On the reciprocal sums of higher-order sequences

Abstract

Let $\{u_n\}$ 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.

1 Introduction

The so-called Fibonacci zeta function and Lucas zeta function defined by

$${\zeta }_F(s)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{F_n^s}\text{and}{\zeta }_L(s)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{L_n^s},$$

where the $F_n$ and $L_n$ denote the Fibonacci numbers and Lucas numbers, have been considered in several different ways. Navas [1] discussed the analytic continuation of these series. Elsner et al. [2] obtained that for any positive distinct integer $s_1$, $s_2$, $s_3$, the numbers ${\zeta }_F(2s_1)$, ${\zeta }_F(2s_2)$, and ${\zeta }_F(2s_3)$ are algebraically independent if and only if at least one of $s_1$, $s_2$, $s_3$ is even.

Ohtsuka and Nakamura [3] studied the partial infinite sums of reciprocal Fibonacci numbers and proved the following conclusions:

$$\begin{array}{c}\lfloor {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{F_k}\right)}^{-1}\rfloor =\{\begin{array}{ll}{F}_{n-2}& \text{if }n\text{ is even and }n\ge 2;\\ F_{n-2}-1& \text{if }n\text{ is odd and }n\ge 1.\end{array}\hfill \\ \lfloor {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{F_k^2}\right)}^{-1}\rfloor =\{\begin{array}{ll}{F}_{n-1}{F}_n-1& \text{if }n\text{ is even and }n\ge 2;\\ F_{n-1}{F}_n& \text{if }n\text{ is odd and }n\ge 1.\end{array}\hfill \end{array}$$

Where $\lfloor \cdot \rfloor$ 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 [6–8]. Related properties of the Fibonacci polynomials and Lucas polynomials can be found in [9–12].

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 [13] and [14]. Kilic and Arikan [15] defined a k th-order linear recursive sequence $\{u_n\}$ 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

$$\parallel {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{u_k}\right)}^{-1}\parallel =u_n-u_{n-1}(n\ge n_0),$$

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

In this paper, we unify the above results by proving some theorems that include all the results, [3–8] and [13–15], 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 $\{u_n\}$ 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 $0\le i<m$ 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 $\{u_n\}$ 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

$$\parallel {\left(\sum _{k=n}^{\lfloor \beta n\rfloor }\frac{1}{u_k}\right)}^{-1}\parallel =u_n-u_{n-1}(n\ge n_1).$$

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

Corollary 1 Let $\{u_n\}$ 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

$$\parallel {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{u_k}\right)}^{-1}\parallel =u_n-u_{n-1}(n\ge n_2).$$

For a positive real number $1<\beta \le 2$, whether there exits an identity for

$$\parallel {\left(\sum _{k=n}^{\lfloor \beta n\rfloor }\frac{1}{u_k}\right)}^{-1}\parallel$$

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

$$f(x)=x^m-a_1{x}^{m-1}-a_2{x}^{m-2}-\cdots -a_{m-1}x-a_m,$$

we have

• (I) Polynomial $f(x)$ has exactly one positive real zero α with $a_1<\alpha <a_1+1$.

• (II) Other $m-1$ zeros of $f(x)$ 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

$$\begin{array}{rl}f(a_1)& =a_1^m-a_1^m-a_2{a}_1^{m-2}-\cdots -a_{m-1}{a}_1-a_m\\ =-a_2{a}_1^{m-2}-\cdots -a_{m-1}{a}_1-a_m<0,\end{array}$$

and

$$\begin{array}{rl}f(a_1+1)& ={(a_1+1)}^m-a_1{(a_1+1)}^{m-1}-a_2{(a_1+1)}^{m-2}-\cdots -a_m\\ >{(a_1+1)}^m-a_1({(a_1+1)}^{m-1}+{(a_1+1)}^{m-2}+\cdots +1)\\ ={(a_1+1)}^m-a_1\cdot \frac{{(a_1+1)}^m-1}{a_1}=1>0.\end{array}$$

Thus there exits a positive real zero α of $f(x)$ with $a_1<\alpha <a_1+1$. According to Descarte’s rule of signs, $f(x)=0$ has at most one positive real root. So, $f(x)$ 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

$$\text{if }x\in \mathbb{R}\text{ such that }x>\alpha ,\text{ then }f(x)>0,$$

(2)

$$\text{if }x\in \mathbb{R}\text{ such that }0<x<\alpha ,\text{ then }f(x)<0.$$

(3)

Let

$$\begin{array}{rl}g(x)& =(x-1)f(x)\\ =x^{m+1}-(a_1+1)x^m+(a_1-a_2)x^{m-1}+(a_2-a_3)x^{m-2}+\cdots +(a_{m-1}-a_m)x+a_m.\end{array}$$

Since $f(x)$ has exactly one positive real zero α, $g(x)$ has two positive real zeros α and 1. Observe that

$$\text{if }x\in \mathbb{R}\text{ such that }x>\alpha ,\text{ then }g(x)>0,$$

(4)

$$\text{if }x\in \mathbb{R}\text{ such that }1<x<\alpha ,\text{ then }g(x)<0.$$

(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 $f(x)$ has no complex zero $z_1$ with $|z_1|>\alpha$.

Proof Assume that there exits such a zero. So, we have

$$f(z_1)=z_1^m-a_1{z}_1^{m-1}-a_2{z}_1^{m-2}-\cdots -a_{m-1}{z}_1-a_m=0,$$

then we obtain

$$\begin{array}{r}\left|z_1^m\right|\le a_1\left|z_1^{m-1}\right|+a_2\left|z_1^{m-2}\right|+\cdots +a_{m-1}|z_1|+a_m,\\ f(|z_1|)=\left|z_1^m\right|-a_1\left|z_1^{m-1}\right|-a_2\left|z_1^{m-2}\right|-\cdots -a_{m-1}|z_1|-a_m\le 0.\end{array}$$

This contradicts with (2). □

Claim 2 $f(x)$ has no complex zero $z_2$ with $1<|z_2|<\alpha$.

Proof Assume that there exits such a zero. Since $f(z_2)=0$,

$$g(z_2)=z_2^{m+1}-(a_1+1)z_2^m+(a_1-a_2)z_2^{m-1}+\cdots +(a_{m-1}-a_m)z_2+a_m=0,$$

then we obtain

$$(a_1+1)|z_2{|}^m\le |z_2{|}^{m+1}+(a_1-a_2)|z_2{|}^{m-1}+\cdots +(a_{m-1}-a_m)|z_2|+a_m.$$

So, we have $g(|z_2|)\ge 0$, which contradicts with (5). □

Claim 3 On the circle $|z_3|=\alpha$ and $|z_3|=1$, $f(x)$ has the unique zero α.

Proof If $f(z_3)=0$, then

$$g(z_3)=z_3^{m+1}-(a_1+1)z_3^m+(a_1-a_2)z_3^{m-1}+\cdots +(a_{m-1}-a_m)z_3+a_m=0,$$

then we obtain

$$(a_1+1)|z_3{|}^m\le |z_3{|}^{m+1}+(a_1-a_2)|z_3{|}^{m-1}+\cdots +(a_{m-1}-a_m)|z_3|+a_m.$$

(6)

If $z_3=\alpha$ or $z_3=1$, then $g(z_3)=0$, so (6) must be an equality. Therefore, $z_3^{m+1},(a_1-a_2)z_3^{m-1},(a_2-a_3)z_3^{m-2},\dots ,(a_{m-1}-a_m)z_3$ and $a_m$ all lie on the same ray issuing from the origin. Since $(a_1-a_2),(a_2-a_3),\dots ,a_m$, 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 $f(z_3)\in {\mathbb{R}}^{+}$. On the circle $|z_3|=\alpha$ and $|z_3|=1$, there are two conditions $z_3=1$ or $z_3=\alpha$. Since $f(1)\ne 0$, α is the unique zero of $f(x)$, Claim 3 holds.

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

Lemma 2 Let $m\ge 2$ and let ${\{u_n\}}_{n\ge 0}$ be an integer sequence satisfying the recurrence formula (1). Then the closed formula of $u_n$ is given by

$$u_n=c{\alpha }^n+\mathcal{O}\left(d^{-n}\right)(n\to \mathrm{\infty }),$$

where $c>0$, $d>1$, and $a_1<\alpha <a_1+1$ is the positive real zero of $f(x)$.

Proof Let $\alpha ,{\alpha }_1,\dots ,{\alpha }_t$ be the distinct roots of $f(x)=0$, where $f(x)=0$ is the characteristic equation of the recurrence formula (1). From Lemma 1 we know that α is the simple root of $f(x)=0$ , 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:

$$u_n=c{\alpha }^n+\sum _{i=1}^t{P}_i(n){\alpha }_i^n,$$

(7)

where

$$P_i(n)\in \mathbb{R}[n],deg{P}_i(n)=r_i-1,r_1+r_2+\cdots +r_t=m-1,\text{and}c\in \mathbb{R}.$$

For example, for positive integers $1\le u,v,w\le t$, if ${\alpha }_u$ is the simple root of $f(x)$, then $P_u(n)=g_1$, where $g_1\in \mathbb{R}$, and $deg{P}_u(n)=0$; if ${\alpha }_v$ is the double root of $f(x)$, then $P_v(n)=g_{2}n+g_3$, where $g_2,g_3\in \mathbb{R}$, and $deg{P}_v(n)=1$; if ${\alpha }_w$ is the multiple root of $f(x)$ with the multiplicity $r_w$, then $P_w(n)=b_1{n}^{r_w-1}+b_2{n}^{r_w-2}+\cdots +b_{r_w-1}n+b_{r_w}$, where $b_1,b_2,\dots ,b_{r_w}\in \mathbb{R}$, and $deg{P}_w(n)=r_w-1$.

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

$$|\sum _{i=1}^t{P}_i(n){\alpha }_i^n|\le \sum _{i=1}^t|P_i(n){\alpha }_i^n|\le M{d}^{-n},$$

which completes the proof (note that if all the roots of $f(x)$ are distinct, we can choose $d^{-1}=max\{|{\alpha }_1|,|{\alpha }_2|,\dots ,|{\alpha }_{m-1}|\}$ 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

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

Using Lemma 2, we have

$$\begin{array}{rl}\frac{1}{u_k}& =\frac{1}{c{\alpha }^k+\mathcal{O}(d^{-k})}=\frac{1}{c{\alpha }^k(1+\mathcal{O}({(\alpha d)}^{-k}))}\\ =\frac{1}{c{\alpha }^k}(1+\mathcal{O}\left({(\alpha d)}^{-k}\right))\\ =\frac{1}{c{\alpha }^k}+\mathcal{O}\left({\left({\alpha }^{2}d\right)}^{-k}\right).\end{array}$$

(8)

Thus

$$\begin{array}{rl}\sum _{k=n}^{\lfloor \beta n\rfloor }\frac{1}{u_k}& =\frac{1}{c}\sum _{k=n}^{\lfloor \beta n\rfloor }\frac{1}{{\alpha }^k}+\mathcal{O}\left(\sum _{k=n}^{\lfloor \beta n\rfloor }{\left({\alpha }^{2}d\right)}^{-k}\right)\\ =\frac{\alpha }{c(\alpha -1)}\cdot {\alpha }^{-n}-\frac{1}{c(\alpha -1)}\cdot {\alpha }^{-\lfloor \beta n\rfloor }+\mathcal{O}\left({\alpha }^{-2n}{d}^{-n}\right)\\ =\frac{\alpha }{c(\alpha -1)}\cdot {\alpha }^{-n}+\mathcal{O}\left({\alpha }^{-2n}{\alpha }^{-\lfloor \beta n\rfloor +2n}\right)+\mathcal{O}\left({\alpha }^{-2n}{d}^{-n}\right)\\ =\frac{\alpha }{c(\alpha -1)}{\alpha }^{-n}+\mathcal{O}\left({\alpha }^{-2n}h\right),\end{array}$$

where $h=max\{{\alpha }^{-\lfloor \beta n\rfloor +2n},d^{-n}\}$.

Taking reciprocal, we get

$$\begin{array}{rl}{\left(\sum _{k=n}^{\lfloor \beta n\rfloor }\frac{1}{u_k}\right)}^{-1}& =\frac{1}{\frac{\alpha }{c(\alpha -1)}{\alpha }^{-n}(1+\mathcal{O}({\alpha }^{-n}h))}\\ =\frac{\alpha -1}{\alpha }c{\alpha }^n(1+\mathcal{O}\left({\alpha }^{-n}h\right))\\ =\frac{\alpha -1}{\alpha }c{\alpha }^n+\mathcal{O}(h)\\ =u_n-u_{n-1}+\mathcal{O}(h).\end{array}$$

Since $h=max\{{\alpha }^{-\lfloor \beta n\rfloor +2n},d^{-n}\}<1$, 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

$$\sum _{k=n}^{\mathrm{\infty }}\frac{1}{u_k}=\frac{1}{c}\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{\alpha }^k}+\mathcal{O}\left(\sum _{k=n}^{\mathrm{\infty }}{\left({\alpha }^{2}d\right)}^{-k}\right)=\frac{\alpha }{c(\alpha -1)}{\alpha }^{-n}+\mathcal{O}\left({\left({\alpha }^{2}d\right)}^{-n}\right).$$

Taking reciprocal, we get

$$\begin{array}{rl}{\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{u_k}\right)}^{-1}& =\frac{1}{\frac{\alpha }{c(\alpha -1)}{\alpha }^{-n}(1+\mathcal{O}({(\alpha d)}^{-n}))}\\ =\frac{\alpha -1}{\alpha }c{\alpha }^n(1+\mathcal{O}\left({(\alpha d)}^{-n}\right))\\ =\frac{\alpha -1}{\alpha }c{\alpha }^n+\mathcal{O}\left(d^{-n}\right)\\ =u_n-u_{n-1}+\mathcal{O}\left(d^{-n}\right).\end{array}$$

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 $\{u_n\}$ 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 $0\le q<p$. 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

$$\begin{array}{rl}\text{(a)}& \parallel {\left(\sum _{k=n}^{\lfloor \beta n\rfloor }\frac{{(-1)}^k}{u_k}\right)}^{-1}\parallel ={(-1)}^n(u_n+u_{n-1})(n\ge n_3),\\ \text{(b)}& \parallel {\left(\sum _{k=n}^{\lfloor \beta n\rfloor }\frac{1}{u_{pk+q}}\right)}^{-1}\parallel =u_{pn+q}-u_{pn-p+q}(n\ge n_4),\\ \text{(c)}& \parallel {\left(\sum _{k=n}^{\lfloor \beta n\rfloor }\frac{{(-1)}^k}{u_{pk+q}}\right)}^{-1}\parallel ={(-1)}^n(u_{pn+q}+u_{pn-p+q})(n\ge n_5).\end{array}$$

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

Corollary 2 Let $\{u_n\}$ 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 $0\le q<p$. Then there exist positive integers $n_6$, $n_7$ and $n_8$ depending on $a_1,a_2,\dots$ , and $a_m$ such that

$$\begin{array}{rl}\text{(e)}& \parallel {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{{(-1)}^k}{u_k}\right)}^{-1}\parallel ={(-1)}^n(u_n+u_{n-1})(n\ge n_6),\\ \text{(f)}& \parallel {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{u_{pk+q}}\right)}^{-1}\parallel =u_{pn+q}-u_{pn-p+q}(n\ge n_7),\\ \text{(g)}& \parallel {\left(\sum _{k=n}^{\mathrm{\infty }}\frac{{(-1)}^k}{u_{pk+q}}\right)}^{-1}\parallel ={(-1)}^n(u_{pn+q}+u_{pn-p+q})(n\ge n_8).\end{array}$$

Proof We shall prove only (c) in Theorem 2 and other identities are proved similarly. From Lemma 2 we have

$$\frac{{(-1)}^k}{u_{pk+q}}=\frac{{(-1)}^k}{c{\alpha }^{pk+q}+\mathcal{O}(d^{-pk-q})}=\frac{{(-1)}^k}{c{\alpha }^{pk+q}}(1+\mathcal{O}\left({(\alpha d)}^{-pk-q}\right)).$$

Thus

$$\begin{array}{rl}\sum _{k=n}^{\lfloor \beta n\rfloor }\frac{{(-1)}^k}{u_{pk+q}}& =\frac{{(-1)}^n{\alpha }^p}{c{\alpha }^{pn+q}({\alpha }^p+1)}+\frac{{(-1)}^n{\alpha }^p}{c{\alpha }^{p\lfloor \beta n\rfloor +q}({\alpha }^p+1)}+\mathcal{O}\left({\left({\alpha }^{2}d\right)}^{-pn-q}\right)\\ =\frac{{(-1)}^n{\alpha }^p}{c{\alpha }^{pn+q}({\alpha }^p+1)}+\mathcal{O}\left({\alpha }^{-p\lfloor \beta n\rfloor -q}\right)+\mathcal{O}\left({\alpha }^{-2pn-2q}{d}^{-pn-q}\right)\\ =\frac{{(-1)}^n{\alpha }^p}{c{\alpha }^{pn+q}({\alpha }^p+1)}+\mathcal{O}\left({\alpha }^{-2pn}{\alpha }^{-p\lfloor \beta n\rfloor +2pn}\right)+\mathcal{O}\left({\alpha }^{-2pn}{d}^{-pn}\right)\\ =\frac{{(-1)}^n{\alpha }^p}{c{\alpha }^{pn+q}({\alpha }^p+1)}+\mathcal{O}\left({\alpha }^{-2pn}{h}^p\right),\end{array}$$

where $h=max\{{\alpha }^{-\lfloor \beta n\rfloor +2n},d^{-n}\}$.

Taking reciprocal, we get

$$\begin{array}{rl}{\left(\sum _{k=n}^{\lfloor \beta n\rfloor }\frac{{(-1)}^k}{u_{pk+q}}\right)}^{-1}& ={(-1)}^n(c{\alpha }^{pn+q}+c{\alpha }^{pn-p+q})(1+\mathcal{O}\left({\alpha }^{-pn}{h}^p\right))\\ ={(-1)}^n(u_{pn+q}+u_{pn-p+q})+\mathcal{O}\left(h^p\right).\end{array}$$

Since $h=max\{{\alpha }^{-\lfloor \beta n\rfloor +2n},d^{-n}\}<1$, 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. □