# 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 $\{c_{n}\}$?

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You function will be discontinuous and unbounded in every $x=c_n$. – Pedro Tamaroff Mar 8 '12 at 2:39
As an example:$$\frac{1}{z} + \sum\limits_{k = 1}^\infty {{{\left( { - 1} \right)}^k}\left( {\frac{1}{{z + k}} + \frac{1}{{z - k}}} \right)} = \frac{\pi }{{\sin \pi z}}$$ But at every $z \in \mathbb{N}$ we're in trouble. – Pedro Tamaroff Mar 8 '12 at 3:19
This is not quite an example of the specific sum Terra M is considering. – Robert Israel Mar 8 '12 at 3:47
@Robert I just wanted to provide a concrete example of a similar looking function, but, yes, you're right. – Pedro Tamaroff Mar 8 '12 at 4:15

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.

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That sum is not of the form given in the original question. – Gerry Myerson Mar 8 '12 at 5:13
@GerryMyerson, just replace $x$ with $\sqrt{x}$ and consider $x > 0$. – Antonio Vargas Mar 8 '12 at 5:18
On second thought, you don't even need to restrict yourself to positive $x$ after making the substitution $x \mapsto \sqrt{x}$; after making a suitable branch cut, the function is real-valued for all real $x \neq 0,1,2,\ldots$ – Antonio Vargas Mar 8 '12 at 5:33

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 $c_n$.

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