The uniform endpoint MLE has rate n, not square-root n

On the probability-one event $0<X_{(n)}\le \theta$, the likelihood at a candidate $u>0$ is $L_n(u)={\prod }_{i=1}^n{p}_u(X_i)=u^{-n}\mathbf{1}\{X_{(n)}\le u\}$. It is zero for $u<X_{(n)}$, while for $u\ge X_{(n)}$ it equals $u^{-n}\le X_{(n)}^{-n}=L_n(X_{(n)})$, with strict inequality if $u>X_{(n)}$. Thus $X_{(n)}$ maximizes the likelihood almost surely; on the null event $X_{(n)}=0$, use the arbitrary positive assignment specified above.

For $0\le t<n\theta$, independence and the uniform CDF give $P_{\theta }\{n(\theta -X_{(n)})>t\}=P_{\theta }\{X_i<\theta -t\mathrm{/}n\text{ for every }i\}=(1-t\mathrm{/}(n\theta ){)}^n$. For fixed $t\ge 0$ this formula applies for all sufficiently large $n$, and its limit is $e^{-t\mathrm{/}\theta }$.

For every $h>0$ small enough that $\theta +h>0$, the gained support interval $(\theta ,\theta +h]$ has $p_{\theta }=0$ and $p_{\theta +h}=1\mathrm{/}(\theta +h)$. If the model were QMD at $\theta$, the score term in [F2] would vanish on this interval and the squared remainder integral would be at least ${\int }_{\theta }^{\theta +h}{p}_{\theta +h}(x)dx=h\mathrm{/}(\theta +h)$. But $[h\mathrm{/}(\theta +h)]\mathrm{/}{h}^2=1\mathrm{/}[h(\theta +h)]\to \mathrm{\infty }$, contradicting the required $o(h^2)$ remainder. Thus the model is not QMD at $\theta$.

Therefore the nonnegative variable $n(\theta -X_{(n)})$ converges in distribution to the exponential law with rate $1\mathrm{/}\theta$, whose CDF is $0$ for $t<0$ and $1-e^{-t\mathrm{/}\theta }$ for $t\ge 0$. The limiting CDF is continuous, so [F3] applies at every real $t$.

For any $\epsilon >0$ and all $n>{\epsilon }^2\mathrm{/}{\theta }^2$, the exact maximum formula gives $P_{\theta }\{\sqrt{n}(\theta -X_{(n)})>\epsilon \}=(1-\epsilon \mathrm{/}(\theta \sqrt{n}){)}^n\to 0$. Thus $\sqrt{n}(\theta -X_{(n)}){\to }_{P}0$ by [F4], refuting the stated nonzero root-$n$ fluctuation and confirming the faster $n$ rate in step 2.1.