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\vskip 1cm{\LARGE\bf 
Series of Error Terms for Rational \\
\vskip .1in
Approximations of Irrational Numbers
}
\vskip 1cm
\large
Carsten Elsner\\
Fachhochschule f{\"u}r die Wirtschaft Hannover\\
Freundallee 15\\
D-30173 Hannover\\
Germany\\
\href{mailto:Carsten.Elsner@fhdw.de}{\tt Carsten.Elsner@fhdw.de} \\
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\vskip .2 in
\begin{abstract}
Let \(p_n/q_n \) be the \(n\)-th convergent of a real irrational
number  \(\alpha \), and let \(\varepsilon_n = \alpha q_n-p_n \). In
this paper we investigate various sums of the type \(\sum_{m}
\varepsilon_m \), \(\sum_{m} |\varepsilon_m| \), and \(\sum_{m}
\varepsilon_m x^m \). The main subject of the paper is bounds for
these sums. In particular, we investigate the behaviour of such sums
when \(\alpha \) is a quadratic surd.  The most significant properties of
the error sums depend essentially on Fibonacci numbers or on related
numbers.
\end{abstract}


\theoremstyle{plain}
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\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}

\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{conjecture}[theorem]{Conjecture}

\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}



\section{Statement of results for arbitrary irrationals} \label{S3}

Given a real irrational number \(\alpha \) and its regular continued fraction expansion
\[\alpha \,=\, \langle \,a_0;a_1,a_2,\ldots \,\rangle \qquad (a_0 \in {\mathbb Z} \,,\, a_{\nu} \in {\mathbb N} \,\,\mbox{for} \,\,\nu \geq 1) \,,\]
the convergents \(p_n/q_n \) of \(\alpha \) form a sequence of best approximating rationals in the 
following sense: 
for any rational \(p/q \) satisfying \(1\leq q &lt; q_n \) we have
\[\left| \,\alpha - \frac{p_n}{q_n} \,\right| \,&lt;\, 
\left| \,\alpha - \frac{p}{q} \,\right| \,.\]
The convergents \(p_n/q_n \) of \(\alpha \) are defined by finite continued fractions
\[\frac{p_n}{q_n} \,=\, \langle \,a_0;a_1,\ldots ,a_n\,\rangle \,.\]
The integers \(p_n \) and \(q_n \) can be computed recursively using the initial values \(p_{-1} =1 \), \(p_0 = a_0 \), \(q_{-1} = 0 \), 
\(q_0 = 1 \), and the recurrence formulae 
\begin{equation}
p_n \,=\, a_np_{n-1} + p_{n-2} \,,\qquad q_n \,=\, a_nq_{n-1} + q_{n-2} 
\label{4}
\end{equation}
with \(n\geq 1 \). Then \(p_n/q_n \) is a rational number in lowest terms satisfying the inequalities
\begin{equation} 
\frac{1}{q_n + q_{n+1}} \,&lt;\, |q_n \alpha - p_n| \,&lt;\, \frac{1}{q_{n+1}} \qquad (n \geq 0) \,.
\label{5}  
\end{equation}
The error terms \(q_n \alpha - p_n \) alternate, i.e., \(\mbox{sgn\,} (q_n \alpha - p_n) = {(-1)}^n \). 
For basic facts on continued fractions and convergents see \cite{Hardy, Khintchine, Perron}. \\
Throughout this paper let 
\[\rho \,=\, \frac{1+\sqrt{5}}{2} \quad \mbox{and} \quad  \overline{\rho} \,=\, -\frac{1}{\rho} \,=\, \frac{1-\sqrt{5}}{2} \,.\]
The Fibonacci numbers \(F_n \) are defined recursively by \(F_{-1} =1 \), \(F_0 = 0 \), and \(F_n = F_{n-1} + F_{n-2} \) for \(n\geq 1 \). In this
paper we shall often apply {\em Binet's formula\/}, 
\begin{equation}
F_n \,=\, \frac{1}{\sqrt{5}} \,\Big( \rho^n - {\Big( -\frac{1}{\rho} \Big)}^n \Big) \qquad (n\geq 0) \,.
\label{720}
\end{equation}   
While preparing a talk on the subject of so-called {\em leaping convergents\/} relying on the papers \cite{E1, K3, K2}, the author applied results 
for convergents to the number \(\alpha = e = \exp (1) \). He found two identities which are based on formulas given by Cohn \cite{Cohn}:
\[\sum_{n=0}^{\infty} (q_n e - p_n) \,=\, 2\int_0^1 \exp (t^2)\,dt -2e + 3\,=\, 0.4887398 \ldots \,,\]
\[\sum_{n=0}^{\infty} |q_n e - p_n| \,=\, 2e\int_0^1 \exp (-t^2) \,dt - e \,=\, 1.3418751\ldots \,.\]
These identities are the starting points of more generalized questions concerning error series of real numbers \(\alpha \). 
\begin{itemize} 
\item[1.)] What is the maximum size \(M\) of the series \(\sum_{m=0}^{\infty} |q_m \alpha - p_m| \) ? One easily concludes that \(M \geq 
(1+\sqrt{5})/2 \), because \(\sum_{m=0}^{\infty} \big| q_m(1+\sqrt{5})/2 -p_m\big| = (1+\sqrt{5})/2 \). 
\item[2.)] Is there a method to compute \(\sum_{m=0}^{\infty} |q_m \alpha - p_m| \) explicitly for arbitrary real quadratic irrationals~? 
\end{itemize}
The series \(\sum_{m=0}^{\infty} |q_m \alpha - p_m| \in [0,M] \) measures the approximation properties of \(\alpha \) on average. The smaller
this series is, the better rational approximations \(\alpha \) has. Nevertheless, \(\alpha \) can be a Liouville number and \(\sum_{m=0}^{\infty} 
|q_m \alpha - p_m| \) takes a value close to \(M\). For example, let us consider the numbers
\[\alpha_n \,=\, \langle \,1;\underbrace{1,\ldots ,1}_{n},a_{n+1},a_{n+2}, \ldots \,\rangle \]
for even positive integers \(n\), where the elements \(a_{n+1}, a_{n+2},\dots \) are defined recursively in the following way. Let \(p_k/q_k = 
\langle 1;\underbrace{1,\ldots ,1\, }_k \rangle \)
for \(k=0,1,\dots ,n \) and set 
\[\begin{array}{cclcclcl}
a_{n+1} &amp;:=&amp; q_n^n \,,\quad &amp; q_{n+1} &amp;=&amp; a_{n+1}q_n + q_{n-1} &amp;=&amp; q_n^n \big( q_n + q_{n-1} \big) \,,\\
a_{n+2} &amp;:=&amp; q_{n+1}^{n+1} \,,\quad &amp; q_{n+2} &amp;=&amp; a_{n+2}q_{n+1} + q_n &amp;=&amp; q_{n+1}^{n+1}\big( q_n^{n+1} + q_{n-1} \big) \,,\\
a_{n+3} &amp;:=&amp; q_{n+2}^{n+2} \,,\quad &amp; q_{n+3} &amp;=&amp; a_{n+3}q_{n+2} + q_{n+1} &amp;=&amp; \dots
\end{array} \]
and so on. In the general case we define \(a_{k+1} \) by \(a_{k+1} = q_k^k \) for \(k=n,n+1,\dots \). Then we have with (\ref{4}) and (\ref{5}) that
\[0 \,&lt;\, \Big| \,\alpha_n - \frac{p_k}{q_k}\,\Big| \,&lt;\, \frac{1}{q_kq_{k+1}} \,&lt;\, \frac{1}{a_{k+1}q_k^2} \,=\, \frac{1}{q_k^{k+2}} \quad (k\geq n) \,.\]
Hence \(\alpha_n \) is a Liouville number. Now it follows from (\ref{H60}) in Theorem~\ref{HS2} below with \(2k=n \) and \(n_0 =(n/2)-1 \) that 
\[\sum_{m=0}^{\infty} |q_m \alpha_n - p_m| \,&gt;\, \sum_{m=0}^{n-1} |q_m \alpha_n - p_m| \,=\, (F_{n-1} -1)(\rho - \alpha_n) + \rho - \rho^{1-n} 
\,\geq \, \rho - \frac{1}{\rho^{n-1}} \,.\]
We shall show by Theorem~\ref{HS2} that \(M=\rho \), such that the error sums of the Liouville numbers \(\alpha_n \) tend to this maximum value \(\rho \)
for increasing \(n\). \\ 
We first treat infinite sums of the form \(\sum_n |q_n \alpha - p_n| \) for arbitrary real irrational numbers \(\alpha = \langle 1;a_1,a_2,\ldots 
\rangle \), when we may assume without loss of generality that \(1&lt; \alpha &lt;2 \). 
\begin{proposition}
Let \(\alpha = \langle 1;a_1,a_2,\ldots \rangle \) be a real irrational number. Then for every integer \(m\geq 0 \), the
following two inequalities hold: Firstly,
\begin{equation}
|q_{2m} \alpha - p_{2m}| + |q_{2m+1} \alpha - p_{2m+1}| \,&lt;\, \frac{1}{\rho^{2m}} \,,
\label{H10}
\end{equation}
provided that either
\begin{equation}
a_{2m}a_{2m+1} \,&gt;\, 1 \qquad \mbox{or} \qquad \big( a_{2m} \,=\, a_{2m+1} \,=\, 1 \,\quad \mbox{and} \quad \,a_1a_2 \cdots a_{2m-1} &gt;1 \big) \,.
\label{H20}
\end{equation}
Secondly,
\begin{equation}
|q_{2m} \alpha - p_{2m}| + |q_{2m+1} \alpha - p_{2m+1}| \,=\, \frac{1}{\rho^{2m}} + F_{2m} (\rho - \alpha) \qquad (0\leq m\leq k) \,,
\label{H30}
\end{equation} 
provided that
\begin{equation}
a_1 \,=\, a_2 \,= \ldots =\, a_{2k+1} \,=\, 1 \,.
\label{H40}
\end{equation}
\label{HS1} 
\end{proposition}
In the second term on the right-hand side of (\ref{H30}), \(\rho - \alpha \) takes positive or negative values according to the
parity of the smallest subscript \(r \geq 1 \) with \(a_r &gt;1 \): For odd \(r\) we have \(\rho &gt; \alpha \), otherwise, \(\rho &lt; \alpha \). \\ 
Next, we introduce a set \(\mathcal{M} \) of irrational numbers, namely
\[\mathcal{M} \,:=\, \left\{ \,\alpha \in {\mathbb R} \setminus {\mathbb Q} \,\,\big| \,\, \exists \,k \in {\mathbb N} \,:\, \alpha = \langle 
1;1,\ldots ,1,a_{2k+1},a_{2k+2}, \ldots \rangle \,\wedge \, a_{2k+1} &gt;1 \right\} \,.\]
Note that \(\rho &gt; \alpha \) for \(\alpha \in \mathcal{M} \). Our main result for real irrational numbers is given by the subsequent theorem.
\begin{theorem}
Let \(1 &lt; \alpha &lt; 2 \) be a real irrational number and let \(g,n\geq 0 \) be integers with \(n\geq 2g \). Set \(n_0 := \lfloor n/2 \rfloor \). 
Then the following inequalities hold. \\
{\bf 1.)} \,For \(\alpha \not\in \mathcal{M} \) we have
\begin{equation}
\sum_{\nu =2g}^n |q_{\nu} \alpha - p_{\nu}| \,\leq \, \rho^{1-2g} - \rho^{-2n_0-1} \,,
\label{H50}
\end{equation}
with equality for \(\alpha = \rho \) and every odd \(n\geq 0 \). \\
{\bf 2.)} \,For \(\alpha \in \mathcal{M} \), say \(\alpha = \langle 1;1,\ldots ,1,a_{2k+1},a_{2k+2}, \ldots \rangle \) with \(a_{2k+1} &gt; 1 \), we 
have
\begin{equation}
\sum_{\nu =2g}^n |q_{\nu} \alpha - p_{\nu}| \,\leq \, (F_{2k-1} - F_{2g-1})(\rho - \alpha) + \rho^{1-2g} - \rho^{-2n_0-1} \,,
\label{H60}
\end{equation} 
with equality for \(n=2k-1 \). \\ 
{\bf 3.)} \,We have
\begin{equation}
\sum_{\nu =2g}^{\infty} |q_{\nu} \alpha - p_{\nu}| \,\leq \, {\rho}^{1-2g} \,,
\label{H70}
\end{equation}
with equality for \(\alpha = \rho \). 
\label{HS2}
\end{theorem} 
In particular, for any positive \(\varepsilon \) and any even integer \(n\) satisfying 
\[n \,\geq \,\frac{\log (\rho /\varepsilon)}{\log \rho} \,,\]
it follows that
\[\sum_{\nu =n}^{\infty} |q_{\nu} \alpha - p_{\nu}| \,\leq \, \varepsilon \,.\]
For \(\nu \geq 1 \) we know by \(q_2 \geq 2 \) and by (\ref{5}) that \(|q_{\nu}\alpha -p_{\nu}| &lt; 1/q_{\nu+1} \leq 1/q_2 \leq 1/2 \), which implies 
\(|q_{\nu}\alpha -p_{\nu}| = \| q_{\nu}\alpha \| \), where \(\| \beta \| \) denotes the distance of a real number \(\beta \) to the nearest integer. 
For \(\alpha = \langle a_0;a_1,a_2,\ldots \rangle \), \(|q_0\alpha - p_0| = \alpha -a_0 = \{ \alpha \} \) is the 
fractional part of \(\alpha \). Therefore, we conclude from Theorem~\ref{HS2} that
\[\sum_{\nu =1}^{\infty} \| q_{\nu}\alpha \| \,\leq \, \rho - \{ \alpha \} \,.\]
In particular, we have for \(\alpha = \rho \) that
\[\sum_{\nu =1}^{\infty} \| q_{\nu}\rho \| \,=\, 1 \,.\]
The following theorem gives a simple bound for \(\sum_m (q_m \alpha - p_m) \).
\begin{theorem}
Let \(\alpha \) be a real irrational number. Then the series \(\sum_{m=0}^{\infty} (q_m \alpha - p_m)x^m \) converges
absolutely at least for \(|x| &lt; \rho \), and
\[0 \,&lt;\, \sum_{m=0}^{\infty} (q_m \alpha - p_m) \,&lt;\, 1 \,.\]
Both the upper bound 1 and the lower bound 0 are best possible.  
\label{Prop1}  
\end{theorem}
The proof of this theorem is given in Section~\ref{S5a}. 
We shall prove Proposition \ref{HS1} and Theorem~\ref{HS2} in Section \ref{S7}, using essentially the properties of Fibonacci numbers.

\section{Statement of Results for Quadratic Irrationals} \label{S2} 
In this section we state some results for error sums involving real quadratic irrational numbers \(\alpha \). Any quadratic irrational \(\alpha \)
has a periodic continued fraction expansion,
\[\alpha \,=\,\langle\,a_0;a_1,\ldots ,a_{\omega},T_1,\ldots ,T_r,T_1,\ldots ,T_r,\ldots \,\rangle \,=\, \langle\,a_0;a_1,\ldots ,a_{\omega},
\overline{T_1,\ldots ,T_r}\,\rangle \,, \]
say. Then there is a linear three-term recurrence formula for \(z_n = p_{rn+s} \) 
and \(z_n = q_{rn+s} \) \((s=0,1,\ldots ,r-1) \), \cite[Corollary\,1]{E2}. This recurrence formula has the form
\[z_{n+2} \,=\, Gz_{n+1}  \pm z_n \qquad (rn &gt; \omega) \,.\]
Here, \(G\) denotes a positive integer, which depends on \(\alpha \) and \(r\), but not on \(n\) and \(s\). The number
\(G\) can be computed explicitly from the numbers \(T_1,\ldots ,T_r \) of the continued fraction expansion of \(\alpha \). 
This is the basic idea on which the following theorem relies. 
\begin{theorem} 
Let \(\alpha \) be a real quadratic irrational number. Then
\[\sum_{m=0}^{\infty} (q_m \alpha - p_m)x^m \,\in\, {\mathbb Q}[\alpha](x) \,.\]
\label{Thm1}
\end{theorem}
It is not necessary to explain further technical details of the proof. Thus, the generating function of the sequence \({(q_m \alpha - p_m)}_{m\geq 0} \) 
is a rational function with coefficients from \({\mathbb Q}[\alpha] \). \\

\begin{example}
Let \(\alpha = \sqrt{7} =\langle 2;\overline{1,1,1,4} \rangle \). Then
\begin{equation}
\sum_{m=0}^{\infty} (q_m\sqrt{7} -p_m)x^m \,=\, \frac{x^3 - (2+\sqrt{7})x^2 + (3+\sqrt{7})x - 
(5+2\sqrt{7})}{x^4 - (8 + 3\sqrt{7})} \,. 
\label{30}
\end{equation}
In particular, for \(x=1 \) and \(x=-1 \) we obtain
\begin{eqnarray*} 
\sum_{m=0}^{\infty} (q_m\sqrt{7} -p_m) &amp;\,=\,&amp; \frac{21 - 5\sqrt{7}}{14} \,=\, 0.555088817 \ldots \,, \\ \\
\sum_{m=0}^{\infty} |q_m\sqrt{7} -p_m| &amp;\,=\,&amp; \frac{7 + 5\sqrt{7}}{14} \,=\, 1.444911182 \ldots \,.
\end{eqnarray*}
\end{example}

Next, we consider the particular quadratic surds
\[\alpha \,=\, \frac{n+\sqrt{4+n^2}}{2} \,=\, \langle \,n;n,n,n,\ldots \,\rangle \]
and compute the generating function of the error terms \(q_m\alpha -p_m \).
\begin{corollary}
Let \(n \geq 1 \) and \(\alpha = (n+\sqrt{4+n^2})/2 \). Then
\[\sum_{m=0}^{\infty} (q_m\alpha -p_m)x^m \,=\, \frac{1}{x+\alpha} \,,\]
particularly
\[\sum_{m=0}^{\infty} (q_m\alpha -p_m) \,=\, \frac{1}{\alpha +1} \,,\quad \sum_{m=0}^{\infty} |q_m\alpha -p_m| \,=\, 
\frac{1}{\alpha -1} \,,\quad \sum_{m=0}^{\infty} \frac{q_m\alpha -p_m}{m+1} \,=\,
\log \Big( 1 + \frac{1}{\alpha} \Big) \,.\] 
\label{Thm2}
\end{corollary}
For the number \(\rho = (1+\sqrt{5})/2 \) we have \(p_m = F_{m+2} \) and \(q_m = F_{m+1} \). Hence,
using \(1/(\rho +1) = (3-\sqrt{5})/2 = 1+\overline{\rho} \), \(1/(\rho - 1) = \rho \),  and \(1 + 1/\rho = \rho \), we get
from Corollary~\ref{Thm2}  
\begin{equation}
\sum_{m=0}^{\infty} (F_{m+1} \rho -F_{m+2}) \,=\, 1+\overline{\rho} \,,\quad \sum_{m=0}^{\infty} |F_{m+1} \rho -F_{m+2}| \,=\, 
\rho \,,\quad \sum_{m=0}^{\infty} \frac{F_{m+1} \rho -F_{m+2}}{m+1} \,=\, \log \rho \,.
\label{40}
\end{equation}
Similarly, we obtain for the number \(\alpha = \sqrt{7} \) from (\ref{30}):
\[\sum_{m=0}^{\infty} \frac{q_m\sqrt{7} -p_m}{m+1} \,=\, \int_0^1 \frac{x^3 - (2+\sqrt{7})x^2 + (3+\sqrt{7})x - 
(5+2\sqrt{7})}{x^4 - (8 + 3\sqrt{7})}\,dx \,=\, 0.5568649708 \ldots \]

\section{Proof of Theorem \ref{Prop1}} \label{S5a} 
Throughout this paper we shall use the abbreviations \(\varepsilon_m (\alpha) = \varepsilon_m :=  q_m \alpha - p_m \) and 
\(\varepsilon(\alpha)  = \sum_{m=0}^{\infty} |\varepsilon_m(\alpha)| \).  
The sequence \({(|\varepsilon_m|)}_{m\geq 0} \) converges strictly decreasing to zero. Since \(\varepsilon_0 &gt;0 \) and 
\(\varepsilon_m \varepsilon_{m+1} &lt;0 \), we have
\[\varepsilon_0 + \varepsilon_1 \,&lt;\, \sum_{m=0}^{\infty} \varepsilon_m \,&lt;\, \varepsilon_0 \,.\]
Put \(a_0 = \lfloor \alpha \rfloor \), \(\theta := \varepsilon_0 = \alpha - a_0 \), so that \(0&lt;\theta &lt; 1 \). Moreover, 
\[\varepsilon_0 + \varepsilon_1 \,=\, \theta + a_1\alpha - (a_0a_1 +1) \,=\, \theta + 
a_1\theta -1 \,=\, \theta + \Big\lfloor \,\frac{1}{\theta}\,\Big\rfloor \,\theta -1 \,.\]
Choosing an integer \(k\geq 1 \) satisfying
\[\frac{1}{k+1} \,&lt;\, \theta \,&lt;\, \frac{1}{k} \,,\]
we get
\[\theta + \Big\lfloor \,\frac{1}{\theta}\,\Big\rfloor \,\theta -1 \,&gt;\, \frac{1}{k+1} + \frac{k}{k+1} - 1 \,=\, 0 \,,\]
which proves the lower bound for \(\sum \varepsilon_m \). \\
In order to estimate the radius of convergence for the series \(\sum \varepsilon_m x^m \) we first prove the inequality
\begin{equation}
q_m \,\geq \, F_{m+1} \qquad (m\geq 0) \,,
\label{50}
\end{equation}
which follows inductively. We have \(q_0 = 1 = F_1 \), \(q_1 = a_1 \geq 1 = F_2 \), and
\[q_m \,=\, a_mq_{m-1} + q_{m-2} \,\geq \, q_{m-1} + q_{m-2} \,\geq \, F_m + F_{m-1} \,=\, F_{m+1} \qquad (m\geq 2) \,,\]
provided that (\ref{50}) is already proven for \(q_{m-1} \) and \(q_{m-2} \). With Binet's formula (\ref{720}) and (\ref{50}) we 
conclude that
\begin{equation}
q_{m+1} \,\geq \, \frac{1}{\sqrt{5}} \Big( \rho^{m+2} - {\Big( -\frac{1}{\rho} \Big)}^{m+2} \Big) \,\geq \, \frac{1}{\sqrt{5}}
\rho^m \qquad (m\geq 0) \,.
\label{45}
\end{equation}
Hence, we have 
\[|\varepsilon_m| x^m \,=\, |q_m\alpha - p_m| x^m  \,&lt;\, \frac{x^m}{q_{m+1}} \,\leq \, \sqrt{5} {\left( \frac{x}{\rho}
\right)}^m \qquad (m\geq 0) \,.\]
It follows that the series \(\sum \varepsilon_m x^m \) converges absolutely at least for \(|x| &lt; \rho \). 
In order to prove that the upper bound 1 is best possible, we choose \(0&lt;\varepsilon &lt; 1 \) and a positive integer \(n\)
satisfying
\[\frac{1}{n} \,\Big( 1 + \frac{\rho \sqrt{5}}{\rho -1} \Big) \,&lt;\, \varepsilon \,.\]
Put
\[\alpha_n \,:=\, \langle \,0;1,\overline{n}\,\rangle \,=\, \frac{1}{2} - \frac{1}{n} + \frac{1}{2} \sqrt{1+\frac{4}{n^2}} 
\,&gt;\, 1 - \frac{1}{n} \,.\]
With \(p_0 = 0 \) and \(q_0 =1 \) we have by (\ref{4}), (\ref{5}), and (\ref{45}),
\begin{eqnarray*}
\sum_{m=0}^{\infty} (q_m\alpha_n - p_m) &amp;\,\geq\,&amp; \alpha_n - \sum_{m=1}^{\infty} |q_m\alpha_n - p_m| \\
&amp;\,&gt;\,&amp; 1 - \frac{1}{n} - \sum_{m=1}^{\infty} \frac{1}{q_{m+1}} \,\geq \, 1 - \frac{1}{n} - \sum_{m=1}^{\infty} \frac{1}{nq_m} \\
&amp;\,\geq \,&amp; 1 - \frac{1}{n} - \frac{\sqrt{5}}{n} \sum_{m=1}^{\infty} \frac{1}{\rho^{m-1}} \\ 
&amp;\,=\,&amp; 1 - \frac{1}{n} \,\Big( 1 + \frac{\rho \sqrt{5}}{\rho -1} \Big) \,&gt;\, 1 - \varepsilon \,.
\end{eqnarray*}
For the lower bound 0 we construct quadratic irrational numbers \(\beta_n \,:=\, \langle \,0;\overline{n}\,\rangle \) and 
complete the proof of the theorem by similar arguments. \hfill \qed 

\section{Proofs of Proposition \ref{HS1} and Theorem \ref{HS2}} \label{S7}
\begin{lemma}
Let \(\alpha = \langle a_0;a_1,a_2,\ldots \rangle \) be a real irrational number with convergents \(p_m/q_m \). Let \(n\geq 
1 \) be a subscript satisfying \(a_n &gt;1 \). Then
\begin{equation}
q_{n+k} \,\geq \, F_{n+k+1} + F_{k+1}F_n \qquad (k\geq 0)\,.
\label{700}
\end{equation}
In the case \(n \equiv k+1 \equiv 0 \pmod 2 \) we additionally assume that \(n\geq 4 \), \(k\geq 3 \). Then
\begin{equation}
F_{n+k+1} + F_{k+1}F_n \,&gt;\, \rho^{n+k} \,.
\label{710}
\end{equation}
\label{L1}
\end{lemma}
When \(\alpha - \rho \not\in {\mathbb Z}  \), the inequality (\ref{700}) with \(m=n+k \) is stronger than (\ref{50}). \\
\begin{proof}
We prove (\ref{700}) by induction on \(k\). Using (\ref{4}) and (\ref{50}), we obtain for \(k=0 \) and \(k=1\), respectively,
\[q_n \,=\, a_nq_{n-1} + q_{n-2} \,\geq \, 2F_n + F_{n-1} \,=\, (F_n + F_{n-1}) + F_n \,=\, F_{n+1} + F_1F_n \,,\]
\[q_{n+1} \,=\, a_{n+1}q_n + q_{n-1} \,\geq \, q_n + q_{n-1} \,\geq \, (F_{n+1} + F_n) + F_n \,=\, F_{n+2} + F_2F_n \,.\]
Now, let \(k\geq 0 \) and assume that (\ref{700}) is already proven for \(q_{n+k} \) and \(q_{n+k+1} \). Then
\begin{eqnarray*}
q_{n+k+2} &amp;\,\geq\,&amp;  q_{n+k+1} + q_{n+k} \\
&amp;\,\geq \,&amp; (F_{n+k+2} + F_{k+2}F_n) + (F_{n+k+1} + F_{k+1}F_n) \\
&amp;\,=\,&amp; F_{n+k+3} + F_{k+3}F_n \,.
\end{eqnarray*}
This corresponds to (\ref{700}) with \(k\) replaced by \(k+2 \). In order to prove (\ref{710}) we express the Fibonacci numbers \(F_m \) by
Binet's formula (\ref{720}). Hence, we have
\begin{eqnarray*}
&amp;&amp; F_{n+k+1} + F_{k+1}F_n \\ 
&amp;\,=\,&amp; \rho^{n+k} \left( \rho \,\Big( \frac{1}{5} + \frac{1}{\sqrt{5}} \Big) + {(-1)}^{n+k} \Big( \frac{1}{\sqrt{5}} - \frac{1}{5} \Big)
\,\frac{1}{\rho^{2n+2k+1}} + \frac{1}{5} \,\Big( \frac{{(-1)}^{n+1}}{\rho^{2n-1}} + \frac{{(-1)}^k}{\rho^{2k+1}} \Big) \right) \,.
\end{eqnarray*} 

\noindent{\em Case~1:\/} \,Let \(n \equiv k \equiv 1 \,\pmod 2 \). \\
In particular, we have \(k\geq 1 \). Then
\begin{eqnarray*}
&amp;&amp; F_{n+k+1} + F_{k+1}F_n \\ 
&amp;\,=\,&amp; \rho^{n+k} \left( \rho \,\Big( \frac{1}{5} + \frac{1}{\sqrt{5}} \Big) + \Big( \frac{1}{\sqrt{5}} - \frac{1}{5} \Big)
\,\frac{1}{\rho^{2n+2k+1}} + \frac{1}{5} \,\Big( \frac{1}{\rho^{2n-1}} - \frac{1}{\rho^{2k+1}} \Big) \right) \\
&amp;\,&gt;\,&amp; \rho^{n+k} \left( \rho \,\Big( \frac{1}{5} + \frac{1}{\sqrt{5}} \Big) - \frac{1}{5\rho^3} \right) \,=\, \rho^{n+k} \,.
\end{eqnarray*} 

\noindent{\em Case~2:\/} \, Let \(n \equiv 1 \,\pmod 2 \),  \(k \equiv 0 \,\pmod 2 \). \\
In particular, we have \(n\geq 1 \) and \(k\geq 0 \). First, we assume that \(k\geq 2 \). Then, by similar computations as in Case~1, we obtain
\begin{eqnarray*}
&amp;&amp; F_{n+k+1} + F_{k+1}F_n \\ 
&amp;\,=\,&amp; \rho^{n+k} \left( \rho \,\Big( \frac{1}{5} + \frac{1}{\sqrt{5}} \Big) - \Big( \frac{1}{\sqrt{5}} - \frac{1}{5} \Big)
\,\frac{1}{\rho^{2n+2k+1}} + \frac{1}{5} \,\Big( \frac{1}{\rho^{2n-1}} + \frac{1}{\rho^{2k+1}} \Big) \,\right) \,&gt;\, \rho^{n+k} \,.
\end{eqnarray*}
For \(k=0 \) and some odd \(n\geq 1 \) we get
\[F_{n+k+1} + F_{k+1}F_n \,&gt;\, \rho^n \left( \rho \,\Big( \frac{1}{5} + \frac{1}{\sqrt{5}} \Big) - \Big( \frac{1}{\sqrt{5}} - \frac{1}{5} \Big)
\,\frac{1}{\rho^3} + \frac{1}{5\rho} \right) \,&gt;\, \rho^n \,.\]

\noindent{\em Case~3:\/} \, Let \(n \equiv 0 \,\pmod 2 \),  \(k \equiv 1 \,\pmod 2 \). \\
By the assumption of the lemma, we have \(n\geq 4 \) and \(k\geq 3 \). Then
\begin{eqnarray*}
&amp;&amp; F_{n+k+1} + F_{k+1}F_n \\ 
&amp;\,=\,&amp; \rho^{n+k} \left( \rho \,\Big( \frac{1}{5} + \frac{1}{\sqrt{5}} \Big) - \Big( \frac{1}{\sqrt{5}} - \frac{1}{5} \Big)
\,\frac{1}{\rho^{2n+2k+1}} + \frac{1}{5} \,\Big( -\frac{1}{\rho^{2n-1}} - \frac{1}{\rho^{2k+1}} \Big) \right) \,&gt;\, \rho^{n+k} \,.
\end{eqnarray*}

\noindent {\em Case~4:\/} \, Let \(n \equiv k \equiv 0 \,\pmod 2 \). \\
In particular, we have \(n\geq 2 \). Then
\begin{eqnarray*}
&amp;&amp; F_{n+k+1} + F_{k+1}F_n \\ 
&amp;\,=\,&amp; \rho^{n+k} \left( \rho \,\Big( \frac{1}{5} + \frac{1}{\sqrt{5}} \Big) + \Big( \frac{1}{\sqrt{5}} - \frac{1}{5} \Big)
\,\frac{1}{\rho^{2n+2k+1}} + \frac{1}{5} \,\Big( -\frac{1}{\rho^{2n-1}} + \frac{1}{\rho^{2k+1}} \Big) \right) \,&gt;\, \rho^{n+k} \,.
\end{eqnarray*} 
This completes the proof of Lemma \ref{L1}. 
\end{proof}
\begin{lemma}
Let \(m\) be an integer. Then
\begin{eqnarray}
\frac{\rho^{2m}}{F_{2m+2}} &amp;\,&lt;\,&amp; 1 \qquad (m\geq 1) \,,
\label{740} \\
\rho^{2m} \Big( \frac{1}{F_{2m+3}} + \frac{1}{F_{2m+3} + F_{2m+1}} \Big) &amp;\,&lt;\,&amp; 1 \qquad (m\geq 0) \,.
\label{730} 
\end{eqnarray} 
\label{L2}
\end{lemma}
\begin{proof} For \(m\geq 1 \) we estimate Binet's formula (\ref{720}) for \(F_{2m+2} \) using \(4m+2 \geq 6 \):
\[F_{2m+2} \,=\, \frac{\rho^{2m}}{\sqrt{5}} \,\Big( \rho^2 - \frac{1}{\rho^{4m+2}} \Big) \,\geq \, \frac{\rho^{2m}}{\sqrt{5}} \Big( \rho^2 -
\frac{1}{\rho^6} \Big) \,&gt;\, \rho^{2m} \,.\] 
Similarly, we prove (\ref{730}) by  
\[F_{2n+1} \,=\, \frac{1}{\sqrt{5}} \,\Big( \rho^{2n+1} + \frac{1}{\rho^{2n+1}} \Big) \,&gt;\, \frac{\rho^{2n+1}}{\sqrt{5}} \qquad (n\geq 0) \,.\]
Hence,
\[\rho^{2m} \Big( \frac{1}{F_{2m+3}} + \frac{1}{F_{2m+3} + F_{2m+1}} \Big) \,&lt;\, \rho^{2m} \,\Big( \frac{\sqrt{5}}{\rho^{2m+3}} + 
\frac{\sqrt{5}}{\rho^{2m+3} + \rho^{2m+1}} \Big) \,&lt;\, 1 \,.\]
The lemma is proven. 
\end{proof}  

{\em Proof of Proposition \ref{HS1}:\/} \,Firstly, we assume the hypotheses in (\ref{H20}) and prove (\ref{H10}). 
As in the proof of Theorem~\ref{Prop1}, put \(a_0 = \lfloor \alpha \rfloor \), \(\theta := \alpha - a_0 \), \(a_1 = \lfloor 1/\theta \rfloor \) with 
\(0&lt;\theta &lt; 1 \) and \(\varepsilon_0 = \theta &lt; 1 \). Then
\[|\varepsilon_0| + |\varepsilon_1| \,=\, \theta + (a_0a_1 +1) - a_1\alpha \,=\, \theta + 1 - a_1\theta \,=\, \theta +1 - \Big\lfloor 
\,\frac{1}{\theta}\, \Big\rfloor \,\theta \,.\]
We have \(0&lt;\theta &lt; 1/2 \), since otherwise for \(\theta &gt; 1/2 \), we obtain \(a_1 = \lfloor 1/\theta \rfloor = 1 \). With \(a_0 = a_1 =1 \) the 
conditions in
(\ref{H20}) are unrealizable both. Hence, there is an integer \(k\geq 2 \) with 
\[\frac{1}{k+1} \,&lt;\,\theta \,&lt;\,\frac{1}{k} \,.\]
Obviously, it follows that \([1/\theta] = k \), and therefore
\[\theta + 1 - \Big\lfloor \,\frac{1}{\theta}\,\Big\rfloor \,\theta \,&lt;\, \frac{1}{k} + 1 - \,\frac{k}{k+1} \,=\, \frac{2k+1}{k(k+1)} 
\,\leq \, \frac{5}{6} \qquad (k\geq 2) \,.\]
Altogether, we have proven that
\begin{equation}
|\varepsilon_0| + |\varepsilon_1| \,\leq \, \frac{5}{6} \,&lt;\, 1 \,.
\label{80}
\end{equation} 
Therefore we already know that the inequality (\ref{H10}) holds for \(m=0 \). Thus, we assume \(m\geq 1 \) in the sequel. Noting that 
\(\varepsilon_{2m} &gt;0 \) and \(\varepsilon_{2m+1} &lt; 0 \) hold for every integer \(m \geq 0 \), we may rewrite (\ref{H10}) as follows:
\begin{equation}
\big( 0 \,&lt;\, \big) \quad (p_{2m+1} - p_{2m}) - \alpha (q_{2m+1} - q_{2m}) \,&lt;\, \frac{1}{\rho^{2m}} \qquad (m\geq 0) \,.
\label{750} 
\end{equation} 
We distinguish three cases according to the conditions in (\ref{H20}). \\

\noindent{\em Case~1:\/} \, Let \(a_{2m+1} \geq 2 \). \\ 
Additionally, we apply the trivial inequality \(a_{2m+2} \geq 1 \). Then, using (\ref{5}), (\ref{50}), and (\ref{730}),
\begin{eqnarray*}
|\varepsilon_{2m}| + |\varepsilon_{2m+1}| &amp;\,&lt;\,&amp; \frac{1}{q_{2m+1}} + \frac{1}{q_{2m+2}} \\
&amp;\,\leq \,&amp; \frac{1}{2q_{2m} + q_{2m-1}} + \frac{1}{q_{2m+1} + q_{2m}} \\
&amp;\,\leq \,&amp; \frac{1}{2q_{2m} + q_{2m-1}} + \frac{1}{3q_{2m} + q_{2m-1}} \\
&amp;\,\leq \,&amp; \frac{1}{2F_{2m+1} + F_{2m}} + \frac{1}{3F_{2m+1} + F_{2m}} \\
&amp;\,&lt;\,&amp; \frac{1}{\rho^{2m}} \qquad (m\geq 0) \,.
\end{eqnarray*}

\noindent{\em Case~2:\/} \, Let \(a_{2m+1} =1 \) and \(a_{2m} \geq 2 \). \\
Here, we have \(p_{2m+1} - p_{2m} = p_{2m} + p_{2m-1} - p_{2m} = p_{2m-1} \), and similarly \(q_{2m+1} - q_{2m} = q_{2m-1} \). 
Therefore, by (\ref{750}), it suffices to show that \(0 &lt; p_{2m-1} - \alpha q_{2m-1} &lt; \rho^{-2m} \) for \(m\geq 1 \). This follows with
(\ref{5}), (\ref{50}), and (\ref{740}) from 
\begin{eqnarray*}
0 &amp;\,&lt;\,&amp; p_{2m-1} - \alpha q_{2m-1} \,&lt;\, \frac{1}{q_{2m}} \\
&amp;\,\leq \,&amp; \frac{1}{2q_{2m-1} + q_{2m-2}} \,\leq \, \frac{1}{2F_{2m} + F_{2m-1}} \\
&amp;\,=\,&amp; \frac{1}{F_{2m+2}} \,&lt;\, \frac{1}{\rho^{2m}} \qquad (m\geq 1) \,.
\end{eqnarray*}

\noindent{\em Case~3:\/} \, Let \(a_{2m} = a_{2m+1} =1 \,\wedge \,a_1a_2 \cdots a_{2m-1} &gt; 1 \). \\
Since \(a_{2m+1} =1 \), we again have (as in Case~2):
\begin{equation}
0 \,&lt;\, |\varepsilon_{2m}| + |\varepsilon_{2m+1}| \,=\, p_{2m-1} - \alpha q_{2m-1} \,&lt;\, \frac{1}{q_{2m}} \,.
\label{760}
\end{equation}
By the hypothesis of Case~3, there is an integer \(n\) satisfying \(1\leq n \leq 2m-1 \) and \(a_n \geq 2 \). 
We define an integer \(k\geq 1 \) by setting \(2m = n+k \). Then we obtain using (\ref{700}) and (\ref{710}),
\[q_{2m} \,=\, q_{n+k} \,\geq \, F_{n+k+1} + F_{k+1}F_n \,&gt;\, \rho^{n+k} \,=\, \rho^{2m} \,.\]
 From the identity \(n+k = 2m \) it follows that the particular condition \(n \equiv k+1 \equiv 0 \pmod 2 \) in Lemma \ref{L1} does not occur. Thus,
by (\ref{760}), we conclude that the desired inequality (\ref{H10}). \\
In order to prove (\ref{H30}), we now assume the hypothesis (\ref{H40}), i.e., \(a_1a_2 \cdots a_{2k+1} =1 \) and \(0\leq m \leq k \). From \(2m-1 \leq
2k-1 \) and \(a_0=a_1=\ldots =a_{2k-1} =1 \) it is clear that \(p_{2m-1} = F_{2m+1} \) and \(q_{2m-1} = F_{2m} \).
Since \(a_{2k+1} =1 \) and \(0\leq m\leq k \), we have 
\begin{eqnarray*}
&amp; &amp; |q_{2m}\alpha - p_{2m}| + |q_{2m+1}\alpha - p_{2m+1}| \\
&amp;\,=\,&amp; p_{2m-1} - \alpha q_{2m-1} \,=\, F_{2m+1} - \alpha F_{2m} \,=\, F_{2m+1} - \rho F_{2m} + (\rho - \alpha)F_{2m} \,.
\end{eqnarray*}
 From Binet's formula (\ref{720}) we conclude that 
\[F_{2m+1} - \rho F_{2m} \,=\, \frac{1}{\sqrt{5}} \,\Big( \frac{1}{\rho^{2m+1}} + \frac{1}{\rho^{2m-1}} \Big) \,=\, \rho^{-2m} \,,\] 
which finally proves the desired identity (\ref{H30}) in Proposition \ref{HS1}. \hfill \qed
\begin{lemma}
Let \(k\geq 1 \) be an integer, and let \(\alpha \,:=\, \langle \,1;1,\ldots ,1,a_{2k+1},a_{2k+2} \ldots \,\rangle \) be a real irrational number with 
partial quotients \(a_{2k+1} &gt;1 \) and \(a_{\mu} \geq 1 \) for \(\mu \geq 2k+2 \). Then we have the inequalities
\begin{equation}
(F_{2k-1}-1)(\rho - \alpha) \,&lt;\, \frac{1}{\rho^{2k}} - |\varepsilon_{2k}| - |\varepsilon_{2k+1}|   
\label{770} 
\end{equation}
for \(a_{2k+1} \geq 3 \), and  
\begin{equation}
(F_{2k-1}-1)(\rho - \alpha) \,&lt;\, \frac{1}{\rho^{2k}} + \frac{1}{\rho^{2k+2}} - |\varepsilon_{2k}| - |\varepsilon_{2k+1}| - |\varepsilon_{2k+2}| 
- |\varepsilon_{2k+3}| 
\label{780} 
\end{equation}
for \(a_{2k+1} =2 \).
\label{L4}
\end{lemma}
One may conjecture that (\ref{770}) also holds for \(a_{2k+1} =2 \). \\

\begin{example}
Let \(\alpha = \langle 1;1,1,1,1,2,\overline{1} \rangle = (21\rho +8)/(13\rho +5) = (257 - \sqrt{5})/158 \). With \(k=2 \) and 
\(a_5 =2 \), we have on the one side 
\[\rho - \alpha \,=\, F_2 (\rho - \alpha) \,=\, \frac{40\sqrt{5} -89}{79} \,=\, 0.005604 \ldots \,,\]
on the other side,
\[\frac{1}{\rho^4} - |\varepsilon_4| - |\varepsilon_5| \,=\, \frac{1}{\rho^4} - \frac{4\sqrt{5} -1}{79} \,=\, 0.045337 \ldots \]
\end{example}

\noindent {\em Proof of Lemma \ref{L4}:\/} \\ 

\noindent {\em Case~1:\/} \,Let \(n:=a_{2k+1} \geq 3 \). \\
Then there is a real number \(\eta \) satisfying \(0&lt;\eta &lt;1 \) and 
\[r_{2k+1} \,:=\, \langle a_{2k+1}; a_{2k+2}, \ldots \rangle \,=\, n +\eta =: 1 + \beta \,. \]
It is clear that \(n-1 &lt; \beta &lt; n \). From the theory of regular continued fractions (see \cite[formula\,(16)]{Khintchine}) it follows that
\begin{eqnarray*}
\alpha &amp;\,=\,&amp; \langle \,1;1,\ldots ,1,a_{2k+1},a_{2k+2} \ldots \,\rangle \,=\, \frac{F_{2k+2} r_{2k+1} + F_{2k+1}}{F_{2k+1}r_{2k+1} + F_{2k}} \\
&amp;\,=\,&amp; \frac{F_{2k+2} (1+\beta) + F_{2k+1}}{F_{2k+1}(1+\beta) + F_{2k}} \,=\, \frac{\beta F_{2k+2} + F_{2k+3}}{\beta F_{2k+1} + F_{2k+2}} \,.
\end{eqnarray*}
Similarly, we have
\[\rho \,=\, \frac{F_{2k+2} \rho + F_{2k+1}}{F_{2k+1}\rho + F_{2k}} \,,\]
hence, by some straightforward computations,
\begin{equation}
\rho - \alpha \,=\, \frac{1+\beta -\rho}{(\rho F_{2k+1} +F_{2k})(\beta F_{2k+1} + F_{2k+2})} \,&lt;\, \frac{n}{(\rho F_{2k+1} +F_{2k})(\beta F_{2k+1} 
+ F_{2k+2})} \,.
\label{785}
\end{equation}
Here, we have applied the identities
\[F^2_{2k+2} - F_{2k+1}F_{2k+3} \,=\, -1 \,,\qquad F^2_{2k+1} - F_{2k}F_{2k+2} \,=\, 1 \,,\]
and the inequality \(1+\beta -\rho &lt; 1+n-\rho &lt; n \). Since \(\beta &gt; n-1 \) and, by (\ref{5}),
\begin{eqnarray*}
|\varepsilon_{2k}| &amp;\,&lt;\,&amp; \frac{1}{q_{2k+1}} \,=\, \frac{1}{nF_{2k+1} + F_{2k}} \,, \\
|\varepsilon_{2k+1}| &amp;\,&lt;\,&amp; \frac{1}{q_{2k+2}} \,=\, \frac{1}{a_{2k+2}q_{2k+1} + F_{2k+1}} \,\leq \, \frac{1}{(n+1)F_{2k+1} + F_{2k}} \,. 
\end{eqnarray*}
(\ref{770}) follows from 
\begin{equation}
\frac{n(F_{2k-1} -1)}{\big( \rho F_{2k+1} + F_{2k} \big) \big( (n-1)F_{2k+1} + F_{2k+2} \big)} \,&lt;\, \frac{1}{\rho^{2k}} - \frac{1}{nF_{2k+1} + 
F_{2k}} - \frac{1}{(n+1)F_{2k+1} + F_{2k}} \,.
\label{790}
\end{equation}
In order to prove (\ref{790}), we need three inequalities for Fibonacci numbers, which rely on Binet's formula. Let \(\delta 
:=1/\rho^4 \). Then, for all integers \(s\geq 1 \), we have
\begin{equation}
\frac{\rho^{2s+1}}{\sqrt{5}} \,&lt;\, F_{2s+1} \,&lt;\, \frac{(1+\delta )\rho^{2s+1}}{\sqrt{5}} \qquad \mbox{and} \qquad \frac{(1-\delta) \rho^{2s}}
{\sqrt{5}} \,\leq \, F_{2s} \,. 
\label{800}
\end{equation}
We start to prove (\ref{790}) by observing that 
\[\sqrt{5} \left( \frac{1+\delta}{\rho^2 (\rho^2 + 1 - \delta )} + \frac{1}{3\rho +1-\delta} + \frac{1}{4\rho + 1 - \delta} \right) \,&lt;\, 1 \,.\]
Here, the left-hand side can be diminished by noting that
\[\frac{1}{\rho} \,&gt;\, \frac{n}{(n-1)\rho + (1-\delta)\rho^2} \,.\]
By \(n\geq 3 \) we get
\[\sqrt{5} \left( \frac{(1+\delta)n}{\rho \big( \rho^2 + 1 - \delta \big) \big( (n-1)\rho + (1-\delta)\rho^2 \big) } + \frac{1}{n\rho +1-\delta} + 
\frac{1}{(n+1)\rho + 1 - \delta} \right) \,&lt;\, 1 \,,\]
or, equivalently, 
\begin{eqnarray*}
&amp; &amp; \frac{(1+\delta)n\rho^{2k-1}/\sqrt{5}}{\big( \rho \cdot \rho^{2k+1}/\sqrt{5} + (1-\delta)\rho^{2k}/\sqrt{5} \big) \big( (n-1)\rho^{2k+1}/\sqrt{5} 
+ (1-\delta)\rho^{2k+2}/\sqrt{5} \big)} \\
&amp;\,&lt;\,&amp; \frac{1}{\rho^{2k}} - \frac{1}{n \rho^{2k+1}/\sqrt{5} + (1-\delta)\rho^{2k}/\sqrt{5}} - \frac{1}{(n+1) \rho^{2k+1}/\sqrt{5} + 
(1-\delta)\rho^{2k}/\sqrt{5}} \,.
\end{eqnarray*}
 From this inequality, (\ref{790}) follows easily by applications of (\ref{800}) with \(s \in \{ 2k-1,2k,2k+1,2k+2 \} \). \\

\noindent {\em Case~2:\/} \,Let \(a_{2k+1} =2 \). \\

\noindent {\em Case~2.1:\/} \, Let \(k\geq 2 \). \\
We first consider the function
\[f(\beta) \,:=\, \frac{1-\rho + \beta}{\beta F_{2k+1} +F_{2k+2}} \qquad (1\leq \beta \leq 2) \,.\]
The function \(f\) increases monotonically with \(\beta \), therefore we have
\[f(\beta) \,\leq \, f(2) \,=\, \frac{3-\rho}{2F_{2k+1} +F_{2k+2}} \,,\]
and consequently we conclude from the identity stated in (\ref{785}) that
\[\rho - \alpha \,\leq \, \frac{3-\rho}{(\rho F_{2k+1} +F_{2k})(2F_{2k+1} +F_{2k+2})} \,.\]
Hence, (\ref{780}) follows from the inequality
\begin{equation}
\frac{(3-\rho)F_{2k-1}}{(\rho F_{2k+1} +F_{2k})(2F_{2k+1} +F_{2k+2})} + \frac{1}{q_{2k+1}} + \frac{1}{q_{2k+2}} + \frac{1}{q_{2k+3}} + 
\frac{1}{q_{2k+4}} \,&lt;\, \frac{1}{\rho^{2k}} + \frac{1}{\rho^{2k+2}} \,.
\label{810}
\end{equation}
On the left-hand side we now replace the \(q\)'s by certain smaller terms in Fibonacci numbers. For \(q_{2k+2} \), \(q_{2k+3} \), and \(q_{2k+4} \),
we find lower bounds by (\ref{700}) in Lemma~\ref{L1}:
\begin{eqnarray*}
q_{2k+1} &amp;\,=\,&amp; a_{2k+1}q_{2k} + q_{2k-1} \,=\, 2F_{2k+1} + F_{2k} \,, \\
q_{2k+2} &amp;\,\geq \,&amp; F_{2k+3} + F_2F_{2k+1} \,=\, F_{2k+3} + F_{2k+1} \,, \\
q_{2k+3} &amp;\,\geq \,&amp; F_{2k+4} + F_3F_{2k+1} \,=\, F_{2k+4} + 2F_{2k+1} \,, \\
q_{2k+4} &amp;\,\geq \,&amp; F_{2k+5} + F_4F_{2k+1} \,=\, F_{2k+5} + 3F_{2k+1} \,. \\
\end{eqnarray*}
Substituting these expressions into (\ref{810}), we then conclude that (\ref{780}) from
\[\frac{(3-\rho)F_{2k-1}}{(\rho F_{2k+1} + F_{2k})(2F_{2k+1} +F_{2k+2})} + \frac{1}{2F_{2k+1} + F_{2k}} + \frac{1}{F_{2k+3} + F_{2k+1}} \]
\begin{equation}
+ \,\frac{1}{F_{2k+4} + 2F_{2k+1}} + \frac{1}{F_{2k+5} + 3F_{2k+1}} \,&lt;\, \frac{1}{\rho^{2k}} \,\Big( 1 + \frac{1}{\rho^2} \Big) \,.
\label{820}
\end{equation}
We apply the inequalities in (\ref{800}) for all \(s\geq 2 \) when \(\delta \) is replaced by \(\delta := 1/\rho^8 \). Using this redefined 
number \(\delta \), we have
\[\sqrt{5} \,\Big( \frac{(3-\rho)(1+\delta)}{\rho \big( \rho^2 +1-\delta \big) \big( 2\rho + (1-\delta)\rho^2 \big)} + \frac{1}{2\rho +1-\delta}
+ \frac{1}{\rho^3 + \rho} + \frac{1}{(1-\delta) \rho^4 + 2\rho} + \frac{1}{\rho^5 + 3\rho} \Big) - \frac{1}{\rho^2} \]
\[&lt;\, 1 \,,\]  
or, equivalently,
\begin{eqnarray*}
&amp; &amp; \frac{(3-\rho)(1+\delta)\rho^{2k-1}/\sqrt{5}}{\big( \rho \cdot \rho^{2k+1}/\sqrt{5} + (1-\delta)\rho^{2k}/\sqrt{5} \big) \big( 2\rho^{2k+1}/
\sqrt{5} + (1-\delta)\rho^{2k+2}/\sqrt{5} \big)} \\
&amp;+&amp; \frac{1}{2\rho^{2k+1}/\sqrt{5} + (1-\delta)\rho^{2k}/\sqrt{5}} + \frac{1}{\rho^{2k+3}/\sqrt{5} + \rho^{2k+1}/\sqrt{5}} \\
&amp;+&amp; \frac{1}{(1-\delta)\rho^{2k+4}/\sqrt{5} + 2\rho^{2k+1}/\sqrt{5}} +  \frac{1}{\rho^{2k+5}/\sqrt{5} + 3\rho^{2k+1}/\sqrt{5}} \\
&amp;\,&lt;\,&amp; \frac{1}{\rho^{2k}} \,\Big( 1 + \frac{1}{\rho^2} \Big) \,.
\end{eqnarray*}
 From this inequality, (\ref{820}) follows by applications of (\ref{800}) with \(s \in \{ 2k-1,2k,2k+1,2k+2,2k+3,2k+4,2k+5 \} \) for \(k\geq 2 \)
(which implies \(s\geq 3 \)). \\

\noindent{\em Case~2.2:\/} \, Let \(k=1 \). \\
 From the hypotheses we have \(a_{2k+1} = a_3 = 2 \). To prove (\ref{780}) it suffices to check the inequality in (\ref{820}) for \(k=1 \). We have
\[\begin{array}{ccccccccclccccc} 
F_{2k-1} &amp;=&amp; F_1 &amp;=&amp; 1 \,,\quad &amp; F_{2k} &amp;=&amp; F_2 &amp;=&amp; 1 \,,\quad &amp; F_{2k+1} &amp;=&amp; F_3 &amp;=&amp; 2 \,, \\
F_{2k+2} &amp;=&amp; F_4 &amp;=&amp; 3 \,,\quad &amp; F_{2k+3} &amp;=&amp; F_5 &amp;=&amp; 5 \,, \\
F_{2k+4} &amp;=&amp; F_6 &amp;=&amp; 8 \,,\quad &amp; F_{2k+5} &amp;=&amp; F_7 &amp;=&amp; 13 \,.
\end{array} \]
Then (\ref{820}) is satisfied because
\[\rho^2 \,\Big( \frac{3-\rho}{7(1+2\rho)} + \frac{1}{5} + \frac{1}{7} + \frac{1}{12} + \frac{1}{19} \Big) - \frac{1}{\rho^2} \,&lt;\, 1 \,.\]
This completes the proof of Lemma {\ref{L4}}. \hfill \qed 
\[\]
{\em Proof of Theorem \ref{HS2}:\/} \,In the sequel we shall use the identity
\begin{equation}
F_{2g} + F_{2g+2} + F_{2g+4} + \ldots + F_{2n} \,=\, F_{2n+1} - F_{2g-1} \qquad (n\geq g \geq 0) \,,
\label{1000}
\end{equation}
which can be proven by induction by applying the recurrence formula of Fibonacci numbers. Note that \(F_{-1} =1 \). Next, we prove (\ref{H50}). \\

\noindent{\em Case~1:\/} \, Let \(\alpha \not\in \mathcal{M} \), \(\alpha = \langle 1;a_1,a_2,\ldots \rangle = \langle 1;1,
\ldots ,1,a_{2k},a_{2k+1}, \ldots \rangle \) with \(a_{2k} &gt; 1 \) for some subscript \(k\geq 1 \). This implies \(\alpha &gt; \rho \). \\

\noindent {\em Case~1.1:\/} \, Let \(0\leq n &lt;2k \). \\
Then \(n_0 = \lfloor n/2 \rfloor \leq k-1 \). In order to treat \(|\varepsilon_{2m}| + |\varepsilon_{2m+1}| \), we apply (\ref{H30}) with \(k\) replaced by 
\(k-1 \) in Proposition \ref{HS1}. For \(\alpha \) the condition (\ref{H40}) with \(k\) replaced by \(k-1 \) is fulfilled. Note that the
term \(F_{2m}(\rho - \alpha) \) in (\ref{H30}) is negative. Therefore, we have
\begin{eqnarray*}
S(n) &amp;\,:=\,&amp; \sum_{\nu =2g}^n |\varepsilon_{\nu}| \,\leq \, \sum_{m=g}^{\lfloor n/2 \rfloor} \big( |\varepsilon_{2m}| + |\varepsilon_{2m+1}| \big) \\
&amp;\,&lt;\,&amp; \sum_{m=g}^{n_0} \frac{1}{\rho^{2m}} \,=\, \frac{\rho^{2-2g} - \rho^{-2n_0}}{\rho^2 -1} \,=\, \rho^{1-2g} - \rho^{-2n_0-1} \,. 
\end{eqnarray*} 

\noindent{\em Case~1.2:\/} \, Let \(n \geq 2k \). \\

\noindent{\em Case~1.2.1:\/} \, Let \(k \geq g \). \\
Here, we get
\begin{equation}
S(n) \,\leq \, \sum_{m=g}^{k-1} \big( |\varepsilon_{2m}| + |\varepsilon_{2m+1}| \big) + \big( |\varepsilon_{2k}| + |\varepsilon_{2k+1}| \big)    
+ \sum_{m=k+1}^{n_0} \big( |\varepsilon_{2m}| + |\varepsilon_{2m+1}| \big) \,.
\label{1010}
\end{equation} 
When \(n_0 \leq k \), the right-hand sum is empty and becomes zero. The same holds for the left-hand sum for \(k=g \). \\
{\bf a)} \,We estimate the left-hand sum as in the preceding case applying (\ref{H30}), \(\rho - \alpha &lt; 0 \), and the hypothesis \(a_1a_2\cdots a_{2k-1} 
=1 \):  
\[\sum_{m=g}^{k-1} \big( |\varepsilon_{2m}| + |\varepsilon_{2m+1}| \big) \,&lt;\, \sum_{m=g}^{k-1} \frac{1}{\rho^{2m}} \,.\]
{\bf b)} \,Since \(a_{2k} &gt;1 \), the left-hand condition in (\ref{H20}) allows us to apply (\ref{H10}) for \(m=k \):
\[|\varepsilon_{2k}| + |\varepsilon_{2k+1}| \,&lt;\, \frac{1}{\rho^{2k}} \,.\]
{\bf c)} \,We estimate the right-hand sum in (\ref{1010}) again by (\ref{H10}). To check the conditions in (\ref{H20}), we use
\(a_1a_2\cdots a_{2m-1} &gt;1 \), which holds by \(m\geq k+1 \) and \(a_{2k} &gt;1 \). Hence,
\[\sum_{m=k+1}^{n_0} \big( |\varepsilon_{2m}| + |\varepsilon_{2m+1}| \big) \,&lt;\, \sum_{m=k+1}^{n_0} \frac{1}{\rho^{2m}} \,.\]
Altogether, we find with (\ref{1010}) that
\begin{equation}
S(n) \,&lt;\, \sum_{m=g}^{n_0} \frac{1}{\rho^{2m}} \,=\, \rho^{1-2g} - \rho^{-2n_0-1} \,.
\label{1015}
\end{equation}

\noindent{\em Case~1.2.2:\/} \, Let \(k&lt;g \). \\
In order to estimate \(|\varepsilon_{2m}| + |\varepsilon_{2m+1}| \) for \(g\leq m \leq n_0 \), we use \(k+1 \leq g \) and the arguments from c)
in Case~1.2.1. Again, we obtain the inequality (\ref{1015}).  
The results from Case~1.1 and Case~1.2 prove (\ref{H50}) for \(a_{2k} &gt;1 \) with \(k\geq 1 \). It remains to investigate the following case. \\

\noindent{\em Case~2:\/} \,Let \(\alpha \not\in \mathcal{M} \), \(\alpha = \langle 1;a_1,a_2,\ldots \rangle \) with \(a_1 &gt;1 \). \\
For \(m=0 \) (provided that \(g=0 \)) the first condition in (\ref{H20}) is fulfilled by \(a_{2m}a_{2m+1} = a_0a_1 = a_1 &gt;1 \). For \(m\geq 1 \) 
we know that \(a_1a_2 \cdots a_{2m-1} &gt;1 \) always satisfies one part of the second condition. Therefore, we apply the inequality from (\ref{H10}):
\[S(n) \,&lt;\, \sum_{m=g}^{n_0} \frac{1}{\rho^{2m}} \,=\, \rho^{1-2g} - \rho^{-2n_0-1} \,. \]
Next, we prove (\ref{H60}). Let \(\alpha \in \mathcal{M} \), \(\alpha = \langle 1;a_1,a_2,\ldots \rangle = \langle 1;1,\ldots ,1,a_{2k+1},a_{2k+2},
\ldots \rangle \) with \(a_{2k+1} &gt; 1 \) for some subscript \(k\geq 1 \). This implies \(\rho &gt; \alpha \). \\

\noindent{\em Case~3.1:\/} \, Let \(0\leq n &lt;2k \). \\
Then \(n_0 = \lfloor n/2 \rfloor \leq k-1 \). In order to treat \(|\varepsilon_{2m}| + |\varepsilon_{2m+1}| \), we apply (\ref{H30}) with \(k\) replaced by 
\(k-1 \) in Proposition \ref{HS1}. For \(\alpha \) the condition (\ref{H40}) with \(k\) replaced by \(k-1 \) is fulfilled. Note that the term 
\(F_{2m}(\rho - \alpha) \) in (\ref{H30}) is positive. Therefore we have, using (\ref{1000}),
\begin{eqnarray*}
S(n) &amp;\,\leq \,&amp; \sum_{m=g}^{n_0} \Big( \frac{1}{\rho^{2m}} + (\rho - \alpha) F_{2m} \Big) \\
&amp;\,=\,&amp; \rho^{1-2g} - \rho^{-2n_0-1} + (\rho - \alpha) \sum_{m=g}^{n_0} F_{2m} \\
&amp;\,=\,&amp; \rho^{1-2g} - \rho^{-2n_0-1} + (\rho - \alpha) (F_{2n_0+1} - F_{2g-1}) \\
&amp;\,\leq \,&amp; \rho^{1-2g} - \rho^{-2n_0-1} + (F_{2k-1} - F_{2g-1})(\rho - \alpha) \,.
\end{eqnarray*}
Here we have used that \(2n_0 +1 \leq 2k-1 \). \\

\noindent{\em Case~3.2:\/} \, Let \(n \geq 2k \). \\ 
Our arguments are similar to the proof given in Case~1.2, using \(a_1a_2 \cdots a_{2k-1} =1 \) and \(a_{2k+1} &gt;1 \). \\

\noindent{\em Case~3.2.1:\/} \, Let \(k \geq g \). \\
Applying (\ref{1000}) again, we obtain 
\begin{eqnarray*}
S(n) &amp;\,\leq \,&amp; \sum_{m=g}^{k-1} \big( |\varepsilon_{2m}| + |\varepsilon_{2m+1}| \big) + \big( |\varepsilon_{2k}| + |\varepsilon_{2k+1}| \big)    
+ \sum_{m=k+1}^{n_0} \big( |\varepsilon_{2m}| + |\varepsilon_{2m+1}| \big) \\
&amp;\,&lt;\,&amp; \sum_{m=g}^{k-1} \Big( \frac{1}{\rho^{2m}} + (\rho - \alpha)F_{2m} \Big) + \frac{1}{\rho^{2k}} + \sum_{m=k+1}^{n_0} \frac{1}{\rho^{2m}} \\
&amp;\,=\,&amp; \sum_{m=g}^{n_0} \frac{1}{\rho^{2m}} + (\rho - \alpha) \sum_{m=g}^{k-1} F_{2m} \\
&amp;\,=\,&amp; \rho^{1-2g} - \rho^{-2n_0-1} + (F_{2k-1} - F_{2g-1})(\rho - \alpha) \,.
\end{eqnarray*} 

\noindent{\em Case~3.2.2:\/} \, Let \(k&lt;g \). \\
 From \(g\geq k+1 \) we get
\[S(n) \,\leq \, \sum_{m=g}^{n_0} \frac{1}{\rho^{2m}} \,=\, \rho^{1-2g} - \rho^{-2n_0-1} \,.\]
The results of Case~3.1 and Case~3.2 complete the proof of (\ref{H60}). \\
For the inequality (\ref{H70}) we distinguish whether \(\alpha \) belongs to \(\mathcal{M} \) or not. \\

\noindent{\em Case~4.1:\/} \, Let \(\alpha \not\in \mathcal{M} \). \\
Then (\ref{H70}) is a consequence of the inequality in (\ref{H50}): 
\[\sum_{\nu =2g}^{\infty} |\varepsilon_{\nu}| \,\leq \, \lim_{n_0 \to \infty} \big( \rho^{1-2g} - \rho^{-2n_0-1} \big) \,=\, {\rho}^{1-2g} \,.\]

\noindent{\em Case~4.2:\/} \, Let \(\alpha \in \mathcal{M} \). \\
There is a subscript \(k\geq 1 \) satisfying \(\alpha = \langle 1;1,\ldots ,1,a_{2k+1},a_{2k+2},\ldots \rangle \) and \(a_{2k+1} &gt;1 \). To simplify
arguments, we introduce the function \(\chi (k,g) \) defined by \(\chi (k,g) = 1 \) (if \(k&gt;g \)), and \(\chi (k,g) = 0 \) (if \(k \leq g \)).
We have 
\begin{eqnarray}
S &amp;\,:=\,&amp; \sum_{\nu =2g}^{\infty} |\varepsilon_{\nu}| \,=\, \sum_{\nu =2g}^{2k-1} |\varepsilon_{\nu}| + \sum_{\nu = \max \{ 2k,2g \}}^{\infty} 
|\varepsilon_{\nu}| \nonumber \\ 
&amp;\,=\,&amp; \chi (k,g) \left( (F_{2k-1} - F_{2g-1})(\rho - \alpha) + \rho^{1-2g} - \rho^{-2k+1} \right) +  
\sum_{m =\max \{ k,g \}}^{\infty} \big( |\varepsilon_{2m}| + |\varepsilon_{2m+1}| \big) \nonumber \\
&amp;\leq \,&amp; (F_{2k-1} - F_{2g-1})(\rho - \alpha) + \rho^{1-2g} - \rho^{-2k+1} +  
\sum_{m =k}^{\infty} \big( |\varepsilon_{2m}| + |\varepsilon_{2m+1}| \big)  \nonumber \\
&amp;\leq \,&amp; (F_{2k-1} - 1)(\rho - \alpha) + \rho^{1-2g} - \rho^{-2k+1} +  
\sum_{m =k}^{\infty} \big( |\varepsilon_{2m}| + |\varepsilon_{2m+1}| \big) \,, 
\label{1020}   
\end{eqnarray}
where we have used (\ref{H60}) with \(n=2k-1 \) and \(n_0 = \lfloor n/2 \rfloor = k-1 \). \\

\noindent{\em Case~4.2.1:\/} \, Let \(a_{2k+1} \geq 3 \). \\
The conditions in Lemma \ref{L4} for (\ref{770}) are satisfied. Moreover, the terms \(|\varepsilon_{2m}| + |\varepsilon_{2m+1}| \) of the series in 
(\ref{1020}) for \(m \geq k+1 \) can be estimated using (\ref{H10}), since \(a_1a_2 \cdots a_{2k+1} &gt;1 \). Therefore, we obtain
\begin{eqnarray*}
S &amp;\,&lt;\,&amp; \frac{1}{\rho^{2k}} - \frac{1}{\rho^{2k-1}} + \rho^{1-2g} + \sum_{m =k+1}^{\infty} \big( |\varepsilon_{2m}| + 
|\varepsilon_{2m+1}| \big) \\
&amp;\,&lt;\,&amp; \frac{1}{\rho^{2k}} - \frac{1}{\rho^{2k-1}} + \rho^{1-2g} + \frac{1}{\rho^{2k+1}} \,=\, {\rho}^{1-2g} \,.
\end{eqnarray*}

\noindent {\em Case~4.2.2:\/} \, Let \(a_{2k+1} = 2 \). \\
Now the conditions in Lemma \ref{L4} for (\ref{780}) are satisfied. Thus, from (\ref{1020}) and (\ref{H10}) we have
\begin{eqnarray*}
S &amp;\,&lt;\,&amp; \frac{1}{\rho^{2k}} + \frac{1}{\rho^{2k+2}} - \frac{1}{\rho^{2k-1}} + \rho^{1-2g} + \sum_{m =k+2}^{\infty} 
\big( |\varepsilon_{2m}| + |\varepsilon_{2m+1}| \big) \\
&amp;\,&lt;\,&amp; \frac{1}{\rho^{2k}} + \frac{1}{\rho^{2k+2}} - \frac{1}{\rho^{2k-1}} + \rho^{1-2g} + \sum_{m =k+2}^{\infty} \frac{1}{\rho^{2m}} \\
&amp;\,=\,&amp; \frac{1}{\rho^{2k}} + \frac{1}{\rho^{2k+2}} - \frac{1}{\rho^{2k-1}} + \rho^{1-2g} + \frac{1}{\rho^{2k+3}} \,=\, {\rho
}^{1-2g} \,.
\end{eqnarray*}
This completes the proof of Theorem \ref{HS2}. \hfill \qed 

\section{Concluding remarks} \label{S6}  
In this section we state some additional identities for error sums \(\varepsilon (\alpha) \). For this purpose let \(\alpha = \langle a_0;a_1,a_2,
\dots  \rangle \) be the continued fraction expansion of a real irrational number. Then
the numbers \(\alpha_n \) are defined by 
\[\alpha \,=\, \langle a_0;a_1,a_2,\dots ,a_{n-1}, \alpha_n \rangle \qquad (n=0,1,2,\dots) \,.\]
\begin{proposition}
For every real irrational number \(\alpha \) we have
\[\varepsilon (\alpha) \,=\, \sum_{n=1}^{\infty} \,\prod_{k=1}^n \frac{1}{\alpha_k} \]
and
\[\left( \begin{array}{c} \varepsilon (\alpha) \\ \cdot \end{array} \right) \,=\, \sum_{n=0}^{\infty} {(-1)}^n \left( \begin{array}{cc} a_n &amp; 1 \\ 
1 &amp; 0 \end{array} \right) \left( \begin{array}{cc} a_{n-1} &amp; 1 \\ 1 &amp; 0 \end{array} \right) \cdots \left( \begin{array}{cc} a_0 &amp; 1 \\ 
1 &amp; 0 \end{array} \right) \left( \begin{array}{r} -1 \\ \alpha \end{array} \right) \,.\] 
\label{Prop2}
\end{proposition}
Next, let \(\alpha = \langle a_0;a_1,a_2,\dots  \rangle \) with \(a_0 \geq 1 \) be a real number with convergents \(p_m/q_m \) \((m\geq 0) \), where 
\(p_{-1}=1 \), \(q_{-1}=0 \). Then the convergents \(\overline{p}_m/\overline{q}_m \) of the number \(1/\alpha = \langle 0;a_0,a_1,a_2,\dots  \rangle \)
satisfy the equations \(\overline{q}_m = p_{m-1} \) and \(\overline{p}_m = q_{m-1} \) for \(m\geq 0 \), since we know that \(\overline{p}_{-1} =1 \),
\(\overline{p}_0 =0 \) and \(\overline{q}_{-1} =0 \), \(\overline{q}_0 =1 \). Therefore we obtain a relation between \(\varepsilon (\alpha) \) and
\(\varepsilon (1/\alpha) \):
\begin{eqnarray*}
\varepsilon (1/\alpha) &amp;\,=\,&amp; \sum_{m=0}^{\infty} \Big| \frac{ \overline{q}_m }{\alpha} - \overline{p}_m \Big| \,=\, \sum_{m=0}^{\infty} \Big| 
\frac{p_{m-1} }{\alpha} - q_{m-1} \Big| \,=\, \frac{1}{\alpha} \sum_{m=0}^{\infty} |q_{m-1}\alpha - p_{m-1}| \\
&amp;\,=\,&amp; \frac{1}{\alpha} \Big( |q_{-1}\alpha - p_{-1}| + \sum_{m=0}^{\infty} |q_m\alpha - p_m| \Big) \,=\, \frac{1}{\alpha} \big( 1 + \varepsilon (\alpha) 
\big) \,.
\end{eqnarray*}
This proves
\begin{proposition}
For every real number \(\alpha &gt;1 \) we have
\[\varepsilon (1/\alpha) \,=\, \frac{1 + \varepsilon (\alpha)}{\alpha} \,.\]
\label{Prop3}
\end{proposition}

\section{Acknowledgment}
The author would like to thank Professor Iekata Shiokawa for helpful comments and useful hints in organizing the paper. I am very much obliged
to the anonymous referee, who suggested the results stated in Section\,\ref{S6}. Moreover, the presentation of the paper was improved by
following the remarks of the referee. 
 
 
 
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\bibitem{Cohn}
H.\,Cohn, A short proof of the simple continued fraction expansion of \(e\),
{\em Amer. Math. Monthly\/} {\bf 113} (2006), 57---62.

\bibitem{E1}
C.\,Elsner, On arithmetic properties of the convergents of Euler's number,
{\em Colloq. Math.\/} {\bf 79} (1999), 133--145.
 
\bibitem{E2}
C.\,Elsner and T.\,Komatsu, On the residue classes of integer sequences satisfying a linear three term recurrence formula,
{\em  Linear Algebra Appl.} {\bf 429} (2008), 933--947.

\bibitem{Hardy} 
G.\,H.\,Hardy and E.\,M.\,Wright, {\em An Introduction to the Theory of
Numbers\/}, Clarendon Press, Oxford, 1984.

\bibitem{Khintchine} 
A.\,Khintchine, {\em Kettenbr{\"u}che\/}, Teubner, Leipzig, 1949.

\bibitem{K3}
T.\,Komatsu, Arithmetical properties of the leaping convergents of $e^{1/s}$,
{\em Tokyo J. Math.\/} {\bf 27} (2004), 1--12.

\bibitem{K2}
T.\,Komatsu, Some combinatorial properties of the leaping convergents,
{\em Integers} {\bf 7} (2) (2007),
\(\sharp \)A21.  Available electronically at
\href{http://www.integers-ejcnt.org/vol7-2.html}{\tt http://www.integers-ejcnt.org/vol7-2.html}.

\bibitem{Perron} 
O.\,Perron, {\em Die Lehre von den Kettenbr{\"u}chen\/}, Chelsea Publishing Company, New York, 1929. 

\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11J04; Secondary 11J70, 11B39.

\noindent \emph{Keywords: } 
continued fractions, convergents, approximation of real numbers, error
terms.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000045},
\seqnum{A001519},
\seqnum{A007676},
\seqnum{A007677},
\seqnum{A041008}, and
\seqnum{A041009}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received July 12 2010;
revised version received  January 10 2011.
Published in {\it Journal of Integer Sequences}, January 28 2011.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}

                                                                                



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