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# On the reciprocal sums of higher-order sequences

Advances in Difference Equations20132013:189

https://doi.org/10.1186/1687-1847-2013-189

• Received: 22 May 2013
• Accepted: 4 June 2013
• Published:

## Abstract

Let $\left\{{u}_{n}\right\}$ be a higher-order recursive sequence. Several identities are obtained for the infinite sums and finite sums of the reciprocals of higher-order recursive sequences.

MSC:11B39.

## Keywords

• infinite sums
• finite sums
• reciprocal
• higher-order recurrences

## 1 Introduction

The so-called Fibonacci zeta function and Lucas zeta function defined by
${\zeta }_{F}\left(s\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{F}_{n}^{s}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\zeta }_{L}\left(s\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{L}_{n}^{s}},$

where the ${F}_{n}$ and ${L}_{n}$ denote the Fibonacci numbers and Lucas numbers, have been considered in several different ways. Navas  discussed the analytic continuation of these series. Elsner et al.  obtained that for any positive distinct integer ${s}_{1}$, ${s}_{2}$, ${s}_{3}$, the numbers ${\zeta }_{F}\left(2{s}_{1}\right)$, ${\zeta }_{F}\left(2{s}_{2}\right)$, and ${\zeta }_{F}\left(2{s}_{3}\right)$ are algebraically independent if and only if at least one of ${s}_{1}$, ${s}_{2}$, ${s}_{3}$ is even.

Ohtsuka and Nakamura  studied the partial infinite sums of reciprocal Fibonacci numbers and proved the following conclusions:

Where $⌊\cdot ⌋$ denotes the floor function.

Further, Wu and Zhang [4, 5] generalized these identities to the Fibonacci polynomials and Lucas polynomials. Similar properties were also investigated in . Related properties of the Fibonacci polynomials and Lucas polynomials can be found in .

Recently, some authors considered the nearest integer of the sums of reciprocal Fibonacci numbers and other famous sequences and obtained several new interesting identities, see  and . Kilic and Arikan  defined a k th-order linear recursive sequence $\left\{{u}_{n}\right\}$ for any positive integer $p\ge q$ and $n>k$ as follows:
${u}_{n}=p{u}_{n-1}+q{u}_{n-2}+{u}_{n-3}+\cdots +{u}_{n-k},$
and they proved that there exists a positive integer ${n}_{0}$ such that
$\parallel {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{u}_{k}}\right)}^{-1}\parallel ={u}_{n}-{u}_{n-1}\phantom{\rule{1em}{0ex}}\left(n\ge {n}_{0}\right),$

where $\parallel \cdot \parallel$ denotes the nearest integer. (Clearly, $\parallel x\parallel =⌊x+\frac{1}{2}⌋$.)

In this paper, we unify the above results by proving some theorems that include all the results,  and , as special cases. We consider the following type of higher-order recurrence sequences. For any positive integer ${a}_{1},{a}_{2},\dots ,{a}_{m}$, we define m th-order linear recursive sequences $\left\{{u}_{n}\right\}$ for $n>m$ as follows:
${u}_{n}={a}_{1}{u}_{n-1}+{a}_{2}{u}_{n-2}+\cdots +{a}_{m-1}{u}_{n-m+1}+{a}_{m}{u}_{n-m},$
(1)

with initial values ${u}_{i}\in \mathbb{N}$ for $0\le i and at least one of them is not zero. If $m=2$, ${a}_{1}={a}_{2}=1$, then ${u}_{n}={F}_{n}$ are the Fibonacci numbers. If $m=2$, ${a}_{1}=2$, ${a}_{2}=1$, then ${u}_{n}={P}_{n}$ are the Pell numbers. Our main results are the following.

Theorem 1 Let $\left\{{u}_{n}\right\}$ be an mth-order sequence defined by (1) with the restriction ${a}_{1}\ge {a}_{2}\ge \cdots \ge {a}_{m}\ge 1$. For any positive real number $\beta >2$, there exists a positive integer ${n}_{1}$ such that
$\parallel {\left(\sum _{k=n}^{⌊\beta n⌋}\frac{1}{{u}_{k}}\right)}^{-1}\parallel ={u}_{n}-{u}_{n-1}\phantom{\rule{1em}{0ex}}\left(n\ge {n}_{1}\right).$

Taking $\beta \to +\mathrm{\infty }$, from Theorem 1 we may immediately deduce the following.

Corollary 1 Let $\left\{{u}_{n}\right\}$ be an mth-order sequence defined by (1) with the restriction ${a}_{1}\ge {a}_{2}\ge \cdots \ge {a}_{m}\ge 1$. Then there exists a positive integer ${n}_{2}$ such that
$\parallel {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{u}_{k}}\right)}^{-1}\parallel ={u}_{n}-{u}_{n-1}\phantom{\rule{1em}{0ex}}\left(n\ge {n}_{2}\right).$
For a positive real number $1<\beta \le 2$, whether there exits an identity for
$\parallel {\left(\sum _{k=n}^{⌊\beta n⌋}\frac{1}{{u}_{k}}\right)}^{-1}\parallel$

is an interesting open problem.

## 2 Several lemmas

To complete the proof of our theorem, we need the following.

Lemma 1 Let ${a}_{1},{a}_{2},\dots ,{a}_{m}$ be positive integers with ${a}_{1}\ge {a}_{2}\ge \cdots \ge {a}_{m}\ge 1$ and $m\in \mathbb{N}$ with $m\ge 2$. Then, for the polynomial
$f\left(x\right)={x}^{m}-{a}_{1}{x}^{m-1}-{a}_{2}{x}^{m-2}-\cdots -{a}_{m-1}x-{a}_{m},$
we have
1. (I)

Polynomial $f\left(x\right)$ has exactly one positive real zero α with ${a}_{1}<\alpha <{a}_{1}+1$.

2. (II)

Other $m-1$ zeros of $f\left(x\right)$ lie within the unit circle in the complex plane.

Proof For any positive integer ${a}_{1}\ge {a}_{2}\ge \cdots \ge {a}_{m}\ge 1$ and $m\ge 2$, we have
$\begin{array}{rl}f\left({a}_{1}\right)& ={a}_{1}^{m}-{a}_{1}^{m}-{a}_{2}{a}_{1}^{m-2}-\cdots -{a}_{m-1}{a}_{1}-{a}_{m}\\ =-{a}_{2}{a}_{1}^{m-2}-\cdots -{a}_{m-1}{a}_{1}-{a}_{m}<0,\end{array}$
and
$\begin{array}{rl}f\left({a}_{1}+1\right)& ={\left({a}_{1}+1\right)}^{m}-{a}_{1}{\left({a}_{1}+1\right)}^{m-1}-{a}_{2}{\left({a}_{1}+1\right)}^{m-2}-\cdots -{a}_{m}\\ >{\left({a}_{1}+1\right)}^{m}-{a}_{1}\left({\left({a}_{1}+1\right)}^{m-1}+{\left({a}_{1}+1\right)}^{m-2}+\cdots +1\right)\\ ={\left({a}_{1}+1\right)}^{m}-{a}_{1}\cdot \frac{{\left({a}_{1}+1\right)}^{m}-1}{{a}_{1}}=1>0.\end{array}$

Thus there exits a positive real zero α of $f\left(x\right)$ with ${a}_{1}<\alpha <{a}_{1}+1$. According to Descarte’s rule of signs, $f\left(x\right)=0$ has at most one positive real root. So, $f\left(x\right)$ has exactly one positive real zero α with ${a}_{1}<\alpha <{a}_{1}+1$. This completes the proof of (I) in Lemma 1.

Observe that from (I) in Lemma 1 we have
(2)
(3)
Let
$\begin{array}{rl}g\left(x\right)& =\left(x-1\right)f\left(x\right)\\ ={x}^{m+1}-\left({a}_{1}+1\right){x}^{m}+\left({a}_{1}-{a}_{2}\right){x}^{m-1}+\left({a}_{2}-{a}_{3}\right){x}^{m-2}+\cdots +\left({a}_{m-1}-{a}_{m}\right)x+{a}_{m}.\end{array}$
Since $f\left(x\right)$ has exactly one positive real zero α, $g\left(x\right)$ has two positive real zeros α and 1. Observe that
(4)
(5)

To complete the proof of (II) in Lemma 1, it is sufficient to show that there is no zero on and outside of the unit circle. □

Claim 1 $f\left(x\right)$ has no complex zero ${z}_{1}$ with $|{z}_{1}|>\alpha$.

Proof Assume that there exits such a zero. So, we have
$f\left({z}_{1}\right)={z}_{1}^{m}-{a}_{1}{z}_{1}^{m-1}-{a}_{2}{z}_{1}^{m-2}-\cdots -{a}_{m-1}{z}_{1}-{a}_{m}=0,$
then we obtain
$\begin{array}{r}|{z}_{1}^{m}|\le {a}_{1}|{z}_{1}^{m-1}|+{a}_{2}|{z}_{1}^{m-2}|+\cdots +{a}_{m-1}|{z}_{1}|+{a}_{m},\\ f\left(|{z}_{1}|\right)=|{z}_{1}^{m}|-{a}_{1}|{z}_{1}^{m-1}|-{a}_{2}|{z}_{1}^{m-2}|-\cdots -{a}_{m-1}|{z}_{1}|-{a}_{m}\le 0.\end{array}$

This contradicts with (2). □

Claim 2 $f\left(x\right)$ has no complex zero ${z}_{2}$ with $1<|{z}_{2}|<\alpha$.

Proof Assume that there exits such a zero. Since $f\left({z}_{2}\right)=0$,
$g\left({z}_{2}\right)={z}_{2}^{m+1}-\left({a}_{1}+1\right){z}_{2}^{m}+\left({a}_{1}-{a}_{2}\right){z}_{2}^{m-1}+\cdots +\left({a}_{m-1}-{a}_{m}\right){z}_{2}+{a}_{m}=0,$
then we obtain
$\left({a}_{1}+1\right)|{z}_{2}{|}^{m}\le |{z}_{2}{|}^{m+1}+\left({a}_{1}-{a}_{2}\right)|{z}_{2}{|}^{m-1}+\cdots +\left({a}_{m-1}-{a}_{m}\right)|{z}_{2}|+{a}_{m}.$

So, we have $g\left(|{z}_{2}|\right)\ge 0$, which contradicts with (5). □

Claim 3 On the circle $|{z}_{3}|=\alpha$ and $|{z}_{3}|=1$, $f\left(x\right)$ has the unique zero α.

Proof If $f\left({z}_{3}\right)=0$, then
$g\left({z}_{3}\right)={z}_{3}^{m+1}-\left({a}_{1}+1\right){z}_{3}^{m}+\left({a}_{1}-{a}_{2}\right){z}_{3}^{m-1}+\cdots +\left({a}_{m-1}-{a}_{m}\right){z}_{3}+{a}_{m}=0,$
then we obtain
$\left({a}_{1}+1\right)|{z}_{3}{|}^{m}\le |{z}_{3}{|}^{m+1}+\left({a}_{1}-{a}_{2}\right)|{z}_{3}{|}^{m-1}+\cdots +\left({a}_{m-1}-{a}_{m}\right)|{z}_{3}|+{a}_{m}.$
(6)

If ${z}_{3}=\alpha$ or ${z}_{3}=1$, then $g\left({z}_{3}\right)=0$, so (6) must be an equality. Therefore, ${z}_{3}^{m+1},\left({a}_{1}-{a}_{2}\right){z}_{3}^{m-1},\left({a}_{2}-{a}_{3}\right){z}_{3}^{m-2},\dots ,\left({a}_{m-1}-{a}_{m}\right){z}_{3}$ and ${a}_{m}$ all lie on the same ray issuing from the origin. Since $\left({a}_{1}-{a}_{2}\right),\left({a}_{2}-{a}_{3}\right),\dots ,{a}_{m}$, are all the elements of ${\mathbb{R}}^{+}$, ${z}_{3}^{m+1},{z}_{3}^{m-1},{z}_{3}^{m-2},\dots ,{z}_{3}$ must be the elements of ${\mathbb{R}}^{+}$. Therefore we obtain $f\left({z}_{3}\right)\in {\mathbb{R}}^{+}$. On the circle $|{z}_{3}|=\alpha$ and $|{z}_{3}|=1$, there are two conditions ${z}_{3}=1$ or ${z}_{3}=\alpha$. Since $f\left(1\right)\ne 0$, α is the unique zero of $f\left(x\right)$, Claim 3 holds.

From the three claims, (II) in Lemma 1 is proven. □

Lemma 2 Let $m\ge 2$ and let ${\left\{{u}_{n}\right\}}_{n\ge 0}$ be an integer sequence satisfying the recurrence formula (1). Then the closed formula of ${u}_{n}$ is given by
${u}_{n}=c{\alpha }^{n}+\mathcal{O}\left({d}^{-n}\right)\phantom{\rule{1em}{0ex}}\left(n\to \mathrm{\infty }\right),$

where $c>0$, $d>1$, and ${a}_{1}<\alpha <{a}_{1}+1$ is the positive real zero of $f\left(x\right)$.

Proof Let $\alpha ,{\alpha }_{1},\dots ,{\alpha }_{t}$ be the distinct roots of $f\left(x\right)=0$, where $f\left(x\right)=0$ is the characteristic equation of the recurrence formula (1). From Lemma 1 we know that α is the simple root of $f\left(x\right)=0$ , then let ${r}_{j}$, for $j=1,2,\dots ,t$, denote the multiplicity of the root ${\alpha }_{j}$. From the properties of m th-order linear recursive sequences, ${u}_{n}$ can be expressed as follows:
${u}_{n}=c{\alpha }^{n}+\sum _{i=1}^{t}{P}_{i}\left(n\right){\alpha }_{i}^{n},$
(7)
where
${P}_{i}\left(n\right)\in \mathbb{R}\left[n\right],\phantom{\rule{1em}{0ex}}deg{P}_{i}\left(n\right)={r}_{i}-1,{r}_{1}+{r}_{2}+\cdots +{r}_{t}=m-1,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}c\in \mathbb{R}.$

For example, for positive integers $1\le u,v,w\le t$, if ${\alpha }_{u}$ is the simple root of $f\left(x\right)$, then ${P}_{u}\left(n\right)={g}_{1}$, where ${g}_{1}\in \mathbb{R}$, and $deg{P}_{u}\left(n\right)=0$; if ${\alpha }_{v}$ is the double root of $f\left(x\right)$, then ${P}_{v}\left(n\right)={g}_{2}n+{g}_{3}$, where ${g}_{2},{g}_{3}\in \mathbb{R}$, and $deg{P}_{v}\left(n\right)=1$; if ${\alpha }_{w}$ is the multiple root of $f\left(x\right)$ with the multiplicity ${r}_{w}$, then ${P}_{w}\left(n\right)={b}_{1}{n}^{{r}_{w}-1}+{b}_{2}{n}^{{r}_{w}-2}+\cdots +{b}_{{r}_{w}-1}n+{b}_{{r}_{w}}$, where ${b}_{1},{b}_{2},\dots ,{b}_{{r}_{w}}\in \mathbb{R}$, and $deg{P}_{w}\left(n\right)={r}_{w}-1$.

From Lemma 1 we have $|{\alpha }_{i}|<1$ for $1\le i\le t$. Since each term of tail in (7) goes to 0 as $n\to \mathrm{\infty }$, we can find the constant $M\in \mathbb{R}$ and $d\in \mathbb{R}$ with $d>1$ for $n>{n}_{0}$ such that
$|\sum _{i=1}^{t}{P}_{i}\left(n\right){\alpha }_{i}^{n}|\le \sum _{i=1}^{t}|{P}_{i}\left(n\right){\alpha }_{i}^{n}|\le M{d}^{-n},$

which completes the proof (note that if all the roots of $f\left(x\right)$ are distinct, we can choose ${d}^{-1}=max\left\{|{\alpha }_{1}|,|{\alpha }_{2}|,\dots ,|{\alpha }_{m-1}|\right\}$ and $M=m-1$). □

## 3 Proof of Theorem 1

In this section, we shall complete the proof of Theorem 1. From the geometric series as $ϵ\to 0$, we have
$\frac{1}{1±ϵ}=1\mp ϵ+\mathcal{O}\left({ϵ}^{2}\right)=1+\mathcal{O}\left(ϵ\right).$
Using Lemma 2, we have
$\begin{array}{rl}\frac{1}{{u}_{k}}& =\frac{1}{c{\alpha }^{k}+\mathcal{O}\left({d}^{-k}\right)}=\frac{1}{c{\alpha }^{k}\left(1+\mathcal{O}\left({\left(\alpha d\right)}^{-k}\right)\right)}\\ =\frac{1}{c{\alpha }^{k}}\left(1+\mathcal{O}\left({\left(\alpha d\right)}^{-k}\right)\right)\\ =\frac{1}{c{\alpha }^{k}}+\mathcal{O}\left({\left({\alpha }^{2}d\right)}^{-k}\right).\end{array}$
(8)
Thus
$\begin{array}{rl}\sum _{k=n}^{⌊\beta n⌋}\frac{1}{{u}_{k}}& =\frac{1}{c}\sum _{k=n}^{⌊\beta n⌋}\frac{1}{{\alpha }^{k}}+\mathcal{O}\left(\sum _{k=n}^{⌊\beta n⌋}{\left({\alpha }^{2}d\right)}^{-k}\right)\\ =\frac{\alpha }{c\left(\alpha -1\right)}\cdot {\alpha }^{-n}-\frac{1}{c\left(\alpha -1\right)}\cdot {\alpha }^{-⌊\beta n⌋}+\mathcal{O}\left({\alpha }^{-2n}{d}^{-n}\right)\\ =\frac{\alpha }{c\left(\alpha -1\right)}\cdot {\alpha }^{-n}+\mathcal{O}\left({\alpha }^{-2n}{\alpha }^{-⌊\beta n⌋+2n}\right)+\mathcal{O}\left({\alpha }^{-2n}{d}^{-n}\right)\\ =\frac{\alpha }{c\left(\alpha -1\right)}{\alpha }^{-n}+\mathcal{O}\left({\alpha }^{-2n}h\right),\end{array}$

where $h=max\left\{{\alpha }^{-⌊\beta n⌋+2n},{d}^{-n}\right\}$.

Taking reciprocal, we get
$\begin{array}{rl}{\left(\sum _{k=n}^{⌊\beta n⌋}\frac{1}{{u}_{k}}\right)}^{-1}& =\frac{1}{\frac{\alpha }{c\left(\alpha -1\right)}{\alpha }^{-n}\left(1+\mathcal{O}\left({\alpha }^{-n}h\right)\right)}\\ =\frac{\alpha -1}{\alpha }c{\alpha }^{n}\left(1+\mathcal{O}\left({\alpha }^{-n}h\right)\right)\\ =\frac{\alpha -1}{\alpha }c{\alpha }^{n}+\mathcal{O}\left(h\right)\\ ={u}_{n}-{u}_{n-1}+\mathcal{O}\left(h\right).\end{array}$

Since $h=max\left\{{\alpha }^{-⌊\beta n⌋+2n},{d}^{-n}\right\}<1$, there exists $n\ge {n}_{1}$ sufficient large so that the modulus of the last error term becomes less than $1/2$, which completes the proof.

Proof of Corollary 1 From identity (8), we have
$\frac{1}{{u}_{k}}=\frac{1}{c{\alpha }^{k}}+\mathcal{O}\left({\left({\alpha }^{2}d\right)}^{-k}\right).$
Thus
$\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{u}_{k}}=\frac{1}{c}\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{\alpha }^{k}}+\mathcal{O}\left(\sum _{k=n}^{\mathrm{\infty }}{\left({\alpha }^{2}d\right)}^{-k}\right)=\frac{\alpha }{c\left(\alpha -1\right)}{\alpha }^{-n}+\mathcal{O}\left({\left({\alpha }^{2}d\right)}^{-n}\right).$
Taking reciprocal, we get
$\begin{array}{rl}{\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{u}_{k}}\right)}^{-1}& =\frac{1}{\frac{\alpha }{c\left(\alpha -1\right)}{\alpha }^{-n}\left(1+\mathcal{O}\left({\left(\alpha d\right)}^{-n}\right)\right)}\\ =\frac{\alpha -1}{\alpha }c{\alpha }^{n}\left(1+\mathcal{O}\left({\left(\alpha d\right)}^{-n}\right)\right)\\ =\frac{\alpha -1}{\alpha }c{\alpha }^{n}+\mathcal{O}\left({d}^{-n}\right)\\ ={u}_{n}-{u}_{n-1}+\mathcal{O}\left({d}^{-n}\right).\end{array}$

So, there exists $n\ge {n}_{2}$ sufficiently large so that the modulus of the last error term becomes less than $1/2$, which completes the proof. □

## 4 Related results

The following results are obtained similarly.

Theorem 2 Let $\left\{{u}_{n}\right\}$ be an mth-order sequence defined by (1) with the restriction ${a}_{1}\ge {a}_{2}\ge \cdots \ge {a}_{m}\ge 1$. Let p and q be positive integers with $0\le q. For any real number $\beta >2$, there exist positive integers ${n}_{3}$, ${n}_{4}$ and ${n}_{5}$ depending on ${a}_{1},{a}_{2},\dots$ , and ${a}_{m}$ such that
$\begin{array}{rl}\text{(a)}& \phantom{\rule{1em}{0ex}}\parallel {\left(\sum _{k=n}^{⌊\beta n⌋}\frac{{\left(-1\right)}^{k}}{{u}_{k}}\right)}^{-1}\parallel ={\left(-1\right)}^{n}\left({u}_{n}+{u}_{n-1}\right)\phantom{\rule{1em}{0ex}}\left(n\ge {n}_{3}\right),\\ \text{(b)}& \phantom{\rule{1em}{0ex}}\parallel {\left(\sum _{k=n}^{⌊\beta n⌋}\frac{1}{{u}_{pk+q}}\right)}^{-1}\parallel ={u}_{pn+q}-{u}_{pn-p+q}\phantom{\rule{1em}{0ex}}\left(n\ge {n}_{4}\right),\\ \text{(c)}& \phantom{\rule{1em}{0ex}}\parallel {\left(\sum _{k=n}^{⌊\beta n⌋}\frac{{\left(-1\right)}^{k}}{{u}_{pk+q}}\right)}^{-1}\parallel ={\left(-1\right)}^{n}\left({u}_{pn+q}+{u}_{pn-p+q}\right)\phantom{\rule{1em}{0ex}}\left(n\ge {n}_{5}\right).\end{array}$

For $\beta \to +\mathrm{\infty }$, we deduce the following identity of infinite sum as a special case of Theorem  2.

Corollary 2 Let $\left\{{u}_{n}\right\}$ be an mth-order sequence defined by (1) with the restriction ${a}_{1}\ge {a}_{2}\ge \cdots \ge {a}_{m}\ge 1$. Let p and q be positive integers with $0\le q. Then there exist positive integers ${n}_{6}$, ${n}_{7}$ and ${n}_{8}$ depending on ${a}_{1},{a}_{2},\dots$ , and ${a}_{m}$ such that
$\begin{array}{rl}\text{(e)}& \phantom{\rule{1em}{0ex}}\parallel {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k}}{{u}_{k}}\right)}^{-1}\parallel ={\left(-1\right)}^{n}\left({u}_{n}+{u}_{n-1}\right)\phantom{\rule{1em}{0ex}}\left(n\ge {n}_{6}\right),\\ \text{(f)}& \phantom{\rule{1em}{0ex}}\parallel {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{u}_{pk+q}}\right)}^{-1}\parallel ={u}_{pn+q}-{u}_{pn-p+q}\phantom{\rule{1em}{0ex}}\left(n\ge {n}_{7}\right),\\ \text{(g)}& \phantom{\rule{1em}{0ex}}\parallel {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k}}{{u}_{pk+q}}\right)}^{-1}\parallel ={\left(-1\right)}^{n}\left({u}_{pn+q}+{u}_{pn-p+q}\right)\phantom{\rule{1em}{0ex}}\left(n\ge {n}_{8}\right).\end{array}$
Proof We shall prove only (c) in Theorem 2 and other identities are proved similarly. From Lemma 2 we have
$\frac{{\left(-1\right)}^{k}}{{u}_{pk+q}}=\frac{{\left(-1\right)}^{k}}{c{\alpha }^{pk+q}+\mathcal{O}\left({d}^{-pk-q}\right)}=\frac{{\left(-1\right)}^{k}}{c{\alpha }^{pk+q}}\left(1+\mathcal{O}\left({\left(\alpha d\right)}^{-pk-q}\right)\right).$
Thus
$\begin{array}{rl}\sum _{k=n}^{⌊\beta n⌋}\frac{{\left(-1\right)}^{k}}{{u}_{pk+q}}& =\frac{{\left(-1\right)}^{n}{\alpha }^{p}}{c{\alpha }^{pn+q}\left({\alpha }^{p}+1\right)}+\frac{{\left(-1\right)}^{n}{\alpha }^{p}}{c{\alpha }^{p⌊\beta n⌋+q}\left({\alpha }^{p}+1\right)}+\mathcal{O}\left({\left({\alpha }^{2}d\right)}^{-pn-q}\right)\\ =\frac{{\left(-1\right)}^{n}{\alpha }^{p}}{c{\alpha }^{pn+q}\left({\alpha }^{p}+1\right)}+\mathcal{O}\left({\alpha }^{-p⌊\beta n⌋-q}\right)+\mathcal{O}\left({\alpha }^{-2pn-2q}{d}^{-pn-q}\right)\\ =\frac{{\left(-1\right)}^{n}{\alpha }^{p}}{c{\alpha }^{pn+q}\left({\alpha }^{p}+1\right)}+\mathcal{O}\left({\alpha }^{-2pn}{\alpha }^{-p⌊\beta n⌋+2pn}\right)+\mathcal{O}\left({\alpha }^{-2pn}{d}^{-pn}\right)\\ =\frac{{\left(-1\right)}^{n}{\alpha }^{p}}{c{\alpha }^{pn+q}\left({\alpha }^{p}+1\right)}+\mathcal{O}\left({\alpha }^{-2pn}{h}^{p}\right),\end{array}$

where $h=max\left\{{\alpha }^{-⌊\beta n⌋+2n},{d}^{-n}\right\}$.

Taking reciprocal, we get
$\begin{array}{rl}{\left(\sum _{k=n}^{⌊\beta n⌋}\frac{{\left(-1\right)}^{k}}{{u}_{pk+q}}\right)}^{-1}& ={\left(-1\right)}^{n}\left(c{\alpha }^{pn+q}+c{\alpha }^{pn-p+q}\right)\left(1+\mathcal{O}\left({\alpha }^{-pn}{h}^{p}\right)\right)\\ ={\left(-1\right)}^{n}\left({u}_{pn+q}+{u}_{pn-p+q}\right)+\mathcal{O}\left({h}^{p}\right).\end{array}$

Since $h=max\left\{{\alpha }^{-⌊\beta n⌋+2n},{d}^{-n}\right\}<1$, there exists $n\ge {n}_{5}$ sufficiently large so that the modulus of the last error term becomes less than $1/2$, which completes the proof. □

## Declarations

### Acknowledgements

The authors express their gratitude to the referee for very helpful and detailed comments. This work is supported by the N.S.F. (11071194, 11001218) of P.R. China and G.I.C.F. (YZZ12062) of NWU.

## Authors’ Affiliations

(1)
Department of Mathematics, Northwest University, Xi’an, Shaanxi, P.R. China

## References

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