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\begin{center}
		
\vskip 1cm{\LARGE\bf   
On the Reciprocal Sums and Products of\\
\vskip 0.1in
$m$th-order Linear Recursive Sequences
}
		
\vskip 1cm
\large  
Tingting Du\\
Research Center for Number Theory and Its Application\\ 
Northwest University \\
Xi'an, Shaanxi 710127 \\
China\\
\href{mailto:dutingtingaaa@163.com}{\tt dutingtingaaa@163.com} \\
\end{center}    
	
\vskip .2 in
	
	
\begin{abstract}
The $m$th-order linear recursive sequence of $(w_{n})_{n\ge 0}$ is defined by
the recursion $w_{n}=a_{1}w_{n-1}+a_{2}w_{n-2}+\cdots+ a_{m}w_{n-m}$ for $n&gt;m$. In previous discussions of the reciprocal sums and products of $(w_{n})_{n\ge 0}$, the condition $a_{1}\ge a_{2}\ge \cdots \ge a_{m}\ge 1$ was typically
imposed. In this paper, we extend previous research and allow the
coefficients to be arbitrary positive integers.
\end{abstract}
	
	\section{Introduction}
	
	For positive integers $a_{1}$, $a_{2}$,$\ldots$, $a_{m}$ with $a_{m} \ne 0$, the $m$th-order linear recursive sequence $(w_{n})_{n\ge 0}$ is defined by 
	\begin{equation}\label{eq1}
		w_{n}=a_{1}w_{n-1}+a_{2}w_{n-2}+\cdots+a_{m}w_{n-m}, \quad n&gt;m,
	\end{equation}
	where the initial values $w_{i}\in \Enn$, and at least one is not zero.
	The characteristic polynomial of the sequence $(w_{n})_{n\ge0}$ is
	\begin{equation}\label{eq2}
		\varphi(x)=x^{m}-a_{1}x^{m-1}-\cdots-a_{m-1}x-a_{m}=(x-\lambda_{1})^{m_{1}}\cdots(x-\lambda_{l})^{m_{l}}, 
	\end{equation} 
	where the $\lambda_{i}$ are called the {\it roots\/} of the sequence.
If the absolute value of a root of the sequence $(w_{n})_{n\ge 0}$ is strictly largest, the root is called the {\it dominant root}. 
	
	If $m=2$, $a_{1}=a_{2}=1$, and $w_{0}=0$,  $w_{1}=1$,
the resulting sequence $(w_{n})_{n\ge 0}$ is the famous Fibonacci sequence $(f_{n})_{n\ge 0}$. Ohtsuka and Nakamura   \cite{Caceres07} considered the 
reciprocal sums of the Fibonacci sequence and obtained  the following result:
 \begin{displaymath}
		\left \lfloor \left({ \sum_{k=n}^{\infty }\frac{1}{f_{k}}}\right)^{-1}   \right \rfloor =
		\begin{cases}
			f_{n-2},     &amp; \text{if $n$ is even and $n \geq 2$;} \\
			f_{n-2}-1,   &amp; \text{if $n$ is odd and $n \geq 1$,}
		\end{cases} 
	\end{displaymath}
where  $\left \lfloor z \right \rfloor$  denotes the floor function.

 Computing the floor function of  $({\sum\limits_{k=n}^{\infty}\frac{1}{w_{k}}})^{-1}$ is a difficult problem.
	  Some researchers have studied the nearest integer to $({\sum\limits_{k=n}^{\infty }\frac{1}{w_{k}}})^{-1}$. For example,  Wu and Zhang  \cite{Caceres08}  proved  that there exists a positive integer $n_{1}$ such that  
	\begin{equation*}
		\left \|(\sum_{k=n}^{\infty} \frac{1}{w_{k}})^{-1}\right \|=w_{n}-w_{n-1},   \quad  n\ge n_{1}, 
	\end{equation*}
	where  $a_{1} \ge a_{2} \ge \dots \ge a_{m}\ge 1$ and $\left \| z  \right \|$ denotes the nearest integer function,
	defined by $\left \| z \right \|=\left \lfloor z+\frac{1}{2}  \right \rfloor$.
	
If two sequences  $(u_{n})_{n\ge 0}$ and $(v_{n})_{n\ge 0}$  satisfy
the condition that $(u_{n}/v_{n})_{n\ge 0}$  tends to $1$ as $n \rightarrow  \infty$, we call them  {\it asymptotically equivalent}.
	Some researchers have continued the study of
reciprocal sums
	by finding a sequence that is  asymptotically equivalent  to  $\left((\sum\limits_{k=n}^{\infty} \frac{1}{w_{k}}) ^{-1}\right)$. Specifically,  Trojovsk\'y  \cite{3}    proved that the sequences
	\begin{equation*}
\left((\sum_{k=n}^{\infty}\frac{1}{P(w_{k})})^{-1}\right)_n
\quad \text{and}  \quad \left(P(w_{n})-P( w_{n-1})\right)_n
	\end{equation*}
	are  asymptotically equivalent, where $P(z)$ is  a non-constant polynomial with  $P(z)\in  \Complexes[z]$.
For more on reciprocal sums and  products, see  \cite{4,9,10,11,12,13,16,17,18,19}.
	
	In previous research concerning reciprocal sums and  products of $(w_{n})_{n\ge 0}$,  the condition $a_{1}\ge a_{2}\ge \cdots \ge a_{m}\ge 1$ was
	typically imposed. In this paper, we allow the coefficients to be
	arbitrary positive integers,
	and obtain a series of sequences that are   asymptotically  equivalent  to $\left((\sum\limits_{k=n}^{\infty } \frac{1}{w_{k}})^{-1}   \right)$ and  
	$\left(1-\prod\limits_{k=n}^{\infty}(1-\frac{1}{w_{k}}))^{-1}   \right)$. The main results are summarized in the following theorem.
	\begin{theorem} \label{th1}
		Let $(w_{n})_{n\ge 0}$ be an  $m$th-order linear recursive sequence defined by \eqref{eq1}. The sequences
		\begin{equation} \label{eq3}
			\left ((\sum_{k=n}^{\infty} \frac{1}{w_{k}})^{-1}   \right) 	  \quad \text{ and} \quad \left(w_{n}-w_{n-1}\right)  
		\end{equation}
		are asymptotically equivalent,  and the sequences
		\begin{equation} \label{eq4}
			\left ((1-\prod_{k=n}^{\infty }(1-\frac{1}{w_{k}}))^{-1}   \right)_n
			\quad \text{ and} \quad \left(w_{n}-w_{n-1}\right)_n
		\end{equation}
		are asymptotically equivalent.
	\end{theorem}
	
	\section{Some lemmas}
	
	In this section, we shall give several lemmas that are useful for the proofs of the theorem.
	\begin{lemma}\label{lem1}
		(Descartes rule of signs). Let $\phi(x) =a_{n_{1}}x^{n_{1}}+\cdots+a_{n_{k} }x^{n_{k}}$ be a    polynomial 
		where the $n_i$ are integers and
		$n_{1}&gt;n_{2}&gt;\cdots &gt;n_{k} \ge 0$. The number of positive roots of $\phi(x)$ is at most the number of sign changes 
		of adjacent nonzero coefficients.
	\end{lemma}
	
	\begin{lemma}(\cite{5,6} Enestr\"om-Kakeya theorem). \label{lem2}
		Let $\phi ^{'}(x)=a_{0}+a_{1}x+\cdots+a_{n}x^{n}$ be a polynomial of order $n$ with real coefficients. If $0\le a_{0} \le a_{1} \le \cdots \le a_{n}$, then all complex roots $z$ of $\phi ^{'}(x)$ 
satisfy  $\left |z\right | \le 1$.
	\end{lemma}
	
	\begin{lemma}\label{lem3}
		Let $\varphi(x)=x^{m}-a_{1}x^{m-1}-\cdots-a_{m-1}x-a_{m}$ be the characteristic polynomial of  $(w_{n})_{n\ge 0}$,  where   the coefficients  $a_{1}$, $a_{2}$, $\ldots$, $a_{m}$ are positive integers. 
Then the following hold:
		\begin{itemize}
			\item	$\varphi (x) $ has only one positive root $\lambda _{1}$, called $\lambda$, and  $\lambda &gt; 1$;
			
			\item The other $m-1$ roots of $\varphi(x)$ lie in the circle $\left | z \right |&lt;\lambda$. Therefore $\lambda$ is the dominant root of $\varphi(x)$.
		\end{itemize}
	\end{lemma}
	
	
	\begin{proof}
		By Lemma \ref{lem1}, the  characteristic  polynomial $\varphi(x ) =x^{m}-a_{1} x^{m-1}-\cdots-a_{m-1} x-a_{m}$ has at most one positive root $\lambda _{1}$, called $\lambda$. In addition, $\lim\limits_{x \to \infty} \varphi(x)=+\infty$, and, by $\varphi(1)&lt; 0$, we obtain that $\lambda&gt;1$ and $\lambda$ is the only positive root of $\varphi(x)$.
		
		By using the relation
		\begin{equation*}
			\lambda ^{m}=a_{1}\lambda ^{m-1}+a_{2}\lambda ^{m-2}+\cdots + a_{m-1}\lambda+a_{m},
		\end{equation*}
		we obtain that $\varphi(x) =( x-\lambda)\psi (x)$, where 
		\begin{equation*}
			\begin{split}
				\psi(x)&amp;=x^{m-1}+(\lambda-a_{1})x^{m-2}+(\lambda ^{2}-a_{1}\lambda -a_{2} )x^{m-3}+\cdots\\
				&amp;+ (\lambda ^{m-2}-a_{1}\lambda^{m-3}-\cdots-a_{m-2} )x+\lambda ^{m-1}-a_{1}\lambda^{m-2}-\cdots-a_{m-1}.
			\end{split}
		\end{equation*}
		\begin{description}
			\item[Claim 1.] \label{cla1}
			If $z$ is a root of $\psi(x)$, then $\left |z \right | \le \lambda$. 
			
			In fact, we prove that all the roots of $\Psi(x):=\psi(\lambda x)$ are in the closed unit ball.  The polynomial
			\begin{equation*}
				\begin{split}
					\Psi (x)&amp;=\lambda^{m-1}x^{m-1}+(\lambda-a_{1}  )\lambda^{m-2} x^{m-2}+(\lambda^{2}-a_{1}\lambda-a_{2})\lambda^{m-3}x^{m-3}+\cdots\\
					&amp;+(\lambda^{m-2}-a_{1}\lambda^{m-3}-\cdots-a_{m-2}  )\lambda x+\lambda ^{m-1}-a_{1}\lambda ^{m-2}-\cdots-a_{m-1}  \\
					&amp;=\lambda^{m-1}x^{m-1}+(\lambda ^{m-1}-a_{1}\lambda ^{m-2}  )x^{m-2}+(\lambda^{m-1}-a_{1}\lambda ^{m-2}-a_{2}\lambda^{m-3})x^{m-3}+\cdots \\
					&amp;+(\lambda^{m-1}-a_{1} \lambda ^{m-2}-\cdots-a_{m-2}\lambda)x+\lambda^{m-1}-a_{1} \lambda ^{m-2}-\cdots-a_{m-1}.     
				\end{split}
			\end{equation*}
			Since 
			\begin{equation*}
				\begin{split}
					&amp;\lambda^{m-1}&gt;\lambda^{m-1}-a_{1}\lambda ^{m-2}&gt; \lambda^{m-1}-a_{1}\lambda ^{m-2}-a_{2}\lambda ^{m-3} &gt;\cdots\\
					&amp;&gt; \lambda^{m-1}-a_{1}\lambda ^{m-2}-\cdots-a_{m-2}\lambda&gt;\lambda ^{m-1}-a_{1}\lambda^{m-2}-\cdots-a_{m-1}=a_{m}/\lambda&gt;0,
				\end{split}
			\end{equation*}
			which $\lambda$ is positive root of $\varphi (x )$. By Lemma \ref{lem2}, this completes the proof of Claim \ref{cla1}.\\
			\item[Claim 2.] \label{cla2}  On the circle $\left |z\right |=\lambda$, the polynomial $\varphi(x )$ has the unique root $\lambda$. 
			
			If $\varphi(z)=0$, then
			\begin{equation*}
				z^{m}=a_{1}z^{m-1}+a_{2}z^{m-2}+\cdots+a_{m-1}z+a_{m}.
			\end{equation*}
			The triangle inequality is satisfied:
			\begin{equation} \label{eq5}
				\left |z\right| ^{m}\le a_{1}\left |z\right |^{m-1}+a_{2}\left|z\right|^{m-2}+\cdots+ a_{m-1}\left|z\right|+a_{m}.
			\end{equation}
			If $z =\lambda$, then $\varphi(z)=0$. So \eqref{eq5} 
must be an equality. Therefore, $a_{1}z^{m-1}$, $a_{2}z^{m-2}$, $\ldots$, $a_{m-1}z$, $a_{m}$ all on the same ray leaving the origin. Since $a_{1}$, $a_{2}$, $\ldots$, $a_{m}$ are all the elements of $\Reals^{+}$, $z^{m-1}$, $z^{m-2}$, $\ldots$, $z$ must  be  elements  of $\Reals^{+}$. Therefore we obtain $\varphi(z)\in \Reals^{+}$. On the circle $\left | z\right |=\lambda$, we obtain $z=\lambda$. This completes the proof of Claim \ref{cla2}.
		\end{description}
	\end{proof} 
	
		
We define $f(x)=\mathcal{O}(g(x))$ to mean that the quotient $f(x)/g(x )$ is bounded for $x\ge a$.
	\begin{lemma}\label{lem4}
		Let $(w_{n})_{n\ge 0}$ be  $m$th-order linear recursive sequence defined by \eqref{eq1}. Then the asymptotic formula of $(w_{n})_{n\ge 0}$  is as follows:
		\begin{equation} \label{eq6}
			w_{n}=c\lambda ^{n} +\mathcal{O}((\lambda \mu )^{\frac{n}{2}})  
		\end{equation}
		where $c$ is a constant. The $\lambda$ defined by Lemma \ref{lem3} is the dominant root of $\varphi (x)$, and    
		$\mu$ is the largest absolute value of the remaining $m-1$ roots of $\varphi(x)$. 
	\end{lemma}
	
	
	\begin{proof} 
		By \cite{7}, there exist unique nonzero polynomials  $\ell _{1}$, $\ldots$, $\ell _{l}\in \Que(\left \{\lambda _{i}\right \} _{i=1}^{l})\left [ x \right ]$, with $\deg \ell _{i} \le m_{i}-1$ (where $m_{i}$ is the multiplicity of $\lambda_{i}$ as a root of the characteristic polynomial  $\varphi (x)$) for $1\le i\le l$, such that
		\begin{equation*}
			w_{n}=\ell _{1} (n) \lambda_{1} ^{n}+\cdots +\ell _{l}(n )\lambda_{l}^{n}, \quad  \text{for all $n$}.  
		\end{equation*}
		By  Lemma \ref{lem3}, we get 
		\begin{equation*}
			w_{n}=c\lambda ^{n}+\sum_{i=2}^{l}\ell_{i}(n)\lambda _{i}^{n}, \quad \ell_{i}(n)\in \Reals[ n ],         
		\end{equation*}
		where 	$\lambda$ is the  dominant root of  $\varphi (x)$, $c$ is a nonzero constant, and 
		\begin{equation*}
			\deg \ell_{i}(n )\le m_{i}-1, \quad \text{for} \quad   i=2, \ldots , l, \quad  m_{2}+\cdots+ m_{l}=m-1, 
		\end{equation*}
		Therefore, we obtain that
		\begin{equation*}
			w_{n}=c\lambda ^{n}+\mathcal{O}(n^{m}\mu ^{n})=c\lambda ^{n}+\mathcal{O}((\lambda \mu )^{\frac{n}{2}}),   
		\end{equation*}
		where $n^{m}&lt; \left | \frac{\lambda }{\mu }  \right |^{\frac{n}{2} }$ for all sufficiently large $n$.
	\end{proof}
	
	
	\begin{remark}  For discussing the reciprocal product of   $m$th-order linear recursive sequences, we assume that $\mu&gt;1$. When  $\mu &lt; 1$, 
	as in \cite{4}, we write $w_{n} =c\lambda^{n} +\mathcal{O}(c^{-n})$, for some $c&gt;1$).
	\end{remark}
	
	\begin{lemma}\label{lem5}
		Let $\lambda&gt; \left | \mu\right |&gt; 1$. We get  
		\begin{equation} \label{eq7}
			\prod_{k=n}^{\infty }(1-\frac{1}{c\lambda ^{k}}+ \mathcal{O}((\frac{\mu}{\lambda ^{3}})^{\frac{k}{2}}))=1-\sum_{k=n}^{\infty }\frac{1}{c\lambda^{k}}+\mathcal{O}((\frac{\mu}{\lambda ^{3}})^{\frac{n}{2}}),  
		\end{equation}
		where $c$ is a constant.
	\end{lemma}
	
	\begin{proof}
		First we prove the following equality:
		\begin{equation} \label{eq8}
			\prod_{k=n}^{n+m}(1-\frac{1}{c\lambda^{k} }+ \mathcal{O}((\frac{\mu }{\lambda^{3}})^{\frac{k}{2}}))=1-\sum_{k=n}^{n+m}\frac{1}{c\lambda ^{k} }+\mathcal{O}((\frac{\mu }{\lambda ^{3}})^{\frac{n}{2}}),
		\end{equation}
		We prove \eqref{eq8} by mathematical induction. When $m=1$,
		we have
		\begin{equation*}
			\begin{split}
				&amp;\prod_{k=n}^{n+1} ( 1-\frac{1}{c\lambda^{k} }+\mathcal{O}((\frac{\mu }{\lambda ^{3}})^{\frac{k}{2}}))\\
				&amp;=( 1-\frac{1}{c\lambda^{n}}+\mathcal{O}((\frac{\mu }{\lambda ^{3}})^{\frac{n}{2}}))\times ( 1-\frac{1}{c\lambda^{n+1} }+\mathcal{O}((\frac{\mu }{\lambda   ^{3}})^{\frac{n+1}{2} }))\\
				&amp;=1-\frac{1}{c\lambda ^{n} }-\frac{1}{c\lambda ^{n+1} }+\frac{1}{c^{2}\lambda ^{2n+1}}+\mathcal{O}((\frac{\mu }{\lambda ^{3}})^{\frac{n}{2}}) \\
				&amp;+ 
				\mathcal{O}((\frac{\mu }{\lambda ^{3}})^{\frac{n+1}{2}}) 
				+\mathcal{O}((\frac{\mu}{\lambda ^{5}})^{\frac{n}{2}} )+\mathcal{O}((\frac{\mu}{\lambda ^{3} }  )^{\frac{2n+1}{2} }  ) \\
				&amp;=1-\frac{1}{c\lambda ^{n}}-\frac{1}{c\lambda ^{n+1} } +\mathcal{O}((\frac{\mu}{\lambda ^{3}  }  )^{\frac{n}{2}}) \\
				&amp;=1-\sum_{k=n}^{n+1 }\frac{1}{c\lambda^{k}}+\mathcal{O}((\frac{\mu }{\lambda ^{3}})^{\frac{n}{2}}).
			\end{split}
		\end{equation*}
		
		That is, \eqref{eq8} holds when $m=1$. Now suppose that for every integer $m$, we have
		\begin{equation*}
			\prod_{k=n}^{n+m}( 1-\frac{1}{c\lambda^{k} }+ \mathcal{O}((\frac{\mu}{\lambda^{3}})^{\frac{k}{2} }))=1-\sum_{k=n}^{n+m }\frac{1}{c\lambda^{k}}+\mathcal{O}((\frac{\mu }{\lambda ^{3}})^{\frac{n}{2}}).
		\end{equation*}
		Then for $m+1$, we have 
		\begin{equation} \label{eq9}
			\begin{split}
				&amp;\prod_{k=n}^{n+m+1}(1-\frac{1}{c\lambda ^{k}}+ \mathcal{O}((\frac{\mu }{\lambda^{3}})^{\frac{k}{2}})) \\
				&amp;=(1-\sum_{k=n}^{n+m }\frac{1}{c\lambda^{k} }+\mathcal{O}((\frac{\mu }{\lambda^{3}})^{\frac{n}{2} })) \times( 1-\frac{1}{c\lambda^{n+m+1} }+\mathcal{O}(( \frac{\mu}{\lambda ^{3}})^{\frac{n+m+1}{2}}))\\
				&amp;=1-\sum_{k=n}^{n+m}\frac{1}{c\lambda^{k}}-\frac{1}{c\lambda^{n+m+1}}+\frac{1}{c\lambda ^{n+m+1}}( \sum_{k=n}^{n+m}\frac{1}{c\lambda^{k}})+\mathcal{O}(( \frac{\mu}{\lambda^{3}})^{\frac{n}{2}}) \\
				&amp;~~~+\mathcal{O}((\frac{\mu }{\lambda^{3}})^{\frac{n+m+1}{2}})+\mathcal{O}((\frac{\mu}{\lambda^{5}})^{\frac{n}{2}})    +\mathcal{O}((\frac{\mu}{\lambda^{3}})^{\frac{2n+m+1}{2}}) \\
				&amp;=1-\sum_{k=n}^{n+m+1}\frac{1}{c\lambda^{k} }+\mathcal{O}((\frac{\mu}{\lambda^{3}})^{\frac{n}{2}}).
			\end{split}
		\end{equation}
		Now \eqref{eq9} follows from \eqref{eq9} and mathematical induction.

Taking $m \rightarrow  \infty$, we have 
		\begin{equation*}
			\prod_{k=n}^{\infty}(1-\frac{1}{c\lambda ^{k}}+\mathcal{O}(\frac{\mu}{\lambda^{3}})^{\frac{k}{2}})=1-\sum_{k=n}^{\infty }\frac{1}{c\lambda ^{k}}+\mathcal{O}(  (\frac{\mu}{\lambda^{3}})^{\frac{n}{2}}),
		\end{equation*}
		which completes the proof.
	\end{proof}
	
	\section{Proof of the theorem}
	
	\begin{proof}[Proof of Theorem \ref{th1}]
		Using  the geometric series, we get
		\begin{equation} \label{eq10}
			\frac{1}{1\pm \epsilon}=1+ \mathcal{O}(\epsilon),
		\end{equation}
		where $\epsilon \rightarrow 0$.
		Using Lemma \ref{lem4}, we obtain
		\begin{equation} \label{eq11}
			\begin{split}
				\frac{1}{w_{k}}&amp;=\frac{1}{c\lambda ^{k}+\mathcal{O}((\lambda\mu )^{\frac{k}{2}})}=\frac{1}{c\lambda ^{k}( 1+\mathcal{O}((\frac{\mu }{\lambda})^{\frac{k}{2}}))}\\
				&amp;=\frac{1}{c\lambda^{k}} (1+\mathcal{O}(\frac{\mu}{\lambda})^{\frac{k}{2}}) =\frac{1}{c\lambda^{k}}+ \mathcal{O}(\frac{\mu}{\lambda ^{3}})^{\frac{k}{2}}.
			\end{split}
		\end{equation}
		Thus 
		\begin{equation*}
			\sum_{k=n}^{\infty }\frac{1}{w_{k}}=\frac{1}{c}\sum_{k=n}^{\infty}\frac{1}{\lambda^{k}}+\mathcal{O}( \sum_{k=n}^{\infty}(\frac{\mu}{\lambda^{3}})^{\frac{k}{2}})=\frac{\lambda}{c(\lambda-1)\lambda^{n}}+\mathcal{O}((\frac{\mu}{\lambda^{3}})^{\frac{n}{2}}).     
		\end{equation*}
		Taking the reciprocal, we get
		\begin{equation*}
			\begin{split}
				(\sum_{k=n}^{\infty }\frac{1}{w_{k}})^{-1}&amp;=\frac{1}{\frac{\lambda}{c(\lambda-1)\lambda^{n}} +\mathcal{O}((\frac{\mu }{\lambda^{3}})^{\frac{n}{2}})}=\frac{1}{\frac{\lambda }{c(\lambda-1)\lambda^{n}}(1+\mathcal{O}((\frac{\mu}{\lambda})^{\frac{n}{2}}))} \\
				&amp;=\frac{c(\lambda -1)\lambda ^{n}}{\lambda}(1+\mathcal{O}((\frac{\mu }{\lambda})^{\frac{n}{2}}  ) )=(c\lambda^{n}-c\lambda ^{n-1})(1+\mathcal{O}((\frac{\mu}{\lambda})^{\frac{n}{2}})),  
			\end{split}
		\end{equation*}
		which yields that
		\begin{equation} \label{eq12}
			\begin{split}
				\frac{(\sum\limits_{k=n}^{\infty }\frac{1}{w_{k}})^{-1}}{w_{n}-w_{n-1}} &amp;=\frac{(c\lambda^{n}-c\lambda ^{n-1})(1+\mathcal{O}((\frac{\mu}{\lambda})^{\frac{n}{2}}))}{(c\lambda^{n}+\mathcal{O}((\lambda\mu )^{\frac{n}{2}} ))-(c\lambda ^{n-1}+\mathcal{O}((\lambda \mu)^{\frac{n-1}{2}}))} \\
				&amp;=\frac{(c\lambda^{n}-c\lambda ^{n-1})(1+\mathcal{O}(  (\frac{\mu}{\lambda})^{\frac{n}{2}}))}{(c\lambda^{n}-c\lambda ^{n-1})(1+\mathcal{O}(\frac{(\lambda \mu)^{\frac{n}{2}}}{\lambda^{n}-\lambda ^{n-1}})+\mathcal{O}(\frac{(\lambda \mu)^{\frac{n-1}{2}} }{\lambda^{n}-\lambda ^{n-1}}))}\\
				&amp;=\frac{(c\lambda^{n}-c\lambda^{n-1})(1+\mathcal{O}((\frac{\mu}{\lambda})^{\frac{n}{2}}))}{(c\lambda^{n}-c\lambda^{n-1})(1+\mathcal{O}((\frac{\mu}{\lambda})^{\frac{n}{2}})+\mathcal{O}((\frac{\mu}{\lambda})^{\frac{n-1}{2}}))}.
			\end{split}
		\end{equation}
		We obtain that
		\begin{equation*}
			\frac{(\sum\limits_{k=n}^{\infty }\frac{1}{w_{k} })^{-1}}{ w_{n} -w_{n-1}}  \quad \text{tends to }  1, \text{as } n \rightarrow  \infty.
		\end{equation*}
		In addition, by Lemma \ref{lem5} and identity \eqref{eq11}, we obtain
		\begin{equation*}
			\begin{split}
				&amp;\prod\limits_{k=n}^{\infty }(1-\frac{1}{w_{k}}) =\prod\limits_{k=n}^{\infty }(1-\frac{1}{c\lambda^{k} }+\mathcal{O}(\frac{\mu}{\lambda ^{3}})^{\frac{k}{2}})  \\
				&amp;=1-\sum\limits_{k=n}^{\infty}\frac{1}{c\lambda^{k} }+\mathcal{O}(\frac{\mu}{\lambda^{3}})^{n}=1-\frac{\lambda}{c(\lambda-1)\lambda^{n}}+\mathcal{O}( \frac{\mu}{\lambda^{3}})^{n}.
			\end{split}
		\end{equation*}
		Taking the reciprocal, we get
		\begin{equation*}
			\begin{split}
				&amp;(1-\prod\limits_{k=n}^{\infty}(1-\frac{1}{w_{k}}))^{-1}=\frac{1}{\frac{\lambda}{c(\lambda-1  )\lambda^{n}}+\mathcal{O}((\frac{\mu }{\lambda ^{3}})^{\frac{n}{2}} )}=\frac{1}{\frac{\lambda}{c(\lambda-1)\lambda^{n}}(1+\mathcal{O}((\frac{\mu }{\lambda})^{\frac{n}{2}}))} \\
				&amp;=\frac{c(\lambda-1)\lambda^{n}}{\lambda}(1+\mathcal{O}((\frac{\mu}{\lambda})^{\frac{n}{2}}))=(c\lambda^{n}-c\lambda^{n-1})(1+\mathcal{O}((\frac{\mu}{\lambda })^{\frac{n}{2}})),  
			\end{split}
		\end{equation*}
		which yields that
		\begin{equation} \label{eq13}
			\begin{split}
				\frac{(1-\prod\limits_{k=n}^{\infty }(1-\frac{1}{w_{k}}))^{-1} }{w_{n}-w_{n-1}}&amp;=\frac{(c\lambda^{n}-c\lambda ^{n-1})( 1+\mathcal{O}((\frac{\mu }{\lambda})^{\frac{n}{2}}))}{(c\lambda^{n}+\mathcal{O}((\lambda\mu )^{\frac{n}{2}}))-(c\lambda^{n-1}+\mathcal{O}((\lambda \mu   )^{\frac{n-1}{2}}))} \\
				&amp;=\frac{(c\lambda^{n}-c\lambda^{n-1})(1+\mathcal{O}((\frac{\mu}{\lambda})^{\frac{n}{2}}))}{(c\lambda ^{n}-c\lambda^{n-1})(1+\mathcal{O}(\frac{(\lambda \mu)^{\frac{n}{2}}}{\lambda^{n}-\lambda ^{n-1}})+\mathcal{O}(\frac{ (\lambda \mu)^{\frac{n-1}{2}} }{\lambda^{n}-\lambda ^{n-1}}))}\\
				&amp;=\frac{(c\lambda^{n}-c\lambda ^{n-1})(1+\mathcal{O}(  (\frac{\mu}{\lambda})^{\frac{n}{2}}))}{(c\lambda^{n}-c\lambda ^{n-1})(1+\mathcal{O}((\frac{\mu}{\lambda}   )^{\frac{n}{2}})+\mathcal{O}((\frac{\mu}{\lambda}   )^{\frac{n-1}{2}}))}.
			\end{split}
		\end{equation}
		We obtain that
		\begin{equation*}
			\frac{(1-\prod\limits_{k=n}^{\infty }(1-\frac{1}{w_{k}})) ^{-1} }{ w_{n}-w_{n-1}}   \quad 
\text{tends to } 1, \text{as } n \rightarrow  \infty,
		\end{equation*}
		which completes the proof.
	\end{proof} 
	
	
	\begin{remark}
		We now discuss the relative error of the asymptotic behavior of the result in  Theorem \ref{th1}.   By identities \eqref{eq12} and \eqref{eq13},  we obtain 
		\begin{equation}  \label{eq14}
			\begin{split} 
				\left |\frac{(\sum\limits_{k=n}^{\infty }\frac{1}{w_{k}})^{-1}}{w_{n}-w_{n-1}}-1 \right |&amp;=\left |\frac{1+\mathcal{O}((\frac{\mu}{\lambda})^{\frac{n}{2}})}{1+\mathcal{O}((\frac{\mu}{\lambda })^{\frac{n}{2}})+\mathcal{O}((\frac{\mu}{\lambda})^{\frac{n-1}{2}})}-1\right | \\
				&amp;=\mathcal{O}((\frac{\mu}{\lambda})^{\frac{n-1}{2}}  )+\mathcal{O}((\frac{\mu}{\lambda})^{\frac{2n-1}{2}}  )=\mathcal{O}((\frac{\mu }{\lambda})^{\frac{n-1}{2}}),
			\end{split} 
		\end{equation}
		and 
		\begin{equation}  \label{eq15}
			\begin{split} 
				\left |\frac{(1-\prod\limits_{k=n}^{\infty }(1-\frac{1}{w_{k}}))^{-1}}{w_{n}-w_{n-1}}-1 \right |&amp;=\left |\frac{1+\mathcal{O}((\frac{\mu}{\lambda})^{\frac{n}{2}})}{1+\mathcal{O}((\frac{\mu}{\lambda})^{\frac{n}{2}})+\mathcal{O}((\frac{\mu}{\lambda})^{\frac{n-1}{2}})}-1 \right | \\
				&amp;=\mathcal{O}((\frac{\mu }{\lambda })^{\frac{n-1}{2}})+\mathcal{O}((\frac{\mu }{\lambda })^{\frac{2n-1}{2}})=\mathcal{O}((\frac{\mu}{\lambda })^{\frac{n-1}{2}}).
			\end{split} 
		\end{equation}
		
		When $n=101$, by the identities \eqref{eq14} and \eqref{eq15}, we can determine the magnitude of the relative error of the asymptotic behavior for the following two sequences $(w_{n})_{n_{1}\ge 0}$  and $(w_{n})_{n_{2}\ge 0}$:
		
		
		
		
		\begin{equation*} 
			w_{n_{1}}=2w_{n-1}+3w_{n-2}+7w_{n-3}+w_{n-4}+w_{n-5},
		\end{equation*}
		\begin{equation*} 
			w_{n_{2}}=3w_{n-1}+5w_{n-2}+w_{n-3}+6w_{n-4}+w_{n-5}.
		\end{equation*}
		
		
		
		
		
		
		\begin{table}[H]  
			\begin{center}
				\begin{tabular}{c c c c}
					\hline
					$w_{n}$ &amp; $\lambda$ &amp; $\mu$ &amp; $\left | \frac{\mu }{\lambda } \right |^{\frac{n-1}{2}}$ \\
					\hline
					$w_{n_{1}}$ &amp; $\approx 3.2421$   &amp;  $\approx -1.1509$   &amp; $\approx 3.2401\times 10^{-23}$
					\\
					$w_{n_{2}}$ &amp; $\approx 4.2965$ &amp; $\approx -1.4946$ &amp; $\approx 1.1762\times 10^{-23}$  \\
					\hline
				\end{tabular} 
				\caption{The higher-order linear recurrences.} \label{tab1}
			\end{center}
		\end{table}
		
		The computations are given in Table \ref{tab1}.  We used the
		software {\tt Mathematica}.
	\end{remark}
	
	
	\section{Acknowledgments} 
	The authors would like to thank the referees for their very helpful and detailed comments. This
	work is supported by the N. S. F. (12126357) of China.
	
	
	
	
	
	
	
	
	
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	\noindent 2020 {\it Mathematics Subject Classification}: Primary 11B39; Secondary 11B37.
	
	\noindent \emph{Keywords:} $m$th-order linear recursive sequence, reciprocal sum, reciprocal product,  characteristic polynomial.


\vspace*{+.1in}
\noindent
Received October 25 2023;
revised versions received  
October 27 2023; January 11 2024;
January 17 2024.
Published in {\it Journal of Integer Sequences},
January 17 2024.

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