## Convergence and integration

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.