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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 $x=c_n$.2012-03-08
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    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.2012-03-08
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    This is not quite an example of the specific sum Terra M is considering.2012-03-08
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    @Robert I just wanted to provide a concrete example of a similar looking function, but, yes, you're right.2012-03-08

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