What does it mean for a cumulative density function of a standardized variable to tend to a limiting value? This is main part of the problem I don't understand too well. Below is the problem I am trying to solve, and my current progress in doing so:
Problem
Let $U_1,\dots,U_n$ be independent uniform random variables on $[0,1]$, and let $U_{(n)}$ be the maximum. Find the cdf of $U_{(n)}$ and a standardized $U_{(n)}$, and show that the cdf of the standardized variable tends to a limiting value.
Solution so far
The density function of a uniform random variable on $[0,1]$ is $f(x) = 1$ and the cdf is $F(x) = x$ for $x \in [0,1]$, so for $U_{(n)}$, the density function is \begin{equation*} f_n(x) = \frac{n!}{(n-1)!(n-n)!}f(x)F^{n-1}(x)[1-F(x)]^{n-n} = nx^{n-1}, \end{equation*} and the cdf is \begin{equation*} F_n(x) = \int_0^x nt^{n-1}\;dt = x^n. \end{equation*} Let $S_n = \frac{U_{(n)} - E(U_{(n)})}{\sqrt{Var(U_{(n)})}}$ be the standardized variable of $U_{(n)}$. \begin{equation} E(U_{(n)}) = \int_0^1xnx^{n-1}\;dx = \frac{n}{n+1} \end{equation} \begin{equation} E(U_{(n)}^2) = \int_0^1x^2nx^{n-1}\;dx = \frac{n}{n+2} \end{equation} \begin{equation} Var(U_{(n)}) = E(U_{(n)}^2)- E(U_{(n)})^2 = \frac{n}{n+2}-\frac{n^2}{(n+1)^2} = \frac{n}{(n+1)^2(n+2)} \end{equation} \begin{equation} S_n = -\sqrt{n(n+2)} + \frac{(n+1)\sqrt{n+2}}{\sqrt{n}}U_{(n)} \end{equation} \begin{equation} E(S_n) = 0, \qquad Var(S_n) = 1 \end{equation}
Idea?
Should I be finding the moment generating function of $U_{(n)}$ and then use it to find the moment generating function of $S$, $M_S(t)$, then take $\lim_{n\rightarrow\infty}M_S(t)$ and see if the result matches any of the moment generating functions for the most common distributions?
Let $\mu_n$ be the mean of $S_n$, and $\sigma_n$ its standard deviation. Then $\Pr(S_n \le s)=\Pr(U_{(n)}\le \sigma_n s+\mu_n)$. This is $(\sigma_n s+\mu_n)^n$. Now we calculate. A little manipulation shows that $$(\sigma_n s+\mu_n)^n=\left(\frac{n}{n+1}\right)^n \left(1+\frac{s}{\sqrt{n}\sqrt{n+2}}\right)^n.$$ The term $\sqrt{n}\sqrt{n+2}$ behaves essentially like $n$, more precisely like $n+1$, but it doesn't matter. The limit is $e^{-1}e^s$.
Added: Please note the comment by Stephen Herschkorn that the limit of the cdf is given by the above expression only for a certain range of values of $s$, since $\Pr(U_{(n)}\le \sigma_n s+\mu_n)=(\sigma_n s+\mu_n)^n$ only when $0\le \sigma_n s+\mu_n\le 1$.