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\begin{center}
\vskip 1cm{\LARGE\bf 
Generalized Alternating Sums of \\
\vskip .1in
Multiplicative Arithmetic Functions
}
\vskip 1cm
\large
Rimer Zurita \\
Carrera de Matem\'atica\\
Universidad Mayor de San Andr\'es\\
Avenida Villaz\'on 1995\\
Planta Baja del Edificio Viejo\\
Monoblock Predio Central\\
La Paz\\
Bolivia\\
\href{mailto:rzuritao@fcpn.edu.bo}{\tt rzuritao@fcpn.edu.bo}
\end{center}

\vskip .2 in

\begin{abstract}
In this paper, given a finite set of primes $Q$, we derive asymptotic
formulas for generalized alternating sums of the form $\sum_{n\leq
x}t_Q(n)f(n)$ and $\sum_{n\leq x}t_Q(n)\frac{1}{f(n)}$, where $f$ is
a multiplicative arithmetic function, and $t_Q(n)$ equals $-1$ if $n$
is divisible by some prime $q\in Q$, and $1$ otherwise. In particular,
these results are applicable to known functions, such as Euler's
totient function, the sum of divisors function, the divisor function,
and others. In the particular case of $Q=\{2\}$, we generalize various
results obtained by  T\'oth, even improving one of his results proposed
as an open problem.
\end{abstract}

\section{Introduction}
Throughout this paper, we let $\mathbb{P}$ denote the set of all prime numbers, and $Q=\{q_1,q_2,\ldots,q_r\}$ be any finite set of prime numbers. We use the notation $q_{\min}:=\min\{q_i\}$ and $q_{\max}:=\max\{q_i\}$. The letter $p$ will always stand for a prime number.

Alternating sums appear in various topics of mathematics, including number theory. For example, Bordell\`es and Cloitre \cite{borde} established asymptotic formulas with error terms for alternating sums of the form
\begin{equation*}
\sum_{n\leq x}(-1)^{n-1}\frac{1}{g(n)},
\end{equation*}
where $g$ belongs to a class of multiplicative functions, including Euler's totient function $\varphi$, the sum-of-divisors function $\sigma$ and the Dedekind function~$\psi$.

T\'oth \cite{lazlo} established some general results for alternating sums of the form
\begin{equation*}
\sum_{n\leq x}(-1)^{n-1}f(n)\quad  \text{ or }\quad \sum_{n\leq x}(-1)^{n-1}\frac{1}{f(n)},
\end{equation*}
where $f$ belongs to a broader class of multiplicative arithmetic functions than those considered by Bordell\`es and Cloitre \cite{borde}, extending their results to a whole new kind of multiplicative functions, such as the divisor function $\tau$, the gcd-sum function $P$, the square free kernel $\kappa$, the square free numbers $\mu^2$ function, the number of abelian groups $a(n)$, the sum-of-unitary-divisor function $\sigma^{*}$, the unitary-Euler function $\varphi^{*}$, the unitary-squarefree kernel $\kappa^{*}$, the powerful part of a number, and the sum-of-bi-unitary-divisors function.

In the last part of his paper, T\'oth proposes a generalization for alternating sums, and also finds an asymptotic result for the generalized sum
\begin{equation*}
\sum_{n\leq x}t_Q(n)\sigma(n),
\end{equation*}
where
\begin{equation*}
t_Q(n)=
\begin{cases}
1, & \text{ if } q\nmid n \text{ for all } q\in Q;\\
-1, &\text{ otherwise},
\end{cases}
\end{equation*}
is defined for a finite set of prime numbers $Q$,
and $\sigma(n)=\sum_{d|n}d$ is the sum of divisors function.

Let
\begin{equation*}
D_Q(f,s):=\sum_{n=1}^\infty t_Q(n)\frac{f(n)}{n^s},
\end{equation*}
be the Dirichlet series for the multiplicative function $f$, with generalized alternating signs depending on $Q$. For example, if $Q=\{2,3\}$, we have
\begin{equation*}
D_{\{2,3\}}(f,s)=\frac{f(1)}{1^s}-\frac{f(2)}{2^s}-\frac{f(3)}{3^s}-\frac{f(4)}{4^s}+\frac{f(5)}{5^s}-\frac{f(6)}{6^s}+\frac{f(7)}{7^s}+\cdots
\end{equation*}
T\'oth \cite[Prop.\ 56]{lazlo} proved that
\begin{equation}\label{sda}
D_{Q}(f,s)=D(f,s)\left( 2\prod_{q\in Q}\left( \sum_{v=0}^\infty \frac{f(q^v)}{q^{vs}}\right)^{-1}-1\right),
\end{equation}
where $D(f,s)$ is the Dirichlet series associated with the multiplicative arithmetic function $f$.

For $q\in Q$, let us consider the formal power series
\begin{equation*}
S_{f,q}(x):=1+\sum_{v=1}^\infty f(q^v)x^v,
\end{equation*}
and its inverse formal series 
\begin{equation*}
\overline{S}_{f,q}(x):=1+\sum_{v=1}^\infty b_{v,q}x^v.
\end{equation*}
From (\ref{sda}) we have, by convolution, that
\begin{equation*}
\sum_{n\leq x}t_Q(n)f(n)=\sum_{d\leq x}h_{f,Q}(d)\sum_{j\leq x/d}f(j),
\end{equation*}
where
\begin{equation*}
h_{f,Q}(n)=
\begin{cases}
2b_{v_1,q_1}\cdots b_{v_r,q_r},&\text{ if } n=q_1^{v_1}\cdots q_r^{v_r}; \\
1,&\text{ if } n=1;\\
0,&\text{ otherwise}.   
\end{cases}
\end{equation*}
From this expression we can derive asymptotic formulas for
\begin{equation}\label{gen}
\sum_{n\leq x}t_Q(n)f(n),
\end{equation}
as long as asymptotic formulas are known for $\sum_{n\leq x}f(n)$, and the coefficients $b_{v,q}$ are adequately estimated.

For example, T\'oth \cite[Teo.\ 57]{lazlo} proved that
{\small
\begin{equation}\label{totheuler}
\sum_{n\leq x}t_Q(n)\sigma(n)=\frac{\pi^2}{12}\left(2\prod_{q\in Q}\left(1-\frac{1}{q}\right)\left(1-\frac{1}{q^2}\right)-1 \right)x^2+O(x(\log x)^{2/3}).
\end{equation}
}
In this paper, we obtain asymptotic expressions for (\ref{gen}), among whose applications we obtain the result (\ref{totheuler}) and others.

\section{Main results}
\begin{theorem}\label{sumf}
Let $f$ be a multiplicative function and consider the following four conditions
\begin{itemize}
\item[(i)] there exists a constant $C_f$ such that
$$\sum_{n\leq x}f(n)=C_fx^2+O(xR_f(x)),$$
where $1\ll R_f(x)$ when $x\rightarrow \infty$ and $R_f(x)$ is an increasing function;
\item[(ii)] $S_{f,q_i}\left(\frac{1}{q_i^2}\right)$ converges for all $i$;
\item[(iii)] 
the sequence $(b_{v,q_i})_{v\geq 0}$ satisfies $b_{v,q_i}\ll 1$ for all $i$;
 \item[(iv)]
the sequence $(b_{v,q_i})_{v\geq 0}$ satisfies $|b_{v,q_i}|\ll (r_i)^v$ with $1\leq r_i\leq \frac{q_i^2}{q_{\max}}$ for all $i$.
\end{itemize}
Assume that conditions (i) and (ii) hold and that one of the two conditions (iii) or (iv) also holds.
Then
\begin{equation*}
\sum_{n\leq x} t_Q(n)f(n)=C_fx^2\left(\frac{2}{S_{f,q_1}(1/q_1^2)\cdots S_{f,q_n}(1/q_n^2)}-1\right)+O(xR_f(x)).
\end{equation*}
\end{theorem}
\begin{theorem}\label{sum1f}
Let $f$ be a multiplicative function, and let us suppose that
\begin{itemize}
\item[(i)] there exist constants $D_f$ and $E_f$ such that
\begin{equation*}
\sum_{n\leq x}\frac{1}{f(n)}=D_f(\log x+E_f)+O(x^{-1}R_{1/f}(x)),
\end{equation*}
where $1\ll R_{1/f}(x)=o(x)$ if $x\rightarrow \infty$ and $R_{1/f}(x)$ is an increasing function;
\item[(ii)] the radius of convergence of $S_{1/f,q_i}(x)$ is $r_{1/f,q_i}>1$, for all $i$;
\item[(iii)] the coefficients of $b_{v,q_i}$ satisfy $b_{v,q_i}\ll M_i^v$ if $v\rightarrow +\infty$, for all $i$ and where $M_i<\frac{q_{\min}}{q_i}$.
\end{itemize}  
Then
{\small
\begin{align*}
&\sum_{n\leq x}t_Q(n)\frac{1}{f(n)}\\
&=D_f\left(\left(\frac{2}{\prod_i S_{1/f,q_i}(1)}-1\right)(\log x+E_f)+\frac{2}{\prod_i S_{1/f,q_i}(1)}\cdot \sum_{i=1}^r \frac{\log(q_i)S'_{1/f,q_i}(1)}{S_{1/f,q_i}(1)} \right)+O(T_{1/f,Q}(x)),
\end{align*}
}
where
{\small
\begin{align*}
&T_{1/f,Q}(x)=\\
&
\begin{cases}
x^{-1}R_{1/f}(x),&\text{ if } \max(q_iM_i)<1;\\
x^{-1}R_{1/f}(x)(\log x)^r,&  \text{ if } \max(q_iM_i)=1;\\
(\log x)^{r-1}\max \{\log x\cdot x^{\log M_{\max}/\log q_{\min}},x^{\log(\max(q_iM_i))/\log q_{\min} }\cdot x^{-1}R_{1/f}(x)\},&\text{ if } \max(q_iM_i)>1.
\end{cases}
\end{align*}
}
\end{theorem}
\section{Proofs of the main results}
\subsection{Proof of Theorem \ref{sumf}}

\begin{proof}
Under the hypothesis of the theorem, we have that
\begin{align*}
&\sum_{n\leq x}t_Q(n)f(n)=\sum_{d\leq x}h_{f,Q}(d)\sum_{j\leq x/d}f(j)
=\sum_{d\leq x}h_{f,Q}(d)\left( C_f\frac{x^2}{d^2}+O\left(\frac{x}{d}R_f(x/d)\right)\right)\\
&=C_fx^2\cdot \sum_{d\leq x}\frac{h_{f,Q}(d)}{d^2}+O\left(xR_f(x)\sum_{d\leq x}\frac{|h_{f,Q}(d)|}{d}\right).
\end{align*}
On one hand, for some $\delta'<1$,
\begin{align*}
& \sum_{d\leq x}\frac{|h_{f,Q}(d)|}{d}\leq \sum_{q_1^{v_1}\cdots q_r^{v_r}\leq x}\frac{2|b_{v_1,q_1}\cdots b_{v_r,q_r}|}{q_1^{v_1}\cdots q_r^{v_r}}+1\\
& \ll \sum_{q_1^{v_1}\cdots q_r^{v_r}\leq x}\frac{r_1^{v_1}\cdots r_r^{v_r}}{q_1^{v_1}\cdots q_r^{v_r}}\ll \sum_{v_1+\cdots +v_r\leq \frac{\log x}{\log q_{\min}}} (\delta')^{v_1+\cdots+v_r}\\
&\ll \sum_{n\leq \frac{\log x}{ \log q_{\min}}}(\delta')^n{n+r-1 \choose r-1}= \sum_{n\leq \frac{\log x}{ \log q_{\min}}}(\delta')^n\frac{(n+r-1)\cdots(n+2)\cdot(n+1)}{(r-1)!}\\
&\ll \sum_{n\leq \frac{\log x}{\log q_{\min}}}(\delta')^n(n+r-1)^{r-1}\ll 1.
\end{align*}
On the other hand, setting $s=2$ in (\ref{sda}),
\begin{align*}
\sum_{d\leq x}\frac{|h_{f,Q}(d)|}{d^2}&=\sum_{d=1}^{\infty} \frac{|h_{f,Q}(d)|}{d^2}-\sum_{d>x}\frac{|h_{f,Q}(d)|}{d^2}
=\frac{2}{S_{f,q_1}(1/q_1^2)\cdots S_{f,q_r}(1/q_r^2)}-1,
\end{align*}
and
\begin{align*}
&x^2\sum_{d>x}\frac{|h_{f,Q}(d)|}{d^2}\ll x^2\sum_{q_1^{v_1}\cdots q_r^{v_r}>x}\frac{b_{v_1,q_1}\cdots b_{v_r,q_r}}{q_1^{2v_1}\cdots q_r^{2v_r}}.
\end{align*}
\textit{Case (iii) of Theorem \ref{sumf}:}
\begin{equation*}
x^2\sum_{d>x}\frac{|h_{f,Q}(d)|}{d^2}\ll x^2\sum_{d>x}\frac{1}{d^2}\ll x \ll xR_f(x).
\end{equation*}
\textit{Case (iv) of Theorem \ref{sumf}:}

Let us define $\delta:=\max_i\left( \frac{r_i}{q_i^2}\right)<\frac{1}{q_{\max}}$. Then
\begin{align*}
&x^2\sum_{d>x}\frac{|h_{f,Q}(d)|}{d^2}\ll x^2\!\!\!\!\!\!\!\!\!\!\!\!\sum_{v_1+\cdots +v_r>\frac{\log x}{\log q_{\max}}}\left( \frac{r_1}{q_1^2}\right)^{v_1}\!\!\!\cdots \left( \frac{r_r}{q_r^2}\right)^{v_r}
\ll x^2\!\!\!\!\!\!\!\!\!\!\!\sum_{v_1+\cdots+v_r>\frac{\log x}{\log q_{\max}}}\!\!\!\!\!\!\!(\delta)^{v_1+\cdots+v_r}\\
& \ll x^2\sum_{n>\frac{\log x}{\log q_{\max}}}\delta^nn^{r-1}
\ll x^2\delta^{\frac{\log x}{\log q_{\max}}}(\log x)^{r-1}=x^{2+\frac{\log \delta}{\log q_{\max}}}(\log x)^{r-1}\ll xR_f(x).
\end{align*}
\end{proof}

\subsection{Proof of Theorem \ref{sum1f}}

\begin{proof}
From the hypothesis of the theorem, we have that
\begin{align*}
&\sum_{n\leq x}t_Q(n)\frac{1}{f(n)}=\sum_{d\leq x}h_{1/f,Q}(d)\sum_{j\leq x/d}\frac{1}{f(j)}\\
&=\sum_{d\leq x}h_{1/f,Q}(d)\left( D_f\left(\log \frac{x}{d}+E_f\right)+O\left( \left(\frac{x}{d}\right)^{-1}R_{1/f}(x/d)\right)\right)\\
&=D_f(\log x+E_f)\sum_{d\leq x}h_{1/f,Q}(d)-D_f\sum_{d\leq x}h_{1/f,Q}(d)\log d
+O\left(x^{-1}R_{1/f}(x)\cdot
\sum_{d\leq x}d|h_{1/f,Q}(d)| \right)\\
&=D_f(\log x+E_f)\sum_{d=1}^{\infty}h_{1/f,Q}(d)+O\left( \log x \sum_{d>x}|h_{1/f,Q}(d)|\right)\\
&-D_f\sum_{d=1}^{\infty}h_{1/f,Q}(d)\log d+O\left( \sum_{d>x}|h_{1/f,Q}(d)|\log d\right)+O\left(x^{-1}R_{1/f}(x)\sum_{d\leq x}d\cdot|h_{1/f,Q}(d)|\right).
\end{align*}
In particular, by (\ref{sda}),
$$
\sum_{d=1}^{\infty}\frac{h_{1/f,Q}(d)}{d^s}=\frac{2}{S_{1/f,q_1}(1/q_1^s)\cdots S_{1/f,q_r}(1/q_r^s)}-1,
$$
then we see that
$$
\sum_{d=1}^{\infty}h_{1/f,Q}(d)=\frac{2}{S_{1/f,q_1}(1)\cdots S_{1/f,q_r}(1)}-1,
$$
and
\begin{align*}
&\sum_{d=1}^{\infty}h_{1/f,Q}(d)\log d=-2\cdot \left(\frac{\log q_1\cdot S'_{1/f,q_1}(1)}{S^2_{1/f,q_1}(1)\cdots S_{1/f,q_r}(1)}+\cdots+\frac{\log q_r\cdot S'_{1/f,q_r}(1)}{S_{1/f,q_1}(1)\cdots S^2_{1/f,q_r}(1)}\right).
\end{align*}
We also have that
\begin{align*}
&\sum_{d>x}|h_{1/f,Q}(d)|\leq 1+\sum_{q_1^{v_1}\cdots q_r^{v_r}>x}2|b_{v_1,q_1}\cdots b_{v_r,q_r}|\ll \sum_{q_1^{v_1}\cdots q_r^{v_r}>x}M_1^{v_1}\cdots M_r^{v_r}
\ll \sum_{q_1^{v_1}\cdots q_r^{v_r}>x}M_{\max}^{v_1+\cdots +v_r}\\
&\ll\sum_{v_1+\cdots +v_r>\frac{\log x}{\log q_{\max}}}M_{\max}^{v_1+\cdots+v_r}\ll \sum_{n>\frac{\log x}{\log q_{\max}}}M_{\max}^n(n+r-1)^{r-1}\ll \left( \frac{\log x}{\log q_{\max}}\right)^{r-1}M_{\max}^{\frac{\log x}{\log q_{\max}}}\\
&\ll(\log x)^{r-1}x^{\frac{\log M_{\max}}{\log q_{\max}}}.
\end{align*}
Similarly,
\begin{align*}
&\sum_{d>x}|h_{1/f,Q}(d)|\log d\ll\sum_{q_1^{v_1}\cdots q_r^{v_r}>x}|b_{v_1,q_1}\cdots b_{v_r,q_r}|(v_1\log q_1+\cdots + v_r\log q_r)\\
&\ll \sum_{q_1^{v_1}\cdots q_r^{v_r}>x}(M_{\max})^{v_1+\cdots+v_r}\log q_{\max}\cdot(v_1+\cdots+v_r)\ll \sum_{n>\frac{\log x}{\log q_{\max}}}(M_{\max})^n(n+r-1)^{r-1}\cdot n\\
&\ll\sum_{n>\frac{\log x}{\log q_{\max}}}(M_{\max})^nn^r\ll (\log x)^{r}x^{\log M_{\max}/\log q_{\max}}.
\end{align*}
On the other hand,
\begin{align*}
&\sum_{d\leq x}d|h_{1/f,Q}(d)|\ll\sum_{q_1^{v_1}\cdots q_r^{v_r}\leq x}q_1^{v_1}\cdots q_r^{v_r}M_1^{v_1}\cdots M_r^{v_r}=\sum_{q_1^{v_1}\cdots q_r^{v_r}\leq x}(q_1M_1)^{v_1}\cdots (q_rM_r)^{v_r}\\
&\ll \sum_{v_1+\cdots+v_r\leq \frac{\log x}{\log q_{\min}}}(\max (q_iM_i))^{v_1+\cdots+v_r}\ll
\sum_{n\leq \frac{\log x}{\log q_{\min}}}(\max (q_iM_i))^n\cdot n^{r-1}\\
&\ll 
\begin{cases}
1,&\text{ if } \max(q_iM_i)<1;\\
(\log x)^r,&\text{ if } \max(q_iM_i)=1;\\
x^{\log \max (q_iM_i)/\log q_{\min}}\cdot (\log x)^{r-1},&\text{ if } \max(q_iM_i)>1.
\end{cases}
\end{align*}

\begin{itemize}[leftmargin=.6in]
\item[Case 1:] If $\max (q_iM_i)<1$, then
\begin{align*}
&(\log x)^r\cdot x^{\log M_{\max}/\log q_{\max}}\ll x^{-1}R_{1/f}(x)\\
&\Leftrightarrow (\log x)^rx^{1+\log M_{\max}/\log q_{\max}}\ll R_{1/f}(x),
\end{align*}
since $q_{\max}M_{\max}<1$, which implies that $\log q_{\max}+\log M_{\max}<0$.
\item[Case 2:] If $\max (q_iM_i)=1$, then
\begin{align*}
&(\log x)^r\cdot x^{\log M_{\max}/\log q_{\max}}\ll(\log x)^r x^{-1}R_{1/f}(x)\\
&\Leftrightarrow x^{1+\log M_{\max}/\log q_{\max}}\ll R_{1/f}(x).
\end{align*}
\item[Case 3:] If $\max (q_iM_i)>1$,
\begin{align*}
&(\log x)^rx^{\log M_{\max}/\log q_{\max}}\ll x^{-1}R_{1/f}(x)x^{\frac{\log \max(q_iM_i)}{\log q_{\min}}}(\log x)^{r-1}\\
&\Leftrightarrow (\log x)\cdot x^{1+\frac{\log M_{\max}}{\log q_{\max}}-\frac{\log \max (q_iM_i)}{\log q_{\min}}}\ll R_{1/f}(x).
\end{align*}
Since all three cases have been considered, the proof of Theorem \ref{sum1f} is complete.
\end{itemize}
\end{proof}

\section{Applications}
For the applications to various multiplicative functions shown in this section, it is sufficient to verify that conditions (i), (ii), (iii) and (iv) of Theorems \ref{sumf} and \ref{sum1f} hold.
\subsection{Euler's totient function \texorpdfstring{$\varphi(n)$}{phi(n)}}
Let us consider Euler's totient function $\varphi(n)=n\prod_{p|n}\left(1-\frac{1}{p}\right)$.
\begin{theorem}
\begin{equation}\label{phigen}
\sum_{n\leq x}t_Q(n)\varphi(n)=\frac{3}{\pi^2}x^2\left( \frac{2}{\prod_{q\in Q}\left(1+\frac{1}{q}\right)}-1\right)+O\left(x(\log x)^{2/3}(\log\log x)^{4/3}\right),
\end{equation}
and
\begin{equation}\label{invphigen}
\begin{split}
&\sum_{n\leq x}t_Q(n)\frac{1}{\varphi(n)}=A\Bigg( \Big( 2\prod _{q\in Q}\frac{(q-1)^2}{q^2-q+1}-1\Big)(\log x+\gamma-B)\\
&+2\prod_{q\in Q}\frac{(q-1)^2}{q^2-q+1}\cdot \sum_{q\in Q}\frac{q^2\log q}{(q-1)(q^2-q+1)}\Bigg)+O(R_{1/\varphi}(x)),
\end{split}
\end{equation}
where $R_{1/\varphi}(x)=x^{-1}(\log x)^{2/3}$ if $2\!\not\in\! Q$, and $R_{1/\varphi}(x)=x^{-1}( \log x)^{2/3+r}$
if $2\!\in\! Q$, $\gamma$ is Euler's constant, and constants $A$ and $B$ are defined by
\begin{equation*}
A=\frac{\zeta(2)\zeta(3)}{\zeta(6)},\qquad B=\sum_{p\in \mathbb{P}}\frac{\log p}{p^2-p+1}.
\end{equation*}
\end{theorem}
\begin{proof}
Concerning (\ref{phigen}),
we know from Walfisz \cite[p.\ 144]{walfisz} that 
\begin{equation*}
\sum_{n\leq x}\varphi(n)=\frac{3}{\pi^2}x^2+O(x(\log x)^{2/3}(\log \log x)^{4/3}).
\end{equation*}
Then we have $R_{\varphi}(x)=(\log x)^{2/3}(\log \log x)^{4/3}$, so condition (i) of Theorem \ref{sumf} is satisfied.

On the other hand, we can see that
\begin{align*}
S_{f,q}(x) & =1+\sum_{v=1}^{\infty}\varphi(q^v)x^v=1+\sum_{v=1}^{\infty}(q^v-q^{v-1})x^v\\
&=1+(1-\frac{1}{q})\left(\frac{qx}{1-qx}\right)=\frac{1-x}{1-qx},\quad |x|<1/q,
\end{align*}
and condition (ii) of Theorem \ref{sumf} is satisfied.

Then we conclude that
\begin{align*}
&\overline{S}_{\varphi,q}(x)=\frac{1-qx}{1-x}=1+\frac{(1-q)x}{(1-x)}=1+(1-q)x\sum_{v=0}^{\infty}x^v\qquad (|x|<1),
\end{align*}
so
$b_{v,q}=(1-q)\ll 1$, and condition (iii) of Theorem \ref{sumf} is satisfied.
Furthermore, we see that $S_{\varphi,q}(1/q^2)=\frac{1-1/q^2}{1-q\cdot 1/q^2}=\frac{q+1}{q}$.

Concerning (\ref{invphigen}),
we know from Landau \cite[ Thm.\ 1.1]{landau} and Sitaramachandraro \cite{sita} that
\begin{equation*}
\sum_{n\leq x}\frac{1}{\varphi(n)}=A(\log x+\gamma-B)+O(x^{-1}(\log x)^{2/3}).
\end{equation*}
Then we have that $R_{1/\varphi}(x)=(\log x)^{2/3}$, impliying that condition (i) of Theorem \ref{sum1f} is satisfied.

On the other hand, we see that 
\begin{align*}
S_{1/\varphi,q}(x) & =1+\sum_{v=1}^{\infty}\frac{1}{\varphi(q^v)}x^v=1+\sum_{v=1}^{\infty}\frac{x^v}{(q^v-q^{v-1})}\\
&=1+\frac{q}{(q-1)}\frac{x}{(q-x)}=\frac{x+q(q-1)}{(q-1)(q-x)}\qquad (|x|<q),
\end{align*}
so condition (ii) of Theorem \ref{sum1f} is satisfied.

Finally,
\begin{align*}
&\overline{S}_{1/\varphi,q}(x)=(q-1)\Bigg(-1+\frac{q^2}{x+q(q-1)}\Bigg)=(q-1)\Big( -1+\frac{q}{q-1}\sum_{v=0}^{\infty}\frac{(-1)^v}{q^v(q-1)^v}x^v\Big)\\
&(|x|<q(q-1)),
\end{align*}
so
$b_{v,q}=\frac{(-1)^vq}{q^v(q-1)^v(q-1)}\ll\left( \frac{1}{q(q-1)}\right)^v$, having that $M_q=\frac{1}{q(q-1)}<\frac{q_{\min}}{q}$, and condition (iii) of Theorem \ref{sum1f} is satisfied with $\max(M_iq_i)=1/(q_{\min}-1)=1$ if $q_{\min}=2$, and $\max(M_iq_i)<1$ if $q_{\min}>2$.
Furthermore, we see that
\begin{equation*}
S_{1/\varphi,q}(1)=\frac{q^2-q+1}{(q-1)^2}\quad \text{ and } \quad S'_{1/\varphi,q}(1)=\frac{q^2}{(q-1)^3}.
\end{equation*}
\end{proof}
\subsection{Sum of divisors function}
The sum of divisors functions is defined by $\sigma(n)=\sum_{d|n}d$.
\begin{theorem}
\begin{equation}\label{sumdivgen}
\sum_{n\leq x}t_Q(n)\sigma(n)=\frac{\pi^2}{12}x^2\left( 2\prod_{q\in Q}\frac{(q-1)^2(q+1)}{q^3}-1\right)+O\left(x(\log x)^{2/3}\right),
\end{equation}
and
\begin{equation}\label{invsumdivgen}
\begin{split}
&\sum_{n\leq x}t_Q(n)\frac{1}{\sigma(n)}=E\Bigg( \Big( \frac{2}{\prod_i K_{q_i}}-1\Big)(\log x+\gamma+F)\\
&+\frac{2}{\prod_i K_{q_i}}\cdot \sum_{i=1}^r\frac{\log q_i\cdot K'_{q_i}}{K_{q_i}}\Bigg)+O(x^{-1}(\log x)^{2/3+r}(\log\log x)^{4/3}),
\end{split}
\end{equation}
where $\gamma$ is Euler's constant, the constants $K_q$ and $K'_q$ are defined by
\begin{align*}
K_q=1+(q-1)\sum_{v=1}^{\infty}\frac{1}{q^{v+1}-1}\quad \text{ and }\quad K'_q=(q-1)\sum_{v=1}^{\infty}\frac{v}{q^{v+1}-1},
\end{align*}
(as a particular case, we have that $K_2\doteq 1.606695$ is
the Erd\H{o}s-Borwein constant, which can be seen in the sequence
\seqnum{A065442} of the Sloane's 
{\it On-line Encyclopedia of Integer Sequences}
(OEIS) \cite{oeis}), and the constants $E$ and $F$ are defined by
\begin{equation*}
E=\prod_{p\in \mathbb{P}}\alpha(p),\qquad F=\sum_{p\in \mathbb{P}}\frac{(p-1)^2\beta(p)\log p}{p\alpha(p)},
\end{equation*}
with
\begin{align*}
&\alpha(p)=1-\frac{(p-1)^2}{p}\sum_{j=1}^{\infty} \frac{1}{(p^j-1)(p^{j+1}-1)},\\
&\beta(p)=\sum_{j=1}^{\infty}\frac{j}{(p^j-1)(p^{j+1}-1)}.
\end{align*}
\end{theorem}

\begin{proof}
Concerning (\ref{sumdivgen}),
we know from Walfisz \cite[p.\ 99]{walfisz} that 
\begin{equation*}
\sum_{n\leq x}\sigma(n)=\frac{\pi^2}{12}x^2+O(x(\log x)^{2/3}).
\end{equation*}
Then we have that $R_{\varphi}(x)=(\log x)^{2/3}$, so condition (i) of Theorem \ref{sumf} is satisfied.

On the other hand, we have that
\begin{align*}
&S_{\sigma,q}(x)=1+\sum_{v=1}^{\infty}\sigma(q^v)x^v=1+\sum_{v=1}^{\infty} \frac{q^{v+1}-1}{q-1}x^v\\
&=1+\frac{1}{q-1}\left(\frac{q^2x}{1-qx}-\frac{x}{1-x}\right)=\frac{1}{(1-qx)(1-x)},\quad |x|<1/q,
\end{align*}
so condition (ii) of Theorem \ref{sumf} is satisfied.

Then we see that
\begin{align*}
&\overline{S}_{\sigma,q}(x)=(1-qx)(1-x)=1-(q+1)x+qx^2\qquad (x\in \mathbb{R}),
\end{align*}
so
$b_{0,q}=1,b_{1,q}=-(q+1),b_{2,q}=q$ and $b_{v,q}=0$ if $v\geq 3$, and condition (iii) of Theorem \ref{sumf} is satisfied.
Furthermore, we see that $S_{\sigma,q}(1/q^2)=\frac{1}{(1-1/q)(1-1/q^2)}=\frac{q^3}{(q-1)^2(q+1)}$.

Concerning (\ref{invsumdivgen}),
we know from Sita~Ramaiah and Suryanarayana \cite[Cor.\ 4.1]{ramasurya} that
\begin{equation*}
\sum_{n\leq x}\frac{1}{\sigma(n)}=E(\log x+\gamma+F)+O(x^{-1}(\log x)^{2/3}(\log\log x)^{4/3}).
\end{equation*}
Then we conclude that $R_{1/\varphi}(x)=(\log x)^{2/3}(\log\log x)^{4/3}$, so condition (i) of Theorem \ref{sum1f} is satisfied.

On the other hand,
\begin{align*}
&S_{1/\sigma,q}(x)=1+\sum_{v=1}^{\infty}\frac{1}{\sigma(q^v)}x^v=1+\sum_{v=1}^{\infty}\frac{(q-1)}{q^{v+1}-1}x^v,\qquad (|x|<q),
\end{align*}
so condition (ii) of Theorem \ref{sum1f} is satisfied.

The coefficients $\left(\frac{q-1}{q^{v+1}-1}\right)$ of this last power series form a log-convex sequence. Indeed,
\begin{align*}
&\left(\frac{q-1}{q^{v+1}-1}\right)^2\leq \left( \frac{q-1}{q^v-1}\right)\cdot \left( \frac{q-1}{q^{v+2}-1}\right),v\geq 0\\
&\Leftrightarrow (q^{v}-1)(q^{v+2}-1)\leq (q^{v+1}-1)^2\Leftrightarrow 2 q^{v+1}\leq q^v+q^{v+2}
\Leftrightarrow 2q\leq 1+q^2.
\end{align*}
By Kaluza's theorem (if a power series $\sum_{v=0}^{\infty}a_v$ satisfies the conditions $a_v>0$ and\\ $a_v^2\leq a_{v-1}a_{v+1}(v\geq 1)$, then the coefficients of its reciprocal power series $\sum_{v\geq 0}b_v$ satisfy \\ $-a_v/a_0^2\leq b_v\leq 0,\,\,\, v\geq 1$) \cite[Lem.\ 8]{lazlo}, we have that $-\frac{q-1}{q^{v+1}-1}\leq b_{v,q}\leq 0$, and therefore, $b_{v,q}\ll \left( \frac{1}{q}\right)^v$. We conclude that $M_i=\frac{1}{q_i}<\frac{q_{\min}}{q_i}$, so condition (iii) of Theorem \ref{sum1f} is satisfied with $\max(q_iM_i)=1$.

Furthermore, we see that
\begin{equation*}
S_{1/\sigma,q}(1)=K_q\quad \text{ and } \quad S'_{1/\sigma,q}(1)=K'_q.
\end{equation*}
\end{proof}
\subsection{Unitary divisor function}
A natural number $m$ is a unitary divisor of a number $n$ if $m$ is a divisor of $n$, and $m$ and $n/m$ are coprime. Let us define the arithmetic function $\sigma^*(n)$ as the sum of the unitary divisors of $n$ (analogous to the sum of divisors function).

We have that $\sigma^{*}$ is multiplicative and $\sigma^{*}(p^v)=p^v+1$.
\begin{theorem}
If $q_{\min}\geq q_{\max}^{2/3}$ (in particular, if $Q$ consists of a single prime), then 
{\small
\begin{equation}\label{sumdivunigen}
\sum_{n\leq x}t_Q(n)\sigma^{*}(n)=\frac{\pi^2}{12\zeta(3)}x^2\left( 2\prod_{q\in Q}\frac{(q^2-1)}{(q^2+q+1)}-1\right)+O\left(x(\log x)^{5/3}\right),
\end{equation}
}
and
\begin{equation}\label{invsumdivunigen}
\begin{split}
&\sum_{n\leq x}t_Q(n)\frac{1}{\sigma^{*}(n)}=E_Q^{*}\log x+F_Q^{*}+O(x^{-1}(\log x)^{5/3+r}(\log\log x)^{4/3})
\end{split}
\end{equation}
for some constant $F_Q^{*}$, and for $E_Q^{*}$, the latter being defined by
\begin{equation*}
E_Q^{*}=B^*\left(\frac{2}{\prod_{q\in Q}R_q}-1 \right)\quad \text{ with }\quad B^*=\prod_p \left(R_p\cdot\left( 1-\frac{1}{p}\right) \right)\quad \text{ and }\quad R_q:=1+\sum_{v=1}^{\infty}\frac{1}{q^v+1}.
\end{equation*}
\end{theorem}
\begin{proof}
We first prove (\ref{sumdivunigen}). We know from Sitaramachandrarao and Suryanarayana \cite[Eq.\ 1.4]{sitasurya} that
\begin{equation*}
\sum_{n\leq x}\sigma^{*}(n)=\frac{\pi^2}{12\zeta(3)}x^2+O(x(\log x)^{5/3}),
\end{equation*}
so $R_{\sigma^{*}}(x)=(\log x)^{5/3}$, and condition (i) of Theorem \ref{sumf} is satisfied.

Similarly, we have that
\begin{equation*}
S_{\sigma^{*},q}(x)=1+\sum_{v=1}^{\infty}\sigma^{*}(q^v)x^v=1+\sum_{v=1}^{\infty}(q^v+1)x^v=\frac{1-qx^2}{(1-qx)(1-x)}
\end{equation*}
for $|x|<1/q$, so condition (ii) of Theorem \ref{sumf} is satisfied.

We conclude that the reciprocal of the power series is given by
\begin{align*}
\overline{S}_{\sigma^{*},q}(x)&=\frac{1-(q+1)x+qx^2}{1-qx^2}=-1+\frac{2-(q+1)x}{1-qx^2}\\
&=-1+\left( \frac{1+1/2(q^{1/2}+q^{-1/2})}{1+\sqrt{q}x}+\frac{1-1/2(q^{1/2}+q^{-1/2})}{1-\sqrt{q}x}\right)\\
&=-1+\left(1+1/2(q^{1/2}+q^{-1/2})\right)\sum_{v=0}^{\infty}(-1)^v\sqrt{q}^{v}x^v+\left(1-1/2(q^{1/2}+q^{-1/2})\right)\sum_{v=0}^{\infty}\sqrt{q}^{v}x^v,
\end{align*}
with $|x|<\frac{1}{\sqrt{q}}$. Then
\begin{align*}
b_{v,q}&=\left(1+1/2(q^{1/2}+q^{-1/2})\right)(-1)^v\sqrt{q}^v+\left(1-1/2(q^{1/2}+q^{-1/2})\right)\sqrt{q}^v\ll \sqrt{q}^v.
\end{align*}
We see that $r_i=q_i^{0.5}$ and $r_i\leq q_i^2/q_{\max}$, so condition (iv) of Theorem \ref{sum1f} is satisfied. Furthermore, we see that
\begin{equation*}
S_{\sigma^{*},q}(\frac{1}{q^2})=\frac{1-1/q^3}{(1-1/q)(1-1/q^2)}=\frac{q^2+q+1}{q^2-1}.
\end{equation*}
Now, we prove (\ref{invsumdivunigen}). We know from Sita~Ramaiah and Suryanarayana \cite[p.\ 1352]{rasu} that
\begin{equation*}
\sum_{n\leq x}\frac{1}{\sigma^*(n)}=B^*\log x+D^*+O(x^{-1}(\log x)^{5/3}(\log \log x)^{4/3}).
\end{equation*}
Then we have $R_{1/\sigma^*}=(\log x)^{5/3}(\log \log x)^{4/3}$, so condition (i) of Theorem \ref{sum1f} is satisfied.

On the other hand,
\begin{equation*}
S_{1/\sigma^*,q}(x)=1+\sum_{v=1}^{\infty}\frac{1}{\sigma^*(q^v)}x^v=1+\sum_{v=1}^{\infty}\frac{1}{q^v+1}x^v\qquad (|x|<q),
\end{equation*}
so condition (ii) of Theorem \ref{sum1f} is satisfied.

Let us choose $a_n:=\frac{1}{q^n+1}$, if $n\geq 1$, and $a_0=1$, as the coefficients for this formal power series. We also define $b_n$ as the coefficients of the associated reciprocal series. Then
\begin{align*}
&a_nb_{n+1}=\sum_{k=1}^{n-1}b_k(a_{n+1}a_{n-k}-a_na_{n+1-k})+b_n(a_{n+1}-a_na_1)\qquad (n\geq 2),\\
&\frac{1}{q^n+1}b_{n+1}=\sum_{k=1}^{n-1}b_k\cdot \left( \frac{1}{q^{n+1}+1}\cdot \frac{1}{q^{n-k}+1}-\frac{1}{q^n+1}\cdot \frac{1}{q^{n+1-k}+1}\right)\\
&\hspace{2 cm}+b_n\left( \frac{1}{q^{n+1}+1}-\frac{1}{q^n+1}\frac{1}{q+1}\right)\\
&=\sum_{k=1}^{n-1}b_k\frac{((q^n+1)(q^{n+1-k}+1)-(q^{n+1}+1)(q^{n-k}+1))}{(q^{n+1}+1)(q^{n-k}+1)(q^n+1)(q^{n+1-k}+1)}+b_n\frac{(q^n+1)(q+1)-(q^{n+1}+1)}{(q^{n+1}+1)(q^n+1)(q+1)}.
\end{align*}
Then
\begin{equation}\label{bn1}
\begin{split}
b_{n+1}&=\frac{q^n(q-1)}{q^{n+1}+1}\cdot \sum_{k=1}^{n-1}\frac{b_k(q^{-k}-1)}{(q^{n-k}+1)(q^{n+1-k}+1)}+\frac{b_n}{q+1}\cdot\frac{(q^n+q)}{(q^{n+1}+1)}\\
&=\frac{(q-1)q^n}{q^{n+1}+1}\cdot \sum_{k=1}^{n-1}b_k\frac{q^k-q^{2k}}{(q^n+q^k)(q^{n+1}+q^k)}+\frac{b_n}{q+1}\cdot\frac{(q^n+q)}{(q^{n+1}+1)}.
\end{split}
\end{equation}
Let us suppose that $|b_i|\leq \frac{C}{q^i}$ for some constant $C$ and for all $i=0,1,2,\ldots, n$. We prove by induction that $|b_{n+1}|\leq \frac{C}{q^{n+1}}$.
Indeed, we have that
\begin{equation}
|b_{n+1}|\leq C\cdot \left( \frac{(q-1)q^n}{q^{n+1}+1}\cdot \sum_{k=1}^{n-1}\frac{q^k}{(q^k+q^n)^2}+\frac{q^n+q}{(q+1)\cdot q^n \cdot (q^{n+1}+1)}\right).
\end{equation}
Let us define the function $f(n):=\frac{q^{n+1}\cdot (q^n+q)}{(q+1)\cdot q^n\cdot (q^{n+1}+1)}=\frac{q^{n+1}+q^2}{(q+1)(q^{n+1}+1)}\rightarrow \frac{1}{q+1}$ if $n\rightarrow \infty$.

Furthermore, $f(n)=\frac{1}{q+1}\left( 1+\frac{q^2-1}{q^{n+1}+1}\right) $ is an increasing function on $n$, therefore $f(n)\leq \frac{1.01}{q+1}$ for $n$ large enough.

Similarly, we define the function $g(n):=\frac{(q-1)q^nq^{n+1}}{q^{n+1}+1}\cdot \sum_{k=1}^{n-1}\frac{q^k}{(q^k+q^n)^2}$. We have that
\begin{align*}
g(n)&=\frac{(q-1)q^{2n+1}}{q^{2n+1}+q^n}\cdot \sum_{k=1}^{n-1} \frac{q^k}{(q^{n/2}+q^{k-n/2})^2}=\frac{q-1}{1+q^{-1-n}}\cdot \sum_{k=1}^{n-1}\frac{1}{(q^{(n-k)/2}+q^{-(n-k)/2})^2}\\
&=\frac{1}{4}\frac{(q-1)}{(1+q^{-1-n})}\cdot \sum_{k=1}^{n-1}\frac{1}{\cosh^2(\frac{n-k}{2}\log q)}=
\frac{(q-1)}{4(1+q^{-1-n})}\cdot \sum_{j=1}^{n-1}\frac{1}{\cosh^2(j\log q/2)}\\
&\leq (q-1)\sum_{j=1}^{\infty}\frac{1}{(e^{j\log q/2}+e^{-j\log q/2})^2}=(q-1)\cdot \sum_{j=1}^{\infty}
\frac{1}{(q^{j/2}+q^{-j/2})^2}\\
&=(q-1)\sum_{j=1}^{\infty}\frac{q^{-j}}{(1+q^{-j})^2}\leq (q-1)\sum_{j=1}^{\infty}q^{-j}(1-q^{-j}+q^{-2j})^2\\
&=(x^{-1}-1)\sum_{j=1}^{\infty}x^j\cdot(1-x^j+x^{2j})^2,
\end{align*} 
where we set $x:=q^{-1}$. We have that
\begin{align*}
g(n)&\leq (q-1)\cdot \sum_{j=1}^{\infty}(x^j-2x^{2j}+3x^{3j}-2x^{4j}+x^{5j})\\
&=(q-1)\left(
\frac{x}{1-x}-\frac{2x^2}{1-x^2}+\frac{3x^3}{1-x^3}-\frac{2x^4}{1-x^4}+\frac{x^5}{1-x^5}
\right)\\
&=\frac{q^9+q^8+5q^7+5q^6+7q^5+7q^4+8q^3+4q^2+3q+1}{(q^4+q^3+q^2+q+1)(q^2+q+1)(q^2+1)(q+1)}=:s(q).
\end{align*}
Continuing with (\ref{bn1}), we have that, for all $n$ sufficiently large,
$$
|b_{n+1}|\leq \frac{C}{q^{n+1}}\left(s(q)+\frac{1.01}{q+1}\right)\leq \frac{C}{q^{n+1}},
$$
since $s(q)+1.01/(q+1)<1$ for all primes $q$. Indeed, the real function $s(x)+1.01/(x+1)$ decreases in the interval $[2,\delta]$, for some $\delta$, and increases in $[\delta,\infty)$, but it is always less than $1$. Then condition condition (iii) of Theorem \ref{sum1f} is satisfied with $M_i=\frac{1}{q_i}$.
\end{proof}

The result (\ref{invsumdivunigen}) improves the result $(51)$ by T\'oth \cite{lazlo}, thus solving open problem $41$ of that publication.

\subsection{Dedekind \texorpdfstring{$\psi$}{psi} function}  
Recall that the Dedekind function $\psi(n)$ is defined as $\displaystyle \psi(n)=n\prod_{p|n}(1+\frac{1}{p})$.
\begin{theorem}\label{teodede}
We have that
\begin{equation}\label{sumgended}
\sum_{n\leq x}t_Q(n)\psi(n)=\frac{15}{2\pi^2}x^2\left( 2\prod _{q\in Q}\left(\frac{q(q-1)}{q^2+1}\right)-1\right)+O(x(\log x)^{2/3})
\end{equation}
and
\begin{equation}\label{sumgendedinv}
\begin{split}
&\sum_{n\leq x}t_Q(n)\frac{1}{\psi(n)}=C\Bigg( (\log x+\gamma+D)\cdot \left( 2\prod_{q\in Q}\frac{q^2-1}{q^2+q-1}-1\right)\\
&+2\prod_{q\in Q}\frac{q^2-1}{q^2+q-1}\cdot \sum_{q\in Q}\frac{q^2\log q}{(q-1)(q^2+q-1)}\Bigg)
+O(x^{-1}(\log x)^{2/3}(\log \log x)^{4/3}),
\end{split}
\end{equation}
where
\begin{equation*}
C=\prod_p\left( 1-\frac{1}{p(p+1)}\right)\quad \text{ and }\quad D=\sum_p\frac{\log p}{p^2+p-1}.
\end{equation*}
(The constant $C\doteq 0.704442$ is sometimes called the carefree constant, and its digits
 form the sequence \seqnum{A065463}
in OEIS \cite{oeis}.) % Rimer: a esto le puse paréntesis.
\end{theorem}
\begin{proof}
The proof of (\ref{sumgended}) is quite similar to that of (\ref{phigen}).
We know from Walfisz \cite[p.\ 100]{walfisz} that
\begin{equation*}
\sum_{n\leq x}\psi(n)=\frac{15}{2\pi^2}x^2+O(x(\log x)^{2/3}),
\end{equation*}
and we obtain that
\begin{equation*}
S_{\psi,q}(x):=1+\sum_{v=1}^{\infty}\psi(q^v)x^v=\frac{1+x}{1-qx} \qquad (|x|<\frac{1}{q}),
\end{equation*}
thus concluding that
\begin{equation*}
\overline{S}_{\psi,q}(x)=\frac{1-qx}{1+x}=1+(q+1)x\sum_{v=1}^{\infty}(-1)^vx^v.
\end{equation*}
Therefore, $b_{v,q}=(-1)^v(q+1)\ll 1$.

The proof of (\ref{sumgendedinv}) is quite similar to that of (\ref{invphigen}).
We know from Sita~Ramaiah and Suryanarayana \cite[Cor.\ 4.2]{rasu} that
\begin{equation*}
\sum_{n\leq x}\frac{1}{\psi(n)}=C(\log x+\gamma +D)+O(x^{-1}(\log x)^{2/3}(\log \log x)^{4/3}).
\end{equation*}
and, for the reciprocal power series, we obtain that
\begin{equation*}
S_{1/\psi,q}(x)=1+\sum_{v=1}^{\infty}\frac{1}{\psi(q^v)}x^v=\frac{q^2+q-x}{(q+1)(q-x)},
\end{equation*}
and
\begin{equation*}
\overline{S}_{1/\psi,q}(x)=\frac{(q+1)(q-x)}{q^2+q-x}=1-q\sum_{v=1}^{\infty}(\frac{1}{q^2+q})^vx^v.
\end{equation*}
Then we have that $b_{v,q}=-q\left( \frac{1}{q^2+q}\right)^v$ and $M_i=1/(q_i^2+q_i)$.
\end{proof}
\subsection{Euler's unitary function}
We now consider an analogue of the Euler totient function, namely the multiplicative function $\varphi^*$ defined on the prime powers $p^v$ by $\varphi^*(p^v)=p^v-1$.
\begin{theorem}
If $q_{\min}\geq q_{\max}^{2/3}$, we have that

\begin{equation}\label{sumgenphiun}
\sum_{n\leq x}t_Q(n)\varphi^*(n)=\frac{C}{2}x^2\left( 2\prod _{q\in Q}\left(\frac{q^2-1}{q^2+q-1}\right)-1\right)+O(x(\log x)^{5/3}(\log \log x)^{4/3}),
\end{equation}

where
$C$ is defined as in Theorem \ref{teodede}.
\end{theorem} 
\begin{proof}
The proof is quite similar to that of (\ref{sumdivunigen}).
We know from Sitaramachandrarao and Suryanarayana \cite{sitasurya} that
\begin{equation*}
\sum_{n\leq x}\varphi^*(n)=\frac{C}{2}x^2+O(x(\log x)^{5/3}(\log \log x)^{4/3}),
\end{equation*}
and we obtain that
\begin{equation*}
S_{\varphi^*,q}(x)=1+\sum_{v=1}^{\infty}\varphi^*(q^v)x^v=\frac{1-2x+qx^2}{(1-qx)(1-x)}
\end{equation*}
thus concluding that
\begin{equation*}
\overline{S}_{\varphi^*,q}(x)=\frac{(1-qx)(1-x)}{qx^2-2x+1}=1+\frac{\sqrt{q-1}}{2}i\left(-\frac{1}{\omega}\sum_{v=0}^{\infty}(x/\omega)^{v+1}+\frac{1}{\overline{\omega}}\sum_{v=0}^{\infty}(x/\overline{\omega})^{v+1}\right),
\end{equation*}
with $\omega=\frac{1}{q}\cdot (1+i\sqrt{q-1})$. Consequently, $b_{v,q}=-\frac{\sqrt{q-1}}{2}i(-\frac{1}{\omega}\omega^{-v}-\frac{1}{\overline{\omega}}\overline{\omega}^{-v})\ll (q^{1/2})^v$.
\end{proof}
\textbf{Note:} Knowing, from \cite{sitasub}, that for certain constants $L^*$ and $M^*$,
\begin{equation*}
\sum_{n\leq x}\frac{1}{\varphi^*(n)}=L^*\log x+M^*+O(x^{-1}(\log x)^{5/3}),
\end{equation*}
the author conjectures that a general result can be established for
\begin{equation*}
\sum_{n\leq x}t_Q(n)\frac{1}{\varphi^*(n)}
\end{equation*}
by proceeding in a manner similar to the one used in the proof of (\ref{invsumdivunigen}).
\section{Acknowledgments}
The author wishes to thank Josimar Ramirez for his thoughtful revision of this paper, and Diego Sejas for his suggestions on the writing of this work.

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Mathematics, Vol.~163, American Mathematical Society, 2015.

\bibitem{lazlo} L. T\'oth, 
Alternating sums concerning multiplicative arithmetic functions,
\textit{J. Integer Sequences} {\bf 20} (2017), 
\href{https://cs.uwaterloo.ca/journals/JIS/VOL20/Toth/toth25.html}{Article 17.2.1}.

\bibitem{walfisz} A. Walfisz, 
\textit{Weylsche {E}xponentialsummen in der neueren {Z}ahlentheorie},
 Mathematische Forschungsberichte, XV, VEB Deutscher Verlag der Wissenschaften, Berlin, 1963.


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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11N37; Secondary 11A05, 11A25, 30B10.

\noindent \emph{Keywords: }
multiplicative arithmetical function, generalized alternating sum, asymptotic formula, Euler's totient function, sum of divisors function, divisor function.

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\noindent (Concerned with sequences
\seqnum{A065442} and
\seqnum{A065463}.)

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\noindent
Received April 16 2020;
revised versions received 
May 7 2020; August 24 2020.
Published in {\it Journal of Integer Sequences}, 
October 17 2020.

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