Rate of convergence of $\frac{1}{\sqrt{n\ln n}}(\sum_{k=1}^n 1/\sqrt{X_k}-2n)$, $X_i$ i.i.d. uniform on $[0,1]$? Let $(X_n)$ be a sequence of i.i.d. random variables uniformly distributed in $[0,1]$; and, for $n\geq 1$, set
$$
S_n = \sum_{k=1}^n \frac{1}{\sqrt{X_k}}\,.
$$
It follows from the generalized central limit theorem (as in [1] and [2, Theorem 3.1]) that
$$
\frac{S_n-2n}{\sqrt{n\ln n}}
$$
converges in law to a Gaussian distribution. However, in this case Berry—Esseen fails to give any convergence rate, as the summands are not square integrable.
Moreover, Hall's results for this sort of sum do not apply. (In [3], this would correspond to $\alpha=2$, while the results hold for $0<\alpha<2$.)

What is the rate of convergence of $\frac{S_n-2n}{\sqrt{n\ln n}}$ to Gaussian?


[1] Shapiro, Jesse M., “Domains of Attraction for Reciprocals of Powers of Random Variables.” SIAM Journal on Applied Mathematics, vol. 29, no. 4, 1975, pp. 734–739. JSTOR, www.jstor.org/stable/2100234.
[2] Christopher S. Withers and Saralees Nadarajah, Stable Laws for Sums of Reciprocals. December 2011.
[3] Hall, P. (1981), On the Rate of Convergence to a Stable Law. Journal of the London Mathematical Society, s2-23: 179-192. doi:10.1112/jlms/s2-23.1.179
 A: Update: what is below is seemingly hinting at a $\Theta(n^{-c})$ rate of convergence for some absolute constant $c>0$. However, this appears to be an artifact of the simulation (sample size too small, and $n$ not large enough). See Iosif Pinelis' answer, which establishes a $\tilde{\Theta}(1/\log n)$ rate of convergence in Kolmogorov distance; and the comments below

A small experimental observation: building on the Mathematica code provided by Przemo, I computed the empirical cumulative distribution of $S_n$ (based on $m=500000$ independent samples) for $n$ ranging from $5$ to $399$. 
list = {}
For[ k = 5, k < 400, k++,
    n = k; m = 500000;
    X = RandomReal[{0, 1}, {m, n}];
    ll = (Total[1/Sqrt[#]] & /@ X - 2 n)/Sqrt[n Log[n]];
    emp = EmpiricalDistribution[ll];
    err :=  Max[ Table[ Abs[CDF[emp, x] - CDF[NormalDistribution[0, 1], x]], {x, -4, 6, 0.05}]];
    AppendTo[ list, err ]
]

Computing the log-log plot of the result:
listpairs = {}
For[ k = 1, k < 395, k++, AppendTo[ listpairs, {4 + k, list[[k]]} ] ]
ListLogLogPlot[listpairs]

it does look like the convergence rate is of the form $1/n^{\epsilon}$ for $\epsilon \simeq 0.14$.

Of course, there are at least two sources of error in the code above (the sampling error, which translates to a supremum norm error in the empirical CDF of order $1/\sqrt{m}$; and the computation error in the Max due to the gridding by 0.05. But it seems unlikely this could somehow change the trend from logarithmic to inverse polynomial.)

Update: Here are the results of a more thorough experiment, taking $m=10^7$ and $n$ from $20$ to $2000$ by steps of $20$ (so 100 different values). Further, the distance between Gaussian and empirical CDFs are now computed as the max over the interval $[-5,5]$, discretized by a step of $0.0001$ (not $[-4,4]$ by $0.05$ as before). Both the regular and log-log plots are below:


I may be misinterpreting it, but this still seems to hint at an inverse polynomial rate, in spite of the theoretical guarantee (?).
A: We shall give an asymptotic formula with error term for its character function $$Z_n:=\frac{1}{\sqrt{n\log n}}\sum_{k=1}^n\left(\frac{1}{\sqrt{X_k}}-2\right)$$
as $n$ tends to infinity. 
Let $n$ be sufficiently large. Note that $X_k$ is $i. i. d.$ we can compute its character function that
\begin{align}
\varphi_{Z_n}(z)&=\left(\int_{0}^1\exp\left({\rm i}z\frac{1/\sqrt{x}-2}{\sqrt{n\log n}}\right)\,d x\right)^n\\
&=\left(\int_{0}^1\,d x\left(\cos\left(z\frac{1/\sqrt{x}-2}{\sqrt{n\log n}}\right)+{\rm i}\sin\left(z\frac{1/\sqrt{x}-2}{\sqrt{n\log n}}\right)\right)\right)^n\\
&:=\left(C_n+{\rm i}S_n\right)^n.
\end{align}
Let $\delta>0$ be fixed. We estimate that
\begin{align}
|S_n|&=\left|\left(\int_0^{\frac{\delta}{n\log n}}+\int_{\frac{\delta}{n\log n}}^1\right)\sin\left(z\frac{1/\sqrt{x}-2}{\sqrt{n\log n}}\right)\,dx\right|\\
&\le \frac{\delta}{n\log n}+|z|\left|\int_{\frac{\delta}{n\log n}}^1\frac{1/\sqrt{x}-2}{\sqrt{n\log n}}\,dx\right|+O\left(|z|^3\int_{\frac{\delta}{n\log n}}^1\left|\frac{1/\sqrt{x}-2}{\sqrt{n\log n}}\right|^3\,dx\right)\\
&=O\left(\frac{\delta}{n\log n}+\frac{\sqrt{\delta}|z|}{n\log n}+\frac{\delta^{-1/2}|z|^3}{n\log n}\right)
\end{align}
uniformly for $z\in\mathbb{R}$ with $|z|\le \sqrt{\delta}$. Similarly,
\begin{align}
C_n&=\left(\int_0^{\frac{\delta}{n\log n}}+\int_{\frac{\delta}{n\log n}}^1\right)\cos\left(z\frac{1/\sqrt{x}-2}{\sqrt{n\log n}}\right)\,dx\\
&=1+O\left( \frac{\delta}{n\log n}\right)-\frac{z^2}{2}\int_{\frac{\delta}{n\log n}}^1\left(\frac{1/\sqrt{x}-2}{\sqrt{n\log n}}\right)^2\,dx+O\left(|z|^3\int_{\frac{\delta}{n\log n}}^1\left|\frac{1/\sqrt{x}-2}{\sqrt{n\log n}}\right|^3\,dx\right)\\
&=1+O\left(\frac{\delta}{n\log n}+\frac{\delta^{-1/2}|z|^3}{n\log n}\right)-\frac{z^2}{2}\frac{1}{n\log n}\left(\log((n\log n)/\delta)+O(1)\right)\\
&=1-\frac{z^2}{2n}-\frac{z^2\log \log n}{2n\log n}+O\left(\frac{\delta+\delta^{-1/2}|z|^3+|z|^2(1+|\log \delta|)}{n\log n}\right).
\end{align}
Therefore,
\begin{align}
\varphi_{Z_n}(z)&=\left(1-\frac{z^2}{2n}-\frac{z^2\log \log n}{2n\log n}+O\left(\frac{\delta+\sqrt{\delta}|z|+\delta^{-1/2}|z|^3+|z|^2(1+|\log \delta|)}{n\log n}\right)\right)^n\\
&=\exp\left(n\log\left(1-\frac{z^2}{2n}-\frac{z^2\log \log n}{2n\log n}+O_{\delta}\left(\frac{1}{n\log n}\right)\right)\right)\\
&=\exp\left(-\frac{z^2}{2}-\frac{z^2}{2}\frac{\log\log n}{\log n}+O_{\delta}\left(\frac{1}{\log n}\right)\right).
\end{align}
Namely,
\begin{align}
\varphi_{Z_n}(z)=\exp\left(-\frac{z^2}{2}-\frac{z^2}{2}\frac{\log\log n}{\log n}\right)\left(1+O_{\delta}\left(\frac{1}{\log n}\right)\right)
\end{align}
holds for all $z\in\mathbb{R},|z|\le \sqrt{\delta}$ as $n\rightarrow \infty$. 
A: $\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\ep}{\epsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\thh}{\theta}
\newcommand{\R}{\mathbb{R}}
\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\PP}{\operatorname{\mathsf P}}$ 
Let $V_k:=1/\sqrt{X_k}$, $b_n:=\sqrt{n\ln n}$, 
\begin{equation*}
 Z_n:=\frac{S_n-2n}{\sqrt{n\ln n}}=\frac1{b_n}\sum_1^n (V_k-\E V_1), 
\end{equation*}
\begin{equation*}
 \de_n:=\sup_{x\in\R}|\De_n(x)|, 
\end{equation*}
where 
\begin{equation*}
 \De_n:=F_n-G,\quad 
F_n(x):= P(Z_n<x),\quad G(x):=P(Z<x), 
\end{equation*}
and $Z\sim N(0,1)$. 
We shall show that for all large enough $n$ 
\begin{equation*}
 \frac{\sqrt{\ln\ln n}}{\ln n}\ll\de_n\ll\ep_n:=\frac{\ln\ln n}{\ln n}; \tag{1}
\end{equation*}
here everywhere, the constants associated with $\ll$, $\gg$, $O(\cdot)$ are universal. Thus, we have a rather tight bracketing of $\de_n$. (Conjecture: $\de_n\asymp\frac{\ln\ln n}{\ln n}$.)
Let $c$ denote various complex-valued expressions (possibly different even within the same formula) such that $|c|\ll1$. 
The pdf of $V_k$ is $x\mapsto\frac2{x^3}\,I\{x>1\}$, where $I$ denotes the indicator. 
Note that $|e^{iu}-1-iu|\le2|u|$ and $|e^{iu}-1-iu+u^2/2|\le|u|^3/6$ for real $u$. 
So, for the characteristic function (c.f.) $f_V$ of $V_k$ and $|t|\le1$ we have 
\begin{multline*}
 \frac12\,f_V(t)=\int_1^\infty\frac{e^{itx}}{x^3}\,dx
 =\int_1^\infty\frac{1+itx}{x^3}\,dx
 -\int_1^{1/|t|}\frac{t^2x^2/2}{x^3}\,dx \\ 
 +\int_1^{1/|t|}\frac{e^{itx}-1-itx+t^2x^2/2}{x^3}\,dx
 +\int_{1/|t|}^\infty\frac{e^{itx}-1-itx}{x^3}\,dx \\
 =\frac12+it-\frac{t^2}2\,\ln\frac1{|t|}
 +c\int_1^{1/|t|}\frac{|t|^3x^3}{x^3}\,dx
 +c\int_{1/|t|}^\infty\frac{|t|x}{x^3}\,dx \\
  =\frac12+it-\frac{t^2}2\,\ln\frac1{|t|}+ct^2.  
\end{multline*}
So, for $|t|\le1$ we have $\ln f_V(t)=2it-t^2\,\ln\frac1{|t|}+ct^2$ and hence for the characteristic function $f_n:=f_{Z_n}$ of $Z_n$ we have 
\begin{multline*}
 \ln f_n(t)=-i2nt/b_n+n\ln f_V(t/b_n)
 =-\frac{t^2}{\ln n}\,\ln\frac{\sqrt{n\ln n}}{|t|}+c\frac{t^2}{\ln n} \\ 
 =-\frac{t^2}2-\frac{t^2}{\ln n}\,\Big(\frac12\,\ln\ln n-\ln|t|\Big)+c\frac{t^2}{\ln n} \\ 
 =-\frac{t^2}2+\frac{t^2}{\ln n}\,\ln|t|+c\frac{t^2}{\ln n}\,\ln\ln n  \tag{2}
\end{multline*}
for $|t|\le b_n=\sqrt{n\ln n}$. So, with $\ep_n$ as in (1), 
\begin{equation*}
  \ln f_n(t)= 
 \begin{cases}
 -\frac{t^2}2+c\ep_n|t| & \text{ if }|t|\le1 \\ 
 -\frac{t^2}2+c\ep_n t^2\ & \text{ if }1\le|t|\le\ln n, 
 \end{cases}
\end{equation*}
whence 
\begin{multline*}
 \int_{|t|<\ln n}\frac{|f_n(t)-e^{-t^2/2}|}{|t|}\,dt \\ 
 \le \int_{|t|<1}\frac{|e^{c\ep_n|t|}-1|}{|t|}\,dt
 +\int_{\R}\frac{|e^{-(1-2c\ep_n)t^2/2}-e^{-t^2/2}|}{|t|}\,dt \le c\ep_n;
\end{multline*}
the latter integral was bounded using the identity $\int_0^\infty\frac{e^{-at^2/2}-e^{-t^2/2}}t\,dt=\ln(1/\sqrt a)$ for $a>0$. 
By the Esseen smoothing inequality (see e.g. formula (6.4)),
\begin{equation*}
 \de_n\le c\int_{|t|<\ln n}\frac{|f_n(t)-e^{-t^2/2}|}{|t|}\,dt +c/\ln n. 
\end{equation*}
Now the second inequality in (1) immediately follows. 
It remains to prove the first inequality in (1). For real $t$ and real $A>0$, 
\begin{multline*}
 \int_0^\infty e^{itx}d\De_n(x)=\int_0^A e^{itx}d\De_n(x)+c(1-F_n(A))+c(1-G(A)) \\ 
 =c\De_n(A)+c\De_n(0)-it\int_0^A e^{itx}\De_n(x)dx+2c(1-G(A))+c\de_n \\ 
  =c\de_n+c|t|A\de_n+ce^{-A^2/2} 
  =c\de_n+c|t|\de_n\sqrt{\ln\frac1{\de_n}} 
\end{multline*}
if $A=\sqrt{2\ln\frac1{\de_n}}$. 
Similarly estimating $\int_{-\infty}^0 e^{itx}d\De_n(x)$, we have 
\begin{equation*}
 f_n(t)-e^{-t^2/2}=\int_{-\infty}^\infty e^{itx}d\De_n(x)=c\de_n+c|t|\de_n\sqrt{\ln\frac1{\de_n}}. 
\end{equation*}
Letting $t=1$ here and in (2), we see from the second line in (2) that 
$$\de_n\sqrt{\ln\frac1{\de_n}}\gg\frac{\ln\ln n}{\ln n},$$ 
whence the first inequality in (1) immediately follows. 
It appears that similar techniques should work for a somewhat wide class of distributions, like the ones referenced by the OP. 
