0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ @Shashi my bad. Edited $\endgroup$ – openspace Apr 5 '18 at 19:12
  • $\begingroup$ nice! {} {} {} {} {} $\endgroup$ – Shashi Apr 5 '18 at 19:22
  • $\begingroup$ I also add some edits I hope you find them fine. $\endgroup$ – Shashi Apr 5 '18 at 19:26
  • $\begingroup$ In order to apply the dominated convergence theorem, we need convergence of the sequence. That is one of the requirements of the theorem! $\endgroup$ – p4sch Apr 5 '18 at 20:14
3
$\begingroup$

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 $x \in [0,1]$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.