Convergent of a series for arbitrary sequence 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:
$$\sum_{n=1}^\infty \frac{1}{n^2|x-x_{n}|}.$$
 A: 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 $\gamma<\beta$. 
