\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \begin{center} \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} \\ \end{center} \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} \newtheorem{theorem}{Theorem} \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 \(1\leq q < q_n \) we have \[\left| \,\alpha - \frac{p_n}{q_n} \,\right| \,<\, \left| \,\alpha - \frac{p}{q} \,\right| \,.\] 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}} \,<\, |q_n \alpha - p_n| \,<\, \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 \[\begin{array}{cclcclcl} a_{n+1} &:=& q_n^n \,,\quad & q_{n+1} &=& a_{n+1}q_n + q_{n-1} &=& q_n^n \big( q_n + q_{n-1} \big) \,,\\ a_{n+2} &:=& q_{n+1}^{n+1} \,,\quad & q_{n+2} &=& a_{n+2}q_{n+1} + q_n &=& q_{n+1}^{n+1}\big( q_n^{n+1} + q_{n-1} \big) \,,\\ a_{n+3} &:=& q_{n+2}^{n+2} \,,\quad & q_{n+3} &=& a_{n+3}q_{n+2} + q_{n+1} &=& \dots \end{array} \] 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 \[0 \,<\, \Big| \,\alpha_n - \frac{p_k}{q_k}\,\Big| \,<\, \frac{1}{q_kq_{k+1}} \,<\, \frac{1}{a_{k+1}q_k^2} \,=\, \frac{1}{q_k^{k+2}} \quad (k\geq n) \,.\] 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 \[\sum_{m=0}^{\infty} |q_m \alpha_n - p_m| \,>\, \sum_{m=0}^{n-1} |q_m \alpha_n - p_m| \,=\, (F_{n-1} -1)(\rho - \alpha_n) + \rho - \rho^{1-n} \,\geq \, \rho - \frac{1}{\rho^{n-1}} \,.\] 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 \(1< \alpha <2 \). \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}| \,<\, \frac{1}{\rho^{2m}} \,, \label{H10} \end{equation} provided that either \begin{equation} a_{2m}a_{2m+1} \,>\, 1 \qquad \mbox{or} \qquad \big( a_{2m} \,=\, a_{2m+1} \,=\, 1 \,\quad \mbox{and} \quad \,a_1a_2 \cdots a_{2m-1} >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 \(a_r >1 \): For odd \(r\) we have \(\rho > \alpha \), otherwise, \(\rho < \alpha \). \\ Next, we introduce a set \(\mathcal{M} \) of irrational numbers, namely \[\mathcal{M} \,:=\, \left\{ \,\alpha \in {\mathbb R} \setminus {\mathbb Q} \,\,\big| \,\, \exists \,k \in {\mathbb N} \,:\, \alpha = \langle 1;1,\ldots ,1,a_{2k+1},a_{2k+2}, \ldots \rangle \,\wedge \, a_{2k+1} >1 \right\} \,.\] Note that \(\rho > \alpha \) for \(\alpha \in \mathcal{M} \). Our main result for real irrational numbers is given by the subsequent theorem. \begin{theorem} Let \(1 < \alpha < 2 \) 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 \(a_{2k+1} > 1 \), 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 \(|q_{\nu}\alpha -p_{\nu}| < 1/q_{\nu+1} \leq 1/q_2 \leq 1/2 \), 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 \(|x| < \rho \), and \[0 \,<\, \sum_{m=0}^{\infty} (q_m \alpha - p_m) \,<\, 1 \,.\] 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 \[z_{n+2} \,=\, Gz_{n+1} \pm z_n \qquad (rn > \omega) \,.\] 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) &\,=\,& \frac{21 - 5\sqrt{7}}{14} \,=\, 0.555088817 \ldots \,, \\ \\ \sum_{m=0}^{\infty} |q_m\sqrt{7} -p_m| &\,=\,& \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 \(\varepsilon_0 >0 \) and \(\varepsilon_m \varepsilon_{m+1} <0 \), we have \[\varepsilon_0 + \varepsilon_1 \,<\, \sum_{m=0}^{\infty} \varepsilon_m \,<\, \varepsilon_0 \,.\] Put \(a_0 = \lfloor \alpha \rfloor \), \(\theta := \varepsilon_0 = \alpha - a_0 \), so that \(0<\theta < 1 \). 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 \[\frac{1}{k+1} \,<\, \theta \,<\, \frac{1}{k} \,,\] we get \[\theta + \Big\lfloor \,\frac{1}{\theta}\,\Big\rfloor \,\theta -1 \,>\, \frac{1}{k+1} + \frac{k}{k+1} - 1 \,=\, 0 \,,\] 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 \[|\varepsilon_m| x^m \,=\, |q_m\alpha - p_m| x^m \,<\, \frac{x^m}{q_{m+1}} \,\leq \, \sqrt{5} {\left( \frac{x}{\rho} \right)}^m \qquad (m\geq 0) \,.\] It follows that the series \(\sum \varepsilon_m x^m \) converges absolutely at least for \(|x| < \rho \). In order to prove that the upper bound 1 is best possible, we choose \(0<\varepsilon < 1 \) and a positive integer \(n\) satisfying \[\frac{1}{n} \,\Big( 1 + \frac{\rho \sqrt{5}}{\rho -1} \Big) \,<\, \varepsilon \,.\] Put \[\alpha_n \,:=\, \langle \,0;1,\overline{n}\,\rangle \,=\, \frac{1}{2} - \frac{1}{n} + \frac{1}{2} \sqrt{1+\frac{4}{n^2}} \,>\, 1 - \frac{1}{n} \,.\] 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) &\,\geq\,& \alpha_n - \sum_{m=1}^{\infty} |q_m\alpha_n - p_m| \\ &\,>\,& 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} \\ &\,\geq \,& 1 - \frac{1}{n} - \frac{\sqrt{5}}{n} \sum_{m=1}^{\infty} \frac{1}{\rho^{m-1}} \\ &\,=\,& 1 - \frac{1}{n} \,\Big( 1 + \frac{\rho \sqrt{5}}{\rho -1} \Big) \,>\, 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 \(a_n >1 \). 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 \,>\, \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} &\,\geq\,& q_{n+k+1} + q_{n+k} \\ &\,\geq \,& (F_{n+k+2} + F_{k+2}F_n) + (F_{n+k+1} + F_{k+1}F_n) \\ &\,=\,& 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*} && F_{n+k+1} + F_{k+1}F_n \\ &\,=\,& \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*} && F_{n+k+1} + F_{k+1}F_n \\ &\,=\,& \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) \\ &\,>\,& \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*} && F_{n+k+1} + F_{k+1}F_n \\ &\,=\,& \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) \,>\, \rho^{n+k} \,. \end{eqnarray*} For \(k=0 \) and some odd \(n\geq 1 \) we get \[F_{n+k+1} + F_{k+1}F_n \,>\, \rho^n \left( \rho \,\Big( \frac{1}{5} + \frac{1}{\sqrt{5}} \Big) - \Big( \frac{1}{\sqrt{5}} - \frac{1}{5} \Big) \,\frac{1}{\rho^3} + \frac{1}{5\rho} \right) \,>\, \rho^n \,.\] \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*} && F_{n+k+1} + F_{k+1}F_n \\ &\,=\,& \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) \,>\, \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*} && F_{n+k+1} + F_{k+1}F_n \\ &\,=\,& \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) \,>\, \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}} &\,<\,& 1 \qquad (m\geq 1) \,, \label{740} \\ \rho^{2m} \Big( \frac{1}{F_{2m+3}} + \frac{1}{F_{2m+3} + F_{2m+1}} \Big) &\,<\,& 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 \(4m+2 \geq 6 \): \[F_{2m+2} \,=\, \frac{\rho^{2m}}{\sqrt{5}} \,\Big( \rho^2 - \frac{1}{\rho^{4m+2}} \Big) \,\geq \, \frac{\rho^{2m}}{\sqrt{5}} \Big( \rho^2 - \frac{1}{\rho^6} \Big) \,>\, \rho^{2m} \,.\] Similarly, we prove (\ref{730}) by \[F_{2n+1} \,=\, \frac{1}{\sqrt{5}} \,\Big( \rho^{2n+1} + \frac{1}{\rho^{2n+1}} \Big) \,>\, \frac{\rho^{2n+1}}{\sqrt{5}} \qquad (n\geq 0) \,.\] Hence, \[\rho^{2m} \Big( \frac{1}{F_{2m+3}} + \frac{1}{F_{2m+3} + F_{2m+1}} \Big) \,<\, \rho^{2m} \,\Big( \frac{\sqrt{5}}{\rho^{2m+3}} + \frac{\sqrt{5}}{\rho^{2m+3} + \rho^{2m+1}} \Big) \,<\, 1 \,.\] 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 \(0<\theta < 1 \) and \(\varepsilon_0 = \theta < 1 \). 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 \(0<\theta < 1/2 \), since otherwise for \(\theta > 1/2 \), 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 \[\frac{1}{k+1} \,<\,\theta \,<\,\frac{1}{k} \,.\] Obviously, it follows that \([1/\theta] = k \), and therefore \[\theta + 1 - \Big\lfloor \,\frac{1}{\theta}\,\Big\rfloor \,\theta \,<\, \frac{1}{k} + 1 - \,\frac{k}{k+1} \,=\, \frac{2k+1}{k(k+1)} \,\leq \, \frac{5}{6} \qquad (k\geq 2) \,.\] Altogether, we have proven that \begin{equation} |\varepsilon_0| + |\varepsilon_1| \,\leq \, \frac{5}{6} \,<\, 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 \(\varepsilon_{2m} >0 \) and \(\varepsilon_{2m+1} < 0 \) hold for every integer \(m \geq 0 \), we may rewrite (\ref{H10}) as follows: \begin{equation} \big( 0 \,<\, \big) \quad (p_{2m+1} - p_{2m}) - \alpha (q_{2m+1} - q_{2m}) \,<\, \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}| &\,<\,& \frac{1}{q_{2m+1}} + \frac{1}{q_{2m+2}} \\ &\,\leq \,& \frac{1}{2q_{2m} + q_{2m-1}} + \frac{1}{q_{2m+1} + q_{2m}} \\ &\,\leq \,& \frac{1}{2q_{2m} + q_{2m-1}} + \frac{1}{3q_{2m} + q_{2m-1}} \\ &\,\leq \,& \frac{1}{2F_{2m+1} + F_{2m}} + \frac{1}{3F_{2m+1} + F_{2m}} \\ &\,<\,& \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 \(0 < p_{2m-1} - \alpha q_{2m-1} < \rho^{-2m} \) for \(m\geq 1 \). This follows with (\ref{5}), (\ref{50}), and (\ref{740}) from \begin{eqnarray*} 0 &\,<\,& p_{2m-1} - \alpha q_{2m-1} \,<\, \frac{1}{q_{2m}} \\ &\,\leq \,& \frac{1}{2q_{2m-1} + q_{2m-2}} \,\leq \, \frac{1}{2F_{2m} + F_{2m-1}} \\ &\,=\,& \frac{1}{F_{2m+2}} \,<\, \frac{1}{\rho^{2m}} \qquad (m\geq 1) \,. \end{eqnarray*} \noindent{\em Case~3:\/} \, Let \(a_{2m} = a_{2m+1} =1 \,\wedge \,a_1a_2 \cdots a_{2m-1} > 1 \). \\ Since \(a_{2m+1} =1 \), we again have (as in Case~2): \begin{equation} 0 \,<\, |\varepsilon_{2m}| + |\varepsilon_{2m+1}| \,=\, p_{2m-1} - \alpha q_{2m-1} \,<\, \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}), \[q_{2m} \,=\, q_{n+k} \,\geq \, F_{n+k+1} + F_{k+1}F_n \,>\, \rho^{n+k} \,=\, \rho^{2m} \,.\] 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*} & & |q_{2m}\alpha - p_{2m}| + |q_{2m+1}\alpha - p_{2m+1}| \\ &\,=\,& 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 \(a_{2k+1} >1 \) and \(a_{\mu} \geq 1 \) for \(\mu \geq 2k+2 \). Then we have the inequalities \begin{equation} (F_{2k-1}-1)(\rho - \alpha) \,<\, \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) \,<\, \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 \(0<\eta <1 \) and \[r_{2k+1} \,:=\, \langle a_{2k+1}; a_{2k+2}, \ldots \rangle \,=\, n +\eta =: 1 + \beta \,. \] It is clear that \(n-1 < \beta < n \). From the theory of regular continued fractions (see \cite[formula\,(16)]{Khintchine}) it follows that \begin{eqnarray*} \alpha &\,=\,& \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}} \\ &\,=\,& \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})} \,<\, \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 \(1+\beta -\rho < 1+n-\rho < n \). Since \(\beta > n-1 \) and, by (\ref{5}), \begin{eqnarray*} |\varepsilon_{2k}| &\,<\,& \frac{1}{q_{2k+1}} \,=\, \frac{1}{nF_{2k+1} + F_{2k}} \,, \\ |\varepsilon_{2k+1}| &\,<\,& \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)} \,<\, \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}} \,<\, F_{2s+1} \,<\, \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 \[\sqrt{5} \left( \frac{1+\delta}{\rho^2 (\rho^2 + 1 - \delta )} + \frac{1}{3\rho +1-\delta} + \frac{1}{4\rho + 1 - \delta} \right) \,<\, 1 \,.\] Here, the left-hand side can be diminished by noting that \[\frac{1}{\rho} \,>\, \frac{n}{(n-1)\rho + (1-\delta)\rho^2} \,.\] By \(n\geq 3 \) we get \[\sqrt{5} \left( \frac{(1+\delta)n}{\rho \big( \rho^2 + 1 - \delta \big) \big( (n-1)\rho + (1-\delta)\rho^2 \big) } + \frac{1}{n\rho +1-\delta} + \frac{1}{(n+1)\rho + 1 - \delta} \right) \,<\, 1 \,,\] or, equivalently, \begin{eqnarray*} & & \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)} \\ &\,<\,& \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}} \,<\, \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} &\,=\,& a_{2k+1}q_{2k} + q_{2k-1} \,=\, 2F_{2k+1} + F_{2k} \,, \\ q_{2k+2} &\,\geq \,& F_{2k+3} + F_2F_{2k+1} \,=\, F_{2k+3} + F_{2k+1} \,, \\ q_{2k+3} &\,\geq \,& F_{2k+4} + F_3F_{2k+1} \,=\, F_{2k+4} + 2F_{2k+1} \,, \\ q_{2k+4} &\,\geq \,& 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}} \,<\, \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 \[\sqrt{5} \,\Big( \frac{(3-\rho)(1+\delta)}{\rho \big( \rho^2 +1-\delta \big) \big( 2\rho + (1-\delta)\rho^2 \big)} + \frac{1}{2\rho +1-\delta} + \frac{1}{\rho^3 + \rho} + \frac{1}{(1-\delta) \rho^4 + 2\rho} + \frac{1}{\rho^5 + 3\rho} \Big) - \frac{1}{\rho^2} \] \[<\, 1 \,,\] or, equivalently, \begin{eqnarray*} & & \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)} \\ &+& \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}} \\ &+& \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}} \\ &\,<\,& \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 \[\begin{array}{ccccccccclccccc} F_{2k-1} &=& F_1 &=& 1 \,,\quad & F_{2k} &=& F_2 &=& 1 \,,\quad & F_{2k+1} &=& F_3 &=& 2 \,, \\ F_{2k+2} &=& F_4 &=& 3 \,,\quad & F_{2k+3} &=& F_5 &=& 5 \,, \\ F_{2k+4} &=& F_6 &=& 8 \,,\quad & F_{2k+5} &=& F_7 &=& 13 \,. \end{array} \] Then (\ref{820}) is satisfied because \[\rho^2 \,\Big( \frac{3-\rho}{7(1+2\rho)} + \frac{1}{5} + \frac{1}{7} + \frac{1}{12} + \frac{1}{19} \Big) - \frac{1}{\rho^2} \,<\, 1 \,.\] 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 \(a_{2k} > 1 \) for some subscript \(k\geq 1 \). This implies \(\alpha > \rho \). \\ \noindent {\em Case~1.1:\/} \, Let \(0\leq n <2k \). \\ 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) &\,:=\,& \sum_{\nu =2g}^n |\varepsilon_{\nu}| \,\leq \, \sum_{m=g}^{\lfloor n/2 \rfloor} \big( |\varepsilon_{2m}| + |\varepsilon_{2m+1}| \big) \\ &\,<\,& \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}), \(\rho - \alpha < 0 \), and the hypothesis \(a_1a_2\cdots a_{2k-1} =1 \): \[\sum_{m=g}^{k-1} \big( |\varepsilon_{2m}| + |\varepsilon_{2m+1}| \big) \,<\, \sum_{m=g}^{k-1} \frac{1}{\rho^{2m}} \,.\] {\bf b)} \,Since \(a_{2k} >1 \), the left-hand condition in (\ref{H20}) allows us to apply (\ref{H10}) for \(m=k \): \[|\varepsilon_{2k}| + |\varepsilon_{2k+1}| \,<\, \frac{1}{\rho^{2k}} \,.\] {\bf c)} \,We estimate the right-hand sum in (\ref{1010}) again by (\ref{H10}). To check the conditions in (\ref{H20}), we use \(a_1a_2\cdots a_{2m-1} >1 \), which holds by \(m\geq k+1 \) and \(a_{2k} >1 \). Hence, \[\sum_{m=k+1}^{n_0} \big( |\varepsilon_{2m}| + |\varepsilon_{2m+1}| \big) \,<\, \sum_{m=k+1}^{n_0} \frac{1}{\rho^{2m}} \,.\] Altogether, we find with (\ref{1010}) that \begin{equation} S(n) \,<\, \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 \(k1 \) 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 \(a_1 >1 \). \\ For \(m=0 \) (provided that \(g=0 \)) the first condition in (\ref{H20}) is fulfilled by \(a_{2m}a_{2m+1} = a_0a_1 = a_1 >1 \). For \(m\geq 1 \) we know that \(a_1a_2 \cdots a_{2m-1} >1 \) always satisfies one part of the second condition. Therefore, we apply the inequality from (\ref{H10}): \[S(n) \,<\, \sum_{m=g}^{n_0} \frac{1}{\rho^{2m}} \,=\, \rho^{1-2g} - \rho^{-2n_0-1} \,. \] 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 \(a_{2k+1} > 1 \) for some subscript \(k\geq 1 \). This implies \(\rho > \alpha \). \\ \noindent{\em Case~3.1:\/} \, Let \(0\leq n <2k \). \\ 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) &\,\leq \,& \sum_{m=g}^{n_0} \Big( \frac{1}{\rho^{2m}} + (\rho - \alpha) F_{2m} \Big) \\ &\,=\,& \rho^{1-2g} - \rho^{-2n_0-1} + (\rho - \alpha) \sum_{m=g}^{n_0} F_{2m} \\ &\,=\,& \rho^{1-2g} - \rho^{-2n_0-1} + (\rho - \alpha) (F_{2n_0+1} - F_{2g-1}) \\ &\,\leq \,& \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 \(a_{2k+1} >1 \). \\ \noindent{\em Case~3.2.1:\/} \, Let \(k \geq g \). \\ Applying (\ref{1000}) again, we obtain \begin{eqnarray*} 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) \\ &\,<\,& \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}} \\ &\,=\,& \sum_{m=g}^{n_0} \frac{1}{\rho^{2m}} + (\rho - \alpha) \sum_{m=g}^{k-1} F_{2m} \\ &\,=\,& \rho^{1-2g} - \rho^{-2n_0-1} + (F_{2k-1} - F_{2g-1})(\rho - \alpha) \,. \end{eqnarray*} \noindent{\em Case~3.2.2:\/} \, Let \(k1 \). To simplify arguments, we introduce the function \(\chi (k,g) \) defined by \(\chi (k,g) = 1 \) (if \(k>g \)), and \(\chi (k,g) = 0 \) (if \(k \leq g \)). We have \begin{eqnarray} S &\,:=\,& \sum_{\nu =2g}^{\infty} |\varepsilon_{\nu}| \,=\, \sum_{\nu =2g}^{2k-1} |\varepsilon_{\nu}| + \sum_{\nu = \max \{ 2k,2g \}}^{\infty} |\varepsilon_{\nu}| \nonumber \\ &\,=\,& \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 \\ &\leq \,& (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 \\ &\leq \,& (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 \(a_1a_2 \cdots a_{2k+1} >1 \). Therefore, we obtain \begin{eqnarray*} S &\,<\,& \frac{1}{\rho^{2k}} - \frac{1}{\rho^{2k-1}} + \rho^{1-2g} + \sum_{m =k+1}^{\infty} \big( |\varepsilon_{2m}| + |\varepsilon_{2m+1}| \big) \\ &\,<\,& \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 &\,<\,& \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) \\ &\,<\,& \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}} \\ &\,=\,& \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 \[\left( \begin{array}{c} \varepsilon (\alpha) \\ \cdot \end{array} \right) \,=\, \sum_{n=0}^{\infty} {(-1)}^n \left( \begin{array}{cc} a_n & 1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{cc} a_{n-1} & 1 \\ 1 & 0 \end{array} \right) \cdots \left( \begin{array}{cc} a_0 & 1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{r} -1 \\ \alpha \end{array} \right) \,.\] \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) &\,=\,& \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}| \\ &\,=\,& \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 \(\alpha >1 \) 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. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{thebibliography}{99} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \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}