One point to add, $\gamma>2$. My approach is to show its complement, i.e $x$ where the series is divergent, is measure zero by Borel-Cantelli lemma. Let $A_{j,m}:=\{\sum_{n=1}^m\sum_{k=1}^n\dfrac{1}{n^\gamma}\dfrac{1}{\sqrt{|x-\frac{k}{2^n}|}}\geq j\}$. The complement now becomes, $$\cap_{j\geq1}\cap_{m\geq N_j}A_{j,m}.$$ It is not the form where Borel-Cantelli is applicable.

Thank you!


$\int_0^1 \sum_{n=1}^\infty\sum_{k=1}^n \frac{1}{n^\gamma}\frac{1}{\sqrt{|x-\frac{k}{2^n}|}}dx = \sum_{n=1}^\infty\frac{1}{n^\gamma}\sum_{k=1}^n \int_0^1 \frac{1}{\sqrt{|x-\frac{k}{2^n}|}}dx = \sum_{n=1}^\infty \frac{1}{n^\gamma} \sum_{k=1}^n 2\frac{k}{2^n}[\sqrt{\frac{2^{2n}}{k^2}-\frac{2^n}{k}}+\sqrt{\frac{2^n}{k}}] \le \sum_{n=1}^\infty \frac{1}{n^\gamma}\sum_{k=1}^n 2\frac{k}{2^n}2\frac{2^n}{k} = \sum_{n=1}^\infty \frac{4}{n^{\gamma-1}} < \infty.$

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  • $\begingroup$ Thank you! Never thought this monster-like series is indeed integrable. $\endgroup$ – Leonardo Sep 21 '18 at 7:51

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