---
abstract: 'As early as the 1930s, Pál Erdős conjectured that: [*for any multiplicative function $f:\mathbb{N}\to\{-1,1\}$, the partial sums $\sum_{n\leqslant x}f(n)$ are unbounded.*]{} In this paper, after providing a counterexample to this conjecture, we consider completely multiplicative functions $f:{\mathbb}{N}\to\{-1,1\}$ as well as a class of similar multiplicative functions $f$ satisfying $$\sum_{p\leqslant x}f(p)=c\cdot\frac{x}{\log x}(1+o(1)).$$ We prove that if $c>0$ then the partial sums of $f$ are unbounded, and if $c<0$ then the partial sums of $\mu f$ are unbounded. Extensions of this result are also given.'
address: 'University of Waterloo, Dept. of Pure Math., Waterloo, ON, N2L 3G1, Canada'
author:
- Michael Coons
title: On the multiplicative Erdős discrepancy problem
---

[^1]

Introduction
============

Erdős [@E1957] asked the following question, sometimes known as the Erdős Discrepancy Problem. [*“Let $f(n)=\pm 1$ be an arbitrary number theoretic function. Is it true that to every $c$ there is a $d$ and an $m$ for which $$\label{E8}\left|\sum_{k=1}^n f(kd)\right|>c\ ?$$ Inequality is one of my oldest conjectures.”*]{} (This particular quote is taken from a restatement of the conjecture in [@E1985 p.78]. See also [@E1985b] and [@EG].) Erdős offered 500 dollars for a proof of this conjecture. Erdős [@E1957 p.293] wrote in 1957 that this conjecture is twenty-five years old, placing its origin at least as far back as the early 1930s. In [@E1957; @E1985; @E1985b], Erdős also stated a multiplicative form of his conjecture.

\[Econj\] Let $f(n)=\pm 1$ be a multiplicative function, (i.e., $f(ab)=f(a)f(b)$, when $\gcd(a,b)=1$). Then $$\label{E9}\limsup_{x\to\infty}\left|\sum_{n\leqslant x} f(n)\right|=\infty;$$ that is, the partial sums of $f$ are unbounded.

Erdős added in [@E1985] that [*“clearly would follow from but as far as I know has never been proved. Incidentally was also conjectured by Tchudakoff.”*]{}

Conjecture \[Econj\] as stated is not true, and while this may be known to others in this field, there seems to be no account of it in the literature.

For a counterexample, consider the multiplicative function $g$ defined by $g(1)=1$, and on prime powers by $$\label{g}g(p^k)=\begin{cases} -1 &\mbox{if $p=2$ and $k\geq 1$}\\ 1 &\mbox{if $p\neq 2$ and $k\geq 1$}.\end{cases}$$ Then $g$ is periodic with period $2$ and for all $n\geq 1$ we have $g(2n)=-1$ and $g(2n-1)=1$. Thus $$\sum_{n\leq x}g(n)=\begin{cases} 1 & \mbox{if $[x]$ is odd}\\ 0 & \mbox{if $[x]$ is even},\end{cases}$$ and so $$\limsup_{x\to\infty}\left|\sum_{n\leqslant x} g(n)\right|=1.$$

It may very well be the case that the function $g$ defined above is the only counterexample to Conjecture \[Econj\], but at least at this point, we can say that this is the only known counterexample. Along with Conjecture \[Econj\], Erdős [@E1957] conjectured a result on the mean values of multiplicative functions. A number–theoretic function $f:\mathbb{N}\to\mathbb{C}$ has a [*mean value*]{}, denoted $M(f)$, provided the limit $$\label{MV} M(f):=\lim_{x\to\infty}\frac{1}{x}\sum_{n\leqslant x}f(n)$$ exists. Erdős [@E1957; @E1985] (among others) conjectured that any multiplicative function taking the values $\pm 1$ has a mean value; this is usually called the Erdős–Wintner Conjecture. In 1961, Delange [@Del1] characterized those functions with positive mean value, and in 1967, Wirsing [@Wir1967] gave a complete solution to this conjecture, as well as the extension to all complex–valued multiplicative functions $f$ satisfying $|f|\leqslant 1$. This was later refined by Halász [@Halasz] in 1968. We state the result here only for those functions with which we are directly concerned.

\[TMV\] Let $f:\mathbb{N}\to\{-1,1\}$ be a multiplicative function. If $$\label{1f}\sum_{p\leqslant x}\frac{1-f(p)}{p}$$ is bounded then $M(f)$ exists and is positive, and if is unbounded then $M(f)=0$.

We note that the ideas of Theorem \[TMV\] have been generalized by many authors, including Granville and Soundararajan [@GS2007; @GS2008] and Goldmakher [@Leo]. In these works the authors use properties of a generalization of to give some new results concerning sums of certain types of Dirichlet characters. The generalization of is usually made by considering a special multiplicative function $g$ (e.g., a Dirichlet character) and comparing it to the multiplicative function of interest $f$ (e.g., a Dirichlet character) by means of investigating the asymptotics of $$\sum_{p\leqslant x}\frac{1-\Re(f\overline{g}(p))}{p}.$$ This sum can be thought of as a metric [@GS2007], and in some sense measures how $g$ mimics $f$; this terminology was introduced in [@Leo].

In contrast to this “mimicry metric,” we consider the asymptotics of $$\sum_{p\leqslant x}\frac{c-f(p)}{p}$$ for $c$ not necessarily equal to $1$. By considering sums like like this, we are able to give the following result toward Conjecture \[Econj\].

\[Tmain\] Let $f:{\mathbb}{N}\to\{-1,1\}$ be a multiplicative function such that there is some $k\geq 1$ with $f(2^k)=1$. Suppose that for some $c\in[-1,1]$ we have $$\sum_{p\leqslant x}f(p)=c\cdot\frac{x}{\log x}(1+o(1)).$$ If $c>0$ then the partial sums of $f$ are unbounded, and if $c<0$ the partial sums of $\mu f$ are unbounded.

Some extensions of this theorem are given in Section \[exten\], including some instances of the case $c=0$. In Section \[CMF\], we show that this theorem is true for completely multiplicative functions without the assumption that there is some $k\geq 1$ with $f(2^k)=1$.

Completely multiplicative functions {#CMF}
===================================

If a multiplicative function $f:{\mathbb}{N}\to\{-1,1\}$ has positive mean value, then clearly the partial sums of $f$ are unbounded; they are asymptotic to $M(f)\cdot x$. The triviality leaves when we consider functions with $M(f)=0$.

Let $f:\mathbb{N}\to\{-1,1\}$ be a completely multiplicative function (i.e., $f(ab)=f(a)f(b)$ for all $a,b\in{\mathbb}{N}$) and suppose that $c\in[-1,1)$. If $$\sum_{p}\frac{c-f(p)}{p}<\infty,$$ then the mean value of $f$ exists and is equal to $0$.

This follows from Theorem \[TMV\] in a very straightforward way. We need only note that $$\begin{gathered}
 \sum_{p\leqslant x}\frac{1-f(p)}{p}=\sum_{n\leqslant x}\frac{1-c+c-f(p)}{p}\\ =(1-c)\sum_{n\leqslant x}\frac{1}{p}+\sum_{n\leqslant x}\frac{c-f(p)}{p}=(1-c)\log\log x+O(1).\qedhere\end{gathered}$$

To prove Theorem \[Tmain\], we will first prove the result for completely multiplicative functions $f:{\mathbb}{N}\to\{-1,1\}$. The bulk of the work is taken up by the following lemma.

\[Lcf\] Let $f:\mathbb{N}\to\{-1,1\}$ be a completely multiplicative function. Suppose that $c\in[-1,1]$ is nonzero and $$\sum_{p}\frac{c-f(p)}{p}<\infty.$$ If $c>0$ then the partial sums of $f$ are unbounded, and if $c<0$ the partial sums of $\mu f$ are unbounded.

Suppose firstly that $c>0$. To give the desired result, it is enough to show that $$\lim_{x\to\infty} \sum_{n\leqslant x}\frac{f(n)}{n}=\infty.$$ To this end, note that for ${\sigma}>1$ we have $$\begin{gathered}
\label{logF} \log F({\sigma})=\log \sum_{n\geqslant 1}\frac{f(n)}{n^{\sigma}} =-\sum_p\log\left(1-\frac{f(p)}{p}\right)\\ =\sum_p\sum_{k\geqslant 1}\frac{f(p)^k}{kp^{k{\sigma}}} =\sum_p \frac{f(p)}{p^{\sigma}}+\sum_p\sum_{k\geqslant 2}\frac{f(p)^k}{kp^{k{\sigma}}}=\sum_p \frac{f(p)}{p^{\sigma}}+O(1),\end{gathered}$$ where the $O(1)$ term is valid for ${\sigma}>1/2$. Since $$\sum_p\frac{c-f(p)}{p}<\infty,$$ we have that $$\label{fposc}\sum_{p\leqslant x}\frac{f(p)}{p}=c\log\log x+O(1).$$ The condition that $c>0$ ensures that $$\lim_{s\to 1^{+}}\sum_{p}\frac{f(p)}{p^{\sigma}}=\infty,$$ and so the divergence of $\log F({\sigma})$ at ${\sigma}=1$ occurs because $\lim_{{\sigma}\to 1^+}F({\sigma})=\infty$.

In the light of it must be the case that $$\label{fninfty}\lim_{x\to\infty}\sum_{n\leqslant x}\frac{f(n)}{n}=\infty.$$ Thus we have that $$\limsup_{x\to\infty} \left|\sum_{n\leqslant x}f(n)\right|=\infty.$$ For if not, there is a real number $M>0$ such that $\left|\sum_{n\leqslant x}f(n)\right|<M,$ and by partial summation, we would then have that $$\sum_{n\leqslant x}\frac{f(n)}{n}=\frac{1}{x}\sum_{n\leqslant x}f(n)+\int_1^x\left(\sum_{n\leqslant t}f(n)\right)\frac{dt}{t^2}=O\left(\int_1^x\frac{dt}{t^2}\right)=O(1),$$ which contradicts .

Now suppose that $c<0$. In this case, instead of $F({\sigma})$, we consider the function $1/F({\sigma})$. Running through the above argument gives $$\label{1overF}-\log F({\sigma})=-\sum_p \frac{f(p)}{p^{\sigma}}+O(1),$$ where again the $O(1)$ term is valid for ${\sigma}>1/2.$ Similar to the above, using the assumption of the lemma, we have that $$\label{fnegc}-\sum_{p\leqslant x}\frac{f(p)}{p}=|c|\log\log x+O(1),$$ which in turn gives, due to that $$\lim_{{\sigma}\to 1^+}\frac{1}{F({\sigma})}=\infty.$$ This implies that $$\lim_{x\to\infty}\sum_{n\leqslant x}\frac{\mu(n)f(n)}{n}=\infty,$$ which using a similar argument as the case $c>0$, give that $$\limsup_{x\to\infty}\left|\sum_{n\leqslant x}\mu(n)f(n)\right|=\infty.$$ This completes the proof of the lemma.

Our proof of the main theorem follows from the similar result for completely multiplicative functions. Using partial summation we have the following theorem.

\[Tcmain\] Let $f:{\mathbb}{N}\to\{-1,1\}$ be a completely multiplicative function. Suppose that for some $c\in[-1,1]$ we have $$\sum_{p\leqslant x}f(p)=c\cdot\frac{x}{\log x}(1+o(1)).$$ If $c>0$ then the partial sums of $f$ are unbounded, and if $c<0$ the partial sums of $\mu f$ are unbounded.

This follows directly from Lemma \[Lcf\]. The condition $$\sum_{p\leqslant x}f(p)=c\cdot\frac{x}{\log x}(1+o(1))$$ gives by partial summation that $$\begin{aligned}
\nonumber \sum_{p\leqslant x}\frac{f(p)}{p} &=\frac{1}{x}\sum_{p\leqslant x}f(p)+\int_1^x\left(\sum_{p\leqslant t}f(p)\right)\frac{dt}{t^2}\\
\nonumber &=c\cdot\frac{1}{\log x}(1+o(1))+c\int_1^x\frac{1}{t\log t}(1+o(1))dt\\
\label{cdens}&=c\log\log x(1+o(1)).\end{aligned}$$ Note that the proof of the lemma follows from the divergent behavior of $\sum_{p\leqslant x}\frac{f(p)}{p}$ in both and , and that this divergence is satisfied by . Thus using in the place of and is enough to prove Lemma \[Lcf\], and thus the condition implies the result of the theorem.

Extension to multiplicative functions
=====================================

The results of the previous section are extendable to multiplicative functions $f:{\mathbb}{N}\to\{-1,1\}$ with the added condition that there is some $k\geq 1$ with $f(2^k)=1$. In this section, by relating a multiplicative function $f:{\mathbb}{N}\to\{-1,1\}$ to a related completely multiplicative function, we are able to deduce Theorem \[Tmain\] as a corollary to Theorem \[Tcmain\]. This is obtained via the following lemma.

\[multf\] Let $f:{\mathbb}{N}\to\{-1,1\}$ be a multiplicative function such that there is some $k\geq 1$ with $f(2^k)=1$. Then $$F({\sigma})=\sum_{n\geqslant 1}\frac{f(n)}{n^{\sigma}}=\Pi({\sigma})\cdot \prod_p\left(1-\frac{f(p)}{p^{\sigma}}\right)^{-1}\qquad \left({\sigma}>1\right),$$ where $$\Pi({\sigma})=\prod_p\left(1+\sum_{k\geqslant 2}\frac{f(p^k)-f(p^{k-1})f(p)}{p^{k{\sigma}}}\right).$$ Moreover, there is a ${\sigma}_0(f)\in(0,1)$ such that $\Pi({\sigma})$ is absolutely convergent for ${\sigma}>{\sigma}_0(f).$

Note that if $f$ is multiplicative, then for ${\sigma}>1$ we have using the Euler product for its generating Dirichlet series that $$\begin{aligned}
 F({\sigma}):=\sum_{n\geqslant 1}\frac{f(n)}{n^{\sigma}}&=\prod_p\left(1+\frac{f(p^2)}{p^{2{\sigma}}}+\frac{f(p^3)}{p^{3{\sigma}}}+\cdots\right)\\
&=\prod_p\left(1-\frac{f(p)}{p^{{\sigma}}}\right)^{-1}\cdot\prod_p\left(1+\sum_{k\geqslant 2}\frac{f(p^k)-f(p^{k-1})f(p)}{p^{k{\sigma}}}\right).\end{aligned}$$

It remains to show that $\Pi({\sigma})$ is absolutely convergent for ${\sigma}>\frac{\log {\varphi}}{\log 2}$. Firstly, note that for each prime $p$ we have $$\begin{gathered}
\label{Piterm}1+\sum_{k\geqslant 2}\frac{f(p^k)-f(p^{k-1})f(p)}{p^{k{\sigma}}}\\ \geq\max\left\{1+\sum_{k\geqslant 2}\frac{f(2^k)-f(2^{k-1})f(2)}{2^{k{\sigma}}},1-\frac{2}{3^{\sigma}(3^{\sigma}-1)}\right\}.\end{gathered}$$ To ensure that none of the factors of the product $\Pi({\sigma})$ is zero, we will show that the right–hand side of is greater than zero for ${\sigma}>\frac{\log{\varphi}}{\log 2}.$ Now if $f(2^k)=1$ for all $k\geq 1$, then $$1+\sum_{k\geqslant 2}\frac{f(2^k)-f(2^{k-1})f(2)}{2^{k{\sigma}}}=1>0,$$ regardless of the range of ${\sigma}$. Thus we may suppose that $f(2^k)\neq 1$ identically. Then, using our assumption, since $f(2^k)=1$ for at least one $k\geq 1$, we have for some $k\geq 2$ that $f(2)\neq f(2^k)$. Rephrased, this means that there is a $k\geq 3$ such that $f(2^{k-1})f(2)=-1$. Denote $$k_0:=\min\{k\geq 3: f(2^{k-1})f(2)=-1\}.$$ Then $$1+\sum_{k\geqslant 2}\frac{f(2^k)-f(2^{k-1})f(2)}{2^{k{\sigma}}}\geq 1-2\sum_{k\geq 2}\frac{1}{2^{k{\sigma}}}+\frac{2}{2^{k_0{\sigma}}}=1-\frac{2}{2^{\sigma}(2^{\sigma}-1)}+\frac{2}{2^{k_0{\sigma}}}.$$ Note that for ${\sigma}>0$ and $k_0\geq 2$, the function $$1-\frac{2}{2^{\sigma}(2^{\sigma}-1)}+\frac{2}{2^{k_0{\sigma}}}$$ is continuous and increasing. Also at ${\sigma}=1$ we have $$1-\frac{2}{2^1(2^1-1)}+\frac{2}{2^{k_0}}=\frac{2}{2^{k_0}}>0,$$ so that by continuity and the fact that $k_0\geq 3$, there is some minimal ${\alpha}:={\alpha}(k_0)\in(0,1)$ such that for ${\sigma}>{\alpha}$ we have $$1-\frac{2}{2^{\sigma}(2^{\sigma}-1)}+\frac{2}{2^{k_0{\sigma}}}>0.$$

Also, we have that $$3^{2{\sigma}}-3^{\sigma}-2>0$$ for ${\sigma}>\frac{\log 2}{\log3}$ by the quadratic formula. Since $3^{2{\sigma}}-3^{\sigma}-2>0$ precisely when $1-\frac{2}{3^{\sigma}(3^{\sigma}-1)}>0$, combining this with the above, we have that each of the terms of the product $\Pi({\sigma})$ is positive for all $${\sigma}>{\sigma}_0(f):=\max\left\{{\alpha},\frac{\log 2}{\log 3}\right\}.$$ Since this maximum is strictly less that one, the only thing left to show is that the sum $\sum_p\sum_{k\geqslant 2}\frac{f(p^k)-f(p^{k-1})f(p)}{p^{k{\sigma}}}$ is absolutely convergent. We have that $$\begin{aligned}
 \sum_p\left|\sum_{k\geqslant 2}\frac{f(p^k)-f(p^{k-1})f(p)}{p^{k{\sigma}}}\right| 
\leqslant \sum_p\sum_{k\geqslant 2}\frac{2}{p^{k\sigma}}
=2\sum_p\frac{1}{p^\sigma}\cdot\frac{1}{p^\sigma-1},\end{aligned}$$ which is convergent when $\sigma>1/2$, proving the lemma.

It is worth remarking that assuming that $f(2^k)=1$ for some $k\geq 1$ ensures that we are not considering the counterexample $g$ defined in .

We now give the proof of Theorem \[Tmain\] as a corollary to Theorem \[Tcmain\].

Let $f:{\mathbb}{N}\to\{-1,1\}$ be a multiplicative function such that $f(2^k)=1$ for some $k\geq 1$, and denote $F({\sigma})=\sum_{n\geqslant 1}\frac{f(n)}{n^{\sigma}}$. By Lemma \[multf\], we have that $$F({\sigma})=\Pi({\sigma})F_c({\sigma}),$$ where $\Pi({\sigma})$ is defined by as in Lemma \[multf\] and $F_c({\sigma})$ is the generating Dirichlet series for the completely multiplicative function $f_c:{\mathbb}{N}\to\{-1,1\}$ defined by $f_c(p)=f(p)$ for all primes $p$. Similar to , we have that $$\log F({\sigma})=\log\Pi({\sigma})+\log F_c({\sigma})=\sum_p\frac{f(p)}{p^{\sigma}}+O(1),$$ since $\Pi({\sigma})>0$ for ${\sigma}\geq 1$. If $c>0,$ then considering the proof of Theorem \[Tcmain\] for $F_c({\sigma})$ gives that $$\lim_{{\sigma}\to 1^+}F_c({\sigma})=\infty,$$ and so $$\lim_{{\sigma}\to 1^+}F({\sigma})=\infty,$$ which in turn gives that the partial sums $\sum_{n\leqslant x} f(n)$ are unbounded.

If $c<0$ we just consider the proof of Theorem \[Tcmain\] for the function $1/F({\sigma})$, and use the equation $$\log \frac{1}{F({\sigma})}=-\log F({\sigma})=-\log \Pi({\sigma})-\log F_c({\sigma})$$ to yield the result.

Weakening of hypotheses and further extensions {#exten}
==============================================

In Theorem \[Tcmain\] we can replace the condition $$\label{cf}\sum_p\frac{c-f(p)}{p}<\infty$$ with something considerably weaker.

Note that assumption is given so that we may use an asymptotic of the form $$\sum_{p\leqslant x}\frac{f(p)}{p}=c\log\log x+O(1),$$ for nonzero $c\in[-1,1]$. In the case of positive $c$ we can weaken the condition to $$\label{liminf}\lim_{x\to\infty}\sum_{p\leqslant x}\frac{f(p)}{p}=\infty,$$ and in the case of negative $c$ we can weaken the condition to $$\label{limsup}\lim_{x\to\infty} \sum_{p\leqslant x}\frac{f(p)}{p}=-\infty.$$ Then if holds we have that $\sum_{n\leqslant x} f(n)$ is unbounded, and if holds we have that $\sum_{n\leqslant x} \mu(n)f(n)$ is unbounded. As far as “density conditions” the above limits are satisfied when we take $$\sum_{p\leqslant x}f(p) =\frac{c\cdot x}{\log x \log_2 x\cdots \log_k x}(1+o(1)),$$ where $\log_j x$ denotes $\log\log\cdots\log x$ with “$\log$” written $j$ times, $k$ is any nonnegative integer, and $c\neq 0$ is taken to be positive or negative depending on the desired case; this is easily seen via partial summation.

We can do a little in the case that $c=0$. Indeed, all we really need is to have for some ${\sigma}>1/2$ that either $$\label{weakgs}\lim_{x\to\infty} \sum_{p\leqslant x}\frac{f(p)}{p^{\sigma}}=\infty\qquad\mbox{or}\qquad\lim_{x\to\infty} \sum_{p\leqslant x}\frac{f(p)}{p^{\sigma}}=-\infty.$$ In this case, the same method gives the following theorem, though consideration of the limits in directly gives a more exact result.

\[extmain\] Let $f:{\mathbb}{N}\to\{-1,1\}$ be a completely multiplicative function, ${\sigma}>1/2$, $k$ a nonnegative integer, and suppose that $$\sum_{p\leqslant x}f(p)=\frac{c\cdot x^{\sigma}}{\log x \log_2 x\cdots \log_k x}(1+o(1)).$$ If $c>0$ then the partial sums of $f$ are unbounded, and if $c<0$ then the partial sums of $\mu f$ are unbounded.

As discussed above, the proof of Theorem \[extmain\] follows exactly the same as that of Theorem \[Tcmain\], and as such we omit it for fear of sounding redundant. Theorem \[Tmain\] can be generalized similarly, but with the added assumptions that both $f\neq g$ for $g$ as defined in , and that ${\sigma}>{\sigma}_0(f)$ as defined in the proof of Lemma \[multf\].

Concluding remarks
==================

Functions satisfying for positive $c$ are, in some sense, large. In fact, since $F({\sigma})$ are divergent at ${\sigma}=1$, we have, using an obvious abuse of notation, at least that $$\sum_{n\leqslant x}f(n)\gg x^{1-{\varepsilon}}$$ for any ${\varepsilon}>0$. Probably this can be improved, but our original purpose was to just prove the unboundedness of partial sums. Indeed, using the terminology of Goldmakher [@Leo], we should have that the function $f(n)$ mimics the function $c^{\Omega(n)}$ and the partial sums of this function are quite large; we have $$\sum_{n\leqslant x} c^{\Omega(n)}\geqslant \sum_{p\leqslant x}c=c\cdot\pi(x).$$

As an extension of the results for negative $c$, it would be nice if one could show that since $\sum_{n\leqslant x}\mu(n)f(n)$ is unbounded, so is $\sum_{n\leqslant x}f(n)$. We suspect that one may have to consider cases whether or not the Riemann hypothesis holds. Nonetheless, since we have $$\sum_{n\leqslant x}\mu(n)f(n)\gg x^{1-{\varepsilon}}$$ for any ${\varepsilon}>0$ and the partial sums of $\mu$ are not too small, $\sum_{n\leqslant x} \mu(n)\neq O(x^{1/2}),$ something may be able to be done in this case. Indeed, we conjecture that in this case one should have at least that $$\sum_{n\leqslant x}f(n)\gg x^{1/2-{\varepsilon}}$$ for any ${\varepsilon}>0$. Towards something like this we have tried to factor $F(s)$ in an enlightening way (to find a singularity at $s=1/2$, but to no avail. We note that one has for any such series $F(s)$ and $c\in[-1,0)$, that $$F(s)=\left(\frac{{\zeta}(2s)}{{\zeta}(s)}\right)^{|c|}\frac{e^{\frac{P(2s)}{2}}D(s)}{{\zeta}(2s)^{\frac{|c|}{2}}}\cdot\exp\left[-\sum_p\frac{c-f(p)}{p^s}\right],$$ where $P(s)$ is the prime zeta function and the function $D(s)$ is absolutely convergent for $\Re(s)>1/3$.

[10]{}

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[^1]: The research of M. Coons is supported by a Fields-Ontario Fellowship and NSERC.
