%%%% THIS FILE IS AUTOMATICALLY GENERATED %%%% DO NOT EDIT THIS FILE DIRECTLY, %%%% ONLY EDIT THE SOURCE, tom-27/document.tex. %%%% %% Standard package list \documentclass[letterpaper]{article} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage[english]{babel} \usepackage[top=3cm, bottom=3cm, left=3.5cm, right=3.5cm]{geometry} \usepackage[onehalfspacing]{setspace} \usepackage{amsmath,amssymb,amsthm,wasysym} \usepackage{nicefrac,booktabs} \usepackage{mathptmx} \usepackage{cite} \usepackage[colorlinks=true]{hyperref} %% Various helpers for Tom's papers \newcommand{\gs}{\textnormal{gs}} \newcommand{\ord}{\textnormal{ord}} \newcommand{\Exp}{\textnormal{Exp}} \newcommand{\Log}{\textnormal{Log}} \newcommand{\lcm}{\textnormal{lcm}} \newcommand{\range}{\textnormal{range}} \newcommand{\NR}{\textnormal{NR}} \newcommand{\Mod}[1]{\left(\textnormal{mod}~#1\right)} \newcommand{\ap}[2]{\left\langle #1;#2 \right\rangle} \newcommand{\summ}[1]{\sum_{k=1}^m{#1}} \newcommand{\bt}[1]{{{#1}\mathbb{N}}} \newcommand{\fp}[1]{{\left\lbrace{#1}\right\rbrace}} \newcommand{\intv}[1]{{\left[1,{#1}\right]}} %% Lifted from http://stackoverflow.com/questions/2767389/referencing-a-theorem-like-environment-by-its-name %% This lets me do things like "Theorem A" and have the references work properly. \makeatletter \let\@old@begintheorem=\@begintheorem \def\@begintheorem#1#2[#3]{% \gdef\@thm@name{#3}% \@old@begintheorem{#1}{#2}[#3]% } \def\namedthmlabel#1{\begingroup \edef\@currentlabel{\@thm@name}% \label{#1}\endgroup } \makeatother % end lift \newtheoremstyle{namedthrm} {}{}{}{}{}{}{ } % This last space needs to be there {\bf\thmname{#1} \thmnote{#3}.} %% End reference hack %% Document start \date{} \begin{document} %% Content start \newtheorem{thrm}{Theorem} \newtheorem{fact}{Fact} \newtheorem{lmma}{Lemma} \newtheorem*{ackn}{Acknowledgement} \title{Sums of Fractional Parts of Integer Multiples of an Irrational} \author{ Tom C. Brown\footnote{Department of Mathematics and Statistics, ¶¡ÏãÔ°AV, Burnaby, British Columbia, V5A 1S6, Canada. \texttt{tbrown@sfu.ca}} and Peter Jau-shyong Shiue\footnote{Department of Mathematical Sciences, University of Nevada, Las Vegas, NV, USA 89154-4020. \texttt{shiue@nevada.edu}}} \maketitle \begin{center}{\small {\bf Citation data:} T.C. {Brown} and P.J.-S. {Shiue}, \emph{Sums of fractional parts of integer multiples of an irrational}, J. Number Theory \textbf{50} (1995), 181--192.}\bigskip\end{center} \newcommand{\floor}[1]{{\left[{#1}\right]}} \begin{abstract}Let $\alpha$ be a positive irrational real number, and let $C_\alpha(n) = \sum_{1\leq k\leq n}(\{k\alpha\}-\frac12)$, $n\geq1$, where $\{x\}$ denotes the fractional part of $x$. We give an explicit formula for $C_\alpha(n)$ in terms of the simple continued fraction for $\alpha$, and use this formula to give simple proofs of several results of A. Ostrowski, G. H. Hardy and J. E. Littlewood, and V. T. S\'os. We also show that there exist positive constants $d_A$ such that if $\alpha = [a_0,a_1,a_2,\ldots]$ and $(1/t)\sum_{1\leq j\leq t}a_j\leq A$ holds for infinitely many $t$, then $C_\alpha(x) > d_A\log x$ and $C_\alpha(x) < - d_A\log x$ each hold for infinitely many $x$. \end{abstract} \section{Introduction \label{s1}} For real numbers $\beta$, let us agree to write $\{\beta\}$ for the fractional part of $\beta$, that is, $\{\beta\} = \beta-\floor{\beta}$, where $\floor{\beta}$ is the greatest integer less than or equal to $\beta$. For an irrational number $\alpha$, $0<\alpha<1$, and for integers $n\geq1$, we write \[ C_\alpha(n) = \sum_{1\leq k\leq n} (\{k\alpha\} - \frac12). \] There are a number of papers dealing with estimates of $C_\alpha(n)$ as a function of $n$. Most of the known results are contained in two long 1922 papers by G. H. Hardy and J. E. Littlewood \cite{hardy+littlewood1922-1, hardy+littlewood1922-2}, a long 1922 paper by A. Ostrowski \cite{ostrowski1922}, and a 1957 paper by Vera T. S\'os \cite{sos1957}. Further references can be found in \cite{shallit1992}. In the present note we give a simple and self-contained proof of a formula for $C_\alpha(n)$ which is more explicit than those which have appeared before, namely in \cite{ostrowski1922,sos1957}. We then use this formula to derive several of the main results in \cite{hardy+littlewood1922-1, hardy+littlewood1922-2,ostrowski1922} and the main result in \cite{sos1957}. It seems to us that our proofs are far simpler than those given previously. In some cases they lead to improvements. In addition, we extend one of the main results in \cite{hardy+littlewood1922-2, ostrowski1922}. Ostrowski and Hardy and Littlewood showed independently that there exist positive constants $c$ and $c_A$ with the following properties. If $\alpha$ is an arbitrarily positive irrational number, then the inequality $|C_a(x)|>c\log x$ holds for infinitely many $x$. If $\alpha=[a_0,a_1,a_2, \ldots]$ and $a_i\leq A$ for all $i$, then $C_a(x)> c_A\log x$ and $C_a(x) < - c_A\log x$ each hold for infinitely many $x$. (Hardy and Littlewood \cite{hardy+littlewood1922-2} call the proofs of these two results the most difficult in their paper. They give no values for the constants $c$ and $c_A$. Ostrowski gives $c\geq\frac1{720}$ and $c_A\geq1/8(A+1)^6$.) We show that there exist positive constants $d_A$ such that if $\alpha[a_0,a_1,a_2,\ldots]$ and $(1/5)\sum_{1\leq j\leq t}a_j\leq A$ holds for infinitely many $t$, then $C_\alpha(x) > d_A\log x$ and $C_\alpha(x) < - d_A\log x$ each hold for infinitely many $x$. (We show that $d_A\geq 1/(7\cdot64(A+1)^2\log(A+1))$. Since $c_A\geq d_A$, this improves Ostrowski's bound. We also show that $c\geq\frac1{256}$.) \section{Notation\label{s2}} Let $\alpha$ be a positive irrational real number, and let \[ \alpha = a_0 + \frac1{a_1 + \frac1{a_2 + \cdots }} \] be the simple continued fraction for $\alpha$, which we abbreviate as $\alpha = [a_0,a_1,a_2,\ldots]$. We write $p_n/q_n = [a_0,a_1,a_2,\ldots, a_n]$ and $d_{2k} = q_{2k}\alpha-p_{2k}$, $d_{2k+1} = p_{2k+1}-q_{2k+1} \alpha$, $k\geq0$. Then $p_n = a_np_{n-1}+p_{n-2}$, $q_n = a_nq_{n-1}+ q_{n-2}$, $n\geq2$, $p_{2k}/q_{2k}<\alpha\frac1{16}\sum_{j\text{ even}}(a_j-1)$. Thus $C_\alpha(M_2) - C_\alpha (M_1) > \frac1{16}\sum_{1\leq j\leq t}(a_j-1)$, therefore $\frac1{32}\sum_{1\leq j\leq t} (a_j-1)<\max_{00$, choose $n$ so that $(\sqrt2)^{-(n-1)} <\epsilon$, and let $P = a_1 + \cdots + a_n$. Choose $s>n$ so that $P/q_n<\epsilon$. For $m\geq q_s$, let $m=\sum_{1\leq j\leq t + 1}z_jq_{j-1}$; then $t\geq s$ and $z_{t+1}\geq1$. Then \begin{align*} \frac1m|C_\alpha(m)| &< \frac1{2m}\sum_{1\leq j\leq t}z_j \leq \frac12\frac{P}{q_t} + \frac12\frac{\sum_{n+1\leq j \leq t + 1}z_j}{\sum_{n+1\leq j\leq t + 1}z_jq_{j-1}} \\ &< \frac12\epsilon + \frac12\frac{\sum_{n+1\leq j\leq t + 1}z_j}{(\sqrt2)^{n-1}\sum_{n+1\leq j\leq t + 1}z_j} < \epsilon. \end{align*} (Here we used $q_n>(\sqrt2)^{n-1}$.)\end{proof} \begin{fact}\label{f2} (Lerch (without proof) \cite{lerch1904}, Hardy and Littlewood \cite{hardy+littlewood1922-1}, and Ostrowski \cite{ostrowski1922} have this result under the stronger hypothesis that $a_i\leq A$ for all $i$.) If $\alpha=[0,a_1,a_2,\ldots]$ and $(1/t)\sum_{1\leq j \leq t}a_j\leq A$ for all $i$, then $C_\alpha(n) = O(\log n)$. In fact, \[ |C_\alpha(n)| < \frac{A}{2\log\tau}\log n + \left(\frac{\log\sqrt5}{2\log\tau}-\frac12\right)A, \qquad n\geq 2,\text{ where } \tau = \frac{1+\sqrt5}2. \] Consequently, $|C_\alpha(n)|<\frac32A\log n$, $n\geq1$, and for every $\epsilon>0$ there is $N_0$ such that $|C_\alpha(n)| < (1/(2\log\tau) +\epsilon)A\log n$, $n\geq N_0$. In particular, $|C_\alpha(n)|<(1.04) A\log n$, $n\geq N_0$.\end{fact} \begin{proof} Let \[ \frac{P_n}{Q_n} = \overbrace{[1,1,\ldots,1]}^{n+1}. \] Then $Q_0 = Q_1 = 1$, $Q_{n+1} = Q_n + Q_{n-1}$, and $Q_t = (1/\sqrt5) (\tau^{t+1} - (-1/\tau)^{t+1}) > (1/\sqrt5)(\tau^{t+1} - 1)$. Since $|C_\alpha(q_t)|<\frac12$, we can assume without loss of generality that $q_t \sqrt5q_t+1\geq\sqrt5Q_t+1> \tau^{t+1}$, so $t+1<(\log\sqrt5 + \log n)/\log\tau$, $\frac12t < \frac12(\log\sqrt5/\log\tau - 1) + \log n/(2\log\tau)$. Since $|C_\alpha(n)|<\frac12\sum_{1\leq j\leq t}a_j\leq \frac12tA$, the result follows.\end{proof} \begin{fact}\label{f3} If $f(n) = o(n)$, then there exists $\alpha$ such that \[ \limsup_{n\to\infty}\left|\frac{C_\alpha(n)}{f(n)}\right| = \infty. \] \end{fact} \begin{proof} Assume that $f(n)>0$ for all $n$. Suppose $a_1,\ldots,a_t$ have already been chosen. Choose $M$ so that $f(x)/x<1/64(t+1)q_t$, $x>M$. Let $P = \max\{f(x):1\leq x\leq M\}$. Choose $a_{t+1}\geq 3$ so that $\frac1{32}(a_{t+1}-1)>(t+1)P$. Finally, choose $x$, $0 \frac{(a_{t+1}-1)}{32P} > (t+1). \] If $M< x< q_{t+1}$, then \[ \frac{|C_\alpha(x)|}{f(x)} = \frac{|C_\alpha(x)|}{x}\frac{x}{f(x)} > \frac{(a_{t+1}-1)}{32q_{t+1}}\cdot64(t+1)q_t > t + 1, \] since $2(a_{t+1}-1)q_t \geq (a_{t+1} + 1)q_t > q_{t+1}$.\end{proof} \begin{fact}\label{f4} If $A\geq2$ and $f(n) = o(\log n)$, then there exists $\alpha = [0,a_1,a_2,\ldots]$ with $a_i\leq A$ for all $i$ and \[ \limsup_{n\to\infty}\left|\frac{C_\alpha(n)}{f(n)}\right| = \infty. \] In fact, this is true for every $\alpha$ such that $a_i1$.\end{fact} \begin{proof} Assume that $f(n)>0$ for all $n$. Fix $L\geq1$. Choose $M$ so that $f(x)/\log x < (c-1)/32L\log(A+1)$, $x>M$. Let $P = \max\{f(x) : 1 \leq x\leq M\}$. Choose $t$ so that $\frac1{32}\sum_{1\leq i\leq t} (a_j-1) \geq\frac1{32}(c-1)t > LP$. Choose $x$, $0 \frac1{32}\sum_{1\leq i\leq t}(a_j-1) > LP$. If $1\leq x\leq M$ then \[ \frac{|C_\alpha(x)|}{f(x)} \geq \frac{|C_\alpha(x)|}{P} > L. \] If $M < x < q_t$ then \[ \frac{|C_\alpha(x)|}{f(x)} = \frac{|C_\alpha(x)|}{\log x}\frac{\log x}{f(x)} > \frac{(c-1)t}{32\log q_t}\cdot\frac{32L\log(A+1)}{(c-1)} > L, \] where the last inequality holds since $q_{s+1}\leq Aq_s + q_{s-1}<(A+1) q_s$, so $\log q_t< t\log(A+1)$.\end{proof} \section{New Results\label{s6}} \begin{thrm}\label{t3} Let $\alpha = [a_0,a_1,a_2,\ldots]$. \begin{enumerate} \item[(a)] If $\sum_{1\leq i\leq t}a_j\geq(1+\frac17)t$ infinitely often, then the inequality $|C_\alpha(x)| > \frac1{256}\log x$ holds for infinitely many $x$. \item[(b)] If $\sum_{1\leq i\leq t}a_j \leq (1+\frac17)t$ infinitely often, then $C_\alpha(x) > \frac1{56}\log x$ and $C_\alpha(x) < -\frac1{56}\log x$ each hold for infinitely many $x$. \end{enumerate}\end{thrm} \begin{lmma} (\cite{ostrowski1922}). For each $t\geq1$, $\sum_{1\leq i\leq t} a_i>\log q_t$.\end{lmma} \begin{proof}[Proof of Lemma.] $q_t\leq (a_t+1)q_{t-1}\cdots \leq (a_t+1)\cdots (a_2+1)(a_1+1)\leq e^{a_t}\cdots e^{a_2}e^{a_1}$.\end{proof} \begin{proof}[Proof of Theorem.] \begin{enumerate} \item[(a)] For $t$ such that $\sum_{1\leq i\leq t}a_j\leq (1 + \frac17)t$, apply Theorem \ref{t2} to get $x$, $0 \frac1{32} \sum_{1\leq j\leq t}(a_j-1)\geq\frac1{32}(t/7)$. By the Lemma, either $\sum_{1\leq j\leq t}(a_j - 1) < \frac18\log q_t$ or $t>\frac78\log q_t$; in either case we get $|C_\alpha(x)| > (1/(8\cdot32))\log x$. \item[(b)] For $t$ such that $\sum_{1\leq j\leq t}a_j\leq (1 + \frac17)t$, let $G = \{ j : 6j+1\leq t, a_{6j}=a_{6j+1}=1\}$. Since the number of $a_j>1$ among $a_1,\ldots,a_t$ is at most $\floor{t/7}$, it follows that $|G|\geq6t/7$. Let $x = \sum_{j\in G} q_{6j}$. Note that $0 \log q_t > \log x$. According to Theorem \ref{t1}(c), $C_\alpha(x) = -\sum_{j\in G}(\frac12 - d_{6j}(m_{6j} + \frac12q_{6j} + \frac12))$. To estimate $d_{6j}m_{6j}$, we use $d_{6j}<1/q_{6j+1}$, $m_{6j}\leq q_6 + q_{12} + \cdots + q_{6j+1}$ (since $q_s < 2q_{s+2}$), $q_{12}<(1/8^{j-2})q_{6j+1}$, etc., to get \[ d_{6j}m_{6j} < \frac18 + \frac1{8^2} + \cdots + \frac1{8^{j-1}} = \frac17 - \frac17\cdot\frac1{8^{j-1}}. \] Next, for $j\in G$ we have $q_{6j}/q_{6j+1} = (q_{6j-1} + q_{6j-2})/(2q_{6j-1} + q_{6j-2}) < \frac23$. Finally, $d_{6j} < 1/q_{6j+1} < 1/8^j$. Therefore \begin{align*} C_{\alpha}(x) &< -\sum_{j\in G} \left(\frac12 - \left(\frac17 - \frac17\cdot\frac1{8^{j-1}} + \frac13 + \frac12\cdot\frac1{8^j}\right)\right) \\ &< -\sum_{j\in G} \left(\frac12 - \left(\frac17 + \frac13\right)\right) = - |G|\frac1{42}. \end{align*} Since $|G|\geq 6t/7$ and $8t/7>\log x$, this gives $C_\alpha(x) < -\frac1{56}\log x$. \end{enumerate} The proof that $C_\alpha(x)>\frac1{56}\log x$ for infinity many $x$ is essentially the same.\end{proof} \begin{thrm}\label{t4} Let $A$ be a positive constant. Then there exists a positive constant $d_A$ such that if $\alpha = [0,a_1,a_2,\ldots]$ is any irrational such that $\sum_{1\leq i\leq t}a_j\leq At$ for infinitely many $t$, then each of $C_\alpha(x) > d_A\log x$ and $C_\alpha(x) < - d_A\log x$ holds for infinitely many $x$.\end{thrm} \begin{proof} We show that $C_\alpha(x) < -d_A\log x$ holds infinitely often. The proof of the other inequality is essentially the same. Our method is similar to the proof of Fact \ref{f5}. First, choose $L$ so that $1/(2^L-1)<\frac14\cdot(1/(4A+2))$. Let $t>4L$ and $t>12A$, with $\sum_{1\leq i\leq t}a_j\leq At$. From among the even terms $a_2,a_4,\ldots,a_{2k},\ldots$ with $2k 2^La_{j_1}$, etc. (since $j_2-j_1\geq2L$), we get \[ d_{j_k}m_{j_k} < \frac1{2^L}+\frac1{2^{2L}} + \cdots + \frac1{2^{(k-1)L}} = \frac1{2^L-1}\left(1 - \frac1{2^{(k-1)L}}\right). \] Next, we use the (easily verified) fact that if $a_{s+1}\geq2$ then $q_s/q_{s+1} <\frac12$ and if $a_{s+1} = 1$ then $q_s/q_{s+1}<1-1/(a_s+2)$. Since $a_{j_k} \leq4A$ (and $\frac12<1-1/(4A + 2)$), $d_{j_k}q_{j_k} < q_{j_k}/q_{j_k+1} < 1 - 1/(4A + 2)$. Finally, $d_{j_k}<1/q_{j_k+1}<1/2^{kL}$. Putting all these together gives \begin{align*} C_\alpha(x) &< -\sum_{1\leq k\leq w}\left(\frac12 - \frac1{2^L-1}\left(1 - \frac1{2^{(k-1)L}}\right) - \frac12\left(1 - \frac1{4A + 2}\right) - \frac12\frac1{2^{kL}}\right) \\ &< -\sum_{1\leq k\leq w}\left(\frac12\left(\frac1{4A + 2}\right) - \frac1{2^L-1}\right) < -w\cdot\frac14\cdot\frac1{4A+2} < -dt, \end{align*} since $w = \floor{t/4L}$ and $L$ depends only on $A$. This completes the proof. \end{proof} By tracing back through the choice of $L$, one can see that $d_A\geq1/(7\cdot 64(A+1)^2\log(A+1))$. \begin{thrm}\label{t5} Let $\alpha = [0,a_1,a_2,\ldots]$. \begin{enumerate} \item[(a)] If $\liminf_{t\to\infty}(1/t)\sum_{1\leq j\leq t}a_j = 1$, then $C_\alpha(x) > \frac1{56}\log x$, $C_\alpha(x) < -\frac1{56}\log x$ hold infinitely often. \item[(b)] If $\liminf_{t\to\infty}(1/t)\sum_{1\leq j\leq t}a_j < \infty$, then $C_\alpha(x) > d\log x$, $C_\alpha(x) < -d\log x$ hold infinitely often, for some $d>0$. \item[(c)] If $\liminf_{t\to\infty}(1/t)\sum_{1\leq j\leq t}a_j = \infty$, then such a $d$ may fail to exist. (See the theorem of Vera T. S\'os below.) However, for every $\epsilon>0$, $|C_\alpha(x)| > (\frac1{32} - \epsilon) \log x$ holds infinitely often. \end{enumerate}\end{thrm} \begin{proof} Part (a) follows from Theorem \ref{t3}(b). Part (b) follows from Theorem \ref{t4}. Part (c) is proved in the same way as Theorem \ref{t3}(a). \end{proof} \section{The Vera T. S\'os Theorem} \setcounter{lmma}{0} Our final application of Theorem \ref{t1}(c) will be a simplified proof of the following result of Vera T. S\'os \cite{sos1957}, which answered a question of Ostrowski \cite{ostrowski1922}. \begin{fact}\label{f5} Let $\alpha = [0,a_1,a_2,\ldots]$, where $a_{2n+1}=1$, $a_{2n}=n^2$, $n\geq0$. Then there exists a constant $C$ such that $C_\alpha(n) > C$ for all $n\geq1$.\end{fact} (In S\'os's paper, the $a_i$'s are indexed differently, and $n^3$ appears rather than $n^2$.) \begin{lmma}\label{l61} For $k\geq1$, $\frac12(q_{2k-2}/q_{2k-1} + d_{2k}/d_{2k-1} -2) < k^2(\frac12 - d_{2k-1}q_{2k-1}(1 + \frac12k^2)) < 0$.\end{lmma} \begin{proof} Using $d_{s+1} (1 + \frac12k^2)/(k^2 + 2) = \frac12$, which gives the right-hand inequality. Next, \begin{align*} 0 &> \frac12 - d_{2k-1}q_{2k-1}\left(1 + \frac12k^2\right) \\ &= \frac12 - \frac{1 + \frac12k^2}{k^2 + \frac{q_{2k-2}}{q_{2k-1}} + \frac{d_{2k}}{d_{2k-1}}} \\ &= \frac{\frac{q_{2k-2}}{q_{2k-1}} + \frac{d_{2k}}{d_{2k-1}} - 2}{2\left(k^2 + \frac{q_{2k-2}}{q_{2k-1}} + \frac{d_{2k}}{d_{2k-1}}\right)} \\ &> \frac{\frac{q_{2k-2}}{q_{2k-1}} + \frac{d_{2k}}{d_{2k-1}} - 2}{2k^2}, \end{align*} which gives the left-hand inequality. \end{proof} \begin{lmma}\label{l62}For $k\geq2$, $q_{2k-2}/q_{2k-1} > 1 - 1/(k-1)^2$. For $k\geq1$, $d_{2k}/d_{2k-1} > d_{2k}q_{2k} > 1 - 2/k^2$.\end{lmma} \begin{proof} For $k\geq 2$, \[ \frac{q_{2k-2}}{q_{2k-1}} = \frac{q_{2k-2}}{q_{2k-2} + q_{2k-3}} = 1 - \frac{q_{2k-3}}{q_{2k-2} + q_{2k-3}} > 1 - \frac{q_{2k-3}}{q_{2k-2}} > 1 - \frac1{(k-1)^2}. \] For $k\geq1$, \begin{align*} \frac{d_{2k}}{d_{2k-1}} &> d_{2k}q_{2k} \\ &= \frac1{\frac{q_{2k+1}}{q_{2k}} + \frac{d_{2k+1}}{d_{2k}}} \\ &= \frac1{1 + \frac{q_{2k-1}}{q_{2k}} + \frac{d_{2k+1}}{d_{2k}}} \\ &> \frac1{1 + \frac1{k^2} + \frac1{(k+1)^2}} > 1 - \frac2{k^2} \end{align*} (Here we used $q_{2k} = k^2q_{2k-1} + q_{2k-2}$ and $d_{2k} = (k+1)^2d_{2k+1} + d_{2k+2}$.) \end{proof} \begin{proof}[Proof of Fact 5.] Let $m = \sum_{1\leq j\leq t}z_jq_{j-1}$ be the Zeckendorff representation of $m$. Then \[ C_\alpha(m) = \sum_{1\leq j\leq t} (-1)^j z_j(\frac12 - d_{j-1}(m_{j-1} + \frac12z_jq_{j-1}+\frac12)) = D_0(m) - D_1(m), \] where \[ D_0(m) = \sum_{1\leq 2k\leq t} z_{2k}(\frac12 - d_{2k-1}(m_{2k-1} + \frac12 z_{2k}q_{2k-1} + \frac12)), \] \[ D_1(m) = \sum_{1\leq 2k+1\leq t} z_{2k+1}(\frac12 - d_{2k}(m_{2k} + \frac12 z_{2k+1}q_{2k} + \frac12)). \] We wish to find constants $A$ and $B$ such that $D_0(m)\geq A$, $D_1(m)\geq B$, $m\geq1$, so that $C_\alpha(m)\geq A - B$, $m\geq1$. Since $m_{2k-1}\leq q_{2k-1}$, we have \begin{align*} D_0(m) &> \sum_{1\leq 2k\leq t} z_{2k}(\frac12 - d_{2k-1}(q_{2k-1} + \frac12z_{2k}q_{2k-1})) \\ &= \sum_{1\leq 2k\leq t} z_{2k}(\frac12 - d_{2k-1}q_{2k-1}(1 + \frac12z_{2k})). \end{align*} Using $z_{2k}\leq a_{2k} = k^2$ and part of Lemma \ref{l61}, this gives $D_0(m) \geq \sum_{1\leq 2k\leq t}k^2(\frac12 - d_{2k-1}q_{2k-1}(1 + \frac12k^2))$. The other part of Lemma \ref{l61}, followed by Lemma \ref{l62}, now gives \begin{align*} D_0(m) &> \frac12\sum_{k=1}^\infty \left(\frac{q_{2k-2}}{q_{2k-1}} + \frac{d_{2k}}{d_{2k-1}} - 2\right) &> \frac12\left(\frac{q_0}{q_1} + \frac{d_2}{d_1} - 2\right) + \frac12\sum_{k=1}^\infty\left(-\frac1{(k-1)^2} - \frac2{k^2}\right) \\ &= A. \end{align*} Next, first omitting some negative terms from $D_1(m)$ gives \[ D_1(m) \leq \sum_{1 \leq 2k+1\leq t}z_{2k+1}(\frac12 - \frac12d_{2k}z_{2k+1}q_{2k}). \] Using Lemma \ref{l62} gives \begin{align*} D_1(m) &\leq \frac12z_1(1 - d_0z_1q_0) + \frac12\sum_{3\leq2k+1\leq t}z_{2k+1}\left(1 - z_{2k+1}\left(1 - \frac2{k^2}\right)\right) \\ &\leq \frac12z_1(1 - d_0z_1q_0) + \frac12\sum_{k=1}^\infty\frac{2}{k^2} = B. \end{align*} (For the last inequality, we used $0\leq z_{2k+1}\leq a_{2k+1} = 1$.) \end{proof} \begin{ackn} The authors are grateful to the referee for Refs. \cite{hardy+littlewood1922-1,hardy+littlewood1922-2,ostrowski1922,sos1957}, and for the statement of Theorem \ref{t1}(c) and the statement and proof of Theorem \ref{t2}, which simplifies so many of the subsequent proofs. Indeed, without theorem \ref{t2} it is possible that Theorems \ref{t3}, \ref{t4} and \ref{t5} would never have been noticed.\end{ackn} \bibliographystyle{amsplain} \bibliography{tom-all} \end{document}