Let $ \{x_n\}_{n=1}^\infty \subset \mathbb{R} $ be a sequence. Prove for $\alpha>1$ that $\sum_{n=1}^\infty \frac{1}{n^\alpha\sqrt{n|x-x_n|}}$ converges for almost every $x$ with regard to Lebesgue measure on $\mathbb{R}$.
I tried solving by finding an integrable function such that this series is bounded below the function's integral, but I didn't find a suitable function.
On
EDIT: See @Michael's comment for good reasoning that shows the conjecture is true. Additionally, see here for another look at a more general problem in this vein. I'll leave the answer below up as I believe it is interesting in its own right, but it should not be looked at as anything other than an explicit example of convergence when $x_n$ is dense in the reals.
This is not a full answer so much as a proof that the question is more difficult than at first glance. Obviously, if $\{x_n\}_{n=1}^\infty$ is made up of isolated points then the conjecture is true. Indeed, even if the sequence also includes some limit points (for example $\{0\}\cup \{1/n:n\in\mathbb{N}\}$) then it is still not very much work to show that the conjecture holds. However, it is less obvious whether the conjecture will hold for a sequence that is dense in the reals. It might seem like no, but in fact there are sequences which are dense in the reals for which the conjecture does hold. We present one such sequence here.
First, we have to define some constants. Let $\alpha>1$ be given and define
$$\epsilon=\alpha-1>0$$
Second, let $\{x_n\}_{n=1}^\infty$ be an enumeration of the rational numbers in the following manner: First, set $x_1=0$. Next, set $x_n=\frac{a_n}{b_n}$ where $b_n\in\mathbb{N}$ and $a_n\in\mathbb{Z}$. Additionally,
$$|a_i|+b_i>|a_j|+b_j\Rightarrow i>j$$
$$x_n>0\Rightarrow x_{n+1}=-x_n$$
Finally, the fractions shall be in their most reduced form with no repetitions. Note that there are many sequences that satisfy this definition, but the first few terms of all of them are
$$0,\frac{1}{1},\frac{-1}{1},\frac{1}{2},\frac{-1}{2},\frac{1}{3},\frac{-2}{3},\cdots$$
Having defined our sequence $x_n$ (or at least defined a set of sequences, each of which will work with our overall logic), let $S$ be the set of number with irrationality measure greater than $2$. By previous results, $\overline{S}$ encompasses almost all real numbers (here $\overline{S}=\{x\in\mathbb{R}:x\not\in S\}$). Thus, if we can show the series converges for $x\in S$ then we are done. As such, let $x$ be an arbitrary real in $S$. By definition
$$|x-x_n|=\left|x-\frac{a_n}{b_n}\right|>\frac{1}{b_n^{2+2\epsilon}}$$
for all sufficiently large $b$ (after some $b^{'}$).
$$\sum_{n=1}^\infty \frac{1}{n^{\alpha+1/2}\sqrt{|x-x_n|}}<K+\sum_{n=N_0}^\infty \frac{b_n^{1+\epsilon}}{n^{\alpha+1/2}}$$
where $N_0$ is defined such that $b_{N_0}=b^{'}$ and $K$ is the sum from $1$ to $N_0-1$. Now, let us put a bound on $b_n$. Define
$$\chi_n=\left\lfloor\frac{b_n}{2}\right\rfloor$$
and note that
$$\frac{\pm1}{1}$$
$$\frac{\pm1}{2}$$
$$\frac{\pm1}{3},\frac{\pm2}{3}$$
$$\vdots$$
$$\frac{\pm1}{\chi_n},\cdots,\frac{\pm(\chi_n-1)}{\chi_n}$$
all appear in the sequence before $\frac{a_n}{b_n}$ since
$$\chi_n-1+\chi_n<2\chi_n\leq b_n$$
In total, there are at least
$$2\sum_{i=1}^{\chi_n}\phi(i)<n$$
rationals in the sequence before $b_n$ (where $\phi(n)$ is the Euler Totient Function). However, we know from here that
$$n>2\sum_{i=1}^{\chi_n}\phi(i)\approx \frac{6\chi_n^2}{\pi^2}\approx \frac{3b_n^2}{2\pi^2}=\frac{1}{9}\cdot \frac{27b_n^2}{2\pi^2}$$
Since $\frac{27}{2\pi^2}>1$, we know that for sufficiently large $b_n$ (past some $n\geq N_1$)
$$n>\frac{1}{9}b_n^2$$
Thus, a bound for $b_n$ is
$$b_n<3\sqrt{n}$$
Putting it all together and setting $N=\max\{N_0,N_1\}$ gives us
$$\sum_{n=1}^\infty \frac{1}{n^{\alpha+1/2}\sqrt{|x-x_n|}}<K+3\sum_{n=N}^\infty \frac{n^{\frac{1+\epsilon}{2}}}{n^{\alpha+1/2}}$$
where $K$ is the sum from $1$ to $N-1$. This is
$$=K+3\sum_{n=N}^\infty \frac{1}{n^{\alpha+1/2-1/2-\epsilon/2}}$$
But we defined $\epsilon$ such that
$$\epsilon=\alpha-1$$
Thus
$$=K+3\sum_{n=N}^\infty \frac{1}{n^{\alpha-\frac{\alpha-1}{2}}}=K+3\sum_{n=N}^\infty \frac{1}{n^{\frac{\alpha+1}{2}}}$$
Since $\frac{\alpha+1}{2}>1$, the sum converges and we are done.
Overall, I would believe that the conjecture holds for any sequence of real numbers. The difficulty lies in the fact that if you change the definition of $x_n$ from how I defined it then you open up the possibility that the sum will diverge for certain reals with irrationality measure $2$. However, I think it would be a countable subset of $\overline{S}$ and the conjecture would still hold, but I do not know how to even begin to prove this.
Partial answer: suppose $x_n$ is a sequence such that the closure of $\{x_n\}_{n \in \mathbb{N}}$ is a null set.
Take $x \in \mathbb{R}$ such that $\delta := \inf_{n \in \mathbb{N}}\{|x-x_n|\} > 0$. Then we have.
\begin{align*} \frac{1}{n^{\alpha} \sqrt{n |x-x_n|}} \leq \frac{1}{n^{\alpha + 1/2} \sqrt{\delta}} \end{align*} and hence \begin{align*} \sum_{n=1}^{\infty} \frac{1}{n^{\alpha} \sqrt{n |x-x_n|}} \leq & \sum_{n=1}^{\infty} \frac{1}{n^{\alpha + 1/2} \sqrt{\delta}} \\ =& \frac{1}{\sqrt{\delta}} \sum_{n=1}^{\infty} \frac{1}{n^{\alpha + 1/2}} \\ \leq& \frac{1}{\sqrt{\delta}} \left( \int_1^{\infty} x^{-\alpha-1/2}dx + 1 \right) \end{align*} Which converges because $\alpha > 1$.
Since this holds for any such $x$, we may conclude that the sum converges everywhere except on $\{x_n\}_{n \in \mathbb{N}}$.