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Prove that for arbitrary sequence $x_{1}, x_{2},...$ of $[0,1]$ there is a $x$ in $[0,1]$ such that below series is convergent: $$\sum_{n=1}^\infty \frac{1}{n^2|x-x_{n}|}.$$

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  • $\begingroup$ what if $x_n$ is an enumeration of the rationals; can there really be such an $x$? $\endgroup$ – user335907 Feb 24 '17 at 0:48
  • $\begingroup$ @james.nixon If the sequence isn't dense, the question will be trivial. $\endgroup$ – Soheil Memarian Feb 24 '17 at 2:33
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We will show that the series converges almost everywhere on $[0,1]$.

Let $a_n(x)=\frac{1}{n^2|x-x_n|}$ be the general term of the series.

Fix $\alpha \in (1,2)$ (say $\alpha=\frac 32$). We first consider the series $\sum_{n=1}^\infty a_n (x)^{1/\alpha}$:

$$ \sum_{n=1}^\infty \frac{1}{n^{2/\alpha}|x-x_n|^{1/\alpha}}.$$

Note that $2/\alpha > 1$ and that $1/\alpha <1$.

We need a simple estimate:

$$ \int_0^1 \frac{dx}{|x-x_n|^{1/\alpha}}\le 2 \int_0^1 \frac{dy}{y^{1/\alpha}}=\frac{2\alpha}{\alpha-1} y^{1-1/\alpha}|_0^1=\frac{2\alpha}{\alpha-1}.$$

Now integrate the series:

$$0\le\int_0^1 \sum_{n=1}^\infty a_n^{1/\alpha}(x) dx =\sum_{n=1}^\infty \int_0^1 a_n(x) dx \le \frac{2\alpha}{\alpha-1} \sum_{n=1}^\infty \frac{1}{n^{2/\alpha}}=c_\alpha < \infty.$$

The first equality is due to monotone convergence.

Being Lebesgue integrable, it follows that $\sum_{n=1}^\infty a_n(x)^{1/\alpha}$ converges almost everywhere on $[0,1]$. Let $E$ denote the set of points where it converges.

Fix $x\in E$. Then convergence implies $\lim_{n\to\infty} a_n(x)^{1/\alpha}= 0$. For all $n$ large enough, $a_n(x)^{1/\alpha}<1$, and since $\alpha >1$, we have $a_n(x) = (a_n(x)^{1/\alpha})^{\alpha}\le a_n(x)^{1/\alpha}$. Therefore by comparison, the series $\sum_{n=1}^\infty a_n (x)$ also converges.

Final note. The same technique can be used to show that $\sum_{n=1}^\infty \frac{1}{n^\beta|x-x_n|^\gamma}$ converges a.e. whenever $\beta >1$ and $\gamma<\beta$.

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  • $\begingroup$ Great solution, and very detailed account. Nice job! $\endgroup$ – Fimpellizieri Feb 24 '17 at 4:09
  • $\begingroup$ thank you. great question. $\endgroup$ – Fnacool Feb 24 '17 at 4:10

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