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\begin{center}
\vskip 1cm{\LARGE\bf 
On Certain Reciprocal Sums
}
\vskip 1cm
\large
Soumyadip Sahu\\
School of Mathematics \\
Tata Institute of Fundamental Research\\
Mumbai 400005\\
India\\
\href{mailto:soumyadip.sahu00@gmail.com}{\tt soumyadip.sahu00@gmail.com} \\
\end{center}

\vskip .2 in

\begin{abstract}        
We use tail sums of convergent series of positive real numbers to define a sequence of non-negative integers, and explicitly determine
this sequence for classes of series defined by reciprocal sums of polynomials and rational functions. For this purpose we develop a new difference calculus method to approximate infinite series.         
\end{abstract}     

\section{Introduction} 
\label{sec1} 
Let $(x_m)_{m=1}^\infty$ be a sequence of positive real numbers satisfying $\sum_{m=1}^\infty x_m &lt; \infty$. With $(x_m)_{m \geq 1}$ one associates a sequence of non-negative integers $(a_n)_{n=1}^{\infty}$ by defining 
\[a_n= \Big \lfloor \frac {1} {\sum_{m=n}^\infty x_m}\Big\rfloor \hspace{5mm} (n \geq 1),\] where $\lfloor.\rfloor$ is the floor function.  We call $(a_n)_{n=1}^{\infty}$ the \textit{reciprocal sequence} of $(x_m)_{m=1}^\infty$. The sequence $(a_n)_{n=1}^{\infty}$ is non-decreasing and divergent. Reciprocal sequences 
capture the rate of convergence of their defining series,
and sometimes give rise to nice arithmetic and combinatorial structures. One can replace the floor function
by the ceiling function or nearest integer function, and create variants of this notion.    
   
Many authors have already studied reciprocal sequence of sequences defined by reciprocals of linear recurrence. Ohtsuka and Nakamura \cite{on} derived a formula for reciprocal sequence of \[x_m=\frac {1} {F_m} \hspace{5mm} (m \geq 1),\] where $F_m$ is the $m^{\text{th}}$ Fibonacci number. Ramifications and generalizations of this result appear in several places, e.g., \cite{kl,ak}. Some papers have investigated the problem beyond reciprocals of linear recurrence relation. Reciprocal sequences of $(\frac {1} {m^4})_{m \geq 1}$ and $(\frac {1} {m^5})_{m \geq 1}$ are recorded in sequences \seqnum{A248230},
\seqnum{A248234} respectively. 
Xin \cite{xin} and Xu \cite{xu} studied reciprocal sequences of \[x_m = \frac{1} {m^k} \hspace{5mm} (m \geq 1)\]  for $k = 2,3,4,5$. They showed  that reciprocal sequences of these sequences are given by polynomials or polynomial like functions  (for precise formulas see Section~\ref{sec4.3.1}).  In present article we extend results of Xin and Xu to general classes of sequences arising from reciprocals of polynomials and rational functions.   

Let $\mathbb{K}$ be a subfield of $\mathbb{R}$. It is sufficient to consider $\mathbb{K} = \mathbb{R}, \mathbb{Q}$ for the purposes of this article. Suppose that $P(X) \in \mathbb{K}[X]$ is a polynomial  of degree $k \geq 2$ with positive leading coefficient.  Now $\mathbb{Q}$ is a subfield of $\mathbb{K}$ and $\mathbb{K}$ is dense in $\mathbb{R}$ with respect to Euclidean topology. Hence there exists $M_0 \in \mathbb{K}$ so that $P(x) &gt; 0$  for all real $x \geq M_{0} +1 $. For fixed choice of such $M_0$, define a sequence of positive real numbers by 
\begin{equation}  
\label{1.1} 
x_m = \frac {1} {P(m+M_0)}\hspace{5mm} (m \geq 1).
\end{equation} 
Since $k \geq 2$ we have $\sum_{m \geq 1} x_m &lt; \infty$. To calculate terms of reciprocal sequence of $(x_m)_{m\geq 1}$ one needs to estimate sums  
of the form
\begin{equation}
\label{1.2}
\sum_{m = n}^{\infty} \frac {1} {P(m+M_0)}. 
\end{equation}  
The standard way to approximate sums of this form is to apply summation formulas from analysis (e.g., the Euler-Maclaurin summation formula \cite[p.\ 806]{as}).  For reciprocal power sums, i.e., $P(X) = X^k$ and $M_0 = 0$, a precise estimate of  \eqref{1.2} is readily available from the asymptotic expansion of the polygamma function \cite[p.\ 260]{as}. Though these summation formulas produce estimates up to higher order,
they often lead to complicated computations and problems regarding convergence.  

The goal of this paper is to present an improvised technique based on difference calculus, which bypasses analytic tools and provides a good upper bound as well as a lower bound for \eqref{1.2}. The central idea behind our method is to find a suitable polynomial $f$ so that the rational function $\frac {1} {P(X)}$ is \textit{approximately} equal to $\frac {1} {f(X)} - \frac {1} {f(X+1)}$, i.e., $\frac {1} {f(X)}$ acts like the difference primitive of $- \frac {1} {P(X)}$. Once we have determined $f$, bounds on \eqref{1.2} simply follow by telescoping. This strategy bears resemblance to the methods employed by Xin \cite{xin} and Xu \cite{xu}.    

Though the technique looks naive and insufficient, it turns out to be powerful enough to calculate the first $k-1$ terms in the asymptotic series of \eqref{1.2}. Our main result is as follows:  
\begin{theorem}
\label{thm1} 
Let $(x_m)_{m\geq 1}$ be the sequence defined by \eqref{1.1}. There exists a polynomial $h(X) \in \mathbb{K}[X]$ of degree $k-1$ and a positive integer $N_0$ depending on $h$ such that 
\begin{equation}
\label{1.3} 0 &lt; h(n) \leq \frac {1} {\sum_{m=n}^{\infty} x_{m}} &lt; h(n)+1 \hspace{5mm}(n \geq N_0). 
\end{equation}
Moreover, $h$ is algorithmically computable and uniquely determined by $P(X)$ and $M_0$ up to a constant term.
\end{theorem}
The leftmost inequality in \eqref{1.3} implies that leading coefficient of $h$ is positive. Now \eqref{1.3} can be restated as 
\begin{equation} 
\label{1.4} 0 &lt; \frac {1} {h(n)+1} &lt; \sum_{m = n}^{\infty} \frac {1} {P(m+M_0)} \leq \frac {1} {h(n)} \hspace{5mm} (n \geq N_0).
\end{equation}  
Note that $\frac {1} {h(n)} - \frac {1} {h(n)+1} = \frac {1} {h(n)(h(n)+1)} = O(n^{-2k+2})$ and $\frac {1} {h(n)}, \frac {1} {h(n)+1}$ are of $O(n^{-k+1})$. Hence \[\sum_{m = n}^{\infty} \frac {1} {P(m+M_0)} = \frac {1} {h(n)} + O(n^{-2k+2}) \hspace{5mm} (n \geq N_0). \] Thus we obtain an estimate such that 
the main term is $O(n^{-k+1})$ and error term is $O(n^{-2k+2})$.
This improvement is due to a formulation as a fractional expression that resembles the
classical Pad\'e approximants. Therefore, Theorem~\ref{thm1} turns out to be a more convenient approximation technique than the usual summation formulas,
and one can directly obtain 
the first $k-1$ terms of the asymptotic expansion for suitably large $n$, as
explained in Section~\ref{sec4.2}.   

We can use Theorem~\ref{thm1} to deduce estimates for classical reciprocal power sums. 
\begin{corollary}
\label{corollary2}
Let $k$ be an integer $\geq 2$. There is a polynomial $h(X) \in \mathbb{Q}[X]$ of degree $k-1$ and a positive integer $N_0$ depending on $h$ such that
\[0 &lt; h(n)  \leq \frac {1} {\sum_{m=n}^{\infty} m^{-k}} &lt; h(n)+1 \hspace{5mm}  (n \geq N_0).\]
Moreover, $h$ is algorithmically computable and unique up to a constant term. 
\end{corollary}
Theorem~\ref{thm1} allows us to study sequences defined by rational functions. Let $P(X), Q(X) \in \mathbb{K}[X]$ be two nonzero polynomials with positive leading coefficients such that $\text{deg}_{\mathbb{K}}P - \text{deg}_{\mathbb{K}}Q = k \geq 2$. Set $R(X) = \frac {P(X)} {Q(X)} \in \mathbb{K}(X)$.  Here $R(X)$ determines $k$ and it is independent of presentation. Since $P$ and $Q$ have positive leading coefficients,  there is an $M_0 \in \mathbb{K}$ so that $P(x), Q(x) &gt; 0$ for all real numbers $x \geq M_0 +1$. For fixed choice of $M_0$, consider the sequence 
\begin{equation}
\label{1.5}  
x_m = \frac {1} {R(m+M_0)}\hspace{5mm} (m \geq 1).
\end{equation}
As before $\sum_{m \geq 1} x_m &lt; \infty$. We have  
\begin{theorem}  
\label{thm3} 
Let $(x_m)_{m\geq 1}$ be the sequence defined by \eqref{1.5}. There exists a polynomial $h(X) \in \mathbb{K}[X]$ of degree $k-1$ and a positive integer $N_0$ depending on $h$ such that 
\begin{equation}
\label{1.6} 0 &lt; h(n) \leq \frac {1} {\sum_{m=n}^{\infty} x_{m}} &lt; h(n)+1 \hspace{5mm} (n \geq N_0). 
\end{equation}
Moreover, $h$ is algorithmically computable and uniquely determined by $R(X)$ and $M_0$ up to a constant term.
\end{theorem}
Observe that if $Q(X) = 1$, then Theorem~\ref{thm3} reduces to Theorem~\ref{thm1}. Though Theorem~\ref{thm3} is a generalization of Theorem~\ref{thm1}, it follows easily from the earlier one. We prove both the theorems in Section~\ref{sec3}. Theorem~\ref{thm3} also offers a convenient asymptotic expression,
as described in paragraphs above. More discussion about this point is postponed to Section~\ref{sec4.2}.   

           
It is clear from Theorem~\ref{thm3} that the $n$-th term of the reciprocal
sequence of \eqref{1.5} is either $\lfloor h(n) \rfloor$ or $\lfloor
h(n)\rfloor + 1$. We pin down the exact expression of the
reciprocal sequence
if $P$ and $Q$ have coefficients in $\mathbb{Q}$, by carefully choosing
constant term of $h$. This procedure and its arithmetic aspects are
described in Section~\ref{sec4.1}.
 
\subsection{Notation and conventions} The symbols $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{R}$, $\mathbb{C}$ have their conventional meaning. In our convention $0 \notin \mathbb{N}$ and the set of nonnegative integers is $\mathbb{Z}_{\geq 0}$. We index sequences by $\mathbb{N}$. 

Let $\mathbb{K}$ be a field. Then $\mathbb{K}^{\times} = \mathbb{K} - \{0\}$
and $\mathbb{K}[X]$ is the ring of polynomials with coefficients
in $\mathbb{K}$. Also $\mathbb{K}(X)$ is the field of rational functions.
If there is more than one variable, we use a boldface symbol to denote a tuple, e.g., $\textbf{X} = (X_1, \ldots, X_n)$. For a fixed integer $d \geq 0$, there is a bijection between $\mathbb{K}^{\times} \times \mathbb{K}^{d}$ and $\mathbb{K}[X]_{d}$, subset polynomials of degree $d$, given by $\iota_{d}: (a_0, \ldots, a_d) \to a_0X^d + \cdots + a_d$. Here, by convention,  $\mathbb{K}^{\times} \times \mathbb{K}^0 = \mathbb{K}^{\times}$.  The degree of the zero polynomial is $-\infty$. If $P$ is a polynomial with coefficients in $\mathbb{R}$, then 
\begin{equation*}
\sgn P = \begin{cases}
                      0, &amp; \text{if $P = 0$;}\\
                      1, &amp; \text{if the leading coefficient of $P$ is positive;}\\
                     -1, &amp; \text{if the leading coefficient of $P$ is negative.}
                 \end{cases}
\end{equation*}     
We write $\lfloor.\rfloor, \lceil.\rceil$ for the floor and ceiling functions,
respectively.
A statement $S(n)$ concerning natural numbers holds for `$n \gg 1$' if there exists a real number $C$ (depending on $S$) so that $S(n)$ is true for all $n \geq C$. 
 
      

\section{Approximate difference primitive}
\label{sec2}  
We begin by introducing some preliminary concepts necessary to define approximate difference primitives. The main result of this section is the existence of canonical approximants satisfying definite requirements. Our arguments are algebraic in nature, and most of the conclusions are valid even in a formal situation,
which is briefly mentioned at the end of section.    

Let $\mathbb{K}$ be a field of characteristic $0$ and $k$ be an integer $\geq 2$. Suppose that  $g(X) \in \mathbb{K}[X]_k$. Write 
\begin{equation}
\label{2.1} 
g(X) = a_0X^k + \cdots + a_k.
\end{equation}  
Here $(a_0, \ldots, a_{k})$ is the unique point in $\mathbb{K}^{\times} \times \mathbb{K}^{k}$ which corresponds to $g(X)$ under $\iota_{k}$. Now $f(X) \in \mathbb{K}[X]$ be a nonzero polynomial such that $\frac{1} {f(X)}$ is a difference primitive of $- \frac {1} {g(X)}$, i.e.,  
$\frac {1} {g(X)} = \frac {1} {f(X)} - \frac {1} {f(X+1)}$ holds in $\mathbb{K}(X)$. 
Then
\begin{equation}
\label{2.2} 
f(X)f(X+1) = g(X)(f(X+1) - f(X)).  
\end{equation} 
Suppose $\text{deg}_{\mathbb{K}} f = d$. Since the left-hand side of \eqref{2.2} is not zero, we have $d \geq 1$. 
Comparing the degrees of both sides gives $d = k-1$.

Let $x_0, x_1,\ldots,x_{k-1}$ be $k$ unknowns. Set  
\begin{equation*}
F(X,\textbf{x}) = x_0X^{k-1} + \cdots + x_{k-1} \in \mathbb{Z}[X, x_0,\ldots,x_{k-1}].   
\end{equation*} With each $\textbf{c} = (c_0, \ldots, c_{k-1}) \in \mathbb{K}^{\times}\times \mathbb{K}^{k-1}$ one associates  $F(X,\textbf{c}) \in \mathbb{K}[X]_{k-1}$. Using \eqref{2.2} we see that $\frac {1} {F(X, \textbf{c})}$ is a difference primitive of $- \frac {1} {g(X)}$ if and only if 
\begin{equation*}
F(X+1,\textbf{c})F(X,\textbf{c}) = g(X)(F(X+1,\textbf{c}) - F(X, \textbf{c})).
\end{equation*}  
To solve the equation above,
one defines a collection of polynomials \[\big\{y_i(\textbf{x}), u_j(\textbf{x}), v_l(\textbf{x},g) \mid 1 \leq i \leq k-1, 0 \leq j, l \leq 2k-2\big\} \subseteq \mathbb{K} [x_0,\ldots,x_{k-1}]\]  by the relations
\begin{align}
&amp; F(X+1,\textbf{x})= x_0X^{k-1}+(x_1+y_1(\textbf{x}))X^{k-2}+\cdots \nonumber \\
&amp; \hspace{4cm}\cdots +(x_{k-2}+y_{k-2}(\textbf{x}))X+(x_{k-1}+y_{k-1}(\textbf{x})),\label{2.3}~\\
&amp; H(X,\textbf{x}) = F(X+1,\textbf{x})F(X,\textbf{x}) \nonumber \\
&amp; \hspace{4cm}= u_0(\textbf{x})X^{2k-2}+\cdots+u_{2k-3}(\textbf{x})X+u_{2k-2}(\textbf{x}),\label{2.4}~\\
&amp; G(X,\textbf{x},g)  = g(X)\big(F(X+1,\textbf{x})- F(X,\textbf{x})\big) \nonumber ~\\ 
&amp; \hspace{3cm} = v_0(\textbf{x},g)X^{2k-2}+\cdots+v_{2k-3}(\textbf{x},g)X+v_{2k-2}(\textbf{x},g).\label{2.5} 
\end{align} 

The coefficient of $X^{k-1}$ in $F(X+1,\textbf{x})$ is $x_0$ and the
degrees of $H(X,\textbf{x}), G(X,\textbf{x},g)$ in $X$ are at most
$2k-2$. Therefore, the equations above are justified. Note that the
polynomials $\{y_i(\textbf{x}), u_j(\textbf{x}) \mid 0 \leq i \leq k-1,
0 \leq j \leq 2k-2 \}$ are independent of $g$ and defined over $\mathbb{Z}
[x_0,\ldots,x_{k-1}]$.

As a consequence of the binomial theorem 
\begin{equation}
\label{2.6}
y_i(\textbf{x}) = \binom {k-i} {1}x_{i-1}+\binom {k-i+1} {2}x_{i-2} + \cdots + \binom {k-1} {i}x_0
\end{equation}
for all $1 \leq i \leq k-1$. For convenience put $y_0(\textbf{x}) = y_{k}(\textbf{x}) = 0$.

 Equation \eqref{2.3}, \eqref{2.4}, and \eqref{2.5} together imply that
\begin{gather}
\label{2.7} u_j(\textbf{x})=\sum_{r=0}^{j} x_r(x_{j-r}+y_{j-r}(\textbf{x}))\hspace{5mm} (0 \leq j \leq k-1), \\
\label{2.8} v_l(\textbf{x},g)=\sum_{r=0}^{l} a_{r}y_{l-r+1}(\textbf{x})\hspace{5mm} (0 \leq l \leq k-1).
\end{gather} 

Subtracting \eqref{2.4} from \eqref{2.5} gives
\begin{equation}
\label{2.9} 
G(X,\textbf{x},g) - H(X,\textbf{x}) = \sum_{i = 0}^{2k-2} (v_{i}(\textbf{x},g) - u_{i}(\textbf{x}))X^{2k-2-i}.
\end{equation}
To construct an approximate difference primitive we need to find a point
$(c_0,\ldots, c_{k-1}) \in \mathbb{K}^k$ so that the first few coefficients
in \eqref{2.9} vanish at the point $(c_0,\ldots, c_{k-1})$.  But for
generic $(a_0,\ldots,a_k)$, one cannot expect to find a common zero
of all the coefficients. Moreover, we would like to have $c_0 \neq 0$,
so that $F(X,c_0, \ldots, c_{k-1})$ is actually of degree $k-1$. These
considerations lead to a formal definition.

\subsection{Good approximants}
\label{sec2.1}  
Let $f(X) \in \mathbb{K}[X]$ be a nonzero polynomial. Set 
\begin{gather}
\delta(f,g,X):= \frac {1} {f(X)} - \frac {1} {f(X+1)} - \frac {1} {g(X)} \label{2.10}, \\
N(f,g,X):= g(X)\big(f(X+1) - f(X)\big) - f(X+1)f(X).\label{2.11}
\end{gather}
A \textit{good approximant} of $g$ is a polynomial $f \in \mathbb{K}[X]_{k-1}$ so that $N(f,g,X)$ is a polynomial of degree $\leq k-1$. 

Note that  
\begin{equation}
\delta(f,g,X) = \frac {N(f,g,X)} {f(X)f(X+1)g(X)} \label{2.12} 
\end{equation}
and $N(f,g,X) = 0$ if and only if $\frac{1} {f(X)}$ is a difference primitive of $-\frac {1} {g(X)}$.  

The following lemma ensures existence of a canonical good approximant:
\begin{lemma}
\label{lemma4}
Let $k \geq 2$ and $g \in \mathbb{K}[X]_{k}$. Consider the system of $k$ equations 
\begin{equation}
\label{2.13} 
u_i(\textbf{x}) = v_i(\textbf{x},g) \hspace{5mm}  (0 \leq i \leq k-1)
\end{equation} in $k$ unknowns $x_0,\ldots,x_{k-1}$. The system of equations \eqref{2.13} has a unique solution in $\mathbb{K}^{\times} \times \mathbb{K}^{k-1}$, i.e., there is a unique tuple $(c_0(g),\ldots,c_{k-1}(g)) \in \mathbb{K}^k$ with $c_0(g) \neq 0$, which is a solution to the system of equation \eqref{2.13}.   
\end{lemma} 
\begin{proof}
To fix notation, assume that $g$ is given in the form $\eqref{2.1}$. This
polynomial remains fixed throughout the discussion, and we omit it from the
notation. Now $u_i$ depends on $\{x_0,\ldots,x_i,y_0,\ldots,y_i\}$. Using
\eqref{2.6}, one concludes that the set of variables appearing in $u_i$ is
$\{x_0,\ldots,x_i\}$. Similarly, $v_i$ depends on $\{y_1,\ldots,y_{i+1}\}$,
i.e., on $\{x_0,\ldots,x_i\}$. These statements hold for all $0 \leq
i \leq k-1$. Therefore one can use a recursive approach to solve the
system of equations.

Let $0 \leq i \leq k-1$ and consider subsystems of $i+1$ equations
\begin{equation}
\label{2.14} u_0 = v_0, \ldots, u_{i} = v_{i} 
\end{equation} in $i+1$ variables $x_0, \ldots, x_{i}$. We would like to show that \eqref{2.14} has unique solution in $\mathbb{K}^{\times} \times \mathbb{K}^{i}$ for each $0 \leq i \leq k-1$.     

For the base case, consider the equation $u_0(x_0) = v_0(x_0)$.
We have $u_0 = x_0^2$, and $v_0 = y_1 = a_0\binom {k-1} {1} x_0=a_0(k-1)x_0$. Now $a_0(k-1) \neq 0$ and it is the only nonzero solution to $u_0 = v_0$. Put $c_0 = a_0(k-1)$. 

Assume that the statement holds for some $0 \leq i \leq k-2$, i.e.,
there is a unique solution to \eqref{2.14} in $\mathbb{K}^{\times}
\times \mathbb{K}^{i}$. Note that the first coordinate of this solution is
necessarily $c_0$. Let the unique solution be $(c_0, \ldots, c_{i})$. We
now construct $c_{i+1} \in \mathbb{K}$ so that $(c_0,\ldots,c_i, c_{i+1})$
is the unique solution of  $u_{i+1}=v_{i+1}$ in $\mathbb{K}^{\times}
\times \mathbb{K}^{i+1}$.  In what follows, we consider
$u_{i+1}$ and $v_{i+1}$ as polynomials of $x_{i+1}$ with coefficients
in $\mathbb{K}[x_0, \ldots, x_i]$. Equation \eqref{2.7} and \eqref{2.8}
together imply that $u_{i+1}, v_{i+1}$ are linear in the variable $x_{i+1}$.

There are two possibilities. 

\subsubsection{Case I: $i &lt; k-2$}

Here $i+1 \leq k-2$. From \eqref{2.7} and \eqref{2.8} we deduce that
coefficient of $x_{i+1}$ in $u_{i+1}$ is $2x_0$ (one $x_0$ arises
from term $x_0(x_{i+1}+y_{i+1})$ and other $x_0$ arises from the term
$x_{i+1}(x_0+y_0)$). Similarly coefficient of $x_{i+1}$ in $v_{i+1}$
is $a_0$ times coefficient of $x_{i+1}$ in $y_{i+2}$, i.e., $a_0\binom
{k-i-2} {1}= a_0(k-i-2)$.

Hence $u_{i+1}=v_{i+1}$ can be rewritten as   
\begin{equation}
\label{2.15}\bigl (2x_0 - a_0(k - i -2)\bigr) x_{i+1} = \text{a polynomial in $x_0,\ldots,x_i$ over $\mathbb{K}$}.
\end{equation}  
But $\bigl (2c_0 - a_0(k - i -2) \bigr) = a_0(k+i) \neq 0$.
Therefore we can substitute $x_0 = c_0,\ldots,x_i = c_i$ in \eqref{2.15} and solve for $x_{i+1}$ to get a tuple $(c_0, \ldots, c_{i+1}) \in \mathbb{K}^{i+2}$ 
that is a solution to the system of equations \[u_0 = v_0, \ldots, u_{i+1} = v_{i+1}.\]
If $(C_0,\ldots, C_{i+1})$ is another solution with $C_0 \neq 0$ then by recursion hypothesis $(c_0, \ldots, c_{i}) = (C_0, \ldots, C_i)$. Using \eqref{2.15} we have $C_{i+1} = c_{i+1}$. Hence the uniqueness. So for $i &lt; k-2$ a solution to the first $i+1$ equations of \eqref{2.13} in $\mathbb{K}^{\times} \times \mathbb{K}^{i}$ extends to a unique solution to the first $i+2$ equations in $\mathbb{K}^{\times} \times \mathbb{K}^{i+1}$.  

\subsubsection{Case II:  $i = k-2$}

This case is essentially similar to Case I. Here coefficient of $x_{k-1}$ in $v_{k-1}$ is $0$. Now $u_{k-1}=v_{k-1}$ can be rewritten as \[2x_0x_{k-1}= \text{a polynomial in $x_0,\ldots,x_{k-2}$ over $\mathbb{K}$}.\] Since $c_0 \neq 0$,
the arguments of the previous case go through to yield a unique tuple $(c_0,\ldots,c_{k-1})$ that is a solution to \eqref{2.13}. 

In this way we can recursively construct $\bigl (c_0(g),\ldots,c_{k-1}(g)\bigr) \in \mathbb{K}^{k}$ so that $\bigl (c_0(g),\ldots,c_{k-1}(g)\bigr)$ is the unique solution to \eqref{2.13} in $\mathbb{K}^{\times} \times \mathbb{K}^{k-1}$. Hence the lemma is proved.
\end{proof} 
Lemma~\ref{lemma4} constructs a point $\textbf{c}(g) = \bigl (c_0(g),\ldots,c_{k-1}(g)\bigr) \in \mathbb{K}^k$ such $c_0(g) \neq 0$ and $G\big(X,\textbf{c}(g),g\big) - H\big(X,\textbf{c}(g)\big)$ is  of degree $\leq k-2$. Therefore $F\big(X,\textbf{c}(g)\big)$ is a good approaximant of $g$. The condition that 
the first $k$ coefficients of $G\big(X,\textbf{c}(g),g\big) - H\big(X,\textbf{c}(g)\big)$ vanish is better than expected, but control on one extra term turns out to be useful. For simplicity we frequently omit $g$ from the
notation for the solution, if it is understood from the context. 

\subsection{Consequences of Lemma~\ref{lemma4}}
\label{sec2.2}
The technique used to prove Lemma~\ref{lemma4} has several corollaries that are indispensable for later developments.  

\begin{corollary}
\label{corollary5}
\leavevmode
\begin{itemize}
\item[(i)] Let $0 \leq i_0 \leq k-1$. Consider the subsystem of equations \[u_i(\textbf{x}) = v_i(\textbf{x},g) \hspace{5mm} (0 \leq i \leq i_0).\] Observe that variables appearing in these equations are $x_0, \ldots, x_{i_0}$. This system has a unique solution in $\mathbb{K}^{\times} \times \mathbb{K}^{i_0}$ given by the first $i_0 + 1$ coordinates of $\textbf{c}$, i.e.,  $(c_0, \ldots, c_{i_{0}})$.

\item[(ii)] Let $g_1, g_2 \in \mathbb{K}[X]_{k}$ with $g_1 - g_2 \in \mathbb{K}$. Then $\textbf{c}(g_1) = \textbf{c}(g_2)$.
\end{itemize}
\end{corollary}

\begin{proof}
\ 
\leavevmode
\begin{itemize}
\item[(i)] Follows from the recursion argument in the proof of Lemma~\ref{lemma4}.

\item[(ii)] A consequence of the fact that for any $g \in \mathbb{K}[X]_{k}$, the polynomials \[\{v_{i}(\textbf{X},g) \mid 0 \leq i \leq k-1\}\] do not depend on the  constant term of $g$.  
\end{itemize}
\end{proof}

The next corollary characterizes all good approximants and provides a necessary and sufficient condition for existence of difference primitives.   
\begin{corollary}
\label{corollary6}
\ 
\leavevmode
\begin{itemize}
\item[(i)] For each $c \in \mathbb{K}$ the polynomial $F(X, c_0,\ldots, c_{k-2}, c)$ is a good approximant of $g$ and every good approximant of $g$ is in this form. 

\item[(ii)] There exists a nonzero polynomial $f(X) \in \mathbb{K}[X]$ so that $\frac {1} {f(X)}$ is a difference primitive of $-\frac {1} {g(X)}$   if and only if the tuple $\textbf{c} = (c_0,\ldots,c_{k-1})$ constructed in Lemma~\ref{lemma4} satisfies \[u_i(\textbf{c}) = v_{i}(\textbf{c},g) \hspace{5mm} (k \leq i \leq 2k-2).\]
If this condition holds then $f(X)$ is uniquely determined and equals $F(X,\textbf{c})$.
\end{itemize}       
\end{corollary} 

\begin{proof}
\ \leavevmode
\begin{itemize}
\item[(i)] By definition, all good approximants of $g$ have degree $k-1$. Let $\textbf{C} = (C_0,\ldots, C_{k-1}) \in \mathbb{K}^{\times} \times \mathbb{K}^{k-1}$. Using \eqref{2.9} one sees that $F(X,\textbf{C})$ is good approximant if and only if $u_{i}(\textbf{C}) = v_{i}(\textbf{C},g)$ for each $0 \leq i \leq k-2$. Now the result follows from uniqueness part of Corollary~\ref{corollary5}.   

\item[(ii)] We have already seen that $f$ has to be of degree $k-1$. From \eqref{2.9} it follows that such $f$ exists if and only if the system of $2k-1$ equations \[u_{i}(\textbf{x}) = v_{i}(\textbf{x},g) \hspace{5mm} (0 \leq i \leq 2k-2)\]
has a solution in $\mathbb{K}^{\times} \times \mathbb{K}^{k-1}$.
Lemma~\ref{lemma4} implies that if such solution exists it is unique and given by the tuple $\textbf{c} = (c_0,\ldots,c_{k-1}) \in \mathbb{K}^{\times} \times \mathbb{K}^{k-1}$ constructed in lemma. Thus both parts of assertion are proved.
\end{itemize}
\end{proof}
The following corollary investigates effect of scaling $g$.         
\begin{corollary}
\label{corollary7} 
Let $g_1(X) \in \mathbb{K}[X]$ be a nonzero scalar multiple of $g(X)$, i.e., $g_1(X) = \alpha g(X)$ for some $\alpha \in \mathbb{K}^{\times}$. Then $\big(\alpha c_0(g), \ldots, \alpha c_{k-1}(g)\big) \in \mathbb{K}^{\times} \times \mathbb{K}^{k-1}$ is the unique solution to system of equations \[u_i(\textbf{x}) = v_i(\textbf{x},g_1)\hspace{5mm} (0 \leq i \leq k-1).\]  
\end{corollary}
\begin{proof}  
One writes coefficients as functions of polynomials. By assumption $a_r(g_1) = \alpha a_r(g)$ for all $0 \leq r \leq k-1$. From explicit expressions \eqref{2.6}, \eqref{2.7} and \eqref{2.8} it follows that $u_j$ is quadratic polynomial of $\{x_0,\ldots,x_{k-1}\}$ while $v_l$ is linear in both $\{x_0,\ldots,x_{k-1}\}$ and $\{a_0,\ldots, a_{k-1}\}$. Therefore the system of equations \[u_{i}(\textbf{x}) = v_{i}(\textbf{x},g) \hspace{5mm} (0 \leq i \leq k-1) \] is invariant under transformation $x_{i} \to \alpha x_i$, $0 \leq i \leq k-1$, and $a_r \to \alpha a_r$, $0 \leq r \leq k-1$. Hence $\big(\alpha c_0(g), \ldots, \alpha c_{k-1}(g)\big)$ is a solution to the system corresponding to $g_1$. But $\alpha c_0(g) \neq 0$. The result follows by uniqueness. 
\end{proof} 
Let $c \in \mathbb{K}$. Define $f_g(c,X) \in \mathbb{K}[X]$ by 
\begin{gather}
\label{2.16}
f_g(c,X) : = c_0(g)X^{k-1} + \ldots + c_{k-2}(g)X + c 
\end{gather}
By Corollary~\ref{corollary6} $f_g(c)$ is a good approximant of $g$ and all good approximants are in this form. The leading term of $N\big(f_g(c),g,X\big)$ is $X^{k-1}$ and its coefficient is \[\big(v_{k-1}(c_0,\ldots,c_{k-2},c;g) - u_{k-1}(c_0,\ldots,c_{k-1},c)\big).\] By \eqref{2.8} $x_{k-1}$ does not appear in $v_{k-1}(\textbf{x})$ and from the proof of Lemma~\ref{lemma4}, we know that the term involving $x_{k-1}$ in $u_{k-1}(\textbf{x})$ is $2x_{k-1}x_0$. Now  
\begin{align}
&amp; v_{k-1}(c_0,\ldots,c_{k-2},c;g) - u_{k-1}(c_0,\ldots,c_{k-1},c)\nonumber\\
&amp; = v_{k-1}(c_0,\ldots,c_{k-2},c_{k-1};g) - u_{k-1}(c_0,\ldots,c_{k-1},c)\nonumber\\
&amp; = u_{k-1}(c_0,\ldots,c_{k-2},c_{k-1};g) - u_{k-1}(c_0,\ldots,c_{k-1},c)\nonumber\\
&amp;\label{2.17} = 2c_0(c_{k-1} - c).   
\end{align}
Here in third step one uses Lemma~\ref{lemma4}. Therefore coefficient of $X^{k-1}$ in $N(f_g(c),g,X)$ is $2c_0\big(c_{k-1}-c\big)$. 

\subsection{Formal algebraic version}
\label{sec2.3} 
We conclude the section with a formal version of Lemma~\ref{lemma4}. Let $k \geq 2$ and $a_0, \ldots, a_{k}$ be formal variables. Suppose that $g(X,\textbf{a})$ is an element of $\mathbb{Z}[a_0, \ldots, a_{k}]$ given by 
\begin{equation}
\label{2.18} 
g(X,\textbf{a}) = a_0X^k + \cdots + a_k. 
\end{equation}
 Define auxiliary polynomials \[ \big\{ y_i(\textbf{x}), u_{j}(\textbf{x}), v_{l}(\textbf{x},\textbf{a}) \mid 0 \leq i \leq k-1, 0 \leq j,l \leq 2k-2 \big\} \subseteq \mathbb{Z}[\textbf{x}, \textbf{a}] \] using \eqref{2.3}, \eqref{2.4} and \eqref{2.5}. It is easy to see that these polynomials satisfy \eqref{2.6}, \eqref{2.7} and \eqref{2.8}. Consider the system of equations 
\begin{equation}
\label{2.19} u_i(\textbf{x}) = v_i(\textbf{x},\textbf{a})\hspace{5mm} (0 \leq i \leq k-1) 
\end{equation}  in variables $x_0, \ldots, x_{k-1}$.     
\begin{lemma}
\label{lemma8} 
Let $\mathbb{F} = \mathbb{Q}(a_0, \ldots,a_k)$, the field of rational functions in variables $a_0,  \ldots, a_k$ with coefficients in $\mathbb{Q}$. Then 
there is a unique tuple of rational functions $(c_0(\textbf{a}), \ldots, c_{k-1}(\textbf{a})) \in \mathbb{F}^k$ with $c_0(\textbf{a}) \neq 0$, which is a solution to the system of equations \eqref{2.19}. Further, $c_i(\textbf{a}) \in \mathbb{Q}[a_0, \ldots, a_i][a_0^{-1}]$ for all $0 \leq i \leq k-1$.
\end{lemma}  
\begin{proof} Similar to the proof of Lemma~\ref{lemma4}. The base case holds since $c_0 = a_0(k-1) \neq 0$. Let $0 \leq i \leq k-2$. The recursion step goes through, since $u_{i+1}$ is independent of $\{a_0,\ldots,a_{k}\}$ and
$v_{i+1}$ depends only on $\{a_0, \ldots, a_{i+1}\}$ and to determine $c_{i+1}$ one needs to divide by an element of $\mathbb{Q}^{\times}a_0$. Recursively,
one deduces \[c_i(\textbf{a}) \in \mathbb{Q}[a_0, \ldots, a_i][a_0^{-1}] \hspace{5mm} (0 \leq i \leq k-1). \] 
\end{proof}    
In the formal version of the theory,
it is enough to consider only one polynomial, namely,  the universal polynomial $g(X,\textbf{a})$. Statements analogous to Corollary~\ref{corollary5}(i) and Corollary \ref{corollary7} hold in this situation, i.e., one can restrict to suitable subsystems and scaling of variables results into scaling of solution. The notion of good approximant with coefficients in $\mathbb{F}$ can be introduced in exactly same manner.  Lemma~\ref{lemma8} constructs a canonical good approximant for $g(X, \textbf{a})$  and classification of Corollary~\ref{corollary6} continues to hold.                         
\section{Proofs of the theorems}    
\label{sec3} 
In this section we use the theory developed in Section~\ref{sec2} to prove Theorem~\ref{thm1}. An appropriate application of the same ideas yields Theorem~\ref{thm3}. Corollary~\ref{corollary2} is an easy consequence of Theorem~\ref{thm1}. 

We initiate the discussion with a useful remark. In what follows,
$\mathbb{K}$ is always a subfield of $\mathbb{R}$ unless otherwise specified.     
\begin{remark}
\label{remark9}
\ \leavevmode
\begin{itemize}
\item[(i)] Let $\phi(X) = a_0X^{d} + a_1X^{d-1} + \cdots + a_d \in \mathbb{K}[X]$ is a polynomial with $\sgn \phi = 1$. Then $\phi(x) &gt; 0$ for all real $x$ satisfying  
\begin{equation}
\label{3.1}
x \geq \max \{1, \frac {(d+1)|a_j|} {|a_0|} \mid 1 \leq j \leq d \}.
\end{equation} 
If $d = 0$ then the right-hand side is interpreted as $1$. Note that expression on the right-hand side is an element of $\mathbb{K}$.  

\item[(ii)] The lower bound appearing in the first part is not best possible. 
To determine the best possible bound, we need to locate the real zeroes of $\phi$.

\item[(iii)] Using (i) one can effectively determine the constant $M_0$ appearing in the statement of Theorems~\ref{thm1} and \ref{thm3}.
\end{itemize}       
\end{remark} 
\subsection{Proof of Theorem~\ref{thm1}} 
\label{sec3.1} 
Let $P(X) \in \mathbb{K}[X]$ be of degree $k \geq 2$ with $\sgn P = 1$. Suppose that $M_0$ is an element of $\mathbb{K}$ with property that $P(x) &gt; 0$ for all real $x \geq M _0 +1$. 

With the notation of Section~\ref{sec2} we use Lemma~\ref{lemma4} for $g(X) = P(X)$. Note that $a_0$, the leading coefficient of $P(X)$, is positive. Let $(c_0,\ldots,c_{k-1}) \in \mathbb{K}^{\times} \times \mathbb{K}^{k-1}$ be the unique solution to the system of equations \eqref{2.13} corresponding to $P$. 

Now $(c_{k-1} - 1, c_{k-1}) \cap \mathbb{K}$ is nonempty since $\mathbb{K}$ is dense in $\mathbb{R}$. Let $c$ be an element of $(c_{k-1}-1, c_{k-1}) \cap \mathbb{K}$. Define $f_P(c,X) \in \mathbb{K}[X]$ by \eqref{2.16}. Since $c_0 = a_0(k-1) &gt; 0$ there exists a $M_{f_P(c)} \in \mathbb{R}$ so that $f_P(c,x) &gt; 0$ for all real $x \geq M_{f_P(c)}$. We know that $f_P(c,X)$ is a good approximant of $P(X)$, and the coefficient of $X^{k-1}$ in $N\big(f_P(c),P,X\big)$ is $2c_0(c_{k-1}-c)$. But $2c_0(c_{k-1}-c) &gt; 0$. Hence there is a $M_1 \in \mathbb{R}$ such that $N(f_P(c),P,x) &gt; 0$ for all real $x \geq M_1$.     

Let $M'_1 = \max \{M_{1}, M_{f_P(c)}\}$. Suppose that $m$ is a positive integer $\geq M'_1 - M_0$. Then by \eqref{2.10} and \eqref{2.11}   
\[\delta\big(f_P(c),P, m+M_0\big) &gt; 0\]
i.e., \[\frac {1} {f_P(c,m+M_0)} -\frac {1} {f_P(c,m+1+M_0)} &gt; \frac {1} {P(m+M_0)}. \]

Using telescoping summation we have
\begin{equation}
\label{3.2} \frac {1} {f_P(c,n+M_0)} &gt; \sum_{m=n}^{\infty} \frac {1} {P(m+M_0)}
\end{equation}
for all positive integers $n \geq M'_1 - M_0$. 
 
Now let $C = c+1 \in \mathbb{K}$, and consider $f_P(C,X) \in \mathbb{K}[X]$. It is a good approximant of $P$ and coefficient of $X^{k-1}$ in $N\big(f_P(C),P,X\big)$ is $2c_0(c_{k-1} -C)$. But $c_{k-1} - c - 1 &lt; 0$. Therefore $2c_0(c_{k-1} - C) &lt; 0$ and there exists a $M_2 \in \mathbb{R}$ so that $N(f_P(C),P,x) &lt; 0$ for all real $x \geq M_2$. 

Let $M'_2 = \max \{M_2, M_{f_P(c)}\}$ and $m$ be a positive integer $\geq M'_2 - M_0$. It follows that 
\[\delta\big(f_P(C),P,m+M_0\big) &lt; 0 \]
i.e.,   
\[\frac {1} {f_P(c,m+M_0)+1} -\frac {1} {f_P(c,m+1+M_0)+1}  &lt; \frac {1} {P(m+M_0)}.\]

By telescoping we have 
\begin{equation}
\label{3.3} \frac {1} {f_P(c,n+M_0)+1} &lt; \sum_{m = n}^{\infty} \frac{1}{P(m+M_0)}
\end{equation} 
for all positive integers $n \geq M'_2 - M_0$.  
     
Suppose that $M_3 = \max \{M'_1, M'_2\}$. Using \eqref{3.2} and \eqref{3.3} 
\[0 &lt; \frac {1} {f_P(c,n+M_0)+1} &lt; \sum_{m=n}^{\infty} \frac {1} {P(m+M_0)} &lt; \frac {1} {f_P(c,n+M_0)}\]
for all positive integers $n \geq M_3 - M_0$. The leftmost inequality is a consequence of $M_3 \geq M_{f_P(c)}$.  

Let $h(X) = f_P(c,X+M_0)$. Note that it has degree $k-1$. Since $M_0 \in \mathbb{K}$, the polynomial $h(X) \in \mathbb{K}[X]$. Put $N_0 = \max \{\lceil M_3  - M_0\rceil, 1\}$. Then \[0 &lt; \frac {1} {h(n) + 1} &lt; \sum_{m=n}^{\infty} \frac {1} {P(m+M_0)} &lt; \frac {1} {h(n)} \] for all integers $n \geq N_0$. It is clear that $h$ and $N_0$ so defined have properties required by Theorem~\ref{thm1}. This construction proves the first part of Theorem~\ref{thm1}.    

Further Lemma~\ref{lemma4} algorithmically determines the tuple $(c_0, \ldots, c_{k-1})$ and $c$ is any element of $(c_{k-1}-1, c_{k-1})\cap \mathbb{K}$. Therefore we can determine $f_P(c)$ algorithmically. Since $M_0$ is part of hypothesis the polynomial $h$ is also algorithmically computable. To finish off proof we need to show uniqueness. This part of assertion is a consequence of Lemma~\ref{lemma10}. Thus the proof of Theorem~\ref{thm1} is complete, modulo Lemma~\ref{lemma10}. \hfill $\square$ 

Corollary~\ref{corollary2} is a special case of Theorem~\ref{thm1}.
\subsubsection{Proof of Corollary~\ref{corollary2}} Follows from Theorem~\ref{thm1} with $\mathbb{K} = \mathbb{Q}$, $P(X) = X^k \in \mathbb{Q}[X]$, and $M_0 = 0$. \hfill $\square$ 

\subsection{Proof of Theorem~\ref{thm3}}
\label{sec3.2} 
In this subsection we prove Theorem~\ref{thm3}. Main idea behind the proof is to approximate the rational function by appropriate polynomial. 
  
Let $P(X),Q(X)$ be two nonzero polynomials in $\mathbb{K}[X]$ so that $\text{deg}_{\mathbb{K}} P - \text{deg}_{\mathbb{K}} Q = k \geq 2$ and $\sgn P = \sgn Q = 1$. Suppose that $R(X) = \frac {P(X)} {Q(X)}$ and $M_0 \in \mathbb{K}$ with $P(x), Q(x) &gt; 0$ for all real $x \geq M_0 +1$.   

Using the division algorithm, we construct  polynomials $A(X), B(X) \in \mathbb{K}[X]$ such that \[P(X) = A(X)Q(X) + B(X)\] and $\text{deg}_{\mathbb{K}} B &lt; \text{deg}_{\mathbb{K}} Q$. It is easy to see that $A(X)$ is of degree $k$ and $\sgn A = 1$.  Now 
\begin{equation} 
\label{3.4}  
R(X) = A(X) + \frac {B(X)} {Q(X)}. 
\end{equation}
Note that $A(X)$ is determined by $R(X)$ and does not depend on individual polynomials $P(X)$ and $Q(X)$. It is the unique polynomial so that the difference $R(X) - A(X)$ is either $0$ or is given by a rational function whose denominator has degree strictly larger than degree of numerator. Since $\text{deg}_{\mathbb{K}} B &lt; \text{deg}_{\mathbb{K}} Q$, $\frac {B(X)} {Q(X)} \to 0$ as $x \to \infty$ on real line. If $B = 0$ then $R(X) = A(X)$ and the result is already true by Theorem~\ref{thm1}.  
 
Let $\epsilon \in \mathbb{K} \cap  (\sgn B)\,\mathbb{R}_{&gt;0}$. If $B = 0$ then $\epsilon = 0$. Note that $Q(x) &gt; 0$ for $x \geq M_0+1$. Hence there is a real number $M_{\epsilon}  \geq M_0 + 1&gt; 0$ such that for all real $x \geq M_{\epsilon}$ we have $(\sgn B) B(x) \geq 0$ and $0 \leq |\frac {B(x)} {Q(x)}| \leq |\epsilon|$. 
  
Define $A_{\epsilon}(X) = A(X) + \epsilon \in \mathbb{K}[X]$.  
It is easy to see that $A_{\epsilon}(X)$ is a polynomial of degree $k$ with $\sgn A_{\epsilon} = 1$. Moreover, if $\sgn B \geq 0$ then
\begin{equation}
\label{3.5}
 A(x) \leq R(x) \leq A_{\epsilon}(x) \hspace{5mm} (x \geq M_{\epsilon})
\end{equation} and if $\sgn B = -1$ then
\begin{equation}
\label{3.6} 
A_{\epsilon}(x) \leq R(x) \leq A(x) \hspace{5mm} (x \geq M_{\epsilon}). 
\end{equation}   

Consider the system of equation \eqref{2.13} in Lemma~\ref{lemma4} for the polynomials $A(X)$ and $A_{\epsilon}(X)$. By Corollary~\ref{corollary5}(ii) 
the same tuple $(c_0,\ldots, c_{k-1}) \in \mathbb{K}^{\times} \times \mathbb{K}^{k-1}$ is a solution to \eqref{2.13} for both $A(X)$ and $A_{\epsilon}(X)$. Since $\sgn A = 1$ we have $c_0 &gt; 0$. Let $c \in (c_{k-1}-1,c_{k-1}) \cap \mathbb{K}$ and consider $f_A(c,X) \in \mathbb{K}[X]$ defined in Section~\ref{sec2}. It 
is a good approximant of each of $A(X)$ and $A_{\epsilon}(X)$. The coefficient of $X^{k-1}$ in both $N\big(f_A(c),A,X\big)$ and $N\big(f_A(c), A_{\epsilon},X\big)$ is $2c_0(c_{k-1} - c) &gt; 0$. Similarly, the coefficient of $X^{k-1}$ in each of $N(f_A(c)+1,A,X)$ and $N(f_A(c)+1, A_{\epsilon},X)$ is $2c_0(c_{k-1}-c-1) &lt; 0$.   

Let $M' \in \mathbb{R}$ be such that $A(x), A_{\epsilon}(x) &gt; 0$ for all real $x \geq M'$. Imitating the proof of Theorem~\ref{thm1}, we can find $M'_3$, $M'_{3,\epsilon} \geq M'$ so that  
\begin{align*}
0 &lt; \frac {1} {f_A(c,n+M_0)+1} &lt; \sum_{m=n}^{\infty} \frac {1} {A(m+M_0)} &lt; \frac {1} {f_A(c,n+M_0)},\\
0 &lt; \frac {1} {f_A(c,n+M_0)+1} &lt; \sum_{m=n}^{\infty} \frac {1} {A_{\epsilon}(m+M_0)} &lt; \frac {1} {f_A(c,n+M_0)} 
\end{align*}     
holds for all positive integers $n \geq M'_3 - M_0$ and $n \geq M'_{3,\epsilon} - M_0$ respectively. 

Let $M_4 = \max\{M'_3, M'_{3,\epsilon}, M_{\epsilon}\}$. From inequalities \eqref{3.5}, \eqref{3.6} and choice of $M'_3, M'_{3,\epsilon}$ it follows that
\[0 &lt; \frac {1} {f_A(c,n+M_0)+1} &lt; \sum_{m=n}^{\infty} \frac {1} {R(m+M_0)} &lt; \frac {1} {f_A(c,n+M_0)}\]
for all positive integers $n \geq M_4 - M_0$. Set $h(X) = f_A(c,X+M_0)$ and $N_0 = \max \{\lceil M_4 - M_0\rceil, 1\}$. One easily sees that $h(X) \in \mathbb{K}[X]_{k-1}$ is a polynomial that satisfies the requirements of the statement of the theorem for the prescribed choice of $N_0$. Note that $A, B$ are 
algorithmically computable. Therefore, by reasoning similar to that in the proof of Theorem~\ref{thm1},  $h$ is also algorithmically computable.

Proof of uniqueness is postponed to Lemma~\ref{lemma10}. \hfill $\square$     
\vspace{3mm}
\subsubsection{Effectiveness of constants}  Using Remark~\ref{remark9} we can determine effective choices for the constants $M_{f_P(c)}, M_1, M_2$ appearing 
in the proof of Theorem~\ref{thm1}. Hence $M'_1, M'_2, M_3,$ and in particular, $N_0$ are effective. These constants depend on choice of $c$. In the proof of Theorem~\ref{thm3}, $\epsilon$ is any number in $\mathbb{K} \cap (\sgn B)\mathbb{R}_{&gt;0}$ and the inequalities $(\sgn B) B(x) \geq 0$, $0 \leq |\frac {B(x)} {Q(x)}| \leq |\epsilon|$ effectively determine $M_{\epsilon}$. Now one can use Remark~\ref{remark9} to show that the constants appearing in the proof of Theorem~\ref{thm3} are also effective. These numbers depend on choices of $P$, $Q$, $c$, and $\epsilon$.     

\subsection{Admissible polynomials} 
\label{sec3.3}  
This subsection studies all polynomials which approximate reciprocals of tail sums. First, we prove a lemma which implies uniqueness part of Theorem~\ref{thm1} and \ref{thm3}. Recall that to retrieve Theorem~\ref{thm1} from Theorem~\ref{thm3} one needs to substitute $Q = 1$. The notation is same as in the
statement of Theorem~\ref{thm3}.        
\begin{lemma}
\label{lemma10} 
Suppose that $h_1(X), h_2(X) \in \mathbb{K}[X]$ are two nonzero polynomials which satisfy \[0 &lt; h_{j}(n) \leq \frac {1} {\sum_{m = n}^{\infty} \frac {1} {R(m + M_0)}} &lt; h_{j}(n) + 1 \hspace{5mm} (j = 1,2)\] for infinitely many $n \in \mathbb{N}$.  Then
\begin{itemize}
\item[(i)] $\emph{deg}_{\mathbb{K}} h_1 = \emph{deg}_{\mathbb{K}} h_2 = k-1$, 
\item[(ii)] $h_1(X) - h_2(X) \in \mathbb{K}$, 
\item[(iii)] $|h_1 - h_2| &lt; 2$,  
\item[(iv)] if there is an infinite subset of $\mathbb{N}$ on which hypothesis 
of the lemma hold simultaneously for $h_1$ and $h_2$, then $|h_1 - h_2| &lt; 1$.
\end{itemize}
\end{lemma}  
\begin{proof}
In what follows the statements hold for both $j = 1,2$. By assumption 
\begin{equation}
\label{3.7} 
 0 &lt; \frac {1} {h_j(n) + 1} &lt; \sum_{m = n}^{\infty} \frac {1} {R(m+M_0)} \leq \frac {1} {h_j(n)} 
\end{equation} for infinitely many $n \in \mathbb{N}$. These inequalities imply that  $h_j$ is non-constant and $\sgn h_j = 1$.

Let $h(X)$ be the polynomial constructed in the proof of Theorem~\ref{thm3}. Comparing with \eqref{3.7}    
\begin{gather*}
0 &lt; \frac {1} {h(n) + 1} &lt; \frac {1} {h_{j}(n)},~\\
0 &lt; \frac {1} {h_{j}(n) + 1} &lt; \frac {1} {h(n)} 
\end{gather*}
for infinitely many $n \in \mathbb{N}$. Hence on an infinite subset of $\mathbb{N}$ 
\begin{equation}
\label{3.8} h(n) - 1 &lt; h_{j}(n) &lt; h(n) + 1.
\end{equation}  It follows that $|h(n) - h_{j}(n)| &lt; 1$ on an infinite subset. But this inequality forces the difference to be constant.  

The conclusion above proves both (i) and (ii).
The proof of (iii) follows from \eqref{3.8}. 

Without loss of generality assume $h_1 - h_2 = C \geq 0$. If $C \geq 1$
then \[0 &lt; \frac {1} {h_1(n)} \leq \frac {1} {h_2(n) + 1} \hspace{5mm}(n
\gg 1).\] This contradicts the hypothesis of (iv). Hence (iv) is proved
by way of contradiction.
\end{proof}

\subsubsection{Admissible polynomials} We introduce a terminology for convenience. A polynomial $h(X) \in \mathbb{K}[X]$ is \textit{admissible} for $(R,M_0)$ if 
\begin{equation*}
0 &lt; h(n) \leq \frac {1} {\sum_{m=n}^{\infty} \frac {1} {R(m+M_0)}} &lt; h(n)+1 \hspace{5mm} (n \gg 1). 
\end{equation*}

In rest of the section let $(c_0,\ldots,c_{k-1}) \in \mathbb{K}^{\times} \times \mathbb{K}^{k-1}$ denote the unique solution to system of equations \eqref{2.13} associated with $A(X)$. The polynomial $A$ depends only on $R$ and is determined by \eqref{3.4}. Since $\sgn A = 1$ it follows that $c_0 &gt; 0$. Moreover, the tuple does not depend on the constant term of $A$ (Corollary~\ref{corollary5}).   

Lemma~\ref{lemma10} implies that every admissible polynomial for $(R,M_0)$ is of degree $k-1$ and has positive leading coefficient. Further if $h_1,h_2$ are two admissible polynomials then $h_1 - h_2$ is constant and $|h_1 - h_2| &lt; 1$. We have shown in Section~\ref{sec3.2} that for each $c \in (c_{k-1}-1, c_{k-1}) \cap \mathbb{K}$ the polynomial $f_A(c,X+M_0)$ is admissible for $(R,M_0)$. In fact it satisfies a stronger inequality, namely,
\[0 &lt; f_A(c,n+M_0)  &lt; \frac {1} {\sum_{m=n}^{\infty} \frac {1} {R(m+M_0)}} &lt; f_A(c,n+M_0)+1 \hspace{5mm} (n \geq N_0).\] Constant term of this polynomial is $f_A(c,M_0)$. Now for all $x, y \in \mathbb{K}$
\begin{equation}
\label{3.9} 
f_A(x,X) - f_A(y,X) = x -y.
\end{equation}
Since $c \in (c_{k-1}-1, c_{k-1})\cap \mathbb{K}$ is infinite, there are infinitely many distinct choices for $c$ which give rise to infinitely many admissible polynomials.

\subsubsection{Effect of scaling}  Let $\alpha \in \mathbb{K}^{\times}$ be  positive. Now $\alpha R(X) = \frac {\alpha P(X)} {Q(X)}$. Therefore same constant $M_0$ has desired positivity properties and the quotient polynomial is $A_{\alpha}(X) := \alpha A(X)$. By Corollary~\ref{corollary7} $\textbf{c}(A_{\alpha}) = \alpha \textbf{c}(A)$. Therefore if $h(X)$ and $h_{\alpha}(X)$ are admissible polynomials with respect to $(R,M_0)$ and $(\alpha R, M_0)$ resp., then $h_\alpha(X) - \alpha h(X) \in \mathbb{K}$.     
  
 
\subsubsection{Choice of constant term}   
Let $c \in (c_{k-1}-1,c_{k-1}) \cap \mathbb{K}$. We have seen that $f_A(c,M_0)$ is the constant term of a polynomial admissible for $(R,M_0)$. Let $C \in \mathbb{K}$ be the constant term of some other admissible polynomial $H(X)$. 
\begin{lemma}
\label{lemma11}
\ \leavevmode
\begin{itemize}
 \item[(i)] There exists unique $c \in [c_{k-1}-1, c_{k-1}] \cap \mathbb{K}$ such that 
\begin{gather*}
C = f_A(c,M_0),\\ 
H(X) = f_A(c,X+M_0).
\end{gather*} 
\item[(ii)] Both $f_A(c_{k-1}-1,X+M_0)$ and $f_A(c_{k-1},X+M_0)$ cannot be admissible polynomials for $(R,M_0)$. 
\end{itemize}  
\end{lemma}
\begin{proof}
\ \leavevmode
\begin{itemize}
\item[(i)] For brevity write $\mathcal{I} = (c_{k-1}-1,c_{k-1}) \cap \mathbb{K}$. 

Let $c = C - f_A(0,M_0) \in \mathbb{K}$. Substituting $x = c$, $y=0$ and $X = M_0$ in \eqref{3.9} we have $f_A(c,M_0) = C$. Moreover, if $x \in \mathbb{K}$ satisfies $f_A(x,M_0) = C$ then $x = c$. By Lemma~\ref{lemma10} (iv), $|C - f_A(x,M_0)| &lt; 1$ holds for each $x \in \mathcal{I}$. Hence $|c-x| &lt; 1$ for all $x \in \mathcal{I}$. Therefore $c \in [c_{k-1}-1,c_{k-1}]$.  Now \[H(X) - f_A(c,X+M_0) = H(X) - f_A(x,X+M_0) + f_A(x,X+M_0) - f_A(c,X+M_0)\] for all $x \in \mathbb{K}$. Letting $x \in \mathcal{I}$ and using admissibility of $H$ we see that 
the right-hand side is constant (Lemma~\ref{lemma10}). But the constant term of the
left-hand side is $0$. Therefore $H(X) = f_{A}(c,X+M_0)$.     

\item[(ii)] Since $f_A(c_{k-1},X+M_0) - f_A(c_{k-1}-1,X+M_0) = 1$ the assertion follows from Lemma~\ref{lemma10} (iv).    
\end{itemize}
\end{proof} 

Lemma~\ref{lemma11} characterizes all admissible polynomials with possible exception at boundary of the interval. Analysis of extremal situation is necessary for later development. We need mild improvement over formalism of Section~\ref{sec2} to discuss admissibility of boundary points. 

\subsubsection{Approximate primitive for rational functions} Let $\mathbb{K}$ be a field of characteristic $0$ and $P(X), Q(X), f(X) \in \mathbb{K}[X] - \{0\}$. Suppose that $\text{deg}_{\mathbb{K}} P - \text{deg}_{\mathbb{K}} Q = k \geq 2$. Define   
\begin{gather*}
\delta(f,R,X) := \frac {1} {f(X)} - \frac {1} {f(X+1)} - \frac {1} {R(X)},\\
N(f,R,X) := P(X)\big(f(X+1) - f(X)\big) - Q(X)f(X)f(X+1) 
\end{gather*}
where $R = \frac {P} {Q}$. These expressions depend on $P$ and $Q$. However if $N(f,R,X) = 0$ for one presentation then it is $0$ for all presentations. 
It is clear that 
\begin{equation}
\label{3.10} 
\delta(f,R,X) = \frac {N(f,R,X)} {f(X)f(X+1)P(X)}.
\end{equation}
Further $\delta(f,R,X) = 0$ implies $\text{deg}_{\mathbb{K}}f = k-1$. Let $A(X),B(X) \in \mathbb{K}[X]$ be the polynomials obtained by the division algorithm, i.e., $P(X) = A(X)Q(X) + B(X)$ with $\text{deg}_{\mathbb{K}} A = k$ and $\text{deg}_{\mathbb{K}}B &lt; \text{deg}_{\mathbb{K}}Q$.  Then  \[\delta(f,R,X) = \delta(f,A,X) + \frac {B(X)} {P(X)A(X)},\] 
i.e,
\begin{equation*}
\delta(f,R,X) = \frac {N(f,A,X)} {f(X)f(X+1)A(X)} + \frac {B(X)} {P(X)A(X)}. 
\end{equation*}
We have $\text{deg}_{\mathbb{K}} P(X)A(X) - \text{deg}_{\mathbb{K}} B(X) &gt; 2k$. Suppose that $f \in \mathbb{K}[X]_{k-1}$. If $N(f,A,X)\in \mathbb{K}[X]_{k-1}$ then $\text{deg}_{\mathbb{K}} f(X)f(X+1)A(X) - \text{deg}_{\mathbb{K}} N(f,A,X) = 2k-1$. Here hypothesis amounts to saying that  $f$ is a good approximant of $A$ with constant term $\neq c_{k-1}$. In such situation behavior of $\delta(f,R,X)$ is dominated by $\delta(f,A,X)$. Now let $f$ be an arbitrary element of $\mathbb{K}[X]_{k-1}$. From expression above it follows that $\delta(f,R,X) = 0$ if and only if 
\begin{equation}
\label{3.11} 
N(f,A,X) P(X) = - f(X)f(X+1)B(X).
\end{equation} 
If \eqref{3.11} holds then $\text{deg}_{\mathbb{K}} N(f,A,X) = k- 2 + \text{deg}_{\mathbb{K}} B - \text{deg}_{\mathbb{K}} Q &lt; k-2$. Moreover, we have $N(f,A,X) = 0$ if $B(X) = 0$.       

Notation and assumptions are identical to Section~\ref{sec3.2}. Since $\sgn P = \sgn Q = 1$ it is easy to see that $\sgn N(f, R, X)$ depends only on $R$ and is independent of presentation.       
\begin{lemma}
\label{lemma12}
\ \leavevmode
\begin{itemize}
\item[(i)] Let $N(f_A(c_{k-1}),R,X) = 0$. Then $f_A(c_{k-1},X+M_0)$ is admissible for $(R,M_0)$. 

\item[(ii)] Now suppose $N(f_A(c_{k-1}),R,X) \neq 0$. If $\sgn N(f_A(c_{k-1}),R,X) = 1$ then $f_A(c_{k-1},X+M_0)$ is admissible and  if $\sgn N(f_A(c_{k-1}),R,X) = -1$ then $f_A(c_{k-1}-1,X+M_0)$ is admissible.
\end{itemize}
\end{lemma}  
\begin{proof} 
\begin{itemize}
\item[(i)] By assumption $N(f_A(c_{k-1}),R,X) = \delta(f_A(c_{k-1}),R,X) = 0$.  
Therefore
\begin{equation}
\label{3.12} 
\frac {1} {R(X)}=\frac {1} {f_A(c_{k-1},X)} - \frac {1} {f_A(c_{k-1},X+1)}. 
\end{equation}  

By \eqref{3.12} the  maximum of real zeroes of $f_A(c_{k-1}, X)$ is $&lt; M_0 +1$. Hence $f_A(c_{k-1},n+M_0) &gt; 0$ for all $n \in \mathbb{N}$.  Telescoping 
\begin{equation}
\label{3.13} 
\sum_{m=n}^{\infty} \frac {1} {R(m+M_0)}=\frac {1} {f_A(c_{k-1}, n+M_0)} \hspace{5mm}(n \geq 1).
\end{equation}  
The coefficient of $X^{k-1}$ in both $N(f_A(c_{k-1})+1,A,X)$ and $N(f_A(c_{k-1})+1,A_{\epsilon},X)$   is $2c_0(c_{k-1} - c_{k-1} -1) = -2c_0 &lt; 0$. Using \eqref{3.5} and \eqref{3.6} and imitating the proof of Theorem~\ref{thm3}, we can show  \[\sum_{m=n}^{\infty} \frac {1} {R(m+M_0)} &gt; \frac {1} {f_A(c_{k-1},n+M_0)+1} \hspace{5mm} (n \gg 1).\] Therefore $f_A(c_{k-1},X+M_0)$ is admissible. 

\item[(ii)] Let $N(f_A(c_{k-1}),R,X) \neq 0$. Suppose that $\sgn N(f_A(c_{k-1}),R,X) = 1$. Hence there is $M_1 \in \mathbb{R}$ such that $N(f_A(c_{k-1}),R,x) &gt; 0$ for all real $x \geq M_1$. One can use \eqref{3.10} and telescoping to deduce  \[\sum_{m=n}^{\infty} \frac {1} {R(m+M_0)} &lt; \frac {1} {f_A(c_{k-1}, n+M_0)}\hspace{5mm} (n \gg 1).\] The coefficient of $X^{k-1}$ in each of $N\big(f_A(c_{k-1})+1,A,X\big)$ and $N\big(f_A(c_{k-1})+1,A_{\epsilon},X\big)$ is $2c_0(c_{k-1} - c_{k-1} -1) = -2c_0 &lt; 0$. We can repeat the arguments from the proof of Theorem~\ref{thm3} to conclude (see part (i))  \[\sum_{m=n}^{\infty} \frac{1} {R(m+M_0)} &gt; \frac {1} {f_A(c_{k-1}, n+M_0)+1} \hspace{5mm} (n \gg 1).\] 
Hence $f_A(c_{k-1}, X+M_0)$ is an admissible polynomial. 

Now assume that $\sgn N(f_A(c_{k-1}),R,X) = -1$. As before, we can use \eqref{3.10} and telescoping to conclude \[\sum_{m=n}^{\infty} \frac {1} {R(m+M_0)} &gt; \frac {1} {f_A(c_{k-1}, n+M_0)}\hspace{5mm} (n \gg 1).\] Note that $f_A(c_{k-1})-1$ is a good approximant of $A$ and $A_{\epsilon}$. Further coefficient of $X^{k-1}$ in both $N\big(f_A(c_{k-1})-1,A,X\big)$ and $N\big(f_A(c_{k-1})-1,A_{\epsilon},X\big)$ is $2c_0(c_{k-1} - c_{k-1}+1) = 2c_0 &gt; 0$. Therefore \[\sum_{m=n}^{\infty} \frac {1} {R(m+M_0)} &lt; \frac {1} {f_A(c_{k-1}, n+M_0)-1}\hspace{5mm} (n \gg 1). \] Hence $f_A(c_{k-1}-1,X+M_0)$ is an admissible polynomial in this case.  
\end{itemize}
\end{proof} 

\subsubsection{Summation by telescoping} If there is $f(X) \in \mathbb{K}[X]- \{0\}$ so that $\frac {1} {R(X)} = \frac {1} {f(X)} - \frac {1} {f(X+1)}$ then $N(f,R,X) = 0$. By \eqref{3.10} and \eqref{3.11} we have $\text{deg}_{\mathbb{K}} f = k-1$, $\text{deg}_{\mathbb{K}} N(f,A,X) &lt; k-2$. Now \eqref{2.9} and  Lemma~\ref{lemma4} implies that $f_A(c_{k-1})$ is the only polynomial which can possibly satisfy these two conditions. Therefore $-\frac {1} {R(X)}$ has an exact difference primitive if and only if $N(f_{A}(c_{k-1}),R,X) = 0$. Whenever this criterion holds, we can compute the exact value of the sum by telescoping. (Cf.\  Corollary~\ref{corollary6}.) 
\begin{remark}
\label{remark13} 
One can use Remark~\ref{remark9} to calculate effective lower bounds on $n$ for inequalities of Lemma~\ref{lemma12} to hold. Computations are analogous to Section~\ref{sec3.2} and we omit the details.      
\end{remark}   
 

\section{Explicit calculations}
\label{sec4}  
\subsection{Reciprocal sequence}
\label{sec4.1}    
This subsection is devoted to explicit calculation of reciprocal sequence.  We begin with an elementary observation. The situation is  same as Theorem~\ref{thm3}.
\begin{remark}
\label{rmk14}  
Let $h$ be a polynomial given by Theorem~\ref{thm3} and $N_0$ be the corresponding integer. Suppose that $(a_n)_{n\geq 1}$ is reciprocal sequence of the sequence $(x_m)_{m \geq 1}$ given by $x_m = \frac {1} {R(m+M_0)}$.  Then from \eqref{1.6} it follows that, for all $n \geq N_0$,
\begin{itemize}
\item[(i)] $a_n$ is either $\lfloor h(n)\rfloor$ or $\lfloor h(n)\rfloor + 1$;

\item[(ii)] if $h(n)$ is an integer for some $n$, then $a_n=h(n)$.
\end{itemize}
\end{remark} 
Let $P, Q \in \mathbb{Q}[X]$ be polynomials so that $\text{deg}_{\mathbb{Q}} P - \text{deg}_{\mathbb{Q}} Q = k \geq 2$ and $\sgn P = \sgn Q = 1$. Suppose that $P(x), Q(x) &gt; 0$ for all real $x \geq 1$, i.e., $0$ is a legitimate choice for $M_0$. One can always ensure this property by shifting the polynomials in hypothesis. Set $R(X) = \frac {P(X)} {Q(X)} \in \mathbb{Q}(X)$ and consider the sequence $(x_m)_{m \geq 1}$ given by 
\begin{equation}
\label{4.1} 
x_{m} = \frac {1} {R(m)} \hspace{5mm} (m\geq1). 
\end{equation}  Subsequent paragraphs contain an algorithmic determination of $(a_n)_{n \geq 1}$, the reciprocal sequence of $(x_m)_{m\geq 1}$.

\subsubsection{Algorithm for reciprocal sequence} Let $A(X)$ and $B(X)$ be the polynomials in $\mathbb{Q}[X]$ obtained by the division algorithm,
i.e, $P(X) = A(X)Q(X) + B(X)$ and $\text{deg}_{\mathbb{Q}} B &lt; \text{deg}_{\mathbb{Q}} Q$. Suppose that $(c_0,\ldots,c_{k-1}) \in \mathbb{Q}^{\times} \times \mathbb{Q}^{k-1}$ is the unique tuple  which is solution to the system of equations \eqref{2.13} corresponding to $A(X)$. Write $c_i=\frac {p_i} {q_i}$ where $p_i,q_i$ are integers with $q_i &gt; 0$ and $\gcd(p_i,q_i)=1$ for each $0 \leq i \leq k-1$. Note that if $p_i = 0$ then $q_i = 1$. Let $L = \lcm (q_0,\ldots,q_{k-2})$. Put 
\begin{equation}
\begin{aligned}
\label{4.2} 
&amp; H(X)= c_0X^{k-1}+\cdots+c_{k-2}X, \\
&amp; \hspace{1cm} H_L(X) = LH(X).
\end{aligned}   
\end{equation} 
It is clear that $H_L(X) \in \mathbb{Z}[X]$.  We determine a family of polynomial in $\mathbb{Q}[X]$ parameterized by residue classes modulo $L$ such that for sufficiently large $n$ value of $a_n$ is obtained by evaluating one of these polynomials at $n$, and this polynomial depends only on residue class of $n$ modulo $L$. 

\subsubsection{Case I: $q_{k-1} \nmid L$}

Under this assumption $q_{k-1} \neq 1$. Write  $c_{k-1} = \lfloor c_{k-1} \rfloor + \frac {r_{k-1}} {q_{k-1}}$ where $r_{k-1}$ is a positive integer $\leq q_{k-1} - 1$. Since $\gcd(p_{k-1}, q_{k-1}) = 1$ it follows that $\gcd(r_{k-1}, q_{k-1}) = 1$. Therefore $\frac {r_{k-1}} {q_{k-1}} \notin \{ \frac {n} {L}\mid n \in \mathbb{Z}\}$. 

Suppose that $r \in \{1,\ldots,L\}$. By the argument above,
there is a unique integer $l(r)$ that satisfies $l(r) -\frac {r} {L} &lt; c_{k-1} &lt; l(r)+1 -\frac {r} {L}$.
Now  let $c(r) = l(r) - \frac {r} {L}$ and $h_r(X)= H(X)+c(r)$. Since $c(r) \in (c_{k-1}-1, c_{k-1})$ there is an integer $N(r)$ so that 
\[h_r(n) \leq \frac {1} {\sum_{m=n}^{\infty} \frac {1} {R(m)}} &lt; h_{r}(n)+1 \hspace{5mm} (n \geq N(r)).\] Fix choice of $N(r)$ for each $r \in \{1,\ldots,L\}$. Put $N = \max\,\{N(1),\ldots,N(L)\}$. Let $n \geq N$. Now $r(n)$ be the unique element in $\{1,\ldots,L\}$ with $H_L(n) \equiv r(n)$ (mod $L$). Evidently $r(n)$ depends only on residue class $n\bmod L$. Note that 
\[h_{r(n)}(n) = \frac {H_L(n)} {L} + l\bigl(r(n)\bigr) - \frac {r(n)} {L} \in \mathbb{Z}\hspace{5mm} (n \geq N).\] Remark~\ref{rmk14} (ii) implies $a_n=h_{r(n)}(n)$.

Therefore in this case we have a closed form formula for $a_n$ depending on equivalence class of $n$ modulo $L$ whenever $n \geq N$. 

\subsubsection{Case II: $q_{k-1} \mid L$}

Let $r \in \{1,\ldots,L\}$. If $\frac{r} {L} \neq 1+\lfloor c_{k-1}\rfloor - c_{k-1}$ then there is a unique integer $l(r)$ with $l(r) - \frac {r} {L} &lt; c_{k-1} &lt; l(r) +1 - \frac {r} {L}$. For these residue classes define $c(r) = l(r) - \frac {r} {L} \in (c_{k-1} - 1, c_{k-1})$. 

Now let $r \in \{1,\ldots,L\}$ with $\frac{r} {L} = 1+\lfloor c_{k-1}\rfloor - c_{k-1}$. Such residue class exists and is unique. Set $l(r) = 1 + \lfloor c_{k-1} \rfloor$. Note that $c_{k-1} = l(r) - \frac {r} {L}$. For this residue class define
\begin{equation*}
c(r) = \begin{cases}
          c_{k-1}, &amp; \text{if $\sgn N(f_A(c_{k-1}),P,X) \geq 0$}; \\
          c_{k-1} - 1, &amp; \text{if $\sgn N(f_A(c_{k-1}),P,X) = -1$}.
          \end{cases}
\end{equation*} 

Let $h_r(X)= H(X)+ c(r)$. By the proof of Theorem~\ref{thm3} and Lemma~\ref{lemma12} it follows that for each $r$ there is an integer $N(r)$ so that
\[h_r(n) \leq \frac {1} {\sum_{m=n}^{\infty} \frac {1} {R(m)}} &lt; h_{r}(n)+1 \hspace{5mm} (n \geq N(r)).\] Suppose that $N = \max \{N(1), \ldots, N(L)\}$. Let $n \geq N$ and $r(n)$ be the unique element in $\{1,\ldots,L\}$ such that $H_L(n) \equiv r(n)$ (mod $L$). It follows that there is a positive integer $N$ so that $a_n = h_{r(n)}(n)$ whenever $n \geq N$. 

The discussion above can be summarized as follows:
\begin{theorem}
\label{thm15} 
Let $(x_m)_{m\geq 1}$ be the sequence defined by \eqref{4.1}. There exist a positive integer $L$, algorithmically computable polynomials $h_{r,L}(X) \in \mathbb{Q}[X]$ parameterized by residue classes modulo $L$, a polynomial $H_L(X) \in \mathbb{Z}[X]$ of degree $k-1$ with constant term $0$, and natural number $N$ so that 
\begin{itemize}
\item[(i)] $H_{L}(X) - Lh_{r,L}(X) \in \mathbb{Z}$ for all residue classes \;$r$,  
\item[(ii)] $a_n = h_{r,L}(n) \hspace{2.5mm}\text{if} \;  H_L(n) \equiv r \;($\emph{mod} $L)$   for all $n \geq N$.
\end{itemize}
\end{theorem}
\begin{proof}
The statement holds with $L = \lcm (q_0,\ldots,q_{k-2})$, $H_{L}$ given by \eqref{4.2}, polynomials $(h_{r}(X))_{r \bmod L}$ and the number $N$ constructed above. The tuple $(c_0,\ldots,c_{k-1})$ is algorithmically constructible and once we have this datum, construction of $(h_{r}(X))_{r \bmod L}$ is already given in two possible cases.        
\end{proof}
\begin{remark}
\label{remark16}
The choice of $L$ in the proof of the theorem is algorithmically computable. Moreover, one can choose $N$ effectively since by Section~\ref{sec3.2} and Remark~\ref{remark13}  each of $N(r)$ is effective.
\end{remark}
Theorem~\ref{thm15} is an analogue of Theorem~\ref{thm3} in context of reciprocal sequence. The modulus in the statement of the theorem adds an arithmetic aspect to the theory. 
\subsubsection{Choice of modulus} Let $\mathfrak{m}$ be a natural number and $\phi(X) \in \mathbb{Z}[X]$. A residue class $r$ modulo $\mathfrak{m}$ is \emph{nontrivial} with respect to $\phi$ if there exists one (and hence infinitely many) integer $n$ such that $\phi(n) \equiv r$ (mod $\mathfrak{m}$). Nontrivial residue classes are exactly residue classes of $\{\phi(j) \mid 1 \leq j \leq \mathfrak{m}\}$.
Observe that we need $h_{r,L}(X)$ only for the residue classes modulo $L$ which are nontrivial with respect to $H_{L}$. For other residue classes, the constant term of $h_{r,L}(X)$ can be chosen arbitrarily. Let $L_1$ be another positive integer so that there exists $H_{L_1}(X) \in \mathbb{Z}[X]$ and polynomials $\{h_{r, L_1}(X) \mid r \bmod L_1\} \subseteq \mathbb{Q}[X]$ satisfying conditions (i) and (ii) in the statement of theorem.  Using condition (i) and Lemma~\ref{lemma10} \[H(X) = \frac {H_{L_1}(X)} {L_1} = \frac {H_{L}(X)} {L}.\] Since $L = \lcm (q_0,\ldots,q_{k-2})$ we have $L | L_1$ and $H_{L_1}(X) = \frac {L_1} {L} H_{L}(X)$. Non-constant part of the polynomials  $\{h_{r,L_1}(X) \mid r \bmod L_1\}$ are same and equals $H(X)$. Now  $H_{L}(n_1) \equiv H_{L}(n_2)$ (mod $L$) if and only if $H_{L_1}(n_1) \equiv H_{L_1}(n_2)$ (mod $L_1$) for all $n_1, n_2 \in \mathbb{Z}$. Hence $H_{L}(n) \to H_{L_1}(n)$ is a bijection between residue classes modulo $L$ nontrivial with respect to $H_{L}$ and residue classes modulo $L_1$ nontrivial with respect to $H_{L_1}$. Let $r$, $r_1$ be two nontrivial residue classes modulo $L$, $L_1$ respectively.  By condition (ii) of the theorem, $h_{r,L}(X) = h_{r_1,L_1}(X)$ if $r$ corresponds to $r_1$ under the bijection mentioned above.  

This phenomenon shows that the modulus in Theorem~\ref{thm15} is essentially unique.   
\subsection{Asymptotic of summation}
\label{sec4.2} 
In this subsection we write down the asymptotic form of Theorem~\ref{thm3}. For this purpose one requires a familiar technique from complex analysis. 
\begin{lemma}
\label{lemma17}
Let $\phi(z) = a_0z^{d} + a_1z^{d-1} + \cdots + a_d \in \mathbb{C}[z]$ be a polynomial of degree $d \geq 1$. There exists $M(\phi) &gt; 0$ (depending on $\phi$) so that on an open set containing the region $|z| \geq M(\phi)$    \[\frac {1} {\phi(z)} = \sum_{j \geq d} \frac {A_j} {z^j}\] where the right-hand side converges absolutely on the open set.
Moreover, there are computable polynomials $F_j \in \mathbb{Z}[x_1, \ldots, x_d]$, $j \geq d$, which depend only on the integer $d$ so that
\begin{itemize}
\item[(i)] $A_j = a_0^{-1} F_j(\frac {a_1} {a_0}, \ldots, \frac {a_d} {a_0})$,
\item[(ii)] $F_d$ is identically $1$ and for $d+1 \leq j \leq 2d-1$ the set of variables appearing in $F_{j}$ is $\{x_{1}, \ldots, x_{j-d}\}$.
\end{itemize}    
\end{lemma} 
\begin{proof}
We have
\begin{align*}
\phi(z)^{-1} &amp; = a_0^{-1} z^{-d} (1 + \frac {a_1} {a_0z} + \cdots + \frac {a_d} {a_0z^{d}})^{-1}.
\end{align*}
Write $\Phi(z) = - (\frac {a_1} {a_0}z + \cdots + \frac {a_d} {a_0}z^d)$. It is easy to see that $|\Phi(\frac {1} {z})| \leq \frac {d} {d+1}$ whenever $|z| \geq M = \max \{ 1 , \frac{(d+1)|a_j|} {|a_0|} \;|\; 1 \leq j \leq d\}$. Let $M(\phi) = 1+M$ and the open set be $\{z \in \mathbb{C} \mid |z| &gt; M\}$. Observe that $\phi(z)$ has no zeroes in region $|z| \geq M$. One obtains the first part of assertion by expanding $(1 - \Phi(\frac {1} {z}))^{-1}$ in a geometric series for $|z| &gt; M$. 

For any $p \geq 0$, coefficient of $z^{-p}$ in $(1 - \Phi(\frac {1} {z}))^{-1}$ is the coefficient of $z^{-p}$ in the finite sum $\sum_{r =0}^{p} \Phi(\frac {1} {z})^r$. Hence, the second part of the assertion is a consequence of the
multinomial theorem.  
\end{proof} 
\subsubsection{Asymptotic form of Theorem~\ref{thm3}} With the notation of Theorem~\ref{thm3} we have $\text{deg}_{\mathbb{K}} h(X) = k-1 \geq 1$ and leading coefficient of $h(X)$ of is $c_0 &gt;0$. Further $h(n) &gt; 0$ for $n \geq N_0$.  
The inequality \eqref{1.6} is equivalent to  
\begin{equation}
\label{4.3} 
0 &lt; \frac {1} {h(n)+1} &lt; \sum_{m = n}^{\infty} \frac {1} {R(m+M_0)} \leq \frac {1} {h(n)} \hspace{5mm}
(n \geq N_0).
\end{equation}  
Now $\frac {1} {h(X)} - \frac {1} {h(X) +1} = \frac {1} {h(X)(h(X)+1)}$, and 
the first term in Taylor series of $\frac {1} {h(z)(h(z)+1)}$ is $c_0^{-2} z^{-2(k-1)}$.  Therefore $\frac {1} {h(n)(h(n)+1)} = O(n^{-2k+2})$  and 
\begin{equation}
\label{4.4}  
\sum_{m = n}^{\infty} \frac {1} {R(m+M_0)} = \frac {1} {h(n)} + O(n^{-2k+2}) \hspace{5mm} (n \geq N_0).
\end{equation} 
Lemma~\ref{lemma17} implies that for each $n \geq \max \{M(h), M(h+1),N_0\} $   
\begin{gather*}
0 &lt; \frac {1} {h(n)} = \sum_{j \geq k-1} \frac {A_j^{(1)}} {n^j},\\
0 &lt; \frac {1} {h(n)+1} = \sum_{j \geq k-1} \frac {A_j^{(2)}} {n^j}.
\end{gather*}
By the second part of the lemma, we have $A_j^{(1)}, A_j^{(2)} \in \mathbb{K}$ for all $j \geq k-1$. Since leading term in expansion of $\frac {1} {h(z)(h(z)+1)}$ is $c_0^{-2}z^{-2k+2}$ we have    
\begin{equation*}
\begin{aligned}
A_j^{(1)} &amp;= A_{j}^{(2)} (= \text{say, $A_j$})  \hspace{1.5cm} (k-1 \leq j \leq 2k-3),\\
A_{2k-2}^{(1)}&amp; = A_{2k-2}^{(2)} + c_0^{-2} &gt; A_{2k-2}^{(2)}.
\end{aligned}   
\end{equation*} 
These statements also follow from lemma above.   Moreover, from absolute convergence ensured by the lemma, it is easy to see that   
\begin{align*}          
\frac {1} {h(n)} &amp; = \sum_{j = k-1}^{2k-3} \frac {A_j} {n^j} + O_{h}(n^{-2k+2}),~\\
\frac {1} {h(n) + 1} &amp; = \sum_{j = k-1}^{2k-3} \frac {A_j} {n^j} + O_{h+1}(n^{-2k+2}) \hspace{5mm} (n \gg 1).
\end{align*}
From \eqref{4.3} one deduces  
\begin{align*}
&amp; \Big|\sum_{m = n}^{\infty} \frac {1} {R(m+M_0)} - \sum_{j = k-1}^{2k-3} \frac {A_j} {n^j}\Big|\\
&amp; \leq \max \big\{\Big|\frac {1} {h(n)} - \sum_{j = k-1}^{2k-3} \frac {A_j} {n^j}\Big|, \Big|\frac {1} {h(n)+1}- \sum_{j = k-1}^{2k-3}\frac {A_j} {n^j}\Big|\big\} \hspace{5mm} (n \geq N_0).  
\end{align*}
Hence   
\begin{equation}
\label{4.5}\sum_{m = n}^{\infty} \frac {1} {R(m+M_0)} = \sum_{j = k-1}^{2k-3} \frac {A_j} {n^j} + O(n^{-2k+2})\hspace{5mm} (n \gg 1). 
\end{equation}   
Thus we have calculated the first $k-1$ terms in the asymptotic expansion of the sum in the left-hand side in the traditional sense. (Compare \eqref{4.4} and \eqref{4.5}.)
 
The constants $M$ and $M(\phi)$ appearing in Lemma~\ref{lemma17} are effective. Let $ |z| \geq M$. Then $|\Phi(\frac {1} {z})| \leq \frac {d} {d+1}$ and \[\frac {1} {\phi(z)} = a_0^{-1}z^{-d}\Big(\sum_{r=0}^p \Phi(\frac {1} {z})^r + \sum_{r \geq p+1} \Phi(\frac {1} {z})^r\Big) \hspace{5mm} (p \geq 0).\] Now suppose $j \geq d$ and $p=j-d$. It is clear that the terms $\sum_{r = d}^{j} \frac {A_r} {z^r}$ in expansion of the lemma appear from the first summation, namely, $a_0^{-1}z^{-d}\sum_{r=0}^{j-d} \Phi(\frac {1} {z})^r$. This summation contributes only finitely many terms in orders higher than $(\frac {1} {z})^{j}$. To estimate the
second summation, note that in the region under consideration
$|\Phi(\frac {1} {z})| \leq \frac {\alpha} {|z|}$ where $\alpha = |\frac {a_1} {a_0}| + \frac {d-1} {d+1}$. Thus, if $|z| \geq \alpha +1$ then $|\sum_{r \geq p+1} \Phi(\frac {1} {z})^r| \leq \frac {\alpha^{p+1}(\alpha +1)} {|z|^{p+1}}$.  Hence for fixed $\epsilon &gt; 0$ one can effectively determine $M_{\phi,j, \epsilon} &gt; 0$  such that \[ \left|\sum_{r \geq j+1}  \frac {A_{r}} {z^r} \right| \leq \epsilon |z|^{-j} \quad (|z| \geq M_{\phi,j,\epsilon}).\]  From this argument,
we conclude that the constant and range appearing in $O_{h}$, $O_{h+1}$, and \eqref{4.5} are effective.       
       
     
\subsection{The example: $P(X) = X^k$}
\label{sec4.3} 
Let $k \geq 2$ and $P(X) = X^k$, $Q(X) =1 \in \mathbb{Q}[X]$. With the notation of Section~\ref{sec3}, $R(X) = P(X) = A(X)= X^k$, $B(X) = 0$ and we are in 
the situtation of Theorem~\ref{thm1}. The objective of this section is to write down the
first few coefficients of the admissible polynomial $f_P(X)$ as a
function of $k$. 

We introduce formal binomial coefficients for convenience. Let $X$ be a formal variable and $r \in \mathbb{Z}_{\geq 0}$. Define $\binom{X} {r}$ to be an element of $\mathbb{Q}[X]$ given by the expression \[\binom{X} {r} = \begin{cases}
            \frac {X(X-1)\cdots(X-r+1)} {r !}, &amp; \text{if}\, r \geq 1;~\\
            1, &amp; \text{if}\,  r =0. 
           \end{cases} \]
           
With the notation of Section~\ref{sec2} suppose that $g(X) = X^k$. For this polynomial 
\begin{equation*}
a_r = \begin{cases}
          1, &amp; \text{if $r=0$;}\\
          0, &amp; \text{if $1 \leq r \leq k$.}
       \end{cases} 
\end{equation*} 
By \eqref{2.8}  
\begin{align}
\label{4.6} 
v_{l}(\textbf{x},X^k) &amp; = y_{l+1}(\textbf{x})\nonumber \\
&amp; =  \binom {k-l-1} {1}x_{l}+\binom {k-l} {2}x_{l-1} + \cdots + \binom {k-1} {l+1}x_0
\end{align}
for all $0 \leq l \leq k-1$. If $l = k-1$ then one observes that \eqref{4.6} holds with formal binomial coefficients.

Substituting \eqref{2.6} into \eqref{2.7} we obtain 
\begin{align}
\label{4.7} 
u_{j}(\textbf{x}) &amp; = \sum_{r=0}^{j}x_{j-r}x_r + \sum_{r=1}^{j}\sum_{s=1}^{r}\binom {k-s} {r-s+1} x_{j-r}x_{s-1} \hspace{5mm} (0 \leq j \leq k-1).       
\end{align}   

Now one can proceed to solve the system of equations \[u_{i}(\textbf{x}) = v_{i}(\textbf{x},X^k)\hspace{5mm} (0 \leq i \leq k-1)\] following the algorithm of Lemma~\ref{lemma4}. The first few solutions are as follows:
\begin{equation}
\label{4.8}  
\begin{aligned}
c_0(k) &amp; = k-1,\\
c_1(k) &amp;= -\frac {(k-1)^2} {2},\\
c_2(k) &amp;= \frac {(k-1)^2(2k-3)} {12}\hspace{5mm} (k\geq 3),\\
c_3(k) &amp;= -\frac {(k-1)^3(k-3)} {24} \hspace{5mm} (k \geq 4),\\
c_4(k) &amp;= \frac {(k-1)^2} {720}(6k^3 -47k^2 + 92k -45) \hspace{5mm} (k \geq 5).
\end{aligned}    
\end{equation}

In general one can use \eqref{4.6}, \eqref{4.7}, and recursion argument of Lemma~\ref{2.1} to conclude that $c_{r}(k)$ is a rational function of $k$ whenever $k \geq r+1$.    

\subsubsection{Asymptotic expansion of polygamma \cite[p.\ 260]{as}}   

Let $k \geq 1$. The polygamma function of order $k$ is a holomorphic function on $\mathbb{C} - \{0, -1, -2, \ldots \}$  defined  by the series \[\psi^{(k)}(z) = (-1)^{k+1} k!\sum_{n = 0}^{\infty} (z+n)^{-k-1}.\] Let $0 &lt; \theta &lt; \frac {\pi} {2}$ and $\mathcal{S}_{\theta} = \{z \in \mathbb{C}\mid z \neq 0, |\arg z| \leq \pi - \theta \}$. Finite order asymptotic expansion of polygamma function on $\mathcal{S}_{\theta}$ is  
\begin{align*}
\psi^{(k)}(z) = (-1)^{k+1}\Big(\frac {(k-1)!} {z^k} + \frac {k!} {2z^{k+1}} + \sum_{n = 1}^{p} B_{2n} \frac {(2n+k-1)!} {(2n)!z^{2n+k}} + O_{p,\theta}(\frac {1} {z^{k +2p+1}})\Big)
 \end{align*}   
where $p$ is any positive integer. Hence on $\mathcal{S}_{\theta}$    
 \begin{align}
 \label{4.9} 
 &amp; \frac {1} {\sum_{n = 0}^{\infty} (z+n)^{-k-1}} \nonumber\\
 &amp; = kz^k \Big(1 + \frac {k} {2z} + k\sum_{n = 1}^{p} \frac {B_{2n}} {2n} \binom {k+2n-1} {2n-1}z^{-2n} + O_{p,\theta}(\frac {1} {z^{2p+1}})\Big)^{-1}. 
 \end{align} 
 As $z \to \infty$ on the sector one can expand the right-hand side to get asymptotic expansion up to arbitrary high order by choosing a large $p$. One can put $p = \lceil \frac {k} {2} \rceil$ to retrieve the first $k$ terms in the asymptotic expansion of $\frac {1} {\sum_{n = 0}^{\infty} (z+n)^{-k-1}}$. These formulas give an alternate gateway to results of this subsection as mentioned in introduction.           
    
\subsubsection{Numerical formula}   
\label{sec4.3.1} 
Let $k \geq 2$. Consider the sequence defined by $x_{m} = m^{-k}$, $m \geq 1$. We explicitly determine the reciprocal sequences associated with $(x_m)_{m\geq 1}$ for small values of $k$. Set up is same as beginning of this section, i.e., $R(X) = P(X) = A(X) = X^k$, $Q(X) = 1$ and $B(X) = 0$.  Notation and arguments from the proof of Theorem~\ref{thm1} and  Section~\ref{sec4.1} are frequently used in 
the discussion below. Here $M_0 = 0$, which is compatible with theory in Section~\ref{sec4.1}.    

\subsubsection{$k = 2$} (Xin \cite{xin}) Using \eqref{4.8} we have $(c_0, c_1) = (1, -\frac {1} {2})$. Hence $L = 1$ and one needs to use Case I. It is clear that $c(1) = -1$. From explicit expression of $f_P(-1,X)$ and numerator polynomials it follows that $2$ is a legitimate choice for $N$. Hence
\begin{align*}
a_n = n-1 \hspace{5mm} (n \geq 2).
\end{align*}

\subsubsection{$k = 3$} (Xin \cite{xin})  By \eqref{4.8} one concludes $(c_0, c_1, c_2) = (2, -2, 1)$.  Here $L = 1$ and we are in the situation
of Case II. Therefore, the constant term has to be $c_{k-1}$ or $c_{k-1}-1$. Note that $N(f_P(c_{k-1}),P,X)$ is a degree $0$ polynomial with negative leading coefficient. Hence $c(1) = c_{k-1}-1 = 0$. It is easy to see that $N = 2$ is a legitimate choice. Thus 
\begin{align*}
a_n = 2n(n-1)\hspace{5mm} (n \geq 2).
\end{align*} 

\subsubsection{$k = 4$} (Xu \cite{xu}) From \eqref{4.8} it follows that $(c_0, c_1, c_2, c_3) = (3, -\frac {9} {2}, \frac {15} {4}, -\frac {9} {8})$. Here  $L = 4$ and Case I holds. All residue classes are nontrivial and $c(1) = -\frac {5} {4}$, $c(2) = -\frac {3} {2}$, $c(3) = -\frac {7} {4}$, $c(4) = -2$. The choice of constants $M_{f_P(c)} = 1$, $M_1 = M_2 = 1$ works for $c = c(2), c(3), c(4)$. If the constant term is $c(1)$, then a legitimate choice is $M_{f_P(c)} = M_2 = 0$, $M_1 = 5$. Therefore $N(2), N(3), N(4) = 1$ and $N(1) = 5$ is a legitimate choice of constants. Hence $N=5$ and for all $n \geq 5$
\begin{align*}
 a_n = 
\begin{cases}
3n^3 - \frac {9} {2}n^2 + \frac {15} {4}n  - \frac {5} {4}, &amp; \text{if $ n \equiv 1$ (mod $4$)};~\\
3n^3 - \frac {9} {2}n^2 + \frac {15} {4}n - \frac {3} {2}, &amp; \text{if $n \equiv 2$ (mod $4$)};~\\
3n^3 - \frac {9} {2}n^2 + \frac {15} {4}n - \frac {7} {4}, &amp; \text{if $n \equiv 3$ (mod $4$)};~\\
3n^3 - \frac {9} {2}n^2 + \frac {15} {4}n -2, &amp; \text{if $n\equiv 4$ (mod $4$)}.
\end{cases}
\end{align*}

\subsection{$k = 5$} (Xu \cite{xu})  Using \eqref{4.8} we see that $(c_0, c_1, c_2, c_3, c_4) = (4, -8 , \frac {28} {3}, -\frac {16} {3}, -\frac {2} {9})$. Hence $L = 3$ and Case I holds. Nontrivial residue classes modulo $L$ with respect to $H_{L}(X)$  are $\{2,3\}$ and $c(2) = - \frac {2} {3}, c(3) = -1$. The choice of constants $M_{f_P(c)} =2$, $M_1 = 1$, $M_2 = 5$ works for $c \in \{c(2), c(3)\}$. Therefore $N = 5$ and for all $ n \geq 5$ 
\begin{align*}
a_n =
\begin{cases}
4n^4 - 8n^3 + \frac {28} {3}n^2 - \frac {16} {3}n - 1, &amp; \text{if $n \equiv 1$ (mod $3$)};~\\
4n^4 - 8n^3 + \frac {28} {3}n^2 - \frac {16} {3}n - \frac {2} {3}, &amp; \text{if $n \equiv 2$ (mod $3$)};~\\
4n^4 - 8n^3 + \frac {28} {3}n^2 - \frac {16} {3}n - 1, &amp; \text{if $n \equiv 3$ (mod $3$)}. 
\end{cases}
\end{align*}
 
\begin{remark}
\label{remark18} 
The results for $k = 4, 5$  answer questions of Kotesovec \cite{oeis}. Theorem~\ref{thm15} provides an algorithmically computable answer to the problem of determining reciprocal sequence for a large class.   
\end{remark} 
\section{Complements: Sequence of polynomials}
\label{sec5.1} 
The discussion in Section~\ref{sec4.3} leads to new kind of formalism. Let $\mathbb{K}$ be a field of characteristic $0$ and $(P_{k}(X))_{k \geq 1}$ be  sequence of polynomials in $\mathbb{K}[X]$ such that $d_k = \text{deg}_{\mathbb{K}} P_k \geq 2$ for all $k \geq 1$ and $d_k \to \infty$ as $k \to \infty$. For $r \in \mathbb{Z}_{\geq 0}$ define \[m_r = \min \{k_0 \in \mathbb{N} \mid k \geq k_0 \implies d_k \geq r+1\}.\] Write $\mathcal{I}_r = \{k \in \mathbb{N} \mid k \geq m_r\}$. By Lemma~\ref{lemma4} one has a well defined function $c_r: \mathcal{I}_r  \to \mathbb{K}$  given by $c_r(k) = c_r(P_k)$. Since the system equations \[u_i(\textbf{X})= v_i(\textbf{X}, P_{k}) \hspace{5mm} \textbf{X} = (X_0,\ldots,X_r), (0 \leq i \leq r \leq d_k-1) \] determines $c_r(k)$, it depends only on coefficients of $X^{d_k}, \ldots, X^{d_k-r}$ in $P_k(X)$. These coefficients are denoted by $a_0(k), \ldots, a_r(k)$ respectively.    

We can study behavior of this function for different sequences. The subsequent remark summarizes some examples of interest.
\begin{example}
\label{example19}
\ \leavevmode
\begin{itemize}
\item[(i)]  Let $P_{k}(X) = X^{k+1} \in \mathbb{Q}[X]$. Note that 
\[a_{r}(k) = \begin{cases}
                  1, &amp; r = 0;\\
                  0,  &amp; r \geq 1.
              \end{cases} \] One uses recursion argument of Lemma~\ref{lemma4} to conclude that there is a rational function $C_r(X) \in \mathbb{Q}(X)$ so that $c_r(k) = C_r(X)$ for all $k \in \mathcal{I}_r$ (cf. Section~\ref{sec4.3}). For $r \geq 1$ these rational functions are divisible by $X^2$. Moreover, the
asymptotic expansion in \eqref{4.9} implies that these rational functions are actually \textbf{polynomials}.  

\item[(ii)] Let $a\in \mathbb{K}$, and $P(X)\in \mathbb{K}[X]$ be a non-constant polynomial. Suppose that $P_{k}(X) = P(X)(X+a)^k \in \mathbb{K}[X]$. For this sequence $a_0(k) = a_0(P)$. Suppose that $k \geq r$. Then  $a_r(k)$ is $\mathbb{K}$-linear combination of elements of the form \[\Big\{\binom {k} {j}a^j \mid 0 \leq j \leq r\Big\}\] with coefficients independent of $k$. By recursion argument there exists a rational function $C_r(X) \in \mathbb{K}(X)$ such that $c_r(k) = C_r(k)$ for $k \gg 1$.    

\item[(iii)] Let $P(X), Q(X) \in \mathbb{K}[X]$ be so that $\text{deg}_{\mathbb{K}} P, \text{deg}_{\mathbb{K}} Q \geq 1$. Construct a sequence by 
$P_{k}(X) = P(X)Q(X)^k$.  Assume leading coefficient of $P(X)$ is $a_0(P)$ and \[Q(X) = X^d + A_1X^{d-1} + \cdots + A_d \in \mathbb{K}[X]\] with $d \geq 1$. Therefore $a_0(k) = a_0(P)$. Using multinomial theorem we deduce that if $k \geq dr$ then $a_r(k)$ is a $\mathbb{K}$-linear combination of elements \[\Big\{ \binom{k} {\textbf{j}} \textbf{A}^{\textbf{j}} \mid \textbf{j}= (j_0,\ldots,j_d) \in \mathbb{Z}^{d+1}_{\geq 0}, \; \sum_{s=0}^{d} j_{s} = k,\; 0 \leq \sum_{s = 0}^{d} sj_s \leq r\Big\}\]
whose coefficients are independent of $k$. Observe that constraints on $\textbf{j}$ imply that $j_0 \geq k - rd$ and $j_s \leq r$ for all $1 \leq s \leq r$. For $k \geq dr$ all possible choices for $\textbf{j}$ appear and these depend only on $d,r$. Then each of these multinomial terms is a polynomial in $k$ of degree $\leq dr$. Using recursion, we can construct $C_{r}(X) \in \mathbb{K}(X)$ such that $c_r(k) = C_r(X)$ for $k \gg 1$.     
\end{itemize}
\end{example}  

In special cases one can use recursion to find finer properties of these functions which indicate possibility of richer structure.   

\section{Acknowledgments}
\label{sec6}
I wish to thank Prof. P. Rath (CMI) for useful comments on an earlier draft. Part of this work was developed during my stay in Chennai Mathematical Institute.   

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\bibitem{as}
M. Abramowitz and I. A. Stegun, eds.,
\textit{Handbook of Mathematical Functions With Formulas, Graphs and Mathematical Tables}, National Bureau of Standards, tenth printing, 1972.

\bibitem{ak}
P. Anantakitpaisal and K. Kuhapatanakul,
Reciprocal sums of the Tribonacci numbers,
\textit{J. Integer Sequences} \textbf{19} (2016), 
\href{https://cs.uwaterloo.ca/journals/JIS/VOL19/Kuhapatanakul/kuha7.html}{Article 16.2.1}.

\bibitem{kl}
T. Komatsu and V. Laohakosol, On the sum of reciprocals of numbers
satisfying a recurrence relation of order $s$, \textit{J. Integer Sequences}
\textbf{13} (2010), 
\href{https://cs.uwaterloo.ca/journals/JIS/VOL13/Komatsu/komatsu6.html}{Article 10.5.8}.

\bibitem{on}
H. Ohtsuka and S. Nakamura,
On the sum of reciprocal Fibonacci numbers,
\textit{Fibonacci Quart.} \textbf{46/47} (2008/2009) 153--159.

\bibitem{oeis}
N. J. A. Sloane et al.,
The On-Line Encyclopedia of Integer Sequences, published electronically at  
\url{https://oeis.org},
2020. 

\bibitem{xin}
L. Xin,
Some identities related to Riemann zeta-function,
\textit{J. Inequal. Appl.} (2016), Paper no.~32.

\bibitem{xu}
H. Xu,
Some computational formulas related Riemann zeta-function tails,
\textit{J. Inequal. Appl.} (2016), Paper no.~132.

\end{thebibliography}      

\bigskip\hrule\bigskip

\noindent 2010 {\it Mathematics Subject Classification:} Primary 11B83; Secondary 39A05, 40A25.   

\noindent\textit{Keywords:} reciprocal sum, difference primitive.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A248230} and
\seqnum{A248234}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received June 11 2017;
revised versions received  June 11 2020; August 13 2020;
November 28 2020; November 29 2020.
Published in {\it Journal of Integer Sequences}, December 2 2020.

\bigskip
\hrule
\bigskip

\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
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