\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} \usepackage[np]{numprint} \npdecimalsign{\ensuremath{.}} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \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} \begin{center} \vskip 1cm{\LARGE\bf Some Explicit Estimates for the \\ \vskip .1in M\"{o}bius Function} \vskip 1cm \large Olivier Bordell\`es\\ 2, all\'{e}e de la Combe\\ 43000 Aiguilhe\\ France\\ \href{mailto:borde43@wanadoo.fr}{\tt borde43@wanadoo.fr} \end{center} \vskip .2 in \newcommand{\N}{\mathbb {N}} \newcommand{\Z}{\mathbb {Z}} \newcommand{\Q}{\mathbb {Q}} \newcommand{\R}{\mathbb {R}} \newcommand{\C}{\mathbb {C}} \begin{abstract} In this work, we provide some explicit upper bounds for certain sums involving the M\"{o}bius function. Thanks to recent results proved by Balazard, some of these bounds improve on earlier estimates given by El Marraki. \end{abstract} \section{Introduction and results} It is often a hard task to give completely explicit results in analytic number theory. For instance, it is well-known that if $\chi$ is a quadratic Dirichlet character to the modulus $q$, then, for any $\varepsilon \in \left( 0,\frac{1}{2} \right)$, there exists a non-effectively computable constant $c_\varepsilon > 0$ such that $$L(1,\chi) > \frac{c_\varepsilon}{q^\varepsilon}$$ and all attempts at providing a value to $c_\varepsilon$ for sufficiently small $\varepsilon$ have been unsuccessful. On the other hand, a wide class of functions of prime numbers have been successfully explicitly estimated during the last fifty years, starting with the benchmarking paper of Rosser and Sch\oe nfeld \cite{ros}. For instance, refining an earlier estimate of Dusart \cite{dus}, Trudgian \cite{tru} proved that, for $x \geq 229$ $$\left | \pi(x) - \textrm{Li} (x) \right | < \np{0.2795} \, \frac{x}{(\log x)^{3/4}} \exp \left( - \sqrt{\frac{\log x}{\np{6.455}}} \right)$$ where $\textrm{Li}$ is the usual logarithmic integral function. As for the M\"{o}bius function, the prime number theorem is known to be equivalent to the estimate $$\sum_{n \leq x} \mu(n) = o(x) \quad \left( x \rightarrow \infty \right) $$ and the method of contour integration may lead to bounds of the form $$\sum_{n \leq x} \mu(n) \ll x \exp \left( -c \sqrt{\log x} \right) \quad \left( x \geq 2, \ c> 0 \right)$$ but no explicit result of this form is known. Refining a method of von Sterneck, based itself upon the work of Chebyshev, MacLeod \cite{mcl} showed that \begin{equation} \left | \sum_{n \leq x} \mu(n) \right | \leq \frac{x+1}{80}+ \frac{11}{2} \quad \left( x \geq 1 \right). \label{e1} \end{equation} Bounds of the form $x(\log x)^{-\alpha}$ with $\alpha > 0$ were then obtained by Sch\oe nfeld \cite{sch} and El Marraki \cite{elm1} who, among others, proved that \cite[Th\'{e}or\`{e}me~2]{elm1} \begin{equation} \left | \sum_{n \leq x} \mu(n) \right | < \frac{\np{0.10917} x}{\log x} \quad \left( x \geq 685 \right) \label{e2} \end{equation} and \cite[Th\'{e}or\`{e}me~3]{elm1} \begin{equation} \left | \sum_{n \leq x} \mu(n) \right | < \frac{362.7x}{(\log x)^2} \quad \left( x > 1 \right). \label{e3} \end{equation} Using an inequality coming from a convolution relation and partial summation, El Marraki \cite{elm2} deduced from \eqref{e2} and \eqref{e3} that \begin{equation} \left | \sum_{n \leq x} \frac{\mu(n)}{n} \right | \leq \frac{\np{0.2185}}{\log x} \ \left( x \geq 33 \right) \quad \textrm{and} \quad \left | \sum_{n \leq x} \frac{\mu(n)}{n} \right | \leq \frac{726}{(\log x)^2} \ \left( x > 1 \right). \label{e4} \end{equation} Ramar\'{e} \cite{ram2} refined the first estimate by showing $$\left | \sum_{n \leq x} \frac{\mu(n)}{n} \right | \leq \frac{1}{69 \log x} \quad \left( x \geq \np{96955} \right).$$ Our first result improves on \eqref{e4} in the following way. \begin{theorem} \label{t1} \begin{enumerate} \item[] \item[$(a)$] For all $x \geq 33$ $$\left | \sum_{n \leq x} \frac{\mu(n)}{n} \right| < \frac{0.19}{\log x}.$$ \item[$(b)$] For all $x > 1$ $$\left | \sum_{n \leq x} \frac{\mu(n)}{n} \right| < \frac{546}{(\log x)^2}.$$ \end{enumerate} \end{theorem} It may be interesting to estimate similar sums twisted by additional conditions. For instance, Ramar\'{e} \cite{ram1,ram2} studied sums of the type $$\sum_{\substack{n \leq x \\ (n,k)=1}} \frac{\mu(n)}{n}$$ where $k \in \Z_{\geq 1}$, and showed among others that $$\left | \sum_{\substack{n \leq x \\ (n,k)=1}} \frac{\mu(n)}{n} \right| \leq \frac{k}{\varphi(k)} \, \frac{0.78}{\log(x/k)} \quad \left( 1 \leq k < x \right).$$ Our second result is a complement to Ramar\'{e}'s bound. \begin{theorem} \label{t2} Let $k,m \in \Z_{\geq 1}$. For all $x \geq k^m$ $$\left | \sum_{\substack{n \leq x \\ (n,k)=1}} \frac{\mu(n)}{n} \right| < \frac{k}{\varphi(k)} \, \frac{C_m}{\left( \log \left( exk^{-m} \right)\right)^2}$$ where $$C_m = \np{1100} \left( 1 + 4 e^{-1}\sqrt{\zeta \left ( m + \tfrac{1}{2} \right )} \right)^2.$$ \end{theorem} The first ten values of the ceiling of $C_m$ are given below. \begin{center} \begin{tabular}{ccccccccccc} \hline $m$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$ & $10$ \\ \hline $\left \lceil C_m \right \rceil$ & $\np{12555}$ & $\np{8045}$ & $\np{7221}$ & $\np{6937}$ & $\np{6820}$ & $\np{6768}$ & $\np{6743}$ & $\np{6731}$ & $\np{6725}$ & $\np{6723}$ \\ \hline \end{tabular} \end{center} \bigskip Next, we estimate the logarithmic mean of the M\"{o}bius function twisted by a Dirichlet character. \begin{proposition} \label{pro3} Let $\chi$ be a non-principal Dirichlet character to the modulus $q \geq 37$ and let $k \in \Z_{\geq 1}$. Then for all $x \geq 1$ $$\left | \sum_{\substack{n \leq x \\ (n,k)=1}} \frac{\mu(n) \chi(n)}{n} \right | \leq \frac{k}{\varphi(k)} \, \frac{2 \sqrt{q} \log q}{|L(1,\chi)|}.$$ \end{proposition} Our last result deals with the following rather curious sum which does not seem to have been studied in the literature before. \begin{theorem} \label{t4} For every $x \geq 1$, define $$S(x) = \sum_{n \leq x} \mu (n) \sum_{k=1}^n \mu(k).$$ Then for all $x \geq \np{1664}$ $$\frac{x}{2 \zeta(2)} - 0.067 \sqrt{x} \leq S(x) \leq \np{0.0006} \left( \frac{x}{\log x} \right)^2 + \frac{x}{2 \zeta(2)} + 0.067 \sqrt{x}.$$ Furthermore, the prime number theorem implies that, for $x$ sufficiently large $$S(x) \ll x^2 e^{-\np{0.4196} \, (\log x)^{3/5} (\log \log x)^{-1/5}}.$$ Finally, the Riemann hypothesis is true if and only if, for all $\varepsilon > 0$ and $x$ sufficiently large $$S(x) \ll x^{1+\varepsilon}.$$ \end{theorem} \bigskip In what follows, we define the functions $M(x)$ and $m(x)$ by $$M(x) = \sum_{n \leq x} \mu(n) \quad \textrm{and} \quad m(x) = \sum_{n \leq x} \frac{\mu(n)}{n}.$$ \section{Tools} Our first lemma follows easily from well-known convolution techniques. \begin{lemma} \label{le45} Let $f$ be a completely multiplicative function and $a \in \Z_{\geq 1}$. Then uniformly for any real number $x \geq 1$ $$\sum_{\substack{n \leq x \\ (n,a) = 1}} \frac{\mu(n)f(n)}{n} = \sum_{\substack{k \leq x \\ k \mid a^\infty}} \frac{f(k)}{k} \sum_{m \leq x/k} \frac{\mu(m) f(m)}{m}.$$ \end{lemma} \begin{proof} If $\mathbf{1}_a^\infty$ is the characteristic function of the set of integers $k \geq 1$ such that $k \mid a^\infty$, then one can easily see that, for any $n \in \Z_{\geq 1}$ $$\left( \mathbf{1}_a^\infty \star \mu \right) (n) = \begin{cases} \mu(n), & \textrm{if\ } (n,a) = 1; \\ 0, & \textrm{otherwise}. \end{cases}$$ Now inserting this in the left-hand side, interchanging the summations and taking the complete multiplicativity of $f$ into account achieve the proof. \end{proof} The first result is an inequality coming from the work of Balazard \cite{bal}, depending on M\"{o}bius' inversion formula and some special properties of the Bernoulli functions, and improving on the inequality used in \cite{elm2,ram1} by a factor $\log$. \begin{lemma} \label{le5} For all $x \geq 1$ $$x \left | m(x) \right| \leq \left | M(x) \right| + \frac{1}{x} \int_1^x \left | M(t) \right| \mathrm{d}t + \frac{8}{3}.$$ \end{lemma} In fact, this inequality is a special case of a more general result stating that, for every $k \in \Z_{\geq 1}$, there exist constants $C_k > 0$ and $D_k > 0$ such that $$\left | x m(x) - M(x) \right | \leq C_k x^{2-k} \int_1^x \left |M(t) \right | t^{k-3} \textrm{d}t + D_k$$ which implies Lemma~\ref{le5} by taking $k=3$. The case $k=2$ provides the bound $$x \left | m(x) \right| \leq \left | M(x) \right| + \int_1^x \frac{\left | M(t) \right|}{t} \, \textrm{d}t + 2 - \frac{2}{x}$$ which proves to be slightly weaker than Lemma~\ref{le5}. \bigskip The next tool is an explicit bound for a certain class of integrals. \begin{lemma} \label{le6} Let $a > 1$, $\alpha > 0$ be real numbers. For all $x \geq a$ $$\int_a^x \frac{t \, \mathrm{d}t}{(\log t)^\alpha} \leq \frac{C_\alpha x^2}{(\log x)^\alpha}$$ where $$C_\alpha = \frac{\alpha^{-1}}{2} \left( \frac{\alpha}{(2e \log a)^{\alpha/(\alpha+1)}} + \alpha^{1/(\alpha+1)} \right)^{\alpha+1}.$$ \end{lemma} \begin{proof} For any $b > 1$, we get $$\int_a^x \frac{t \, \mathrm{d}t}{(\log t)^\alpha} = \left( \int_a^{x^{1/b}} + \int_{x^{1/b}}^x \right) \frac{t \, \textrm{d}t}{(\log x)^\alpha} \leq \frac{1}{2} \left( \frac{x^{2/b}}{(\log a)^\alpha} + \frac{b^\alpha x^2}{(\log x)^\alpha} \right).$$ The inequality $\log x \leq Ce^{-1} x^{1/C}$, used with $C = \dfrac{b \alpha}{2(b-1)}$, yields $$(\log x)^\alpha \leq \left( \frac{b \alpha}{2e(b-1)} \right)^\alpha x^{2-2/b} \Longleftrightarrow x^{2/b} \leq \left( \frac{b \alpha}{2e(b-1)} \right)^\alpha \frac{x^2}{(\log x)^\alpha}$$ and hence $$\int_a^x \frac{t \, \textrm{d}t}{(\log t)^\alpha} \leq \frac{x^2}{2} \left( \frac{b}{\log x} \right)^\alpha \left( \left( \frac{\alpha}{2e (b-1) \log a} \right)^\alpha + 1 \right)$$ and choosing $$b = 1 + \left( \frac{\alpha}{2e \log a} \right)^{\alpha / (\alpha + 1)}$$ concludes the proof. \end{proof} The next lemma will be proved to be useful in the proof of Theorem~\ref{t2}. \begin{lemma} \label{le7} Let $k,m \in \Z_{\geq 1}$. For all $x \geq 1$ $$\sum_{\substack{n \leq x \\ (n,k)=1}} \frac{\mu(n)}{n} = \sum_{d_1 \mid k} \frac{\mu(d_1)^2}{d_1} \sum_{d_2 \mid d_1} \frac{\mu(d_2)^2}{d_2} \dotsb \sum_{d_m \mid d_{m-1}} \frac{\mu(d_m)^2}{d_m} \sum_{\substack{h \leq \frac{x}{d_1 \dotsb d_m} \\ h \mid d_m^\infty}} \frac{1}{h} \sum_{j \leq \frac{x}{hd_1 \dotsb d_m}} \frac{\mu(j)}{j}.$$ \end{lemma} \begin{proof} We proceed by induction on $m$. For $m=1$, we have \begin{eqnarray*} \sum_{\substack{n \leq x \\ (n,k)=1}} \frac{\mu(n)}{n} &=& \sum_{d_1 \mid k} \frac{\mu(d_1)}{d_1} \sum_{h \leq x/d_1} \frac{\mu(hd_1)}{h} \\ &=& \sum_{d_1 \mid k} \frac{\mu(d_1)^2}{d_1} \sum_{\substack{h \leq x/d_1 \\ (h,d_1)=1}} \frac{\mu(h)}{h} \\ &=& \sum_{d_1 \mid k} \frac{\mu(d_1)^2}{d_1} \sum_{\substack{h \leq x/d_1 \\ h \mid d_1^\infty}} \frac{1}{h} \sum_{j \leq x/(hd_1)} \frac{\mu(j)}{j} \end{eqnarray*} where we used Lemma~\ref{le45} in the last equality. Now assume that the statement is true for some $m \geq 1$. From Lemma~\ref{le45} again $$\sum_{\substack{h \leq \frac{x}{d_1 \dotsb d_m} \\ h \mid d_m^\infty}} \frac{1}{h} \sum_{j \leq \frac{x}{hd_1 \dotsb d_m}} \frac{\mu(j)}{j} = \sum_{\substack{h \leq \frac{x}{d_1 \dotsb d_m} \\ (h,d_m)=1}} \frac{\mu(h)}{h}$$ so that using the induction hypothesis \begin{eqnarray*} \sum_{\substack{n \leq x \\ (n,k)=1}} \frac{\mu(n)}{n} &=& \sum_{d_1 \mid k} \frac{\mu(d_1)^2}{d_1} \sum_{d_2 \mid d_1} \frac{\mu(d_2)^2}{d_2} \dotsb \sum_{d_m \mid d_{m-1}} \frac{\mu(d_m)^2}{d_m} \sum_{\substack{h \leq \frac{x}{d_1 \dotsb d_m} \\ (h,d_m)=1}} \frac{\mu(h)}{h} \\ &=& \sum_{d_1 \mid k} \frac{\mu(d_1)^2}{d_1} \dotsb \sum_{d_m \mid d_{m-1}} \frac{\mu(d_m)^2}{d_m} \sum_{d_{m+1} \mid d_m} \mu \left( d_{m+1} \right) \sum_{h \leq \frac{x}{d_1 \dotsb d_m d_{m+1}}} \frac{\mu \left( h d_{m+1} \right)}{h d_{m+1}} \\ &=& \sum_{d_1 \mid k} \frac{\mu(d_1)^2}{d_1} \dotsb \sum_{d_m \mid d_{m-1}} \frac{\mu(d_m)^2}{d_m} \sum_{d_{m+1} \mid d_m} \frac{\mu \left( d_{m+1} \right)^2}{d_{m+1}} \sum_{\substack{h \leq \frac{x}{d_1 \dotsb d_m d_{m+1}} \\ (h,d_{m+1}) = 1}} \frac{\mu(h)}{h} \\ &=& \sum_{d_1 \mid k} \frac{\mu(d_1)^2}{d_1} \dotsb \sum_{d_{m+1} \mid d_m} \frac{\mu \left( d_{m+1} \right)^2}{d_{m+1}} \sum_{\substack{h \leq \frac{x}{d_1 \dotsb d_{m+1}} \\ h \mid d_{m+1}^\infty}} \frac{1}{h} \sum_{j \leq \frac{x}{hd_1 \dotsb d_{m+1}}} \frac{\mu(j)}{j} \end{eqnarray*} achieving the proof. \end{proof} \bigskip The identity below may be proved by induction. We leave the details to the reader. \begin{lemma} \label{le8} Let $k,m \in \Z_{\geq 1}$. Then $$\sum_{d_1 \mid k} \frac{\mu(d_1)^2}{d_1} \sum_{d_2 \mid d_1} \frac{\mu(d_2)^2}{d_2} \dotsb \sum_{d_m \mid d_{m-1}} \frac{\mu(d_m)^2}{d_m} \prod_{p \mid d_m} \left( 1 - \frac{1}{p^{1/2}} \right)^{-1} = \frac{k}{\varphi(k)} \prod_{p \mid k} \left( 1 + \frac{1}{p^{m+1/2}} \right).$$ \end{lemma} \section{Proofs of the Theorems} \subsection{Theorem~\ref{t1}} \begin{proof} \begin{enumerate} \item[] \item[$(a)$] We check numerically the inequality for $x \in \left[ 33,\np{6000} \right]$, and we assume $x > \np{6000}$. Let $T \in \left[685,x\right]$ be a parameter at our disposal. From Lemma~\ref{le5} and the bounds \eqref{e1} and \eqref{e2}, we infer \begin{eqnarray*} x |m(x)| & \leq & |M(x)| + \frac{1}{x} \left( \int_1^T + \int_T^x \right) |M(t)| \, \textrm{d}t + \frac{8}{3} \\ & < & \frac{\np{0.10917} x}{\log x} + \frac{T(T+882)}{160x} - \frac{883}{160x} + \frac{\np{0.10917}}{x} \int_T^x \frac{t \, \textrm{d}t}{\log t} + \frac{8}{3} \\ & \leq & \frac{\np{0.10917} x}{\log x} + \frac{T^2(1+882/685)}{160x} + \frac{\np{0.10917}}{x} \int_T^x \frac{t \, \textrm{d}t}{\log t} + \frac{8}{3} \end{eqnarray*} and Lemma~\ref{le6} implies that $$x |m(x)| < \frac{\np{0.10917} x}{\log x} + \frac{\np{1567}}{685} \, \frac{T^2}{160x} + \frac{\np{0.10917}}{2} \left( 1 + \frac{1}{\sqrt{2e \log T}} \right)^2 \frac{x}{\log x} + \frac{8}{3}.$$ We choose $T = \np{0.337} \, x (\log x)^{-1/2}$. Since $x > \np{6000}$, we have $T>685$ and thus \begin{eqnarray*} x |m(x)| &<& \frac{x}{\log x} \biggl( \np{0.10917} + \np{0.00163} + \frac{\np{0.10917}}{2} \left( 1 + \frac{1}{\sqrt{2e \log 685 }} \right)^2 \biggr. \\ & & \biggl. {} + \frac{8}{3} \, \frac{\log \np{6000}}{\np{6000}} \biggr) \\ &<& \frac{0.19x}{\log x} \end{eqnarray*} achieving the proof of the inequality. \item[$(b)$] The inequality is first checked on $\left] 1,2 \right[$ via $$\frac{546}{(\log x)^2} > \frac{546}{(\log 2)^2} > 1 = |m(x)|$$ and then numerically for $x \in \left[ 2,33 \right]$. If $x \in \left[ 33,e^{\np{2873}} \right]$, then $$|m(x)| < \frac{\np{0.19}}{\log x} < \frac{546}{(\log x)^2}$$ so that we may suppose $x > e^{\np{2873}}$. Using Lemma~\ref{le5} as above, Lemma~\ref{le6} with $\alpha = 2$, and \eqref{e3}, we get for any $T > 1$ \begin{eqnarray*} x |m(x)| & \leq & \frac{\np{362.7}x}{(\log x)^2} + \frac{T^2(1+882/T)}{160x} - \frac{883}{160x} + \frac{\np{362.7}}{x} \int_T^x \frac{t \, \textrm{d}t}{(\log t)^2} + \frac{8}{3} \\ & < & \frac{\np{362.7}x}{(\log x)^2} + \frac{T^2(1+882/T)}{160x} \\ & & {} + \frac{\np{362.7}}{4} \left( \frac{2}{(2e \log T)^{2/3}} + 2^{1/3} \right)^3 \frac{x}{(\log x)^2} + \frac{8}{3}. \end{eqnarray*} Choosing $T=x(\log x)^{-1} > \dfrac{e^{\np{2873}}}{\np{2873}}$, we obtain \begin{eqnarray*} x |m(x)| &<& \frac{x}{(\log x)^2} \Biggl( \np{362.7} + \frac{1}{160} \left( 1 + \frac{882 \times \np{2873}}{e^{\np{2873}}} \right) \Biggr. \\ & & \Biggl. {} + \frac{\np{362.7}}{4} \biggl( 2^{1/3} + 2\left( e \log \left( \frac{e^{\np{2873}}}{\np{2873}} \right) \right)^{-2/3} \biggr)^3 + \frac{8}{3} \, \frac{\np{2873}}{e^{\np{2873}}} \Biggr) \\ & < & \frac{x}{(\log x)^2} \left( \np{362.7} + \np{0.00625} + \np{182.73823} + \frac{8}{3} \, \frac{\np{2873}}{e^{\np{2873}}} \right) \\ &<& \frac{546x}{(\log x)^2}. \end{eqnarray*} \end{enumerate} The proof is completed. \end{proof} \subsection{Theorem~\ref{t2}} We first state the following result, which is an easy consequence of Theorem~\ref{t1}. \begin{lemma} \label{le9} For all $N \in \Z_{\geq 1}$ $$\left | \sum_{n=1}^N \frac{\mu(n)}{n} \right | \leq \frac{1}{\log \left( \frac{e}{2}(N+1) \right)} \quad \text{and} \quad \left | \sum_{n=1}^N \frac{\mu(n)}{n} \right | < \frac{550}{\left( \log \left( e(N+1) \right)\right)^2}.$$ \end{lemma} \begin{proof} We check numerically the first inequality for $N \in \{1,\dotsc,32 \}$ and, if $N \geq 33$, then by Theorem~\ref{t1} $$\left | \sum_{n=1}^N \frac{\mu(n)}{n} \right | < \frac{\np{0.19}}{\log N} \leq \frac{1}{\log \left( \frac{e}{2}(N+1) \right)}.$$ Now let us have a look at the second inequality. If $N \in \{1,\dotsc,10^{119}\}$, then $$\left | \sum_{n=1}^N \frac{\mu(n)}{n} \right | \leq \frac{1}{\log \left( \frac{e}{2}(N+1) \right)} < \frac{550}{\left( \log \left( e(N+1) \right)\right)^2}$$ and for $N > 10^{119}$ $$\left | \sum_{n=1}^N \frac{\mu(n)}{n} \right | < \frac{546}{\log^2 N} \leq \frac{550}{\left( \log \left( e(N+1) \right)\right)^2}$$ concluding the proof. \end{proof} We now are in a position to show Theorem~\ref{t2}. \begin{proof}[Proof of Theorem~\ref{t2}] Setting $$T= \left( exk^{-m} \right)^{\frac{4 \sqrt{\zeta(m+1/2)}}{4 \sqrt{\zeta(m+1/2)} + e}}$$ and using Lemmas~\ref{le7} and~\ref{le9}, the sum at the left-hand side does not exceed \begin{eqnarray*} & < & 550 \sum_{d_1 \mid k} \frac{\mu(d_1)^2}{d_1} \sum_{d_2 \mid d_1} \frac{\mu(d_2)^2}{d_2} \dotsb \sum_{d_m \mid d_{m-1}} \frac{\mu(d_m)^2}{d_m} \sum_{\substack{h \leq \frac{x}{d_1 \dotsb d_m} \\ h \mid d_m^\infty}} \frac{1}{h \left( \log \left( \frac{ex}{hd_1 \dotsb d_m}\right) \right)^2} \\ &=& 550 \sum_{d_1 \mid k} \frac{\mu(d_1)^2}{d_1} \dotsb \sum_{d_m \mid d_{m-1}} \frac{\mu(d_m)^2}{d_m} \left( \sum_{\substack{h \leq T \\ h \mid d_m^\infty}} + \sum_{\substack{T < h \leq \frac{x}{d_1 \dotsb d_m} \\ h \mid d_m^\infty}} \right) \frac{1}{h \left( \log \left( \frac{ex}{hd_1 \dotsb d_m}\right) \right)^2} \\ & \leq & 550 \sum_{d_1 \mid k} \frac{\mu(d_1)^2}{d_1} \sum_{d_2 \mid d_1} \frac{\mu(d_2)^2}{d_2} \dotsb \sum_{d_m \mid d_{m-1}} \frac{\mu(d_m)^2}{d_m} \frac{1}{\left( \log \left( \frac{ex}{Td_1 \dotsb d_m}\right) \right)^2} \frac{d_m}{\varphi(d_m)} \\ & & {} + 550 \, T^{-1/2} \sum_{d_1 \mid k} \frac{\mu(d_1)^2}{d_1} \sum_{d_2 \mid d_1} \frac{\mu(d_2)^2}{d_2} \dotsb \sum_{d_m \mid d_{m-1}} \frac{\mu(d_m)^2}{d_m} \sum_{h \mid d_m^\infty} \frac{1}{h^{1/2}} \\ \end{eqnarray*} where in the second sum we used the fact that, if $h > T$, then $h^{-1} < (hT)^{-1/2}$. Now from Lemma~\ref{le8} we get \begin{eqnarray*} \left | \sum_{\substack{n \leq x \\ (n,k)=1}} \frac{\mu(n)}{n} \right| & \leq & \frac{550}{\left( \log \left( \frac{ex}{Tk^m}\right) \right)^2} \sum_{d_1 \mid k} \frac{\mu(d_1)^2}{d_1} \sum_{d_2 \mid d_1} \frac{\mu(d_2)^2}{d_2} \dotsb \sum_{d_m \mid d_{m-1}} \frac{\mu(d_m)^2}{\varphi(d_m)} \\ & & {} + 550 \, T^{-1/2} \sum_{d_1 \mid k} \frac{\mu(d_1)^2}{d_1} \dotsb \sum_{d_m \mid d_{m-1}} \frac{\mu(d_m)^2}{d_m} \prod_{p \mid d_m} \left( 1 - \frac{1}{p^{1/2}} \right)^{-1} \\ &=& \frac{550 k}{\varphi(k)} \left( \frac{1}{\left( \log \left( \frac{ex}{Tk^m}\right) \right)^2} + T^{-1/2} \prod_{p \mid k} \left( 1 + \frac{1}{p^{m+1/2}} \right) \right) \\ & \leq & \frac{550 k}{\varphi(k)} \left( \frac{1}{\left( \log \left( \frac{ex}{Tk^m}\right) \right)^2} + \frac{16 \, \zeta \left( m + \tfrac{1}{2} \right)}{(e \log T)^2} \right) \end{eqnarray*} giving the asserted result if we replace $T$ by its value given above. \end{proof} \subsection{Proposition~\ref{pro3}} \begin{proof} Since a Dirichlet character is a completely multiplicative function, we get from Lemma~\ref{le45} $$\sum_{\substack{n \leq x \\ (n,k)=1}} \frac{\mu(n) \chi(n)}{n} = \sum_{\substack{n \leq x \\ n \mid k^\infty}} \frac{\chi(n)}{n} \sum_{m \leq x/n} \frac{\mu(m) \chi(m)}{m}$$ and we conclude using the inequality \cite[page~4]{bor} $$\left | \sum_{n \leq x} \frac{\mu(n) \chi(n)}{n} \right | \leq \frac{2 \sqrt{q} \log q}{|L(1,\chi)|}$$ valid whenever $q \geq 37$. \end{proof} \subsection{Theorem~\ref{t4}} \begin{proof} The proof will follow from the identity \begin{equation} S(x) = \frac{1}{2} \left( M(x)^2 + \sum_{k \leq x} \mu(k)^2 \right). \label{e5} \end{equation} Indeed \begin{eqnarray*} S(x) &=& \sum_{n \leq x} \mu (n) \sum_{k=1}^n \mu(k) = \sum_{k \leq x} \mu (k) \sum_{k \leq n \leq x} \mu(n) \\ &=& M(x) \sum_{n \leq x} \mu (n)- \sum_{k \leq x} \mu (k) \sum_{n=1}^{k-1} \mu(n) \\ &=& M(x)^2 - \sum_{k \leq x} \mu (k) \left( \sum_{n=1}^{k} \mu(n) - \mu(k) \right) \\ &=& M(x)^2 - S(x) + \sum_{k \leq x} \mu(k)^2 \end{eqnarray*} giving \eqref{e5}. Now from \cite{coh} we know that $$\left | \sum_{k \leq x} \mu(k)^2 - \frac{x}{\zeta(2)} \right | \leq \np{0.1333} \sqrt{x}$$ as soon as $x \geq \np{1664}$. This along with \eqref{e2} leads to the explicit inequality of the theorem. The second inequality follows from the fully explicit bound for the Riemann zeta-function given in \cite{for} providing $$M(x) \ll x e^{-\np{0.2098} (\log x)^{3/5}(\log \log x)^{-1/5}}.$$ Now assume RH. Using Soundararajan's result \cite[Theorem~1]{sou} we infer $$S(x) \ll x e^{2(\log x)^{1/2} (\log \log x)^{14}} \ll x^{1+\varepsilon}.$$ Conversely, if $S(x) \ll x^{1+\varepsilon}$, then $$M(x)^2 = 2 S(x) - \sum_{n \leq x} \mu(n)^2 \ll x^{1+\varepsilon}$$ so that $M(x) \ll x^{1/2+\varepsilon}$, which is known to be equivalent to the Riemann hypothesis. The proof of Theorem~\ref{t4} is complete. \end{proof} \section{Acknowledgments} I would like to express my profound gratitude to the anonymous referee for his careful reading of the manuscript and the many valuable suggestions and corrections he made in it. \begin{thebibliography}{9} \bibitem{bal} M. Balazard, Elementary remarks on M\"{o}bius' function, \textit{Proc. Steklov Inst. Math.} \textbf{276} (2012), 33--39. \bibitem{bor} O. Bordell\`{e}s, An explicit Mertens' type inequality for arithmetic progressions, \textit{J. Inequal. Pure and Appl. Math} \textbf{6} (2005), Art. 67. \bibitem{coh} H. Cohen and F. Dress, Estimations num\'{e}riques du reste de la fonction sommatoire relative aux entiers sans facteur carr\'{e}, \textit{Pub. Math. Orsay: Colloque de Th\'{e}orie Analytique des Nombres, Marseille, 1985} (1988), 73--76. \bibitem{dus} P. Dusart, Autour de la fonction qui compte le nombre de nombres premiers, Th\`{e}se, Universit\'{e} de Limoges, 1998. \bibitem{elm1} M. 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Trudgian, Updating the error term in the prime number theorem, \textit{Ramanujan J.} (2015), 1--10. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11N37; Secondary 11Y35, 11N56. \noindent \emph{Keywords: } M\"{o}bius function, explicit estimate, Balazard's inequality. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A002321} and \seqnum{A008683}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received June 11 2015; revised version received September 25 2015. Published in {\it Journal of Integer Sequences}, November 15 2015. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}