# 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 almost everywhere.

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.

• Your second observation about the limit points provides a clue to why you need to use the condition on $b_n$: Yes there is a subsequence that grows, but if you have to go too far in $n_k$ the multiplying $b_n$ must shrink more rapidly. Of course you can find some subset of points where the whole sequence is "early" but the measure of the set of such points is not going to be non-zero. Commented Apr 13, 2018 at 0:40
• You mentioned the case $a_n=0$ for all $n.$ That one's easy: $$\sum_{n=1}^{\infty}b_n|x|^{-\alpha} = |x|^{-\alpha}\sum_{n=1}^{\infty}b_n.$$ Since $b_n\le b_n^{1/\alpha}$ for large $n$ (because $b_n \to 0),$ we see the above sum is finite for all $x\ne 0.$
– zhw.
Commented Apr 13, 2018 at 0:50