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\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}
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\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 $c_\varepsilon &gt; 0$ such that
$$L(1,\chi) &gt; \frac{c_\varepsilon}{q^\varepsilon}$$ 
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 $x \geq 229$
$$\left | \pi(x) - \textrm{Li} (x) \right | &lt; \np{0.2795} \, \frac{x}{(\log x)^{3/4}} \exp \left( - \sqrt{\frac{\log x}{\np{6.455}}} \right)$$
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  
$$\sum_{n \leq x} \mu(n) \ll x \exp \left( -c \sqrt{\log x} \right) \quad \left( x \geq 2, \ c&gt; 0 \right)$$
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 $\alpha &gt; 0$ 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 | &lt; \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 | &lt; \frac{362.7x}{(\log x)^2} \quad \left( x &gt; 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 &gt; 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 $x \geq 33$
$$\left | \sum_{n \leq x} \frac{\mu(n)}{n} \right| &lt; \frac{0.19}{\log x}.$$
   \item[$(b)$] For all $x &gt; 1$
$$\left | \sum_{n \leq x} \frac{\mu(n)}{n} \right| &lt; \frac{546}{(\log x)^2}.$$
\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
$$\left | \sum_{\substack{n \leq x \\ (n,k)=1}} \frac{\mu(n)}{n} \right| \leq \frac{k}{\varphi(k)} \, \frac{0.78}{\log(x/k)} \quad \left( 1 \leq k &lt; x \right).$$
Our second result is a complement to Ramar\'{e}'s bound.

\begin{theorem}
\label{t2}
Let $k,m \in \Z_{\geq 1}$. For all $x \geq k^m$
$$\left | \sum_{\substack{n \leq x \\ (n,k)=1}} \frac{\mu(n)}{n} \right| &lt; \frac{k}{\varphi(k)} \, \frac{C_m}{\left( \log \left( exk^{-m} \right)\right)^2}$$
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
$m$ &amp; $1$ &amp; $2$ &amp; $3$ &amp; $4$ &amp; $5$ &amp; $6$ &amp; $7$ &amp; $8$ &amp; $9$ &amp; $10$ \\
\hline
$\left \lceil C_m \right \rceil$ &amp; $\np{12555}$ &amp; $\np{8045}$ &amp; $\np{7221}$ &amp; $\np{6937}$ &amp; $\np{6820}$ &amp; $\np{6768}$ &amp; $\np{6743}$ &amp; $\np{6731}$ &amp; $\np{6725}$ &amp; $\np{6723}$ \\
\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 $\varepsilon &gt; 0$ 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 $n \in \Z_{\geq 1}$
$$\left( \mathbf{1}_a^\infty \star \mu \right) (n) = \begin{cases} \mu(n), &amp; \textrm{if\ } (n,a) = 1; \\ 0, &amp; \textrm{otherwise}. \end{cases}$$
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 $C_k &gt; 0$ and $D_k &gt; 0$ 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 $a &gt; 1$, $\alpha &gt; 0$ 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 $b &gt; 1$, 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} &amp;=&amp; \sum_{d_1 \mid k} \frac{\mu(d_1)}{d_1} \sum_{h \leq x/d_1} \frac{\mu(hd_1)}{h} \\
   &amp;=&amp; \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} \\
   &amp;=&amp; \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} &amp;=&amp; \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} \\
   &amp;=&amp; \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}} \\
   &amp;=&amp; \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}  \\
   &amp;=&amp; \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 $x &gt; \np{6000}$. 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)| &amp; \leq &amp; |M(x)| + \frac{1}{x} \left( \int_1^T + \int_T^x \right) |M(t)| \, \textrm{d}t + \frac{8}{3} \\
   &amp; &lt; &amp; \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} \\
   &amp; \leq &amp; \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
$$x |m(x)| &lt; \frac{\np{0.10917} x}{\log x} + \frac{\np{1567}}{685} \, \frac{T^2}{160x} + \frac{\np{0.10917}}{2} \left( 1 + \frac{1}{\sqrt{2e \log T}} \right)^2 \frac{x}{\log x} + \frac{8}{3}.$$
We choose $T = \np{0.337} \, x (\log x)^{-1/2}$. Since $x &gt; \np{6000}$, we have $T&gt;685$ and thus
\begin{eqnarray*}
   x |m(x)| &amp;&lt;&amp; \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. \\
   &amp; &amp; \biggl. {} + \frac{8}{3} \, \frac{\log \np{6000}}{\np{6000}} \biggr) \\
   &amp;&lt;&amp; \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
   $$\frac{546}{(\log x)^2} &gt; \frac{546}{(\log 2)^2} &gt; 1 = |m(x)|$$
   and then numerically for $x \in \left[ 2,33 \right]$. If $x \in \left[ 33,e^{\np{2873}} \right]$, then
   $$|m(x)| &lt; \frac{\np{0.19}}{\log x} &lt; \frac{546}{(\log x)^2}$$
   so that we may suppose $x &gt; e^{\np{2873}}$. Using Lemma~\ref{le5} as above, Lemma~\ref{le6} with $\alpha = 2$, and \eqref{e3}, we get for any $T &gt; 1$
\begin{eqnarray*}
   x |m(x)| &amp; \leq &amp; \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} \\
   &amp; &lt; &amp; \frac{\np{362.7}x}{(\log x)^2} + \frac{T^2(1+882/T)}{160x} \\
   &amp; &amp; {} + \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 $T=x(\log x)^{-1} &gt; \dfrac{e^{\np{2873}}}{\np{2873}}$, we obtain
\begin{eqnarray*}
   x |m(x)| &amp;&lt;&amp; \frac{x}{(\log x)^2} \Biggl( \np{362.7} + \frac{1}{160} \left( 1 + \frac{882 \times \np{2873}}{e^{\np{2873}}} \right)  \Biggr. \\
   &amp; &amp; \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) \\
   &amp; &lt; &amp; \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) \\
   &amp;&lt;&amp; \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 $N \in \Z_{\geq 1}$
$$\left | \sum_{n=1}^N \frac{\mu(n)}{n} \right | \leq \frac{1}{\log \left( \frac{e}{2}(N+1) \right)} \quad \text{and} \quad \left | \sum_{n=1}^N \frac{\mu(n)}{n} \right | &lt; \frac{550}{\left( \log \left( e(N+1) \right)\right)^2}.$$
\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}
$$\left | \sum_{n=1}^N \frac{\mu(n)}{n} \right | &lt; \frac{\np{0.19}}{\log N} \leq \frac{1}{\log \left( \frac{e}{2}(N+1) \right)}.$$
Now let us have a look at the  second inequality. If $N \in \{1,\dotsc,10^{119}\}$, then
$$\left | \sum_{n=1}^N \frac{\mu(n)}{n} \right | \leq \frac{1}{\log \left( \frac{e}{2}(N+1) \right)} &lt; \frac{550}{\left( \log \left( e(N+1) \right)\right)^2}$$
and for $N &gt; 10^{119}$
$$\left | \sum_{n=1}^N \frac{\mu(n)}{n} \right | &lt; \frac{546}{\log^2 N} \leq \frac{550}{\left( \log \left( e(N+1) \right)\right)^2}$$
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*}
   &amp; &lt; &amp; 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} \\
   &amp;=&amp; 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 &lt; 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} \\
   &amp; \leq &amp; 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)} \\
   &amp; &amp; {} + 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 $h &gt; T$, then $h^{-1} &lt; (hT)^{-1/2}$. Now from Lemma~\ref{le8} we get
\begin{eqnarray*}
   \left | \sum_{\substack{n \leq x \\ (n,k)=1}} \frac{\mu(n)}{n} \right| &amp; \leq &amp; \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)} \\
   &amp; &amp; {} + 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} \\
   &amp;=&amp; \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) \\
   &amp; \leq &amp;  \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) &amp;=&amp; \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) \\
  &amp;=&amp; M(x) \sum_{n \leq x} \mu (n)- \sum_{k \leq x} \mu (k) \sum_{n=1}^{k-1} \mu(n) \\
  &amp;=&amp; M(x)^2 - \sum_{k \leq x} \mu (k) \left( \sum_{n=1}^{k} \mu(n) - \mu(k) \right) \\
  &amp;=&amp; 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.

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\end{thebibliography}


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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11N37; Secondary 11Y35, 11N56.

\noindent \emph{Keywords: } M\"{o}bius function, explicit estimate,
Balazard's inequality.

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\noindent (Concerned with sequences
\seqnum{A002321} and
\seqnum{A008683}.)

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\vspace*{+.1in}
\noindent
Received June 11 2015;
revised version received  September 25 2015.
Published in {\it Journal of Integer Sequences}, November 15 2015.

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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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