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\begin{document}


\medskip \begin{center}{{\Large\sc Proposed to The American Mathematical Monthly
\footnote{E-mail addresses: {\tt
vicentiu.radulescu@math.cnrs.fr}\qquad {\tt
teodoraradulescu@yahoo.com} \\ {\tt
http://www.inf.ucv.ro/\~{}radulescu}}}}\end{center}

\bigskip
{\sl
\begin{enumerate} \item[{(\bf{a})\phantom{$^*$}}]  Find a sequence of distinct complex
numbers $(z_n)_{n\geq 1}$ and a sequence of nonzero real numbers
$(\alpha_n)_{n\geq 1}$ such that the series
$\sum_{n=1}^\infty\alpha_n\,|z-z_n|^{-1}$ either converges to a
positive number or diverges to $+\infty$ for almost all complex
numbers $z$, but not all $\alpha_n$ are positive.

\item[{(\bf{b})\phantom{$^*$}}] Let $(z_n)_{n\geq 1}$ be a
sequence of distinct complex numbers. Assume that
$\sum_{n=1}^\infty \alpha_n$ is an absolutely convergent series of
real numbers such that $\sum_{n=1}^\infty\alpha_n\,|z-z_n|^{-1}$
converges to a nonnegative number, for almost all $z\in\CC$. Prove
that $\alpha_n$ are nonnegative for all $n\geq 1$.

\end{enumerate} }
\begin{flushright}{
Teodora-Liliana R\u adulescu, ``Fra\c tii Buze\c sti" College,
Craiova, Romania\\
Vicen\c{t}iu R\u{a}dulescu, University of Craiova,
Romania}\end{flushright}

\bigskip
{\sc Solution}. {(\bf{a})} We prove that the series
\begin{equation}\label{1}-\frac{1}{|z|}+\frac{1}{\left|z+\frac 12\right|}+
\frac{1}{\left|z-\frac 12\right|}+\cdots +\frac{1}{\left|z+\frac
1n\right|}+\frac{1}{\left|z-\frac 1n\right|}+\cdots\end{equation}
diverges to $+\infty$ for all $z\in\CC\setminus\left\{0,\pm\frac
12,\ldots ,\pm\frac 1n,\ldots\right\}$.

Indeed, we first observe that, for any fixed
$z\in\CC\setminus\left\{0,\pm\frac 12,\ldots ,\pm\frac
1n,\ldots\right\}$, the above series has the same nature as the
series $-1+1+1+1+\ldots$, which diverges. Next, we observe that
$$-\frac{1}{|z|}+\frac{1}{\left|z+\frac 1n\right|}\geq 0\quad\mbox{if and only if}
\quad \mbox{Re}\,z\leq -\frac{1}{2n}$$ and
$$-\frac{1}{|z|}+\frac{1}{\left|z-\frac 1n\right|}\geq 0\quad\mbox{if and only if}
\quad \mbox{Re}\,z\geq \frac{1}{2n}\,.$$ The above relations show
that for any $z\in\CC\setminus\left\{0,\pm\frac 12,\ldots ,
\pm\frac 1n,\ldots\right\}$ with $\mbox{Re}\,z\not=0$ there exists
$N\in\NN$ such that
$-\frac{1}{|z|}+\sum_{k=1}^N\left(\frac{1}{\left|z+\frac
1k\right|}+ \frac{1}{\left|z-\frac 1k\right|}\right)>0$. It
remains to prove that this is also true if $z=iy$,
$y\in\RR\setminus\{0\}$. For this purpose we observe that
$$\begin{array}{ll}
\di &\di \frac{1}{\left|z+\frac 12\right|}+\frac{1}{\left|z-\frac
12\right|}+\cdots
+\frac{1}{\left|z+\frac 1n\right|}+\frac{1}{\left|z-\frac 1n\right|}=\\
&\di \frac{2}{\sqrt{y^2+\frac 14}}+\cdots
+\frac{2}{\sqrt{y^2+\frac 1{n^2}}}\geq \frac{2n-2}{\sqrt{y^2+\frac
14}}\geq\frac{1}{|y|}=\frac{1}{|z|}\,,
\end{array}$$
provided $2|y|\geq (4n^2-8n+3)^{-1}$. In conclusion, the series
\eq{1} diverges to $+\infty$ for all
$z\in\CC\setminus\left\{0,\pm\frac 12,\ldots ,\pm\frac
1n,\ldots\right\}$.

Another example of series with the above properties is
$$-\frac{1}{|z|}+\sum_{n=2}^\infty\frac 1n\left(\frac{1}{|z+\ln n|}+
\frac{1}{|z-\ln n|}\right)\,,\qquad z\in\CC\setminus\{0,\pm\ln
2;\pm\ln 3, \ldots\}\,.$$

\smallskip
{(\bf{b})} It is sufficient to focus on an arbitrary term of the
sequence, say $\alpha_1$, and to show that $\alpha_1\geq 0$. We
can assume, without loss of generality, that $z_1=0$. Fix
arbitrarily $\ep\in (0,1)$. Since $\sum_{n=1}^\infty
|\alpha_n|<\infty$, there exists a positive integer $N$ such that
$\sum_{i=N+1}^\infty |\alpha_i|<\ep$. Next, we choose $r>0$ small
enough so that $|a_i|>r/\ep$, for all $i\in\{2,\ldots ,N\}$. Set
$$f(z)=\sum_{n=1}^\infty\frac{\alpha_n}{|z-z_n|}\,.$$
It follows that
\begin{equation}\label{2}
\begin{array}{ll}
0 &\di\leq\intbr f(z)dz=\alpha_1\intbr\frac{dz}{|z|}+\sum_{i=2}^N\alpha_i\intbr\frac{dz}{|z-z_i|}+
\sum_{i=N+1}^\infty\alpha_i\intbr\frac{dz}{|z-z_i|}\\
&\di\leq\alpha_1\intbr\frac{dz}{|z|}+\sum_{i=2}^N|\alpha_i|\intbr\frac{dz}{|z-z_i|}+
\sum_{i=N+1}^\infty |\alpha_i|\intbr\frac{dz}{|z-z_i|}\\
&\di =2\pi r\alpha_1+\sum_{i=2}^N|\alpha_i|\intbr\frac{dz}{|z-z_i|}+
\sum_{i=N+1}^\infty |\alpha_i|\intbr\frac{dz}{|z-z_i|}\\
&\di \leq 2\pi r\alpha_1+\sum_{i=2}^N|\alpha_i|\intbr\frac{dz}{|z-z_i|}+
\ep\sup_{i\geq N+1}\intbr\frac{dz}{|z-z_i|}\,.
\end{array}
\end{equation}
For every $i\in\{2,\ldots N\}$ we have $|z-z_i|\geq |z_i|-|z|\geq \frac r\ep -r=r(1-\ep)/\ep$, so
\begin{equation}\label{3}
\intbr\frac{dz}{|z-z_i|}\leq\frac{\ep}{r(1-\ep)}\intbr dz=\frac{\ep\pi r}{1-\ep}\,.\end{equation}
If $i\geq N+1$ we distinguish two cases: either $|z_i|\geq 2r$ or $|z_i|<2r$. In the first situation
we deduce that $|z-z_i|\geq r$, for any $z\in B_r(0)$. Thus
$$\intbr\frac{dz}{|z-z_i|}\leq \frac 1r \intbr dz=\pi r\,.$$
If  $|z_i|<2r$ then
$$\intbr\frac{dz}{|z-z_i|}\leq \int_{B_{4r}(z_i)}\frac{dz}{|z-z_i|}=8\pi r\,.$$
The above two relations show that
\begin{equation}\label{4}
\sup_{i\geq N+1}\intbr\frac{dz}{|z-z_i|}\leq 8\pi r\,.\end{equation}
Using \eq{2}, \eq{3} and \eq{4} we obtain
$$0\leq 2\pi r\alpha_1+\sum_{i=2}^N |\alpha_i|\cdot\frac{\ep\pi r}{1-\ep}+8\ep\pi r\,.$$
Dividing by $r$ and letting $\ep\ri 0$ we deduce that $\alpha_1\geq 0$. \qed

\medskip
{\bf Remark}. For part {(\bf{a})} of this proposal, we have {\bf
not} been able to find an example of series
$\sum_{n=1}^\infty\alpha_n\,|z-z_n|^{-1}$ which converges to a
positive number for almost all complex numbers $z$, but not all
$\alpha_n$ being positive. It might be possible that such a series
does not exist and the unique situation which can occur is that,
under our assumptions described in {(\bf{a})}, the series
$\sum_{n=1}^\infty\alpha_n\,|z-z_n|^{-1}$ always diverges to
$+\infty$. We let at your choice to decide if this assertion could
be included as an open problem in this proposal.

\end{document}
