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 $c_n$.