Is $f(x)=\sum \limits_{n=1}^{\infty}\frac{1}{x-c_{n}}$ bounded? I have this function which is defined by infinite series
$$f(x)=\sum_{n=1}^{\infty}\frac{1}{x-c_{n}}$$
where $\{c_{n}\}$ is a sequence of nonzero real numbers such that $\sum \frac{1}{c_{n}}<\infty$.
My question is: Is $f$ bounded on $\mathbb R$? i.e. $|f(x)|<\infty$ for all $x\in \mathbb R$?
If not, can we make it bounded by assuming another condition on the sequence $\{c_{n}\}$?
 A: As Peter T.off observed, the limit function will not be bounded.
Indeed, one such example is
$$\sum_{n=1}^{\infty} \frac{1}{x^2-n^2} = \frac{\pi x \cot(\pi x) - 1}{2x^2},$$
where the sum converges uniformly on compact subsets of $\mathbb{C}\setminus\mathbb{Z}$.  The limit function has a pole at every integer.
A: By the way, convergence of $\sum_{n=1}^\infty \frac{1}{c_n}$ doesn't imply that $\sum_{n=1}^\infty \frac{1}{x-c_n}$ converges for $x \notin \{c_n: n=1\ldots\infty\}$.  For example,
take $c_{2n} = d_n$ and $c_{2n+1}=-d_n$ where $d_n$ is any sequence of positive numbers with $d_n \to \infty$.  Then $\sum_{n=1}^\infty \frac{1}{c_n} = 0$.  But since $\frac{1}{x-d_n} + \frac{1}{x+d_n} = \frac{2x}{x^2 - d_n^2}$, $\sum_{n=1}^\infty \frac{1}{x-c_n}$ diverges whenever $x \ne 0$ and $\sum_{n=1}^\infty 1/d_n^2$ diverges. 
On the other hand, if $\sum_{n=1}^\infty \frac{1}{c_n}$ converges absolutely, 
$\sum_{n=1}^\infty \frac{1}{x-c_n}$ will also converge absolutely as long as $x \notin \{c_n: n = 1 \ldots \infty\}$, because $\frac{1}{|x - c_n|} \le \frac{2}{|c_n|}$ when $n$ is large enough that $|c_n| > 2 |x|$.  The sum will then be a meromorphic function of $x$ with a pole at each $c_n$.
