Let $U_1,U_2,...U_n,$ be iid samples from uniform distribution $U(0,1)$. And the order Statistic:
\begin{equation}
U_{n,1}\leq U_{n,2}\leq ...\leq U_{n,n},
\end{equation}
Let $1\leq k_n \leq n$ and $k_n\rightarrow\infty$ when $n\rightarrow\infty$. Proof: $Y_n=nU_{n,{k_n}} \rightarrow \infty$ in probability, i.e. for every $M>0$,
\begin{equation}
lim_{n\rightarrow\infty} \quad P(Y_n>M)=1
\end{equation}
My ideas so far:
The density function of $U_{n,k_n}$ is
\begin{equation}
f_U(x)=\frac{n!}{(k_n-1)!(n-k_n)!}x^{k_n-1}(1-x)^{n-k_n}.
\end{equation}
And the density function of $Y_n=nU_{n,k_n}$ is
\begin{equation}
f_Y(x)=\frac{(n-1)!}{(k_n-1)!(n-k_n)!}(\frac{x}{n})^{k_n-1}(1-\frac{x}{n})^{n-k_n}.
\end{equation}
But the integral
\begin{equation}
P(Y_n>M)=\int_{M}^{\infty} \frac{(n-1)!}{(k_n-1)!(n-k_n)!}(\frac{x}{n})^{k_n-1}(1-\frac{x}{n})^{n-k_n}dx.
\end{equation}
seems very complicated, what can I do next?
Thanks in advance for any help!
Note that $U_{n,k}$ has the same distribution as $S_k/S_{n+1}$, where $$ S_n:=\sum_{i=1}^n v_i, \quad v_i\overset{\text{iid}}{=}\text{Exp}(1). $$ Thus, for any $M<\infty$, $$ \mathsf{P}\!\left(nU_{n,k_n}>M\right)=\mathsf{P}\!\left(\frac{n}{S_{n+1}}\cdot S_{k_n}>M\right)\to 1 $$ as $n\to\infty$ because $S_{n+1}/n\to 1$ a.s. and $S_{k_n}\to \infty$ a.s.