# Convergence almost everywhere of sum.

Consider $\{r_{n}\}_{n \ge 1}$ set of all rational numbers of $[0,1]$.

Now lets define $$\displaystyle S_{k}(x) := \sum_{n=1}^{k} \frac{1}{n^{2} \sqrt{|x-r_{n}|}}$$

My question is : does $S_k(x)$ converge almost everywhere on $[0,1]$\ $\mathbb{Q}$?

I thought about considering $\displaystyle \lim_{k \to \infty} \int_{[0,1]} S_{k}(x)\, dx$ and use Lebesgue theorem about limit under integration. Then if exact result is converges , I could say if my series converges a.e. or not. Am I right?

However , can we say something about this series without Lebesgue integral?

• @Shashi my bad. Edited – openspace Apr 5 '18 at 19:12
• nice! {} {} {} {} {} – Shashi Apr 5 '18 at 19:22
• I also add some edits I hope you find them fine. – Shashi Apr 5 '18 at 19:26
• In order to apply the dominated convergence theorem, we need convergence of the sequence. That is one of the requirements of the theorem! – p4sch Apr 5 '18 at 20:14

## 1 Answer

Since the $S_k(x)$ is pointwise monotone-inscreasing, we may use the montone convergence theorem to conclude that $$\int_0^1 \lim_{n \rightarrow \infty} S_n(x) \mathop{dx} = \lim_{n \rightarrow \infty} \sum_{k=1}^n \frac{1}{k^2} \int_0^1 \frac{1}{\sqrt{|x-r_n|}} \mathop{dx} \leq 4 \sum_{k=1}^\infty \frac{1}{k^2} <\infty.$$ Thus, $S(x) := \lim_{n \rightarrow \infty} S_n(x) < \infty$ for $\lambda$-almost all $x \in [0,1]$.

• I don't get the last estimate. Would you please provide some explanations? – Shashi Apr 5 '18 at 20:51
• We have $\int_0^1 |x-r_n|^{-1/2} \mathop{dx} = \left. -2 \sqrt{r_n -x} \right|_{x=0}^{r-n} + \left. 2 \sqrt{x -r_n} \right|_{x=r_n}^{1} = 2 (\sqrt{1-r_n} + \sqrt{r_n}) \leq 4$. – p4sch Apr 5 '18 at 21:06
• thanks I see it! – Shashi Apr 5 '18 at 21:07