Let $(a_n)$ and $(b_n)$ two sequences of real numbers where $b_n \geq 0$.
If $\sum_{n=1}^\infty b_n^{1/\alpha} < \infty$ for some $\alpha > 1$, then $\sum_{n=1}^\infty b_n|x-a_n|^{-\alpha}$ converges for almost every $x \in \mathbb{R}$.
I'm having trouble convincing myself that this is true, and I'm having trouble using the convergence of $\sum b_n^{1/\alpha}$. Also, if we let $(a_n)$ be some enumeration of $\mathbb{Q}$, then every point $x$ would be a limit point, so there would be a subsequence $|x-a_{n_k}|^{-\alpha}$ which would grow without bound, it seems. I've tried some special cases by letting $a_n = 0$ for all $n$, but still could not prove the result.
$$\begin{align} & Consider\text{ }the\text{ }set \\ & E_{n}^{\varepsilon }=\left\{ x:\ {{\left| x-{{a}_{n}} \right|}^{\alpha }}<\varepsilon b_{n}^{1-\frac{1}{\alpha }} \right\},n=1,2,3,\ldots \\ & Then\text{ }the\text{ }measure\text{ }of\text{ }this\text{ }set \\ & m\left( E_{n}^{\varepsilon } \right)=2{{\left( \varepsilon b_{n}^{1-\frac{1}{\alpha }} \right)}^{\frac{1}{\alpha }}} \\ & Now\text{ }For\text{ }all\,x\in \mathbb{R}/\bigcup\limits_{n=1}^{\infty }{E_{n}^{\varepsilon }}~,\text{ }we\text{ }have \\ & {{\left| x-{{a}_{n}} \right|}^{\alpha }}\ge \varepsilon b_{n}^{1-\frac{1}{\alpha }}\Rightarrow \frac{b_{n}^{1-\frac{1}{\alpha }}}{{{\left| x-{{a}_{n}} \right|}^{\alpha }}}\le \frac{1}{\varepsilon }\Rightarrow \frac{{{b}_{n}}}{{{\left| x-{{a}_{n}} \right|}^{\alpha }}}\le \frac{1}{\varepsilon }b_{n}^{\frac{1}{\alpha }} \\ & \Rightarrow \sum\nolimits_{n=1}^{\infty }{\frac{{{b}_{n}}}{{{\left| x-{{a}_{n}} \right|}^{\alpha }}}}\le \frac{1}{\varepsilon }\sum\nolimits_{n=1}^{\infty }{b_{n}^{\frac{1}{\alpha }}} \\ & As\ \varepsilon \to 0,\ m\left( \bigcup\limits_{n=1}^{\infty }{E_{n}^{\varepsilon }} \right)\to 0.Hence\ \sum\nolimits_{n=1}^{\infty }{\frac{{{b}_{n}}}{{{\left| x-{{a}_{n}} \right|}^{\alpha }}}}\ converges\text{ }for\text{ }almost\text{ }all\ x\in \mathbb{R} \\ \end{align}$$