This is a follow-up question to 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 $[0,1]$? . My motivation is to construct a statistic whose rate convergence to a Gaussian will be very slow and as such to formulate a counterexample to Berry-Esseen' theorem. We therefore define the following statistic: \begin{equation} S_n := \frac{\left(\sum\limits_{k=1}^n f^{-1}(X_k) - \frac{3}{2} e \cdot n\right)}{e\cdot \sqrt{n \log(\log(n))}} \end{equation} where $X_k$ are i.i.d. uniformly distributed in $(0,1)$ and the function $f()$ is defined as follows: \begin{equation} f(x) := \frac{e^2}{2} \cdot \frac{1+\log(x)}{x^2 \log(x)^2} 1_{x \ge e} \end{equation} Now, the probability density of $f(X)$ is as follows: \begin{eqnarray} \rho_{f(X)}(z) &=& \int\limits_0^1 \delta(z - f^{-1}(x)) dx =-\int\limits_{e}^\infty \delta(z-u) f^{'}(u) du= -f^{'}(z) 1_{z \ge e}\\ &=& \frac{e^2}{2} \cdot \frac{2+3 \log(z)+2 \log(z)^2}{z^3 \log(z)^3}1_{z \ge e} \end{eqnarray} From this we readily get the moments: \begin{eqnarray} E\left[ f^{-1}(X) \right] = \frac{3}{2} e\\ E\left[ (f^{-1}(X))^2 \right] = \infty \end{eqnarray} We also get the characteristic function. It reads: \begin{eqnarray} \kappa_{f^{-1}(X)}(k) = e^{\imath k e}+ \imath k \frac{e}{2} e^{\imath k e}-k^2 \frac{e^2}{2}\cdot \int\limits_0^\infty (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) d \delta \end{eqnarray} for $0<k<1$.

Note: The last integral on the right hand side is for me hard to crack. However numerical computations suggest that: \begin{equation} \lim_{k\rightarrow 0} \frac{1}{\log(\log(1/k))} \cdot \int\limits_0^\infty (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) d \delta = 1 \end{equation} Unfortunately for the time being I am unable to prove or disprove that claim above. However if we assume that the above is true then at least we can make sure that our test statistic is properly normalized. Indeed we have: \begin{eqnarray} \log\left( \kappa_{S_n}(k)\right) &=& -\imath k \frac{3}{2} \frac{n}{c_n} + n \log\left[ \kappa_{f^{-1}(X)}(\frac{k}{e c_n})\right] \\ &=& \frac{1+4 \log(2)-4 \log(2-2 \log(k)+\log(n) + \log(\log(\log(n))))}{8\log(\log(n))} k^2 + O(\frac{k^3}{\sqrt{n \cdot \log(\log(n))}}) \end{eqnarray} where in the second line I simply used the Series[] command in Mathematica to obtain the expansion . Now we can clearly see that: \begin{equation} \lim_{n\rightarrow \infty} \log\left( \kappa_{S_n}(k)\right) = -\frac{k^2}{2} \end{equation} as it should be.

Now, I carried out a Monte Carlo simulation and computed the sample Cumulative Distribution Function (CDF) of our statistic and plotted it along with the CDF of a standardized Gaussian distribution with the former and the later being plotted in Blue and Purple respectively. Here I took $n=5,10,15$ and in each case I used $m=1000$ realizations. The figures are below:

CDFs at $n=5$

CDFs at $n=10$

CDFs at $n=15$

I have used the following Mathematica code to produce those figures:

m = 1000; n = 15; delta = 1/10;
bins = Table[-5 + delta/2 + j delta, {j, 1, (10 - delta)/delta}];
limD = CDF[NormalDistribution[0, 1], bins];
X = RandomReal[{0, 1}, {m, n}];
x =.; {t0, Y} = 
 Timing[(x /. 
    Map[First[
       NSolve[(E^2 (1 + Log[x]))/(2 x^2 Log[x]^2) == # && x > E, x, 
        Reals]] &, X, {2}])];
ll = (Total[#] & /@ Y - 3/2 E n)/(E Sqrt[n Log[Log[n]]];
emp = EmpiricalDistribution[ll];
DD = CDF[emp, bins];
pl = ListPlot[Transpose[{bins, #}] & /@ {DD, limD}, ImageSize -> 800, 
   LabelStyle -> {15, FontFamily -> "Arial"}, 
   BaseStyle -> {15, FontFamily -> "Bold"}, 
   PlotLabel -> "n=" <> ToString[n]];
Export["LimitBehavior1_n_" <> ToString[n] <> ".jpg", pl, "JPEG"];
Import["LimitBehavior1_n_" <> ToString[n] <> ".jpg"]

Having said all this my question is the following. What is the rate of convergence of our statistic towards a Gaussian. To be specific we are asking about the behavior of the supremum norm of the difference in CDFs for large values of $n$.