1-\alpha^{l+1})^2} + \sum\limits_{k = n}^{\infty}\frac{R_k}{\alpha^{mk}}\\ & = O\left(\frac{1}{\alpha^{3mn}}\right). \end{aligned}$

Then we have

$\begin{equation} \frac{\alpha^{mn}(\alpha^m-1)}{\alpha^m}C_m = O\left(\frac{1}{\alpha^{2mn}}\right). \end{equation}$

(3.20)

By (3.2a), (3.19), and (3.20), we have

$\begin{aligned} \left(\sum\limits_{k = n}^{\infty}\frac{1}{\sum\limits_{i = 0}^l u_{mk+i}}\right)^{-1} = &\left(\frac{\alpha^m(\alpha-1)(\alpha-\beta)} {\alpha^{mn}(\alpha^{l+1}-1)(\alpha^m-1)}\right)^{-1} \left(1+O\left(\frac{1}{\alpha^{2mn}}\right)\right)^{-1}\\ = &\frac{\alpha^{mn}(\alpha^{l+1}-1)(\alpha^m-1)}{\alpha^m(\alpha-1)(\alpha-\beta)} \left(1+O\left(\frac{1}{\alpha^{2mn}}\right)\right)\\ = &\frac{\alpha^{mn+l+1}-\alpha^{m(n-1)+l+1}-\alpha^{mn}+\alpha^{m(n-1)}} {(\alpha-1)(\alpha-\beta)}+ O\left(\frac{1}{\alpha^{mn}}\right)\\ = &\frac{\alpha^{m(n-1)}-\beta^{m(n-1)}+\beta^{m(n-1)}}{(\alpha-1)(\alpha-\beta)} -\frac{\alpha^{m(n-1)+l+1}-\beta^{m(n-1)+l+1}+\beta^{m(n-1)+l+1}}{(\alpha-1)(\alpha-\beta)} \\ &-\frac{\alpha^{mn}-\beta^{mn}+\beta^{mn}}{(\alpha-1)(\alpha-\beta)} +\frac{\alpha^{mn+l+1}-\beta^{mn+l+1}+\beta^{mn+l+1}}{(\alpha-1)(\alpha-\beta)} +O\left(\frac{1}{\alpha^{mn}}\right)\\ = &\frac{1}{\alpha-1}\left(u_{mn+l+1}-u_{m(n-1)+l+1}-u_{mn}+u_{m(n-1)}\right)\\ &+\frac{\beta^{mn+l+1}-\beta^{mn}+\beta^{m(n-1)}-\beta^{m(n-1)+l+1}}{(\alpha-1)(\alpha-\beta)} +O\left(\frac{1}{\alpha^{mn}}\right). \end{aligned}$

Then

$\begin{aligned} &\underset{n\to\infty}{\lim}\left( \left( \sum\limits_{k = n}^{\infty}\frac{1}{\sum\nolimits_{i = 0}^l u_{mk+i}}\right)^{-1} - \frac{1}{\alpha-1}\left(u_{mn+l+1}-u_{m(n-1)+l+1}-u_{mn}+u_{m(n-1)}\right) \right)\\ & = \underset{n\to\infty}{\lim}\left(\frac{\beta^{mn+l+1}-\beta^{mn}+\beta^{m(n-1)}-\beta^{m(n-1)+l+1}}{(\alpha-1)(\alpha-\beta)}+O\left(\frac{1}{\alpha^{mn}}\right) \right) = 0. \end{aligned}$

So we have

$\left( \sum\limits_{k = n}^{\infty}\frac{1}{\sum\nolimits_{i = 0}^l u_{mk+i}}\right)^{-1} \sim \frac{1}{\alpha-1}\left(u_{mn+l+1}-u_{m(n-1)+l+1}-u_{mn}+u_{m(n-1)}\right).$

□

3.7. The proof of Theorem 2.7

In this subsection, we will provide a proof of Theorem 2.7.

Proof. Let $h$ be a positive integer. By (1.2), we have

$\begin{equation} \begin{aligned} \frac{1}{u_{mk}u_{mk+h}} & = (\alpha-\beta)^2\left( (\alpha^{mk}-\beta^{mk})(\alpha^{mk+h}-\beta^{mk+h})\right)^{-1} = \frac{(\alpha-\beta)^2}{\alpha^{2mk+h}}\left( (1-\frac{B^{mk}}{\alpha^{2mk}}) (1-\frac{B^{mk+h}}{\alpha^{2mk+2h}})\right)^{-1}\\ & = \frac{(\alpha-\beta)^2}{\alpha^{2mk+h}}\left( 1-\frac{B^{mk}}{\alpha^{2mk}} -\frac{B^{mk+h}}{\alpha^{2mk+2h}}+ \frac{B^h}{\alpha^{4mk+2h}}\right)^{-1} = \frac{(\alpha-\beta)^2}{\alpha^{2mk+h}}\left( 1-\eta\right)^{-1}, \end{aligned} \end{equation}$

(3.21)

where

$\eta = \frac{B^{mk}}{\alpha^{2mk}} +\frac{B^{mk+h}}{\alpha^{2mk+2h}}- \frac{B^h}{\alpha^{4mk+2h}} = \frac{B^{mk}}{\alpha^{2mk}} +\frac{B^{mk+h}}{\alpha^{2mk+2h}}+O\left(\frac{1}{\alpha^{4mk}}\right).$

Then we have

$\begin{equation} \eta^2+\frac{\eta^3}{1-\eta} = O\left(\frac{1}{\alpha^{4mk}}\right). \end{equation}$

(3.22)

By (3.2b), (3.21), and (3.22), we have

$\begin{equation} \begin{aligned} \frac{1}{u_{mk}u_{mk+h}} & = \frac{(\alpha-\beta)^2}{\alpha^{2mk+h}}\left(1+\eta+\eta^2+\frac{\eta^3}{1-\eta}\right) = \frac{(\alpha-\beta)^2}{\alpha^{2mk+h}}\left(1+\eta+O\left(\frac{1}{\alpha^{4mk}}\right)\right)\\ & = \frac{(\alpha-\beta)^2}{\alpha^{2mk+h}}\left( 1+\frac{B^{mk}}{\alpha^{2mk}}\left( 1+\frac{B^h}{\alpha^{2h}}\right)+ O\left(\frac{1}{\alpha^{4mk}}\right)\right)\\ & = \frac{(\alpha-\beta)^2}{\alpha^{h}}\left( \frac{1}{\alpha^{2mk}}+\frac{B^{mk}}{\alpha^{4mk}}\left( 1+\frac{B^h}{\alpha^{2h}}\right)+ O\left(\frac{1}{\alpha^{6mk}}\right)\right). \end{aligned} \end{equation}$

(3.23)

Let $n$ be a positive integer. Then it follows from (3.23) that

$\begin{equation} \begin{aligned} \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}u_{mk+h}} & = \frac{(\alpha-\beta)^2}{\alpha^h}\left( \sum\limits_{k = n}^{\infty}\frac{1}{\alpha^{2mk}}+ \left( 1+\frac{B^h}{\alpha^{2h}}\right) \sum\limits_{k = n}^{\infty}\frac{B^{mk}}{\alpha^{4mk}}\right)+ O\left(\frac{1}{\alpha^{6mn}}\right) \\ & = \frac{(\alpha-\beta)^2}{\alpha^h}\left( \frac{\alpha^{2m}}{\alpha^{2mn}(\alpha^{2m}-1)}+\left(1+\frac{B^h}{\alpha^{2h}}\right) \frac{B^{mn}\alpha^{4m}}{\alpha^{4mn}(\alpha^{4m}-B^m)}\right)+ O\left(\frac{1}{\alpha^{6mn}}\right)\\ & = \frac{(\alpha-\beta)^2\alpha^{2m}}{\alpha^{2mn+h}(\alpha^{2m}-1)}\left( 1+\left(1+\frac{B^h}{\alpha^{2h}}\right) \frac{B^{mn}\alpha^{2m}(\alpha^{2m}-1)}{\alpha^{2mn}(\alpha^{4m}-B^m)}+ O\left(\frac{1}{\alpha^{4mn}}\right)\right)\\ & = \frac{(\alpha-\beta)^2\alpha^{2m}}{\alpha^{2mn+h}(\alpha^{2m}-1)}\left( 1+\omega\right), \end{aligned} \end{equation}$

(3.24)

where

$\omega = \left(1+\frac{B^h}{\alpha^{2h}}\right) \frac{B^{mn}\alpha^{2m}(\alpha^{2m}-1)}{\alpha^{2mn}(\alpha^{4m}-B^m)}+ O\left(\frac{1}{\alpha^{4mn}}\right).$

Then we have

$\begin{equation} \omega^2-\frac{\omega^3}{1+\omega} = O\left(\frac{1}{\alpha^{4mn}}\right). \end{equation}$

(3.25)

By (3.2a), (3.24), and (3.25), we have

$\begin{equation} \begin{aligned} \left(\sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}u_{mk+h}}\right)^{-1} & = \left(\frac{(\alpha-\beta)^2\alpha^{2m}}{\alpha^{2mn+h}(\alpha^{2m}-1)}\right)^{-1} \left( 1+\omega\right)^{-1}\\ & = \frac{\alpha^{2mn+h}(\alpha^{2m}-1)}{(\alpha-\beta)^2\alpha^{2m}} \left(1-\omega+\omega^2-\frac{\omega^3}{1+\omega}\right)\\ & = \frac{\alpha^{2mn+h}(\alpha^{2m}-1)}{(\alpha-\beta)^2\alpha^{2m}} \left(1-\omega+O\left(\frac{1}{\alpha^{4mn}}\right)\right)\\ & = \frac{\alpha^{2mn+h}(\alpha^{2m}-1)}{(\alpha-\beta)^2\alpha^{2m}}\left( 1-\left(1+\frac{B^h}{\alpha^{2h}}\right) \frac{B^{mn}\alpha^{2m}(\alpha^{2m}-1)}{\alpha^{2mn}(\alpha^{4m}-B^m)}+ O\left(\frac{1}{\alpha^{4mn}}\right)\right)\\ & = \frac{\alpha^{2mn+h}-\alpha^{2m(n-1)+h}}{(\alpha-\beta)^2}-\left(1+\frac{B^h}{\alpha^{2h}}\right) \frac{B^{mn}(\alpha^{2m}-1)^2\alpha^h}{(\alpha-\beta)^2(\alpha^{4m}-B^m)}+ O\left(\frac{1}{\alpha^{2mn}}\right). \end{aligned} \end{equation}$

(3.26)

(ⅰ) If we take $h = 2l$ , then it follows from (3.26) that

$\begin{aligned} \left(\sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}u_{mk+2l}}\right)^{-1} = &\left(\frac{\alpha^{mn+l}-\beta^{mn+l}+\beta^{mn+l}}{\alpha-\beta}\right)^2- \left(\frac{\alpha^{m(n-1)+l}-\beta^{m(n-1)+l}+\beta^{m(n-1)+l}}{\alpha-\beta}\right)^2\\ &-\left(1+\frac{1}{\alpha^{4l}}\right) \frac{B^{mn}(\alpha^{2m}-1)^2\alpha^{2l}}{(\alpha-\beta)^2(\alpha^{4m}-B^m)}+ O\left(\frac{1}{\alpha^{2mn}}\right)\\ = & u_{mn+l}^2-u_{m(n-1)+l}^2+\delta+O\left(\frac{1}{\alpha^{2mn}}\right), \end{aligned}$

where

$\begin{aligned} \delta& = \frac{2(B^{mn+l}-B^{m(n-1)+l})}{(\alpha-\beta)^2}- \left(1+\frac{1}{\alpha^{4l}}\right) \frac{B^{mn}(\alpha^{2m}-1)^2\alpha^{2l}}{(\alpha-\beta)^2(\alpha^{4m}-B^m)}\\ & = -\frac{B^{mn}(\alpha^{2m}-1)^2}{\alpha^{4m}-B^m} \left(\frac{\alpha^{2l}+\beta^{2l}}{(\alpha-\beta)^2}- \frac{2B^l(1-B^m)(\alpha^{4m}-B^m)}{(\alpha-\beta)^2(\alpha^{2m}-1)^2} \right)\\ & = -\frac{B^{mn}(\alpha^{2m}-1)^2}{\alpha^{4m}-B^m} \left(u_l^2+\frac{2B^l}{(\alpha-\beta)^2}- \frac{2B^l(1-B^m)(\alpha^{4m}-B^m)}{(\alpha-\beta)^2(\alpha^{2m}-1)^2}\right)\\ & = -\frac{B^{mn}(\alpha^{2m}-1)^2}{\alpha^{4m}-B^m} \left(u_l^2+\frac{2B^l}{(\alpha-\beta)^2}\left(1- \frac{(1-B^m)(\alpha^{4m}-B^m)}{(\alpha^{2m}-1)^2}\right)\right). \end{aligned}$

Let $C_{m, l} = u_l^2+\frac{2B^l}{(\alpha-\beta)^2}\left(1- \frac{(1-B^m)(\alpha^{4m}-B^m)}{(\alpha^{2m}-1)^2}\right)$ . Then

$\underset{n\to\infty}{\lim}\left( \left( \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}u_{mk+2l}}\right)^{-1} -\left( u_{mn+l}^2-u_{m(n-1)+l}^2- \frac{B^{mn}(\alpha^{2m}-1)^2}{\alpha^{4m}-B^m}C_{m,l} \right)\right) = 0.$

So we have

$\left( \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}u_{mk+2l}}\right)^{-1} \sim u_{mn+l}^2-u_{m(n-1)+l}^2- \frac{B^{mn}(\alpha^{2m}-1)^2}{\alpha^{4m}-B^m}C_{m,l},$

where $C_{m, l} = u_l^2+\frac{2B^l}{A^2-4B}\left(1- \frac{(1-B^m)(\alpha^{4m}-B^m)}{(\alpha^{2m}-1)^2}\right)$ .

(ⅱ) If we take $h = 2l-1$ , then it follows from (3.26) that

$\begin{aligned} \left(\sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}u_{mk+2l-1}}\right)^{-1} = &\frac{\alpha^{2mn+2l-1}-\alpha^{2m(n-1)+2l-1}}{(\alpha-\beta)^2} -\left(1+\frac{B}{\alpha^{4l-2}}\right) \frac{B^{mn}(\alpha^{2m}-1)^2\alpha^{2l-1}}{(\alpha-\beta)^2(\alpha^{4m}-B^m)}+ O\left(\frac{1}{\alpha^{2mn}}\right)\\ = &(\alpha-\alpha\beta^{2m})\frac{\alpha^{2mn+2l-2}}{(\alpha-\beta)^2}-\left(1+\frac{B}{\alpha^{4l-2}}\right) \frac{B^{mn}(\alpha^{2m}-1)^2\alpha^{2l-1}}{(\alpha-\beta)^2(\alpha^{4m}-B^m)}+ O\left(\frac{1}{\alpha^{2mn}}\right)\\ = &(\alpha-\alpha\beta^{2m}) \left(\frac{\alpha^{2mn+2l-2}-\beta^{2mn+2l-2}+\beta^{2mn+2l-2}}{(\alpha-\beta)^2}\right)\\ &-\left(1+\frac{B}{\alpha^{4l-2}}\right) \frac{B^{mn}(\alpha^{2m}-1)^2\alpha^{2l-1}}{(\alpha-\beta)^2(\alpha^{4m}-B^m)}+ O\left(\frac{1}{\alpha^{2mn}}\right)\\ = &(\alpha-\alpha\beta^{2m})u_{mn+l-1}^2+\delta+O\left(\frac{1}{\alpha^{2mn}}\right), \end{aligned}$

where

$\begin{aligned} \delta& = \frac{2B^{mn+l-1}(\alpha-\alpha\beta^{2m})}{(\alpha-\beta)^2}- \left(1+\frac{B^{2l-1}}{\alpha^{4l-2}}\right) \frac{B^{mn}(\alpha^{2m}-1)^2\alpha^{2l-1}} {(\alpha-\beta)^2(\alpha^{4m}-B^m)}\\ & = -\frac{B^{mn}(\alpha^{2m}-1)^2}{\alpha^{4m}-B^m} \left(\frac{\alpha^l\alpha^{l-1}+\beta^{2l-1}}{(\alpha-\beta)^2} -\frac{2B^{l-1}(\alpha-\alpha\beta^{2m})(\alpha^{4m}-B^m)} {(\alpha-\beta)^2(\alpha^{2m}-1)^2} \right)\\ & = -\frac{B^{mn}(\alpha^{2m}-1)^2}{\alpha^{4m}-B^m} \left(u_l u_{l-1}+\frac{B^{l-1}(\alpha+\beta)}{(\alpha-\beta)^2} -\frac{2B^{l-1}(\alpha-\alpha\beta^{2m})(\alpha^{4m}-B^m)} {(\alpha-\beta)^2(\alpha^{2m}-1)^2}\right)\\ & = -\frac{B^{mn}(\alpha^{2m}-1)^2}{\alpha^{4m}-B^m} \left(u_l u_{l-1}+ \frac{B^{l-1}}{(\alpha-\beta)^2}\left((\alpha+\beta) -\frac{2(\alpha-\alpha\beta^{2m})(\alpha^{4m}-B^m)} {(\alpha^{2m}-1)^2}\right)\right). \end{aligned}$

Let $C_{m, l}^{\prime} = u_l u_{l-1}+ \frac{B^{l-1}}{(\alpha-\beta)^2}\left((\alpha+\beta) -\frac{2(\alpha-\alpha\beta^{2m})(\alpha^{4m}-B^m)} {(\alpha^{2m}-1)^2}\right)$ . Then

$\underset{n\to\infty}{\lim}\left( \left( \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}u_{mk+2l-1}}\right)^{-1} -\left( (\alpha-\alpha\beta^{2m})u_{mn+l-1}^2-\frac{B^{mn}(\alpha^{2m}-1)^2}{\alpha^{4m}-B^m} C_{m,l}^{\prime} \right)\right) = 0.$

So we have

$\left( \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}u_{mk+2l-1}}\right)^{-1} \sim (\alpha-\alpha\beta^{2m})u_{mn+l-1}^2- \frac{B^{mn}(\alpha^{2m}-1)^2}{\alpha^{4m}-B^m}C_{m,l}^{\prime},$

where $C_{m, l}^{\prime} = u_l u_{l-1}+ \frac{B^{l-1}}{A^2-4B}\left(A -\frac{2(\alpha-\alpha\beta^{2m})(\alpha^{4m}-B^m)} {(\alpha^{2m}-1)^2}\right)$ .

□

3.8. The proof of Theorem 2.8

In this subsection, we will provide a proof of Theorem 2.8.

Proof. By (1.2), we have

$\begin{equation} \begin{aligned} u_{mk}+C = \frac{\alpha^{mk}-\beta^{mk}}{\alpha-\beta}+C = \frac{\alpha^{mk}}{\alpha-\beta}\left(1+\frac{C(\alpha-\beta)}{\alpha^{mk}} -\frac{B^{mk}}{\alpha^{2mk}}\right) = \frac{\alpha^{mk}}{\alpha-\beta}\left(1+\eta\right), \end{aligned} \end{equation}$

(3.27)

where

$\eta = \frac{C(\alpha-\beta)}{\alpha^{mk}}-\frac{B^{mk}}{\alpha^{2mk}}.$

Then we have

$\begin{equation} \eta^2-\frac{\eta^3}{1+\eta} = O\left(\frac{1}{\alpha^{2mk}}\right). \end{equation}$

(3.28)

From (3.2a), (3.27), and (3.28), it follows that

$\begin{equation} \begin{aligned} \frac{1}{u_{mk}+C} & = \left(\frac{\alpha^{mk}}{\alpha-\beta}\right)^{-1}\left(1+\eta\right)^{-1} = \frac{\alpha-\beta}{\alpha^{mk}}\left(1-\eta+\eta^2-\frac{\eta^3}{1+\eta}\right) = \frac{\alpha-\beta}{\alpha^{mk}}\left(1-\eta+O\left(\frac{1}{\alpha^{2mk}}\right)\right)\\ & = \frac{\alpha-\beta}{\alpha^{mk}} \left(1-\frac{C(\alpha-\beta)}{\alpha^{mk}} +O\left(\frac{1}{\alpha^{2mk}}\right)\right) = (\alpha-\beta)\left(\frac{1}{\alpha^{mk}}-\frac{C(\alpha-\beta)}{\alpha^{2mk}} +O\left(\frac{1}{\alpha^{3mk}}\right)\right). \end{aligned} \end{equation}$

(3.29)

Let $n$ be a positive integer. Then it follows from (3.29) that

$\begin{equation} \begin{aligned} \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}+C} & = \sum\limits_{k = n}^{\infty}\frac{\alpha-\beta}{\alpha^{mk}}- \sum\limits_{k = n}^{\infty}\frac{C(\alpha-\beta)^2}{\alpha^{2mk}}+ O\left(\frac{1}{\alpha^{3mn}}\right) = \frac{\alpha^m(\alpha-\beta)}{\alpha^{mn}(\alpha^m-1)}- \frac{C\alpha^{2m}(\alpha-\beta)^2}{\alpha^{2mn}(\alpha^{2m}-1)} +O\left(\frac{1}{\alpha^{3mn}}\right)\\ & = \frac{(\alpha-\beta)\alpha^m}{\alpha^{mn}(\alpha^m-1)} \left(1-\frac{C(\alpha-\beta)\alpha^m}{\alpha^{mn}(\alpha^m+1)}+ O\left(\frac{1}{\alpha^{2mn}}\right)\right) = \frac{(\alpha-\beta)\alpha^m}{\alpha^{mn}(\alpha^m-1)} \left(1-\omega\right), \end{aligned} \end{equation}$

(3.30)

where

$\omega = \frac{C(\alpha-\beta)\alpha^m}{\alpha^{mn}(\alpha^m+1)}+ O\left(\frac{1}{\alpha^{2mn}}\right).$

Then we have

$\begin{equation} \omega^2+\frac{\omega^3}{1-\omega} = O\left(\frac{1}{\alpha^{2mn}}\right). \end{equation}$

(3.31)

By (3.2b), (3.30), and (3.31), we have

$\begin{aligned} \left(\sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}+C}\right)^{-1} = &\left(\frac{(\alpha-\beta)\alpha^m}{\alpha^{mn}(\alpha^m-1)}\right)^{-1}\left(1-\omega\right)^{-1} = \frac{\alpha^{mn}(\alpha^m-1)}{(\alpha-\beta)\alpha^m}\left(1+\omega+\omega^2+\frac{\omega^3}{1-\omega}\right)\\ = &\frac{\alpha^{mn}(\alpha^m-1)}{(\alpha-\beta)\alpha^m}\left(1+\omega+O\left(\frac{1}{\alpha^{2mn}}\right)\right) = \frac{\alpha^{mn}(\alpha^m-1)}{(\alpha-\beta)\alpha^m} \left(1+\frac{C(\alpha-\beta)\alpha^m}{\alpha^{mn}(\alpha^m+1)}+ O\left(\frac{1}{\alpha^{2mn}}\right)\right)\\ = &\frac{\alpha^{mn}-\alpha^{m(n-1)}}{\alpha-\beta}+ \frac{C(\alpha^m-1)}{\alpha^m+1}+ O\left(\frac{1}{\alpha^{mn}}\right)\\ = &\frac{\alpha^{mn}-\beta^{mn}+\beta^{mn}}{\alpha-\beta} -\frac{\alpha^{m(n-1)}-\beta^{m(n-1)}+\beta^{m(n-1)}}{\alpha-\beta}+\frac{C(\alpha^m-1)}{\alpha^m+1}+ O\left(\frac{1}{\alpha^{mn}}\right)\\ = &u_{mn}-u_{m(n-1)}+\frac{\beta^{mn}-\beta^{m(n-1)}}{\alpha-\beta} +\frac{C(\alpha^m-1)}{\alpha^m+1} +O\left(\frac{1}{\alpha^{mn}}\right). \end{aligned}$

Then

$\begin{aligned} &\underset{n\to\infty}{\lim}\left( \left( \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}+C}\right)^{-1} -\left( u_{mn}-u_{m(n-1)}+\frac{C(\alpha^m-1)}{\alpha^m+1} \right)\right) = \underset{n\to\infty}{\lim}\left(\frac{\beta^{mn}-\beta^{m(n-1)}}{\alpha-\beta} +O\left(\frac{1}{\alpha^{mn}}\right) \right) = 0. \end{aligned}$

So we have

$\left( \sum\limits_{k = n}^{\infty}\frac{1}{u_{mk}+C}\right)^{-1} \sim u_{mn}-u_{m(n-1)}+C\frac{\alpha^m-1}{\alpha^m+1}.$

□

4. Conclusions

Let $\left(u_n\right)_{n\geq0}$ be the special Lucas $u$ -sequence defined by $u_{n+2} = Au_{n+1}-Bu_n, \, u_0 = 0, \, u_1 = 1$ , where $n\geq0$ , $B = \pm1$ , and $A$ is an integer such that $A^2-4B > 0$ . In this paper, we study the asymptotic behavior of the sequences involving $u_n$ . In Section 1, we give the definition of the asymptotic behavior and introduce the asymptotic behavior of some sequences. In Section 2, we give the asymptotic formulas for $\left(\sum\limits_{k = n}^\infty a_k\right)^{-1}$ , where

$a_k = \frac{1}{u_{mk}^s},\,\frac{1}{u_{mk}+u_{mk+l}},\,\frac{1}{\sum\nolimits_{i = 0}^l u_{mk+i}},\,\frac{1}{u_{mk}u_{mk+2l}},\,\frac{1}{u_{mk}u_{mk+2l-1}},\,\frac{1}{u_{mk}+C},$

$m, \, l$ are positive integers, $s = 1, 2, 3, 4$ , and $C$ is any constant. In Section 3, we give the proof of these results.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

Hongjian Li was supported by the Project of Guangdong University of Foreign Studies (Grant No. 2024RC063). Pingzhi Yuan was supported by the National Natural Science Foundation of China (Grant No. 12171163) and the Basic and Applied Basic Research Foundation of Guangdong Province (Grant No. 2024A1515010589).

Conflict of interest

The authors declare there are no conflicts of interest.

References

[1]	E. Marder, D. Bucher, Central pattern generators and the control of rhythmic movements, Curr. Biol., 11 (2001), R986–R996. https://doi.org/10.1016/S0960-9822(01)00581-4 doi: 10.1016/S0960-9822(01)00581-4
[2]	E. Marder, R. L. Calabrese, Principles of rhythmic motor pattern generation, Physiol. Rev., 76 (1996), 687–717. https://doi.org/10.1152/physrev.1996.76.3.687 doi: 10.1152/physrev.1996.76.3.687
[3]	A. Sakurai, C. A. Gunaratne, P. S. Katz, Two interconnected kernels of reciprocally inhibitory interneurons underlie alternating left-right swim motor pattern generation in the mollusk Melibe leonina, J. Neurophysiol., 112 (2014), 1317–1328. https://doi.org/10.1152/jn.00261.2014 doi: 10.1152/jn.00261.2014
[4]	D. Alaçam, A. Shilnikov, Making a swim central pattern generator out of latent parabolic bursters, Int. J. Bifurcation Chaos, 25 (2015), 1540003. https://doi.org/10.1142/S0218127415400039 doi: 10.1142/S0218127415400039
[5]	A. I. Selverston, Model Neural Networks and Behavior, New York, 1985. https://doi.org/10.1007/978-1-4757-5858-0
[6]	W. N. Frost, P. S. Katz, Single neuron control over a complex motor program, PNAS, 93 (1996), 422–426. https://doi.org/10.1073/pnas.93.1.422 doi: 10.1073/pnas.93.1.422
[7]	P. S. Katz, S. L. Hooper, Invertebrate central pattern generators, Cold Spring Harbor Monogr. Ser., 49 (2007), 251.
[8]	E. Marder, S. Kedia, E. O. Morozova, New insights from small rhythmic circuits, Curr. Opin. Neurobiol., 76 (2022), 102610. https://doi.org/10.1016/j.conb.2022.102610 doi: 10.1016/j.conb.2022.102610
[9]	E. Marder, Neuromodulation of neuronal circuits: back to the future, Neuron, 76 (2012), 1–11. https://doi.org/10.1016/j.neuron.2012.09.010 doi: 10.1016/j.neuron.2012.09.010
[10]	I. Belykh, A. Shilnikov, When weak inhibition synchronizes strongly desynchronizing networks of bursting neurons, Phys. Rev. Lett., 101 (2008), 078102. https://doi.org/10.1103/PhysRevLett.101.078102 doi: 10.1103/PhysRevLett.101.078102
[11]	T. Nowotny, M. I. Rabinovich, Dynamical origin of independent spiking and bursting activity in neural microcircuits, Phys. Rev. Lett., 98 (2007), 128106. https://doi.org/10.1103/PhysRevLett.98.128106 doi: 10.1103/PhysRevLett.98.128106
[12]	A. I. Selverston, Invertebrate central pattern generator circuits, Phil. Trans. R. Soc. B, 365 (2010), 2329–2345. https://doi.org/10.1098/rstb.2009.0270 doi: 10.1098/rstb.2009.0270
[13]	A. I. Selverston, M. I. Rabinovich, H. D Abarbanel, R. Elson, A. Szücs, R. D. Pinto, et al., Reliable circuits from irregular neurons: a dynamical approach to understanding central pattern generators, J. Physiol.-Paris, 94 (2000), 357–374. https://doi.org/10.1016/S0928-4257(00)01101-3 doi: 10.1016/S0928-4257(00)01101-3
[14]	R. Huerta, M. A. Sánchez-Montañés, F. Corbacho, J. A. Sigüenza, A central pattern generator to control a pyloric-based system, Biol. Cybern., 82 (2000), 85–94. https://doi.org/10.1007/PL00007963 doi: 10.1007/PL00007963
[15]	M. Lodi, A. L. Shilnikov, M. Storace, Design principles for central pattern generators with preset rhythms, IEEE Trans. Neural Networks Learn. Syst., 31 (2019), 3658–3669. https://doi.org/10.1109/TNNLS.2019.2945637 doi: 10.1109/TNNLS.2019.2945637
[16]	S. Chen, Y. Liu, T. Chen, J. Lou, Rhythm motion control in bio-inspired fishtail based on central pattern generator, IET Cyber-Syst. Robot., 3 (2021), 53–67. https://doi.org/10.1049/csy2.12007 doi: 10.1049/csy2.12007
[17]	J. Wojcik, J. Schwabedal, R. Clewley, A. L. Shilnikov, Key bifurcations of bursting polyrhythms in 3-cell central pattern generators, PLoS One, 9 (2014), e92918. https://doi.org/10.1371/journal.pone.0092918 doi: 10.1371/journal.pone.0092918
[18]	J. T. C. Schwabedal, A. B. Neiman, A. L. Shilnikov, Robust design of polyrhythmic neural circuits, Phys. Rev. E, 90 (2014), 022715. https://doi.org/10.1103/PhysRevE.90.022715 doi: 10.1103/PhysRevE.90.022715
[19]	R. Azodi-Avval, F. Bahrami, A mathematical model of arm movement during rhythmic motor activity, in 2011 18th Iranian Conference of Biomedical Engineering (ICBME), (2011), 304–308. https://doi.org/10.1109/ICBME.2011.6168578
[20]	M. B. Reyes, P. V. Carelli, J. C. Sartorelli, R. D. Pinto, A modeling approach on why simple central pattern generators are built of irregular neurons, PLoS One, 10 (2015), e0120314. https://doi.org/10.1371/journal.pone.0120314 doi: 10.1371/journal.pone.0120314
[21]	J. Dethier, G. Drion, A. Franci, R. Sepulchre, A positive feedback at the cellular level promotes robustness and modulation at the circuit level, J. Neurophysiol., 114 (2015), 2472–2484. https://doi.org/10.1152/jn.00471.2015 doi: 10.1152/jn.00471.2015
[22]	J. Collens, K. Pusuluri, A. Kelley, D. Knapper, T. Xing, S. Basodi, et al., Dynamics and bifurcations in multistable 3-cell neural networks, Chaos, 30 (2020), 072101. https://doi.org/10.1063/5.0011374 doi: 10.1063/5.0011374
[23]	Q. Lu, X. Wang, J. Tian, A new biological central pattern generator model and its relationship with the motor units, Cognit. Neurodyn., 16 (2022), 135–147. https://doi.org/10.1007/s11571-021-09710-0 doi: 10.1007/s11571-021-09710-0
[24]	Q. Lu, J. Tian, Synchronization and stochastic resonance of the small-world neural network based on the CPG, Cognit. Neurodyn., 8 (2014), 217–226. https://doi.org/10.1007/s11571-013-9275-8 doi: 10.1007/s11571-013-9275-8
[25]	Y. Zang, E. Marder, Neuronal morphology enhances robustness to perturbations of channel densities, PNAS, 120 (2023), e2219049120. https://doi.org/10.1073/pnas.2219049120 doi: 10.1073/pnas.2219049120
[26]	E. M. Izhikevich, Neural excitability, spiking, and bursting, Int. J. Bifurcation Chaos, 10 (2000), 1171–1266. https://doi.org/10.1142/S0218127400000840 doi: 10.1142/S0218127400000840
[27]	B. Lu, X. Jiang, Reduced and bifurcation analysis of intrinsically bursting neuron model, Electron. Res. Arch., 31 (2023), 5928–5945. https://doi.org/10.3934/era.2023301 doi: 10.3934/era.2023301
[28]	F. Zhan, S. Liu, X. Zhang, J. Wang, B. Lu, Mixed-mode oscillations and bifurcation analysis in a pituitary model, Nonlinear Dyn., 94 (2018), 807–826. https://doi.org/10.1007/s11071-018-4395-7 doi: 10.1007/s11071-018-4395-7
[29]	W. B. Kristan, Neuronal decision-making circuits, Curr. Biol., 18 (2008), R928–R932. https://doi.org/10.1016/j.cub.2008.07.081 doi: 10.1016/j.cub.2008.07.081
[30]	K. L. Briggman, W. B. Kristan, Multifunctional pattern-generating circuits, Annu. Rev. Neurosci., 31 (2008), 271–294. https://doi.org/10.1146/annurev.neuro.31.060407.125552 doi: 10.1146/annurev.neuro.31.060407.125552
[31]	A. A. A. Hill, J. Lu, M. A. Masino, O. H. Olsen, R. L. Calabrese, A model of a segmental oscillator in the leech heartbeat neuronal network, J. Comput. Neurosci., 10 (2001), 281–302. https://doi.org/10.1023/A:1011216131638 doi: 10.1023/A:1011216131638
[32]	R. L. Calabrese, Half-center oscillators underlying rhythmic movements, in The Handbook of Brain Theory and Neural Networks, (1998), 444–447.
[33]	Y. Zang, S. Hong, E. D. Schutter, Firing rate-dependent phase responses of Purkinje cells support transient oscillations, eLife, 9 (2020), e60692. https://doi.org/10.7554/eLife.60692 doi: 10.7554/eLife.60692
[34]	M. Liu, L. Duan, In-phase and anti-phase spikes synchronization within mixed Bursters of the pre-Bözinger complex, Electron. Res. Arch., 30 (2022), 961–977. https://doi.org/10.3934/era.2022050 doi: 10.3934/era.2022050
[35]	S. Li, G. Zhang, J. Wang, Y. Chen, B. Deng, Emergent central pattern generator behavior in chemical coupled two-compartment models with time delay, Physica A, 491 (2018), 177–187. https://doi.org/10.1016/j.physa.2017.08.121 doi: 10.1016/j.physa.2017.08.121
[36]	A. Doloc-Mihu, R. L. Calabrese, A database of computational models of a half-center oscillator for analyzing how neuronal parameters influence network activity, J. Biol. Phys., 37 (2011), 263–283. https://doi.org/10.1007/s10867-011-9215-y doi: 10.1007/s10867-011-9215-y
[37]	A. Doloc-Mihu, R. L. Calabrese, Analysis of family structures reveals robustness or sensitivity of bursting activity to parameter variations in a half-center oscillator (HCO) model, eNeuro, 3 (2016). https://doi.org/10.1523/ENEURO.0015-16.2016
[38]	I. Elices, P. Varona, Asymmetry factors shaping regular and irregular bursting rhythms in central pattern generators, Front. Comput. Neurosci., 11 (2017), 9. https://doi.org/10.3389/fncom.2017.00009 doi: 10.3389/fncom.2017.00009
[39]	A. J. White, Sensory feedback expands dynamic complexity and aids in robustness against noise, Biol. Cybern., 116 (2022), 267–269. https://doi.org/10.1007/s00422-021-00917-2 doi: 10.1007/s00422-021-00917-2
[40]	Z. Yu, P. J. Thomas, Dynamical consequences of sensory feedback in a half-center oscillator coupled to a simple motor system, Biol. Cybern., 115 (2021), 135–160. https://doi.org/10.1007/s00422-021-00864-y doi: 10.1007/s00422-021-00864-y
[41]	R. Huerta, P. Varona, M. I. Rabinovich, H. D. I. Abarbanel, Topology selection by chaotic neurons of a pyloric central pattern generator, Biol. Cybern., 84 (2001), L1–L8. https://doi.org/10.1007/PL00007976 doi: 10.1007/PL00007976
[42]	V. In, A. Kho, P. Longhini, J. D. Neff, A. Palacios, P. L. Buono, Meet ANIBOT: the first biologically-inspired animal robot, Int. J. Bifurcation Chaos, 32 (2022), 2230001. https://doi.org/10.1142/S0218127422300014 doi: 10.1142/S0218127422300014
[43]	A. S. Lele, Y. Fang, J. Ting, A. Raychowdhury, Learning to walk: bio-mimetic hexapod locomotion via reinforcement-based spiking central pattern generation, IEEE J. Emerging Sel. Top. Circuits Syst., 10 (2020), 536–545. https://doi.org/10.1109/JETCAS.2020.3033135 doi: 10.1109/JETCAS.2020.3033135
[44]	T. Sun, Z. Dai, P. Manoonpong, Distributed-force-feedback-based reflex with online learning for adaptive quadruped motor control, Neural Networks, 142 (2021), 410–427. https://doi.org/10.1016/j.neunet.2021.06.001 doi: 10.1016/j.neunet.2021.06.001
[45]	A. Espinal, H. Rostro-Gonzalez, M. Carpio, E. I. Guerra-Hernandez, M. Ornelas-Rodriguez, M. Sotelo-Figueroa, Design of spiking central pattern generators for multiple locomotion gaits in hexapod robots by christiansen grammar evolution, Front. Neurorobot., 10 (2016), 6. https://doi.org/10.3389/fnbot.2016.00006 doi: 10.3389/fnbot.2016.00006
[46]	F. Zhan, S. Liu, Response of electrical activity in an improved neuron model under electromagnetic radiation and noise, Front. Comput. Neurosci., 11 (2017), 107. https://doi.org/10.3389/fncom.2017.00107 doi: 10.3389/fncom.2017.00107
[47]	F. Zhan, S. Liu, J. Wang, B. Lu, Bursting patterns and mixed-mode oscillations in reduced Purkinje model, Int. J. Mod. Phys. B, 32 (2018), 1850043. https://doi.org/10.1142/S0217979218500431 doi: 10.1142/S0217979218500431
[48]	D. Terman, J. E. Rubin, A. C. Yew, C. J. Wilson, Activity patterns in a model for the subthalamopallidal network of the basal ganglia, J. Neurosci., 22 (2002), 2963–2976. https://doi.org/10.1523/JNEUROSCI.22-07-02963.2002 doi: 10.1523/JNEUROSCI.22-07-02963.2002
[49]	F. Su, J. Wang, S. Niu, H. Li, B. Deng, C. Liu, et al., Nonlinear predictive control for adaptive adjustments of deep brain stimulation parameters in basal ganglia–thalamic network, Neural Networks, 98 (2018), 283–295. https://doi.org/10.1016/j.neunet.2017.12.001 doi: 10.1016/j.neunet.2017.12.001
[50]	J. Song, S. Liu, H. Lin, Model-based quantitative optimization of deep brain stimulation and prediction of Parkinson's states, Neuroscience, 498 (2022), 105–124. https://doi.org/10.1016/j.neuroscience.2022.05.019 doi: 10.1016/j.neuroscience.2022.05.019
[51]	J. Song, H. Lin, S. Liu, Basal ganglia network dynamics and function: role of direct, indirect and hyper-direct pathways in action selection, Network: Comput. Neural Syst., 34 (2023), 84–121. https://doi.org/10.1080/0954898X.2023.2173816 doi: 10.1080/0954898X.2023.2173816
[52]	M. Valero, I. Zutshi, E. Yoon, G. Buzsáki, Probing subthreshold dynamics of hippocampal neurons by pulsed optogenetics, Science, 375 (2022), 570–574. https://doi.org/10.1126/science.abm1891 doi: 10.1126/science.abm1891
[53]	C. A. Tassinari, G. Cantalupo, B. Hoegl, P. Cortelli, L. Tassi, S. Francione, et al., Neuroethological approach to frontolimbic epileptic seizures and parasomnias: the same central pattern generators for the same behaviours, Rev. Neurol., 165 (2009), 762–768. https://doi.org/10.1016/j.neurol.2009.08.002 doi: 10.1016/j.neurol.2009.08.002
[54]	C. A. Tassinari, E. Gardella, G. Cantalupo, G. Rubboli, Relationship of central pattern generators with parasomnias and sleep-related epileptic seizures, Sleep Med. Clin., 7 (2012), 125–134. https://doi.org/10.1016/j.jsmc.2012.01.003 doi: 10.1016/j.jsmc.2012.01.003

Reader Comments

Your name:*

Email:*
© 2024 the Author(s), licensee AIMS Press. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0)