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\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}
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\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}
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\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 $a&lt;\alpha&lt;a+1$;
\item[(b)] the remaining $s-1$ zeros of $f(x)$ lie within the unit circle in the complex plane.
\end{enumerate}
\end{lemma}
\begin{proof}
The case where $a=1$ can be found in \cite{Mi}, so we assume from now on that $a\ge 2$.

~~~~~By Descarte's rule of signs, we see that $f(x)$ has at most one positive real zero. Since $f(a)&lt;0$ and $f(a+1)&gt;0$, its unique positive real zero, say $\alpha$, satisfies $(2\le)~a&lt;\alpha&lt;a+1$. Since multiple roots are counted separately by Descarte's rule again, part (a) is proved. Observe from part (a) that 
\begin{align} \label{f+}
\textnormal{for real}~~ &amp;x&gt; \alpha, \textnormal{we have}~~ f(x)&gt;0,\\
\textnormal{while for real}~~ &amp;0&lt;x&lt;\alpha, \textnormal{we have}~~ f(x)&lt;0.\label{f-}
\end{align}
Next, let 
$$g(x)=(x-1)f(x)= x^{s+1}-(a+1)x^s+(a-1)x^{s-1}+1.$$ 
Observe further that 
\begin{align}
\textnormal{for real}~~ &amp;x&gt; \alpha, \textnormal{we have}~~ g(x)&gt;0,\label{g+}\\
\textnormal{while for real}~~ &amp;1&lt;x&lt;\alpha, \textnormal{we have}~~ g(x)&lt;0.\label{g-}
\end{align}
To prove part (b), we proceed by establishing several claims.

\bigskip

\noindent{\bf Claim 1.} $f(x)$ has no complex zero $z_1$ with $\left|z_1\right|&gt;\alpha$.\\

\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 $1&lt;\left|z_2\right|&lt; \alpha$.\\

\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, &amp;\text{if $s$ is even}; \\
x^{s+1}+(a+1)x^s+(a-1)x^{s-1}+1, &amp; \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 $f(0)=-1,~f(-1)=a&gt;0$, 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
$$ 2^s\le a^s&lt;\left|\alpha\right|^s=\left|z_3\right|^s=\frac{1}{a+1}\le\frac{1}{3}~~~~\textnormal{or}~~~~
1=\left|z_3\right|^s=\frac{1}{a+1}\le\frac{1}{3}.$$

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&amp;=g(\alpha)=\alpha^{s+1}-(a+1)\alpha^s+(a-1)\alpha^{s-1}+1   \label{11}\\
%0&amp;=G(\alpha)=(s+1)\alpha^2-(a+1)s\alpha+(a-1)(s-1)   \label{12}\\
%0&amp;=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 $(a^2-a+2)\alpha^{s-1}+(a+1)\alpha^{s-2}+(a+1)\alpha^{s-3}+\cdots+(a+1)\alpha&gt;s$, 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 $c&gt;0$, $d&gt;1$, and $\alpha&gt;a$ 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 $\left|\alpha_j\right|&lt;1$ ($2\le t\le s$) 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}&amp;=\frac{1}{c\alpha^k+\mathcal O(d^{-k})} 
=\frac{1}{c\alpha^k(1+\mathcal O((\alpha d)^{-k})))}\\
&amp;=\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}&amp;=\frac{1}{c}\sum_{k=n}^\infty\frac{1}{\alpha^k}+\mathcal O\left(\sum_{k=n}^\infty(\alpha^2 d)^{-k}\right)\\
&amp;=\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}&amp;=\frac{\alpha-1}{\alpha}c\alpha^n+\mathcal O(d^{-n})\\
&amp;=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} 
&amp;\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}\\
&amp;\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}\\
&amp;\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}\\
&amp;\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}\\
&amp;\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}&amp;=\sum_{k=n}^\infty\frac{(-1)^k}{c\alpha^k+\mathcal O(d^{-k})}\\
%&amp;=\sum_{k=n}^\infty\frac{(-1)^k}{c\alpha^k(1+\mathcal O((\alpha d)^{-k}))}\\
&amp;=\sum_{k=n}^\infty\frac{(-1)^k}{c\alpha^k}\left(1+\mathcal O((\alpha d)^{-k})\right)\\
%&amp;=\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)\\
&amp;=\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}%&amp;=\left(\frac{\alpha}{c(-\alpha)^n(\alpha+1)}\left(1+\mathcal O((\alpha d)^{-n})\right)\right)^{-1}\\
&amp;=\frac{c(-\alpha)^n(\alpha+1)}{\alpha}\left(1+\mathcal O((\alpha d)^{-n})\right)\\
&amp;=(-1)^n(c\alpha^n+c\alpha^{n-1})+\mathcal O((-d)^{-n})\\
&amp;=(-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} 
&amp;\left\|\left(\sum_{k=n}^\infty\frac{1}{T_k}\right)^{-1}\right\|
=T_n-T_{n-1}\quad(n\ge 1)\,.\\ 
&amp;\left\|\left(\sum_{k=n}^\infty\frac{1}{T_{2k}}\right)^{-1}\right\|=T_{2n}-T_{2n-2}\quad(n\ge 1)\,.\\
&amp;\left\|\left(\sum_{k=n}^\infty\frac{1}{T_{2k-1}}\right)^{-1}\right\|=T_{2n-1}-T_{2n-3}\quad(n\ge 2)\,.\\
&amp;\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)\,.\\
&amp;\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)\,.\\
&amp;\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} &amp;=(u_{\nu-1}^s+u_\nu^s)A_{\nu -1} -u_{\nu-1}^{2s}A_{\nu -2} &amp; \, &amp;(\nu \ge 2), &amp;\quad A_0&amp;= 0,&amp;\quad A_1&amp;= 1; \\
B_{\nu} &amp;=(u_{\nu-1}^s+u_\nu^s)B_{\nu -1} -u_{\nu-1}^{2s}B_{\nu -2} &amp; \, &amp;(\nu \ge 2), &amp;\quad B_0&amp;= 1,&amp;\quad B_1&amp;= 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 $\zeta_{u^{(m)}}(s)-A_n/B_n$.







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T. Komatsu,  
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\emph{Fibonacci Quart.} \textbf{46/47} (2008/2009), 268--278. 

\bibitem{Kom2} 
T. Komatsu, 
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T. Koshy, 
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M. D. Miller, 
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H. Ohtsuka and S. Nakamura, 
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%%%%%%%%%%%%%%%%%%%%%%%%%
\bibitem{oeis}
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\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
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\noindent (Concerned with sequence
\seqnum{A000073}.)

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\vspace*{+.1in}
\noindent
Received January 20 2010;
revised version received May 19 2010. 
Published in {\it Journal of Integer Sequences}, May 20 2010.

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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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