%%version Dec7/07 %%%%% \documentclass[12pt]{article} \usepackage{amsmath,amssymb,amsbsy,amsfonts,amsthm,latexsym, amsopn,amstext,amsxtra,euscript,amscd,amsthm} \newtheorem{thm}{Theorem} \newtheorem{theorem}[thm]{Theorem} \newtheorem{conj}{Conjecture} \newtheorem{conjecture}[conj]{Conjecture} \textwidth= 6.5in \textheight= 9.0in \topmargin = -20pt \evensidemargin=0pt \oddsidemargin=0pt \headsep=25pt \parskip=10pt \font\smallit=cmti10 \font\smalltt=cmtt10 \font\smallrm=cmr9 \def\cA{{\mathcal A}} \def\cB{{\mathcal B}} \def\cC{{\mathcal C}} \def\cE{{\mathcal E}} \def\cF{{\mathcal F}} \def\cI{{\mathcal I}} \def\cL{{\mathcal L}} \def\cM{{\mathcal M}} \def\cN{{\mathcal N}} \def\cR{{\mathcal R}} \def\cS{{\mathcal S}} \def\cP{{\mathcal P}} \def\cQ{{\mathcal Q}} \def\cT{{\mathcal T}} \def\cX{{\mathcal X}} \def\N{{\mathbb N}} \def\Z{{\mathbb Z}} \def\R{{\mathbb R}} \def\C{{\mathbb C}} \def\eps{\varepsilon} \newcommand{\comm}[1]{\marginpar{\fbox{#1}}} \def\xxx{\vskip5pt\hrule\vskip5pt} \def\yyy{\vskip5pt\hrule\vskip2pt\hrule\vskip5pt} \makeatletter \renewcommand\section{\@startsection {section}{1}{\z@}% % {-3.5ex \@plus -1ex \@minus -.2ex}% % here is the vskip of 30pt: {-30pt \@plus -1ex \@minus -.2ex}% {2.3ex \@plus.2ex}% {\normalfont\normalsize\bfseries}} \renewcommand\subsection{\@startsection{subsection}{2}{\z@}% {-3.25ex\@plus -1ex \@minus -.2ex}% {1.5ex \@plus .2ex}% {\normalfont\normalsize\bfseries}} % add a point after section numbers: \renewcommand{\@seccntformat}[1]{\csname the#1\endcsname. } %\quad} \makeatother \begin{document} \vspace*{-40pt} \centerline{\smalltt INTEGERS: \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 7 (2007), \#A31} \vskip 40pt \begin{center} {\bf ON THE IRRATIONALITY OF A DIVISOR FUNCTION SERIES} \vskip 20pt {\bf John B.~Friedlander\footnote{Research of J.~F.was supported in part by NSERC grant A5123.}}\\ {\smallit Department of Mathematics, University of Toronto, Toronto, Ontario M5S 3G3, Canada}\\ {\tt frdlndr@math.toronto.edu}\\ \vskip 10pt {\bf Florian Luca\footnote{Research of F.~L. was supported in part by grants SEP-CONACYT 46755 and a Guggenheim Fellowship.}}\\ {\smallit Instituto de Matem{\'a}ticas, Universidad Nacional Aut\'onoma de M{\'e}xico, C.P. 58089, Morelia, Michoac{\'a}n, M{\'e}xico}\\ {\tt fluca@matmor.unam.mx}\\ \vskip 10pt {\bf Mihai Stoiciu}\\ {\smallit Department of Mathematics and Statistics, Williams College, Williamstown, MA 01267, USA}\\ {\tt mstoiciu@williams.edu} \end{center} \vskip 30pt \centerline{\smallit Received: 12/8/06, Revised: 3/20/07, Accepted: 6/12/07, Published: 7/3/07} \vskip 30pt \centerline{\bf Abstract} \noindent Here, we show, unconditionally for $k=3$, and on the prime $k$-tuples conjecture for $k\ge 4$, that $ \sum_{n=1}^{\infty} \frac{\sigma_k(n)}{n!} $ is irrational, where $\sigma_k(n)$ denotes the sum of the $k$th powers of the divisors of $n$. \pagestyle{myheadings} \markright{\smalltt INTEGERS: \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 7 (2007), \#A31\hfill} \thispagestyle{empty} \baselineskip=15pt \section{Introduction} For a positive integer $n$ put $$ \sigma_k(n)=\sum_{d\, \mid\, n} d^k $$ for the sum of the $k$th powers of the (positive) divisors of $n$. In \cite{EK}, Erd\H os and Kac showed that the series $$ \sum_{n=1}^{\infty}\frac{\sigma_k(n)}{n!} $$ is irrational for both $k=1$ and $k=2$. The problem is mentioned also in \cite{E}, wherein it is stated that the method does not seem to extend to $k\ge 3$, and it appears as B14 in \cite{RKG}. Let us quickly give proofs of the irrationality of these series when $k=0,1,2$. Assume that the given series is $A/B$, where $A$ and $B$ are positive integers. Multiplying by $(n-1)!$, where $n>B$, we get that $$ \sum_{j=1}^{n-1}\sigma_k(j)\frac{(n-1)!}{j!}+\frac{\sigma_k(n)}{n}+\frac {\sigma_k(n+1)}{n(n+1)}+\sum_{j\ge 2}\frac{\sigma_k(n+j)}{n(n+1)\cdots (n+j)}=\frac{A(n-1)!}{B}. $$ The first of the above sums is an integer and the right hand side is also an integer. The second sum is positive and, for $k\le 2$, its size is $$ \le \frac{\sigma_2(n+2)}{n(n+1)(n+2)}+\sum_{j\ge 3}\frac{\sigma_2(n+j)}{n(n+1)\cdots (n+j)}\ll \frac{1}{n}+\sum_{m\ge n}\frac{1}{m^2}\ll \frac{1}{n}, $$ so that this term belongs to the interval $[0,c/n]$, where $c$ is an absolute constant. We shall take $n$ to be a prime $n\equiv 1\pmod 4$ such that $(n+1)/2$ has no prime factor $y$. Thus, when $k=1$, we get that $\sigma_1(n)/n=1+1/n$, and we conclude that there is an integer in the interval $[1/n,(c+1+c_1\log\log n)/n]$, while when $k=2$ we get that $\sigma_2(n)/n=n+1/n$, and so there exists an integer in the interval $[1/4, 1/4+c/n+c_2/y]$, where $c_2$ is the constant implied by the $O$ symbol in \eqref{eq:c2}. Choosing $n$ (and $y$) to be sufficiently large, we get a contradiction in the case $k=1$ (and $k=2$), which completes the proof. \medskip In this paper, we make a modest contribution to the problem, proving two results about the irrationality of the above series for $k\ge 3$. \begin{thm} \label{thm:3} The sum of the series \begin{equation} \label{eq:ser} \sum_{n=1}^{\infty}\frac{\sigma_3(n)}{n!} \end{equation} is irrational. \end{thm} We recall the statement of the {\it Prime $k$-tuples Conjecture} (see~\cite{Dick,HL,SS}), which is due to Dickson. \begin{conj} \label{conjkprim} For any $k\ge 2$, let $a_1,\ldots,a_k$ and $b_1,\ldots,b_k$ be integers with $a_i>0$ for each $i=1,\ldots, k$. Suppose that for every prime number $p$ there exists an integer $n$ such that $\prod_{i=1}^k(a_in+b_i)$ is not a multiple of $p$. Then there exist infinitely many positive integers $n$ such that $p_i=a_in+b_i$ is prime for all $i=1,\ldots,k$. \end{conj} We have the following result for any positive integer $k$. \begin{thm} \label{thm:k} The Prime $k$-tuples Conjecture implies that $$ \sum_{n= 1}^{\infty} \frac{\sigma_k(n)}{n!} $$ is irrational. \end{thm} Throughout this paper, we use the Landau symbols $O$ and $o$ as well as Vinogradov's symbols $\gg,~\ll$ and $\asymp$ with their regular meaning. The constants implied by these symbols depend at most on $k$. We use $p$ and $q$ to denote prime numbers. \medskip \noindent{\bf Acknowledgements.} We thank the referee for a careful reading of the manuscript and for suggestions which improved the quality of this paper. Most of this paper was written during a very enjoyable visit by the first two authors to Williams College; these authors wish to express their thanks to that institution for the hospitality and support. \section{Proof of Theorem \ref{thm:3}} Our main tool is the following well-known theorem of Chen (see \cite{C1}, \cite{C2}). \begin{thm} \label{thm:chen} Let $a$ be an integer. There exists $x_a$ such that for $x>x_a$, the interval $[x/2,x]$ contains $\gg xa/\varphi (a)^2(\log x)^2$ prime numbers $p\equiv 1\, (\mod\, a)$ such that $q=(p+1)/2$ is either a prime or a product of two primes each one of which exceeds $x^{1/10}$. \end{thm} A proof containing the basic ideas for this theorem appears, for example, in Chapter 11 of \cite{HR}. Chen actually proved that for large even integers $N$ there are $\gg N/(\log N)^2$ primes $p$ such that $N-p$ is either a prime or a product of two large primes. Easy and well-known modifications of Chen's argument yield the above theorem. \medskip To prove our Theorem \ref{thm:3} we assume that the sum of the series shown at \eqref{eq:ser} is rational and deduce a contradiction. We write it, as we did for $k=0,~1$ and $2$ as $A/B$, multiply across by $(n-1)!$ for some large $n$ (with $n>B$), and obtain \begin{equation} \label{eq:1} \frac{\sigma_3(n)}{n}+\frac{\sigma_3(n+1)}{n(n+1)}+\frac{\sigma_3(n+2)}{n (n+1)(n+2)}+ \sum_{j\ge 3}\frac{\sigma_3(n+j)}{n(n+1)\cdots (n+k)}=(n-1)!\left(\frac{A}{B}-\sum_{j=1}^{n-1}\frac{\sigma_k(j)}{j!}\right), \end{equation} where the right hand side is an integer. In what follows, we shall exploit the above relation. Since $\sigma_3(n)\ll n^3$, we have $$ \sum_{j\ge 3}\frac{\sigma_3(n+j)}{n(n+1)\cdots (n+j)}\ll \frac{1}{n}+\sum_{m\ge n}\frac{1}{m^2}\ll \frac{1}{n}. $$ Furthermore, the sum appearing on the left hand side of formula \eqref{eq:1} above is a positive integer. We now let $y$ be a large positive real number which we shall fix later, and take $a=72\prod_{5\le p\le y} p$. Note that, by Chebyshev's bounds, this means that $y\asymp \log a$. We let $x$ be large compared to $a$ and $m\in [x/2a,x/a]$ be such that $p=am+1$ is prime and $q=(p+1)/2=(am+2)/2$ is either a prime or a product of two primes $q_1q_2$ each exceeding $x^{1/10}$. Choose $n$ to be the prime $n=p$. We then note that $$ \frac{\sigma_3(n)}{n}=n^2+\frac{1}{n}. $$ Furthermore, writing $n+2=am+3=3t$, we have that $t$ is coprime to all primes $q\le y$, and so obtain the inequalities \begin{eqnarray*} \frac{28}{27}<\frac{\sigma_3(n+2)}{n(n+1)(n+2)} & = & \frac{28}{27}\frac{\sigma_3(t)}{t^3}\left(1+O\left(\frac{1}{n}\right)\right) \\ & < & \frac{28}{27}\prod_{p>y}\left(1-\frac{1}{p^3}\right)^{-1}\left(1+O\left (\frac{1}{n}\right)\right)\\ & < & 1+\frac{1}{27}+O\left(\frac{1}{y^2}\right)\ , \end{eqnarray*} since $y<\sqrt x$. Combining this with our estimate for the tail of the series we have, for $n$ a prime as above, \begin{equation} \label{eq:****} \frac{\sigma_3(n)}{n}+\frac{\sigma_3(n+2)}{n(n+1)(n+2)}+\sum_{j\ge 3}\frac{\sigma_3(n+j)}{n(n+1)\cdots (n+j)}=A_n+\frac{1}{27}+O\left(\frac{1}{y^2}\right), \end{equation} where $A_n$ is a positive integer. We now consider the remaining term $\sigma_3(n+1)/(n(n+1))$. \medskip Assume first that for some large $x$ there is a prime $p=am+1\in [x/2,x]$ such that $q=(p+1)/2=(a/2)m+1$ is prime. Then \begin{equation*} \begin{split} \frac{\sigma_3(n+1)}{n(n+1)} & =\frac{\sigma_3(2q)}{2q(2q-1)} =\frac{9(q^3+1)}{2q(2q-1)}\\ & = \frac{9}{4} q+\frac{9}{8}+\frac{9q-4}{8q(2q-1)}= B_n+\frac{3}{8}+O\left(\frac{1}{n}\right) \end{split} \end{equation*} with some integer $B_n$, where we have used the fact that $q=(a/2)m+1\equiv 1\pmod 4$, because $8\mid a$. Summing up everything, we find that $$ \frac{1}{27}+\frac{3}{8}+O\left(\frac{1}{y^2}\right)\in \Z, $$ which is impossible if $y$ is chosen to be sufficiently large. \medskip So, we are left with the more interesting part of the problem where $q=q_1q_2$, where $q_1>q_2>x^{1/10}$ holds for all the $\gg xa/\varphi(a)^2(\log x)^2$ choices of primes $p=am+1\in [x/2,x]$ guaranteed by Chen's Theorem \ref{thm:chen}. We put $$ M=\left\lfloor \frac{\sigma_3(n+1)}{n(n+1)}\right\rfloor= \left\lfloor \frac{9\sigma_3(q)}{2q(2q-1)}\right\rfloor. $$ Note that since $q_2\le q^{1/2}$, we have that $$ \frac{\sigma_3(q)}{2q(2q-1)}=\frac{q^3+q_1^3+q_2^3+1}{2q(2q-1)}= \frac{q^3+q_1^3+O(q^{3/2})}{2q(2q-1)}=\frac{q^3+q_1^3}{2q(2q-1)}+O\left(\frac {1}{q^{1/2}}\right). $$ Thus, using also the previous calculations of fractional parts (see \eqref{eq:****}), we find that for large $x$, $$ M+\frac{26}{27}-\frac{c_0}{y^2}\le \frac{9q^3+9q_1^3}{2q(2q-1)}\le M+\frac{26}{27}+\frac{c_0}{y^2} $$ holds for all large $x$ with some constant $c_0>0$, an inequality which can be rewritten as \begin{eqnarray*} & & q^2\left(4M+3-9q+\frac{23}{27}-\frac{2q(m+O(1))}{q^2}-\frac{4c_0}{y^2} \right) \le 9q_1^3\\ & \le & q^2\left(4M+3-9q+\frac{23}{27}-\frac{2q(m+O(1))}{q^2}+\frac{4c_0}{y^2} \right). \end{eqnarray*} From this inequality and the fact that $q_1q_2^2/27$ if $y^2>220 c_0$, so if $y$ satisfies the above inequality and $x$ is large, then $q_2^2/27\le q_1\le x/q_2$, and therefore $q_2\le 3x^{1/3}$. Now fix $q_2$. Then the left hand side of the above inequality is $\ge (L+1/3)q_2^2/27$ and the middle term satisfies $q_1\le x/q_2$. Thus, $L+1/3\le 27x/q_2^3$. Since $L\ge 0$ is an integer, it follows that the number of possibilities for $L$ is \begin{equation} \label{eq:posforL} 1+\left\lfloor \frac{27x}{q_2^3}\right\rfloor\le \frac{81x}{q_2^3}. \end{equation} We now fix also $L\ge 0$. Then $q_1$ is a prime in the interval shown at \eqref{eq:bounds} above such that $q_1q_2\equiv 1\pmod a$ and $2q_1q_2-1=p$ is a prime. By the Brun sieve, the number of such primes is $$ \ll \frac{ c_0q_2^2 }{y^2}\cdot \left(\frac{a}{\varphi(a)^2}\right)\cdot \frac{1}{\left(\log(10c_0q_2^2/(9y^2a)) \right)^2}. $$ Since $q_2>x^{1/10}$, it follows that for large $x$ we have $10c_0q_2^2/(9y^2a) > x^{1/6}$. Thus, the number of possibilities for $q_1$ once $q_2$ and $L$ are fixed is $$ \ll \frac{1}{y^2}\cdot\frac{a}{\varphi(a)^2} \cdot \frac{q_2^2}{(\log x)^2}. $$ Summing up over all the possibilities for $L$ shown at \eqref{eq:posforL}, we get that the total number of possibilities for $q_1$ when $q_2$ is fixed is $$ \ll \frac{1}{y^2}\cdot\frac{a}{\varphi(a)^2} \cdot \frac{x}{(\log x)^2}\cdot\frac{1}{q_2}, $$ and now summing up the above bound over all primes $q_2\in [x^{1/10}, 3x^{1/3}]$, we get that the total number of possibilities for $p$ is \begin{equation} \label{eq:last1} \ll \frac{1}{y^2}\frac{a}{\varphi(a)^2} \frac{x}{(\log x)^2} \sum_{x^{1/10}\le q_2\le 3x^{1/3}}\frac{1}{q_2}\ll \frac{1}{y^2}\frac{a}{\varphi(a)^2} \frac{x}{(\log x)^2}\ , \end{equation} where the fact that the last sum above is $O(1)$ follows from Mertens's estimate for the summatory function of the reciprocal of the primes. Let $c_1$ be the constant implied in the Vinogradov symbol from Chen's Theorem \ref{thm:chen} and $c_2$ be the constant implied in the last Vinogradov symbol in \eqref{eq:last1}. Comparing the estimates for the number of primes $p$ under scrutiny we get $$ \frac{c_1 a x}{\varphi(a)^2(\log x)^2}\le \frac{c_2 ax } {\varphi(a)^2y^2 (\log x)^2} $$ leading to $y^2k$ such that $p_i\nmid \sigma_k(i)Q_i(-i)$. (For any other $i$, for notational purposes, take $p_i=1$.) As we have seen, $Q_k(-k)=1\ne 0$, so that $p_k\nmid \sigma_k(k)Q_k(-k)$ can be arranged simply by choosing $p_k>\sigma_k(k)$. Let $P=\prod_{i=1}^k p_i$ and let $m$ be a positive integer such that $$ \frac{(k!)^2}{i} m+1\equiv p_i\pmod {p_i^2}\qquad i=1,\ldots,k. $$ By the Chinese Remainder Theorem, there are infinitely many such positive integers $m$ and they form the arithmetic progression $m_0\pmod {P^2}$, where $m_0$ is the first positive integer in this progression. Given such an $m$, we choose $n=(k!)^2m$ and write $m=m_0+\ell P^2$ with some nonnegative integer $\ell$. Then $$ n+i=i\left(\frac{(k!)^2}{i}m+1\right)=ip_i\left(\frac{(k!)^2}{i}\frac{P^2} {p_i} \ell+\frac{(k!)^2m_0/i+1}{p_i}\right). $$ Put $$A_i={\displaystyle{\frac{(k!)^2}{i}\frac{P^2}{p_i}}} \quad \text{\rm and}\quad B_i={\displaystyle{\frac{(k!)^2m_0/i+1}{p_i}}} $$ for $i=1,\ldots,k$. One checks easily that we can apply the Prime $k$-tuples Conjecture \ref{conjkprim} to the linear polynomials $A_i\ell+B_i$. Indeed, the prime numbers dividing $A_i$ are exactly the primes $p\le k$ together with the primes $p_i$. Since $B_i\mid (k!)^2m_0/i+1$, we see that $B_i$ is coprime to all primes $p\le k$. Further, $B_i\equiv 1\pmod {p_i}$ from the way $m_0$ was chosen. To see that $B_i$ is coprime to $p_j$ if $j\ne i$ assume otherwise. Then $p_j\mid ((k!)^2/j) m_0+1$ and $p_j\mid B_i\mid ((k!)^2/i)m_0+1$. Hence, $p_j\mid (k!)^2m_0+j$ and $p_j\mid (k!)^2m_0+i$, so $p_j\mid (j-i)$, but this is false because $p_j>k$. Thus, the conditions for the applicability of the Prime $k$-tuples Conjecture \ref{conjkprim} are fulfilled and we can let $\ell$ be large and such that $A_i\ell+B_i=q_i$ are all primes for $i=1,\ldots,k$. Put $a_i=ip_i$ and note that if $m=m_0+\ell P^2$, then $n+i=a_i q_i$ for $i=1,\ldots,k$. Now assume that the series $$ \sum_{j=1}^{\infty} \frac{\sigma_k(j)}{j!}=\frac{A}{B} $$ with $A$ and $B$ coprime integers. Suppose $n$ is sufficiently large, and in particular $n>B$. Multiplying across by $n!$, we get \begin{equation} \label{eq:n} \sum_{j\le n}\sigma_k(j)\frac{n!}{j!}+\sum_{i=1}^k \frac{\sigma_k(n+i)}{(n+1)\cdots (n+i)}+\sum_{j\ge k+1}\frac{\sigma_k(n+j)}{(n+1)\cdots (n+j)}\in \Z. \end{equation} The first sum above is an integer, while the last sum is positive and, since $\sigma_k(n)\ll n^k$, \begin{eqnarray*} \sum_{j\ge k+1}\frac{\sigma_k(n+j)}{(n+1)\cdots (n+j)} & = & \frac{\sigma_k(n+k+1)}{(n+1)\cdots (n+k+1)}+\sum_{j\ge k+2}\frac{\sigma_k(n+j)}{(n+1)\cdots (n+j)}\\ & \ll & \frac{1}{n}+\sum_{j\ge k+2}\frac{1}{(n+j)^2}\ll \frac{1}{n}. \end{eqnarray*} As for the intermediate terms, since $q_i$ is prime we have \begin{eqnarray*} \frac{\sigma_k(n+i)}{(n+1)\cdots (n+i)} & = & \frac{\sigma_k(a_i)(q_i^k+1)}{(n+1)\cdots (n+i)}\\ & = & \frac{\sigma_k(a_i)}{a_i^k}\left(\frac{(n+i)^k+a_i^k}{(n+1)\cdots (n+i)}\right)\\ & = & \frac{\sigma_k(a_i)}{a_i^k}\left(Q_i(n)+\frac{R_i(n)+a_i^k}{(n+1)\cdots(n+i)} \right)\\ & = & \frac{\sigma_k(a_i)}{a_i^k} Q_i(n)+O\left(\frac{1}{n}\right), \end{eqnarray*} where for the above error term we used the fact that ${\text{\rm deg}}\,\, R_i\le i-1$. Since $n\equiv -i\pmod {p_i}$, we get that $Q_i(n)\equiv Q_i(-i)\pmod {p_i}$ so that $Q_i(n)=Q_i(-i)+p_i\ell_i(n)$, for some integers $\ell_i(n)$. Thus, $$ \frac{\sigma_k(n+i)}{(n+1)\cdots (n+i)}=\frac{\sigma_k(ip_i)}{(ip_i)^k}\left(Q_i(-i)+\ell_i(n) p_i\right)+O\left(\frac{1}{n}\right). $$ We add everything together, multiply formula \eqref{eq:n} by $(k!)^kP^{k-1}$ and get $$ \sum_{i=1}^kP_i^{k-1} \sigma_k(ip_i)\frac{(k!)^k}{i^k} \frac{1}{p_i}\left(Q_i(-i)+p_i\ell_i(n)\right)+O\left(\frac{1}{n}\right)\in \Z, $$ where $$ P_i =Pp_i^{-1}=\prod_{\substack{1\le j\le k\\j\ne i}}p_j\ . $$ {From} this it follows that $$ \sum_{i=1}^kP_i^{k-1} \frac{(k!)^k}{i^k} \frac{\sigma_k(ip_i)Q_i(-i)}{p_i}+O\left(\frac{1}{n}\right)\in \Z. $$ At this moment, we observe that the sum on the left does not depend on $n$ and so $$ \sum_{1\le i\le k} P_i^{k-1}\frac{(k!)^k}{i^k} \frac{\sigma_k(ip_i)}{p_i}Q_i(-i)\in \Z. $$ Using also the fact that $\sigma_k(ip_i)=\sigma_k(i)\sigma_k(p_i)= \sigma_k(i)(p_i^k+1)$, we get that the above relation implies $$ \sum_{1\le i\le k} P_i^{k-1}\frac{(k!)^k}{i^k} \frac{\sigma_k(i)}{p_i}Q_i(-i)\in \Z. $$ This is a non-empty (since $Q_k(-k)=1$) sum of non-zero rational numbers whose reduced denominators are distinct primes. 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