\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://www.research.att.com/cgi-bin/access.cgi/as/~njas/sequences/eisA.cgi?Anum=#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \begin{center} \vskip 1cm{\LARGE\bf On the Sum of Reciprocals of Numbers \\ \vskip .2cm Satisfying a Recurrence Relation of Order $s$ } \vskip 1cm \large Takao Komatsu\footnote{ The first author was supported in part by the Grant-in-Aid for Scientific Research (C) (No.\ 22540005), the Japan Society for the Promotion of Science.} \\ Graduate School of Science and Technology\\ Hirosaki University \\ Hirosaki, 036-8561 \\ Japan\\ \href{mailto:komatsu@cc.hirosaki-u.ac.jp}{\tt komatsu@cc.hirosaki-u.ac.jp} \\ \ \\ Vichian Laohakosol\footnote{ The second author was supported in part by the Commission on Higher Education and the Thailand Research Fund RTA5180005. }\\ Department of Mathematics\\ Kasetsart University \\ Bangkok 10900 \\ Thailand\\ \href{mailto:fscivil@ku.ac.th}{\tt fscivil@ku.ac.th} \end{center} \vskip .2 in \begin{abstract} We discuss the partial infinite sum $\sum_{k=n}^{\infty}u_k^{-s}$ for some positive integer $n$, where $u_k$ satisfies a recurrence relation of order $s$, $u_n= a u_{n-1}+u_{n-2}+\cdots+u_{n-s}$ ($n\ge s$), with initial values $u_0\ge 0$, $u_k\in\mathbb N$ ($0\le k\le s-1$), where $a$ and $s(\ge 2)$ are positive integers. If $a=1$, $s=2$, and $u_0=0$, $u_1=1$, then $u_k=F_k$ is the $k$-th Fibonacci number. Our results include some extensions of Ohtsuka and Nakamura. We also consider continued fraction expansions that include such infinite sums. \end{abstract} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{defin}[theorem]{Definition} \newenvironment{definition}{\begin{defin}\normalfont\quad}{\end{defin}} \newtheorem{examp}[theorem]{Example} \newenvironment{example}{\begin{examp}\normalfont\quad}{\end{examp}} \newtheorem{rema}[theorem]{Remark} \newenvironment{remark}{\begin{rema}\normalfont\quad}{\end{rema}} \def\fl#1{\left\lfloor#1\right\rfloor} \def\cl#1{\left\lceil#1\right\rceil} \def\rint#1{\left(\!\!\left(#1\right)\!\!\right)} \def \C{\mathbb{C}} \def \N{\mathbb{N}} \def \P{\mathbb{P}} \def \Q{\mathbb{Q}} \def \R{\mathbb{R}} \def \Z{\mathbb{Z}} \section{Introduction} The so-called {\em Fibonacci zeta function\/} is defined by $$ \zeta_F(s)=\sum_{n=1}^{\infty}\frac{1}{F_n^s}\,, $$ where $F_n$ is the $n$-th Fibonacci number satisfying the recurrence formula $$ F_n=F_{n-1}+F_{n-2}\quad(n\ge 2),\quad F_0=0,\quad F_1=1\,. $$ Ohtsuka and Nakamura \cite{ON} studied the partial infinite sums of reciprocal Fibonacci numbers $\sum_{k\ge n}^{\infty}1/F_k^s$. They gave an explicit formula for the integer part of $(\sum_{k\ge n}^\infty F_k^{-1})^{-1}$ and $(\sum_{k\ge n}^\infty F_k^{-2})^{-1}$. Holliday and Komatsu \cite{HK} generalized these results to the cases of $G_n$ and $H_n$, satisfying $G_n=a G_{n-1}+G_{n-2}$ ($n\ge 2$) with $G_0=0$ and $G_1=1$, and $H_n=H_{n-1}+H_{n-2}$ ($n\ge 2$) with $H_0=c$ and $H_1=1$, where $a\ge 1$ and $c\ge 0$ are integers. In this paper we shall not consider the integer part, but the nearest integer function of $(\sum_{k\ge n}^\infty u_k^{-1})^{-1}$, where $\{u_k\}_{k\ge 0}$ is a sequence of non-negative integers satisfying a linear recurrence formula of the type $$ u_n=a u_{n-1}+u_{n-2}+\cdots+u_{n-s}\,, $$ where $a$ and $s\;(\ge 2)$ are positive integers. Here, $\|\cdot\|$ denotes the nearest integer\footnote{In other contexts, this notation is sometimes used for the distance from the nearest integer.}, namely, $\|x\|=\fl{x+1/2}$. Our main result is the following: \begin{theorem} \label{thm1} Let $\{u_n\}_{n\ge 0}$ be an integer sequence satisfying the recurrence formula \begin{equation} u_n=a u_{n-1}+u_{n-2}+\cdots+u_{n-s}\quad(n\ge s) \label{u1} \end{equation} with initial conditions \begin{equation} u_0\ge 0,\quad u_k\in\mathbb{N}\quad(0\le k\le s-1)\,, \label{u2} \end{equation} where $a$ and $s\;(\ge 2)$ are positive integers. Then there is a positive integer $n_0$ such that \begin{equation} \left\|\left(\sum_{k=n}^\infty\frac{1}{u_k}\right)^{-1}\right\| =u_n-u_{n-1}\quad(n\ge n_0)\,. \label{u3} \end{equation} \label{ut} \end{theorem} If $a=1$, $u_0=0$, $u_1=u_2=1$, $u_3=2$, $\cdots$, $u_{s-1}=2^{s-3}$, then the $u_n$'s are generalized Fibonacci numbers (sometimes called ``Fibonacci $s$-step numbers'' \cite{Eric}). If $s=2$, then $u_k=F_k$ are Fibonacci numbers, while if $s=3$, then $u_k=T_k$ are Tribonacci numbers. We need the following two lemmas in order to prove this theorem. \begin{lemma} \label{lemma1} Let $a,s\in\N,~s\ge 2$ and let $$f(x)=x^s-a x^{s-1}-x^{s-2}-\cdots-x-1.$$ Then \begin{enumerate} \item[(a)] $f(x)$ has exactly one positive simple zero $\alpha\in\R$ with $a<\alpha0$, its unique positive real zero, say $\alpha$, satisfies $(2\le)~a<\alpha \alpha, \textnormal{we have}~~ f(x)>0,\\ \textnormal{while for real}~~ &0 \alpha, \textnormal{we have}~~ g(x)>0,\label{g+}\\ \textnormal{while for real}~~ &1\alpha$.\\ \noindent{\it Proof of Claim 1.} If $0=f(z_1)=z_1^s-az_1^{s-1}-z_1^{s-2}-\cdots-z_1-1$, then $$\left|z_1\right|^s\le a\left|z_1\right|^{s-1}+\left|z_1\right|^{s-2}+\cdots+\left|z_1\right|+1,$$ which implies that $f\left(\left|z_1\right|\right)\le 0$, contradicting \eqref{f+}.\\ \noindent{\bf Claim 2.} $f(x)$ has no complex zero $z_2$ with $1<\left|z_2\right|< \alpha$.\\ \noindent{\it Proof of Claim 2.} If $f(z_2)=0$, then $0=g(z_2)=z_2^{s+1}-(a+1)z_2^s+(a-1)z_2^{s-1}+1$ and so $$(a+1)\left|z_2\right|^s\le \left|z_2\right|^{s+1}+(a-1)\left|z_2\right|^{s-1}+1,$$ i.e., $g\left(\left|z_2\right|\right)\ge 0$, contradicting \eqref{g-}.\\ \noindent{\bf Claim 3.} $f(x)$ has no complex zero $z_3\ne \alpha$, with either $\left|z_3\right|=\alpha$ or $\left|z_3\right|=1$.\\ \noindent{\it Proof of Claim 3.} If $f(z_3)=0$, then \begin{equation} \label{gz3} 0=g(z_3)=z_3^{s+1}-(a+1)z_3^s+(a-1)z_3^{s-1}+1 \end{equation} so that \begin{equation} \label{g=0} (a+1)\left|z_3\right|^s\le \left|z_3\right|^{s+1}+(a-1)\left|z_3\right|^{s-1}+1. \end{equation} If $\left|z_3\right|=\alpha$ or $\left|z_3\right|=1$, then $g\left(\left|z_3\right|\right)=0$ and so \eqref{g=0} must be an equality. Then the two conditions $z_3^{s+1}~\textnormal{and}~z_3^{s-1}\in\R$, or $z_3^{s+1}=-(a-1)z_3^{s-1}$ follow from two applications of the fact that $$ |z_1+z_2|=|z_1|+|z_2|\quad\Longleftrightarrow\quad\frac{z_1}{z_2}\in\R_{\ge 0}\quad(\hbox{for }z_2\ne 0) $$ and from $$ \bigl(z_1+z_2\in\mathbb R\land\frac{z_1}{z_2}\in\R\bigr)\quad\Longrightarrow\quad (z_1, z_2\in\R\lor z_1=-z_2)\,. $$ \\ $\bullet$ If $z_3^{s+1}~\textnormal{and}~z_3^{s-1}\in\R$, then \eqref{gz3} shows that $z_3^s\in\R$, which in turn forces $z_3\in\R$. Thus, $z_3=\pm \;\alpha$ or $z_3=\pm\; 1$. The possibility $z_3=\alpha$ is ruled out by the hypothesis, and the possibility $z_3=1$ is ruled out by \eqref{f-}. To rule out the remaining two possibilities of negative zeros, consider \begin{equation} \label{negroot} g(-x) = \begin{cases} -x^{s+1}-(a+1)x^s-(a-1)x^{s-1}+1, &\text{if $s$ is even}; \\ x^{s+1}+(a+1)x^s+(a-1)x^{s-1}+1, & \text{if $s$ is odd}. \end{cases} \end{equation} By Descarte's rule of signs applied to $g(-x)$, we deduce that $g(x)$ and so also $f(x)$, has at most one real negative zero if $s$ is even and has no real negative zeros if $s$ is odd. When $s$ is even, since $f(0)=-1,~f(-1)=a>0$, should $f(x)$ have a real negative zero, such zero must lie in the interval $(-1,0)$ and so can neither be $-\alpha$ nor $-1$.\\ \noindent $\bullet$ If $z_3^{s+1}=-(a-1)z_3^{s-1}$, then \eqref{gz3} gives $z_3^s=\frac{1}{a+1}$. Thus, either $$ 2^s\le a^s<\left|\alpha\right|^s=\left|z_3\right|^s=\frac{1}{a+1}\le\frac{1}{3}~~~~\textnormal{or}~~~~ 1=\left|z_3\right|^s=\frac{1}{a+1}\le\frac{1}{3}.$$ Both possibilities are untenable and Claim 3 is proved. ~~~~~Part (b) now follows from Claims 1--3.\\ %To prove part (c), assume that $f(x)$ has a multiple root and so does $g(x)$. Thus, $g(x)$ and %$$g'(x)=x^{s-2}\left\{(s+1)x^2-(a+1)sx +(a-1)(s-1)\right\}=:x^{s-2}G(x)$$ %have a common zero. Such a common zero can only be the roots of the quadratic polynomial $G(x)$. It is easily checked that both roots of $G(x)$ are real and positive. Thus, this multiple root of $f(x)$ can only be $\alpha$. Now consider %\begin{align} %0&=g(\alpha)=\alpha^{s+1}-(a+1)\alpha^s+(a-1)\alpha^{s-1}+1 \label{11}\\ %0&=G(\alpha)=(s+1)\alpha^2-(a+1)s\alpha+(a-1)(s-1) \label{12}\\ %0&=f(\alpha)=\alpha^s-a\alpha^{s-1}-\alpha^{s-2}-\cdots-\alpha-1. \label{13} %\end{align} %Using \eqref{11} and \eqref{12}, we get %\begin{equation} \label{14} %0=(a+1)\alpha^s-2(a-1)\alpha^{s-1}-(s+1). %\end{equation} %Using \eqref{13} and \eqref{14}, we get %$$0=(a^2-a+2)\alpha^{s-1}+(a+1)\alpha^{s-2}+(a+1)\alpha^{s-3}+\cdots+(a+1)\alpha +(a-s).%$$ %Since $(a^2-a+2)\alpha^{s-1}+(a+1)\alpha^{s-2}+(a+1)\alpha^{s-3}+\cdots+(a+1)\alpha>s$, this last relation can never hold and part (c) is proved. \end{proof} We shall keep the notation of Lemma~\ref{lemma1} throughout the rest of the paper. \begin{lemma} Let $s\ge 2$ and let $\{u_n\}_{n\ge 0}$ be an integer sequence satisfying the recurrence formula (\ref{u1}) and the initial conditions (\ref{u2}). Then there are real numbers $c>0$, $d>1$, and $\alpha>a$ such that \begin{equation} u_n=c\alpha^n+\mathcal O(d^{-n})\qquad(n\rightarrow\infty)\,. \label{u5} \end{equation} \label{ul} \end{lemma} \begin{proof} Let $\alpha_1=\alpha,~\alpha_2,\ldots,\alpha_t$ with $\left|\alpha_j\right|<1$ ($2\le t\le s$) be the distinct roots of $f(x)$, and let $r_j$ for $j=2,3,\ldots,t$ denote the multiplicity of the root $\alpha_j$. Then the expansion formula (\ref{u5}) follows from the shape of $u_n$, which is given by $$ u_n=c\alpha^n+\sum_{j=2}^t P_j(n)\alpha_J^n\,, $$ where $$ P_j(x)\in\R[x],\qquad\hbox{deg}P_j=r_J-1,\qquad 1+r_2+r_3+\cdots+r_t=s\,. $$ \end{proof} \noindent{\it Proof of Theorem \ref{ut}.} Applying Lemma~\ref{ul} and the expansion formula $$ \frac{1}{1\pm\epsilon}=1\mp\epsilon+\mathcal O(\epsilon^2)=1+\mathcal O(\epsilon)\qquad(\epsilon\to0)\,, $$ we have \begin{align*} \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((\alpha d)^{-k}))) =\frac{1}{c\alpha^k}+\mathcal O((\alpha^2 d)^{-k})\,, \end{align*} Since \begin{align*} \sum_{k=n}^\infty\frac{1}{u_k}&=\frac{1}{c}\sum_{k=n}^\infty\frac{1}{\alpha^k}+\mathcal O\left(\sum_{k=n}^\infty(\alpha^2 d)^{-k}\right)\\ &=\frac{\alpha}{c(\alpha-1)}\alpha^{-n}+\mathcal O((\alpha^2 d)^{-n})\,, \end{align*} we obtain \begin{align*} \left(\sum_{k=n}^\infty\frac{1}{u_k}\right)^{-1}&=\frac{\alpha-1}{\alpha}c\alpha^n+\mathcal O(d^{-n})\\ &=u_n-u_{n-1}+\mathcal O(d^{-n})\,. \end{align*} Theorem \ref{ut} follows by choosing $n\ge n_0$ sufficient large so that the modulus of the last error term becomes less than $1/2$. \qed \section{Related results} The following results are similarly obtained. Here, $n_1$, $n_2$, $n_3$, $n_4$ and $n_5$ are positive integers depending only on $a$. \begin{theorem} \label{thm2} \begin{align} &\left\|\left(\sum_{k=n}^\infty\frac{1}{u_{2k}}\right)^{-1}\right\|=u_{2n}-u_{2n-2}\quad(n\ge n_1)\,. \label{t42}\\ &\left\|\left(\sum_{k=n}^\infty\frac{1}{u_{2k-1}}\right)^{-1}\right\|=u_{2n-1}-u_{2n-3}\quad(n\ge n_2)\,. \label{t43}\\ &\left\|\left(\sum_{k=n}^\infty\frac{(-1)^k}{u_k}\right)^{-1}\right\|= (-1)^n(u_{n}+u_{n-1})\quad(n\ge n_3)\,. \label{t51}\\ &\left\|\left(\sum_{k=n}^\infty\frac{(-1)^k}{u_{2k}}\right)^{-1}\right\|=(-1)^n(u_{2n}+u_{2n-2})\quad(n\ge n_4)\,. \label{t52}\\ &\left\|\left(\sum_{k=n}^\infty\frac{(-1)^k}{u_{2k-1}}\right)^{-1}\right\|=(-1)^n(u_{2n-1}+u_{2n-3})\quad(n\ge n_5)\,. \label{t53} \end{align} \end{theorem} \begin{proof} We shall prove only (\ref{t51}). The other identities are proved similarly. By (\ref{u5}) we get \begin{align*} \sum_{k=n}^\infty\frac{(-1)^k}{u_k}&=\sum_{k=n}^\infty\frac{(-1)^k}{c\alpha^k+\mathcal O(d^{-k})}\\ %&=\sum_{k=n}^\infty\frac{(-1)^k}{c\alpha^k(1+\mathcal O((\alpha d)^{-k}))}\\ &=\sum_{k=n}^\infty\frac{(-1)^k}{c\alpha^k}\left(1+\mathcal O((\alpha d)^{-k})\right)\\ %&=\frac{1}{c}\sum_{k=n}^\infty\left(-\frac{1}{\alpha}\right)^k+\mathcal O\left(\sum_{k=n}^\infty(-\alpha^2 d)^{-k}\right)\\ &=\frac{\alpha}{c(-\alpha)^n(\alpha+1)}+\mathcal O\left((-\alpha^2 d)^{-n}\right)\,. \end{align*} By taking its reciprocal, we have \begin{align*} \left(\sum_{k=n}^\infty\frac{(-1)^k}{u_k}\right)^{-1}%&=\left(\frac{\alpha}{c(-\alpha)^n(\alpha+1)}\left(1+\mathcal O((\alpha d)^{-n})\right)\right)^{-1}\\ &=\frac{c(-\alpha)^n(\alpha+1)}{\alpha}\left(1+\mathcal O((\alpha d)^{-n})\right)\\ &=(-1)^n(c\alpha^n+c\alpha^{n-1})+\mathcal O((-d)^{-n})\\ &=(-1)^n(u_n+u_{n-1})+\mathcal O(d^{-n})\,. \end{align*} The identity \eqref{t51} follows by choosing $n\ge n_3$ sufficiently large so that the modulus of the last error term becomes less than $1/2$. \end{proof} \section{The sum of reciprocal Tribonacci numbers} The so-called Tribonacci numbers $T_n$ (\cite[Ch.\ 46]{Kos}, \cite[sequence \seqnum{A000073}]{oeis}, \cite{Ki}) are defined by $$ T_n=T_{n-1}+T_{n-2}+T_{n-3}\quad(n\ge 3),\quad T_0=0,\quad T_1=T_2=1\,. $$ By setting $a=1$, $s=3$ and $u_k=T_k$ ($k\ge 0$) in Theorem~\ref{thm1} and Theorem~\ref{thm2}, we get some identities about the partial Tribonacci zeta functions. Numerical evidences imply that identities hold for smaller positive integers $n$, as indicated in the identities. The detailed explanations for small $n$ can be seen in \cite{Kom2}. \begin{corollary} \begin{align} &\left\|\left(\sum_{k=n}^\infty\frac{1}{T_k}\right)^{-1}\right\| =T_n-T_{n-1}\quad(n\ge 1)\,.\\ &\left\|\left(\sum_{k=n}^\infty\frac{1}{T_{2k}}\right)^{-1}\right\|=T_{2n}-T_{2n-2}\quad(n\ge 1)\,.\\ &\left\|\left(\sum_{k=n}^\infty\frac{1}{T_{2k-1}}\right)^{-1}\right\|=T_{2n-1}-T_{2n-3}\quad(n\ge 2)\,.\\ &\left\|\left(\sum_{k=n}^\infty\frac{(-1)^k}{T_k}\right)^{-1}\right\|= (-1)^n(T_{n}+T_{n-1})\quad(n\ge 2)\,.\\ &\left\|\left(\sum_{k=n}^\infty\frac{(-1)^k}{T_{2k}}\right)^{-1}\right\|=(-1)^n(T_{2n}+T_{2n-2})\quad(n\ge 1)\,.\\ &\left\|\left(\sum_{k=n}^\infty\frac{(-1)^k}{T_{2k-1}}\right)^{-1}\right\|=(-1)^n(T_{2n-1}+T_{2n-3})\quad(n\ge 2)\,. \end{align} \end{corollary} \section{Continued fraction expansion of generalized $m$-step zeta functions} The first author \cite{Kom} studied several continued fraction expansions of some types of Fibonacci zeta functions $\zeta_F(s):=\sum_{n=1}^{\infty}F_n^{-s}$ and Lucas zeta functions in $\zeta_L(s):=\sum_{n=1}^{\infty}L_n^{-s}$, where $L_n$ is the $n$-th Lucas number defined by $$ L_n=L_{n-1}+L_{n-2}\quad(n\ge 2)\quad L_0=2,\quad L_1=1. $$ A continued fraction expansion of the generalized $m$-step zeta functions defined by $\zeta_{u^{(m)}}(s):=\sum_{n=1}^\infty u_n^{-s}$, where $$ u_n=a u_{n-1}+u_{n-2}+\cdots+u_{n-m}\quad(n\ge m) $$ with initial positive integral values $u_k$ ($0\le k\le m-1$), is given by $$ \zeta_{u^{(m)}}(s) =\frac{1}{u_1^s - \displaystyle \frac{u_1^{2s}}{u_1^s+u_2^s - \displaystyle \frac{u_2^{2s}}{u_2^s+u_3^s - \displaystyle \frac{u_3^{2s}}{u_3^s+u_4^s - {\atop \ddots\displaystyle - ~\frac{u_{n-1}^{2s}}{u_{n-1}^s+u_n^s - \cdots }}}}}}\,. $$ Define $A_n$ (respectively, $B_n$) as the numerator (respectively, denominator) of the $n^{th}$ convergent of the continued fraction expansion given for $\zeta_{u^{(m)}}(s)$: $$ \frac{A_n}{B_n}=\frac{1}{u_1^s - \displaystyle \frac{u_1^{2s}}{u_1^s+u_2^s - \displaystyle \frac{u_2^{2s}}{u_2^s+u_3^s - \displaystyle \frac{u_3^{2s}}{u_3^s+u_4^s - {\atop \ddots\displaystyle - ~\frac{u_{n-1}^{2s}}{u_{n-1}^s+u_n^s}}}}}}. $$ Hence $\{A_\nu\}_{\nu\ge 0}$ and $\{B_\nu\}_{\nu\ge 0}$ satisfy the following recurrence formulas. \begin{align*} A_{\nu} &=(u_{\nu-1}^s+u_\nu^s)A_{\nu -1} -u_{\nu-1}^{2s}A_{\nu -2} & \, &(\nu \ge 2), &\quad A_0&= 0,&\quad A_1&= 1; \\ B_{\nu} &=(u_{\nu-1}^s+u_\nu^s)B_{\nu -1} -u_{\nu-1}^{2s}B_{\nu -2} & \, &(\nu \ge 2), &\quad B_0&= 1,&\quad B_1&= u_1^s\, \end{align*} In fact, $A_\nu$ and $B_\nu$ can be expressed explicitly as follows. \begin{lemma} For $n=1,2,\ldots$ $$ A_n=(u_1 u_2\cdots u_n)^s\sum_{\nu=1}^n\frac{1}{u_\nu^s},\qquad B_n=(u_1 u_2\cdots u_n)^s\,. $$ \label{GL1} \end{lemma} \begin{proof} By induction we have $B_n=(u_1 u_2\cdots u_n)^s$. Thus, $$ A_n=B_n\sum_{\nu=1}^n\frac{1}{u_\nu^s}=(u_1 u_2\cdots u_n)^s\sum_{\nu=1}^n\frac{1}{u_\nu^s}\,. $$ \end{proof} Theorem~\ref{ut} provides us with interesting information about the nearest integer of the reciprocal of $\zeta_{u^{(m)}}(s)-A_n/B_n$. \begin{thebibliography}{99} \bibitem{Eric} Eric Weisstein, World of Mathematics, published electronically at \\ \href{http://mathworld.wolfram.com/Fibonaccin-StepNumber.html}{\tt http://mathworld.wolfram.com/Fibonaccin-StepNumber.html}. \bibitem{HK} S. H. Holliday and T. Komatsu, On the sum of reciprocal generalized Fibonacci numbers, submitted. \bibitem{Ki} E. Kili\c{c}, Tribonacci sequences with certain indices and their sums, \emph{Ars Combin.} \textbf{86} (2008), 13--22. \bibitem{Kom} T. Komatsu, On continued fraction expansions of Fibonacci and Lucas Dirichlet series, \emph{Fibonacci Quart.} \textbf{46/47} (2008/2009), 268--278. \bibitem{Kom2} T. Komatsu, On the sum of reciprocal Tribonacci numbers, \emph{Ars Combin.}, to appear. \bibitem{Kos} T. Koshy, \emph{Fibonacci and Lucas Numbers with Applications}, John Wiley \& Sons, 2001. \bibitem{Mi} M. D. Miller, On generalized Fibonacci numbers, \emph{Amer. Math. Monthly} \textbf{78} (1971), 1108--1109. \bibitem{ON} H. Ohtsuka and S. Nakamura, On the sum of reciprocal Fibonacci numbers, \emph{Fibonacci Quart.} \textbf{46/47} (2008/2009), 153--159. %%%%%%%%%%%%%%%%%%%%%%%%% \bibitem{oeis} N. J. A. Sloane, (2010), The On-Line Encyclopedia of Integer Sequences, published electronically at \href{http://www.research.att.com/~njas/sequences/}{\tt http://www.research.att.com/\char'176njas/sequences/} . \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11A55; Secondary 11B39. \noindent \emph{Keywords: } Fibonacci numbers, recurrence relations of $s$-th order, partial infinite sum. \bigskip \hrule \bigskip \noindent (Concerned with sequence \seqnum{A000073}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received January 20 2010; revised version received May 19 2010. Published in {\it Journal of Integer Sequences}, May 20 2010. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}