I am having some trouble with a textbook problem.
I am given that $U_1, ... U_n$ are independent random variables with uniform distribution over $[0,1]$ and that $U_{(n)}$ is the maximum.
I have to show that the CDF of the standardized $U_{(n)}$ approaches a limiting value.
I know that the CDF of $U_{(n)}$ is $F(U)=(u)^n$ and that the standardized $U_{(n)}$ is $U_{(n)}^*=\frac{U_{(n)}-\frac{n}{n+1}}{\sqrt{\frac{n}{(n+1)^2(n+2)}}}$, where $E[U_{(n)}] = \frac{n}{n+1}$ and $var(U_{(n)})=\frac{n}{(n+1)^2(n+2)}$.
However, I am unsure on how to find the CDF of $U_{(n)}^*$ and then show the limiting value.
Any help is appreciated, thank you in advance.
$$F_{U^*_{(n)}}(t) = F_{U_{(n)}}(t \cdot \text{SD}(U_{(n)}) + E[U_{(n)}]) = \begin{cases} 0 & t \sqrt{\frac{n}{(n+1)^2 (n+2)}} + \frac{n}{n+1} < 0 \\ 1& t \sqrt{\frac{n}{(n+1)^2 (n+2)}} + \frac{n}{n+1} > 1 \\ \left(t \sqrt{\frac{n}{(n+1)^2 (n+2)}} + \frac{n}{n+1}\right)^n & \text{otherwise} \end{cases}$$
I believe that if you take $n \to \infty$ and do some computations (recall $e^x = \left(1+\frac{x}{n}\right)^n$), the limit is $F(t) = e^{t-1}$ for $t \le 1$, and $F(t)=1$ for $t>1$. That is, if $U^*$ is the random variable with the limiting distribution, then $1-U^*$ is an exponential random variable with mean $1$.