\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{amscd} \usepackage{graphicx} \usepackage{enumitem} \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} \usepackage{latexsym} \usepackage{epsf} \usepackage{breakurl} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{https://oeis.org/#1}{\rm \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 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|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|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|