TITLE: convergence in distriubtion of $(\max_{1\leq k\leq n} X_k)/ln(n)$ where $X_k\sim \text{Exp}(1)$.
QUESTION [0 upvotes]: The problem is: Let $X_k$'s be i.i.d. $\text{Exp}(1)$ r.v.'s. define:
$$\xi_n = max_{1\leq k\leq n} X_k$$
$$\eta_n = \xi_n/ln(n)$$.
The question is what is the limit of $\eta_n$ as $n\to \infty$?
Here is my approach:
$$P(\xi_n\leq x) = P(\max_{1\leq k\leq n}X_k\leq x) = \prod_{k=1}^{n}(1-e^{-x}) = (1-e^{-x})^n$$
$$P(\eta_n\leq x) = P(\frac{\xi_{n}}{ln(n)}\leq x) = P(\xi_n\leq ln(n) x) = (1-e^{-ln(n) x})^n = (1-\frac{1}{n^x})^n$$
What is the limit of $(1-e^{-ln(n) x})^n$ as $n\to \infty$?
Thank you for the help.

REPLY [2 votes]: What is the limit of $(1-e^{-ln(n) x})^n$ as $n\to \infty$?

Hint: Let $x_n=(1-a_n)^n$ with $a_n$ in $(0,1)$ for every $n$.

If $na_n\to\alpha$ with $\alpha$ in $(0,+\infty)$ then $x_n\to\mathrm e^{-\alpha}$
If $na_n\to0$ then $x_n\to1$
If $na_n\to\infty$ then $x_n\to0$

In your case, you know that $(1-e^{-\ln(n) x})^n=(1-1/n^x)^n$ hence $a_n=1/n^x$. The result follows.
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