%%%% THIS FILE IS AUTOMATICALLY GENERATED %%%% DO NOT EDIT THIS FILE DIRECTLY, %%%% ONLY EDIT THE SOURCE, tom-26/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{corl}{Corollary} \newtheorem*{ackn}{Acknowledgement} \theoremstyle{namedthrm} \newtheorem{thrm}{Theorem} \title{Irrational Sums} \author{Tom C. Brown, D.-Y. Pei and Peter Jau-Shyong Shiue} \maketitle \begin{center}{\small {\bf Citation data:} T.C. {Brown} and P.J.-S. {Shiue}, \emph{Irrational sums}, Rocky Mountain J. Math. \textbf{25} (1995), 1219--1223.}\bigskip\end{center} \section{Introduction\label{s1}} In this note we give some sufficient conditions for the irrationality of the sum of the series $\sum_{n=1}^\infty 1/H(f(n))$, where $H(k))_{k\geq0}$ is a sequence of integers, positive from some point on, satisfying a homogeneous linear recurrence relation with integer coefficients, and $f$ is a strictly increasing function from the set of positive integers to the set of nonnegative integers. We will refer to such a sequence $(H(k))_{k\geq0}$ simply as a ``recurrent sequence,'' and the symbol $f$ will always denote a strictly increasing function from the set of positive integers to the set of nonnegative integers. Let us agree that the symbol $\sum1/H(f(n))$ denotes the summation of all those terms $1/H(f(n))$ for which $H(f(n))>0$. All of our results are based on the following theorem of C. Badea \cite{badea1987}. \begin{thrm}[A]\namedthmlabel{tA} (Badea \cite{badea1987}). If $(a_k)_{k\geq0}$ is a sequence of positive integers such that $a_{k+1}>a_k^2 - a_k+1$ for all sufficiently large $k$, then $\sum1/a_k$ is irrational.\end{thrm} A simple example to show that the converse of Badea's Theorem \ref{tA} is false is the series $\sum1/n! = e$. Another easy example to see that the converse of Badea's result is false is the following. Let $\{c_n\}$, $n\geq1$, be a nonperiodic sequence of 2's and 5's, and let $a_n = 10^n/c_n$, $n\geq1$. Then $\sum1/a_n$ is irrational, and $a_{n+1} < a_n^2 - a_n + 1$, $n\geq3$. Thus our goal is to find simple conditions on $H(k)$ and $f(n)$ which ensure that $H(f(n+1)) > H(f(n))^2 - H(f(n)) + 1$ for all sufficiently large $n$. To avoid complications, \emph{from now on we will always assume that the characteristic polynomial of the recurrent sequence $H(k)$ has a unique (real) root $\beta>1$ of maximum modulus}. It then follows from standard properties of recurrence relations (see, for example, \cite{liu1985}) that there exist numbers $A>0$ and $c\geq0$ such that $\lim_{k\to\infty} H(k)/(k^c\beta^k) = A$. (If $\beta$ is a root of multiplicity 1, then $c=0$.) \section{Main results.} \begin{thrm}[1]\namedthmlabel{t1} If $f(n+1)-2f(n)\to\infty$ as $n\to\infty$ and $f(n+1)\geq f(n)^2$ for all sufficiently large $n$, then $\sum1/H(f(n))$ is irrational for every recurrent sequence $H(k)$.\end{thrm} \begin{proof} Assume that $H(k)/k^c\beta^k\to A$ as $k\to\infty$ (where $\beta>1$, $A>0$ and $c\geq0$). To apply Badea's result, we need to show that $H(f(n+1))/(H(f(n))^2 - H(f(n))+1)>1$ for sufficiently large $n$. We do this by dividing the numerator and denominator of the lefthand side of this inequality by $f(n+1)^c\beta^{f(n+1)}$. Since $H(f(n+1))/f(n+1)^c\beta^{f(n+1)}\to A > 0$ as $n\to\infty$, then $H(f(n+1))/f(n+1)^c\beta^{f(n+1)} > (2/3)A$ for all sufficiently large $n$. Next, \begin{align*} \frac{H(f(n))^2 - H(f(n)) + 1}{f(n+1)^c\beta^{f(n+1)}} &= \frac{f(n)^{2c}}{f(n+1)^c}\frac1{\beta^q}\left(\frac{H(f(n))^2}{f(n)^{2c}\beta^{2f(n)}} - \frac{H(f(n))}{f(n)^{2c}\beta^{2f(n)}}\right) \\ &\qquad + \frac1{f(n+1)^c\beta^{f(n+1)}}, \end{align*} where $q = f(n+1)-2f(n)$. Since the expression inside the large brackets converges to $A^2$ and the other term converges to 0, for sufficiently large $n$ (using also $f(n)^{2c}/f(n+1)^c\leq1$) \[ \frac{H(f(n))^2 - H(f(n)) + 1}{f(n+1)^c\beta^{f(n+1)}} < \beta^{-q}(A^2 + 1) + (1/3)A. \] Finally, \[ \frac{H(f(n+1))}{H(f(n))^2 - H(f(n)) + 1} > \frac{(2/3)A}{\beta^{-q}(A^2 + 1) + (1/3)A} > 1, \] as required.\end{proof} \begin{corl} For every recurrent sequence $H(k)$, $\sum1/H(2^{2^n})$ is irrational.\end{corl} For the next result, we weaken the condition on $f$ and strengthen the condition on $H(k)$. \begin{thrm}[2]\namedthmlabel{t2} If $f(n+1)-2f(n)\to\infty$ as $n\to\infty$, then $\sum1/H(f(n))$ is irrational for every recurrent sequence $H(k)$ for which $\beta$ has multiplicity 1. (Recall that $\beta>1$ is the unique root of maximum modulus of the characteristic polynomial of $H(k)$.)\end{thrm} \begin{proof} The proof of Theorem \ref{t1}, with $c$ set equal to 0 throughout, gives a proof of Theorem \ref{t2}.\end{proof} \begin{corl}Let $H(k)$ be a recurrent sequence for which $\beta$ has multiplicity 1. Then for every $\epsilon>0$, $\sum1/H([(2+\epsilon)^n])$ is irrational. For every $0<\epsilon<1$, $\sum1/H(2^n - [(2-\epsilon)^2])$ is irrational.\end{corl} \begin{thrm}[3]\namedthmlabel{t3} Let $H(k)$ be a recurrent sequence for which $\beta$ has multiplicity 1. Then there exists an integer $P$ such that for every pair of fixed integers $s,p$ with $s>0$, $-\infty1$ and $A>0$. Let $s,p$ be given with $s>0$ and $p<-\log A/\log\beta$. Let $f(n) = s2^n+p$, $n\geq1$. Since $f(n+1) - 2f(n) = -p$, \[ \frac{H(f(n))^2 - H(f(n)) + 1}{\beta^{f(n+1)}} = \frac{1}{\beta^{-p}}\left(\frac{H(f(n))^2}{\beta^{2f(n)}} - \frac{H(f(n))}{\beta^{2f(n)}}\right) + \frac1{\beta^{f(n+1)}}\to \beta^pA^2. \] Thus, since $H(f(n+1))/\beta^{f(n+1)}\to A$, \[ \frac{H(f(n+1))}{H(f(n))^2 - H(f(n)) + 1} \to \frac1{\beta^pA}. \] Since $\beta^pA < 1$ by the choice of $p$, \[ \frac{H(f(n+1))}{H(f(n))^2 - H(f(n)) + 1} > 1 \] for sufficiently large $n$, and therefore $1/\sum H(f(n)) = \sum1/H(s2^n+p)$ is irrational, by Badea's theorem.\end{proof} \section{Remarks\label{s3}} For the Fibonacci sequence $F(k)$, where \[ F(0) = 0,\qquad F(1) = 1,\qquad F(k+2) = F(k+1) + F(k), ~~ k\geq0 \] \[ F(k) = (1/\sqrt5)(((1 + \sqrt5)/2)^k - ((1 - \sqrt5)/2)^k), \] \[ \beta = (1 + \sqrt5)/2, \qquad A = 1/\sqrt5, \] $-\log A/\log\beta = 1.67\ldots$. Thus, according to the proof of Theorem \ref{t3}, $\sum1/F(s2^n+p)$ is irrational for every fixed pair of integers $s>0$ and $p\leq1$. This is a generalization of a result of C. Badea \cite{badea1987}, who showed, answering a question of Erd\H{o}s and Graham \cite{erdos+graham1980}, that $\sum1/F(2^n+1)$ is irrational. More generally, let $H(0)=0$, $H(1)=1$, $H(k+2) = aH(k+1) + bH(k)$, $k\geq0$, where $a\geq1$, $b\geq1$. Then $H(k) = (1/\sqrt{a^2 + 4b})(((a + \sqrt{a^2+4b}) /2)^k - ((a - \sqrt{a^2 + 4b})/2)^k)$, $\beta = (a + \sqrt{a^2 + 4b})/2$, $A = 1/\sqrt{a^2 + 4b}$, and $\beta^pA < 1$ for $p\leq 1$, so again $\sum 1/H(s2^n+p)$ is irrational for every fixed pair of integers $s>0$ and $p\leq1$. This extends a result of Kuipers \cite{kuipers1977} (see also \cite{laohakosol+roenrom1984}), who showed this in the case $b=1$ and $p=0$. (One can relax the requirement $a\geq1$, $b\geq1$ to $a=1$, $b\geq1$ or $a\geq2$, $a^2 + 4b >0$. In these cases $A<1<\beta$, so that $\beta^pA < 1$ holds for $p\leq0$ and $\sum1/H(s2^n+p)$ is irrational for $s>0$ and $p\leq0$.) If $a^2+4b<0$, so that the characteristic polynomial $x^2-ax-b$ of the sequence $H(k)$ no longer has a unique root of maximum modulus, it is easy to verify that the sequence $H(k)$ has infinitely many negative terms, for any nontrivial initial values $H(0)$, $H(1)$. For such a sequence the present methods give no information about the irrationality of $\sum1/H(f(n))$ for any function $f$. Some examples of polynomials for which $\beta>1$ and $b$ has multiplicity 1 ($\beta$ is the unique root of maximum modulus for the given polynomial) are discussed in Hua and Wang \cite{hua+wang1981}, including the polynomials $x^d-x^{d-1}-\cdots-x-1$, $d\geq2$, (which come from the generalized Fibonacci sequences $F(0) = F(1) = \cdots F(d-2) = 0$, $F(d-1) = 1$, $F(k+d) = F(k+d-1) + F(k+d-2) + \cdots + F(k+1) + F(k)$, $k\geq0$), $x^d - Lx^{d-1}-1$, $d\geq2$, $L\geq2$, and $x^t - t^2r^{t-1}x^{t-1} + (-1)^{t-2} A_{t-2}r^{t-2}x^{t-2} + \cdots - A_1rx - 1 = 0$, $t\geq2$, where \[ A_1 = \binom{2t}1,\qquad A_k = \binom{2t}k - A_1\binom{2t-2}{k-1} - \cdots - A_{k-1}\binom{2t-2k+2}1, \] $t - 2\geq k>1$, and the positive integer $r$ satisfies $t^2>2/r^{t-1} + |A_1|/r^{t-2} + \cdots + |A_{t-2}|/r$. \begin{ackn} We are grateful to the referee for several helpful remarks. \end{ackn} \bibliographystyle{amsplain} \bibliography{tom-all} \end{document}