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

Let $\displaystyle x\in \Omega$ and $\displaystyle s>0$ such that $\displaystyle \Omega _1 := \overline{B_s(x)} \subset \Omega$. Then since $\displaystyle (r_k)$ is dense there are an infinite number of elements of the sequence in $\displaystyle \Omega _1$ and in $\displaystyle \Omega \setminus \Omega _1$. Now if $\displaystyle \alpha=0$ the series obviously diverges for all $\displaystyle x$ so assume first that $\displaystyle \alpha >0$ then there is $\displaystyle (r_{k_l})$ such that $\displaystyle \| x-r_{k_l}\| ^{\alpha } >s^{\alpha }$ (taking the subsequence to be in $\displaystyle \Omega \setminus \Omega _1$) and so $\displaystyle f(x)\geq \sum_{k_l} \| x-r_{k_l}\| ^{\alpha } \geq \sum_{k_l} s^{\alpha } \rightarrow \infty$ so the series diverges at $\displaystyle x$. Using an analogous argument for $\displaystyle \alpha <0$ (taking the subsequence to be in $\displaystyle \Omega _1$ and noting that the exponent is negative and thus reverses inequalities) we see that the series diverges. Since $\displaystyle x$ was arbitrary we get that $\displaystyle f \equiv \infty$ which is nonsense, since an integrable function attains $\displaystyle \infty$ 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 $\displaystyle \| x-r_k\| \rightarrow 0$ for all $\displaystyle k$ which actually might never happen).

Any thoughts and comments are appreciated.