We consider iid random variables $X_1,X_2,\ldots,X_n\sim\mathcal{U}_{[0,\theta]}$ and are interested in the asymptotic behaviour of the corresponding MLE for $\theta$, i.e. $$M_n:=\mathrm{max}_{i\in\{1,2,\ldots,n\}}X_i.$$ More precisely, I would like to confirm explicitly that $M_n$ converges (in some sense) to a reversed-Weibull-distributed rv in distribution and I suppose that I am just missing some computational detail/trick/idea...
I thought I should try studying a standardised version of $M_n$ since that's what seems to be done in extreme value theory and more generally when studying limiting distributions.
We know that $$\mathbb{E}_{\theta}[M_n]=\frac{n}{n+1}\theta,~~\mathbb{Var}_\theta[M_n]=\frac{n}{(n+1)^2(n+2)}\theta^2,$$ so I consider the cdf of $$Z_n=\frac{M_n-\mathbb{E}_\theta[M_n]}{\sqrt{\mathbb{Var}_\theta[M_n]}}~~\mathrm{or}~~Z_n'=\frac{M_n-\theta}{\sqrt{\mathbb{Var}_\theta[M_n]}},$$ which leads to $$\mathbb{P}_\theta(Z_n'\leq x)=\left(1+\frac{\sqrt{n}}{(n+1)\sqrt{n+2}}x\right)^n$$ and $$\mathbb{P}_\theta(Z_n\leq x)=\left(1-\frac{1}{n+1}\left(1-\frac{\sqrt{n}}{\sqrt{n+2}}x\right)\right)^n$$ for $x\leq 0$ and $x\leq \theta/(n+1)$ respectively. However, I don't see how the limit with respect to $n$ takes the form $$\exp\left(-\left(\frac{1}{1+\xi x}\right)^{1/\xi}\right)$$ for some $\xi<0$ here. (I know that $\lim_{n\rightarrow\infty}(1+x/n)^n=\exp(x)$)
How can I resolve this? Is it maybe just a clever alternative for the denominator $\sqrt{\mathbb{Var}_\theta[M_n]}$?
Hint:
Consider $T_n=n(\theta-M_n)$. From the distribution function of $M_n$, find the distribution function of $T_n$. Take the limit of the DF of $T_n$ as $n\to\infty$. Just remember that $\lim_{n\to\infty} \left(1+\frac{a}{n}\right)^n=e^a$. You would be easily able to identify the asymptotic distribution of $M_n$.
A quick search shows that this question has been answered in detail here on CV.