Okay, so I'm being asked to prove that there exists \alpha and n such that f(x)= \sum_{k=0}^{ \infty } \| x-r_k\| ^{\alpha } is (Lebesgue) integrable in \Omega:=B_{\frac{1}{2} } (0) \subset \mathbb{R} ^n where (r_k)_{k\in \mathbb{N} } is dense in \Omega . My problem is I'm pretty sure we need some coefficients to control the series, here's my reasoning:

Let x\in \Omega and s>0 such that \Omega _1 := \overline{B_s(x)} \subset \Omega. Then since (r_k) is dense there are an infinite number of elements of the sequence in \Omega _1 and in \Omega \setminus \Omega _1. Now if \alpha=0 the series obviously diverges for all x so assume first that \alpha >0 then there is (r_{k_l}) such that \| x-r_{k_l}\| ^{\alpha } >s^{\alpha } (taking the subsequence to be in \Omega \setminus \Omega _1) and so f(x)\geq \sum_{k_l} \| x-r_{k_l}\| ^{\alpha } \geq \sum_{k_l} s^{\alpha } \rightarrow \infty so the series diverges at x. Using an analogous argument for \alpha <0 (taking the subsequence to be in \Omega _1 and noting that the exponent is negative and thus reverses inequalities) we see that the series diverges. Since x was arbitrary we get that f \equiv \infty which is nonsense, since an integrable function attains \infty at most in a null set.

Does this look right? I'm not too sure myself since I'm not overly comfortable with my series knowledge, but it feels right (since we don't really know if \| x-r_k\| \rightarrow 0 for all k which actually might never happen).

Any thoughts and comments are appreciated.