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Consider $$\zeta_n=\max_{1\le k \le n} X_k ,$$ where $X_k\sim \exp(1)$. Let $\eta_n=\frac{\zeta_n}{\ln(n)}$. What is the limit of $\eta_n$ as $n\to \infty$? (Convergence in distribution).

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closed as off-topic by Davide Giraudo, Silvia Ghinassi, C. Falcon, Math1000, Daniel W. Farlow Jun 30 at 1:32

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Well, knowing that $\eta_n$ is a maximum, you could try to write the event $[\eta_n\le x]$ for a fixed $x$ using only events $[X_k\le x]$. // Re your approach of MSE, let me venture two guesses: First, the present question is homework. Second, you ask questions here at a pace such that you do not have the time to accept the answers you get, even less to study and understand them. Am I right on both counts? – Did Sep 12 '11 at 9:56
    
I am thinking of $P[\zeta_n\le \ln(n)x]$, so we will have $F(\ln(n) x)$, where $\zeta_n$ has a distribution of $F(x)=(1-\exp(-\lambda x))^n$, then find the distribution for one n and take the limit. i have $\lim_{n\rightarrow\infty} (1-\exp(-x\ln(n)))^n=\lim_{n\rightarrow\infty}(1-(\exp(-\ln(n))^x)^n=\lim_{n\rig‌​htarrow\infty} (1-(\frac{1}{n})^x)^n$, but i am not even sure if this limit exists. for x>0, we have a limit of 1 x=0,we have a limit of 0. can we just definite x<0 to have a limit of 0 as well? – bear Sep 12 '11 at 14:11
    
The logarithm of the quantity of interest is $n\log(1-n^{-x})$. Perhaps you know the limit of $n\log(1-an^{-1})$ for every given $a$. Then you might want to show that the limit of $n\log(1-n^{-x})$ is $-\infty$ if $x<1$ and $0$ if $x>1$. Once you know that, what does that tell you on the limit in distribution you are after? – Did Sep 12 '11 at 14:26
    
we have F(x)=1 for x>1, and F(x)=0, for x<1 – bear Sep 12 '11 at 14:30
    
Sorry? Do you mean the limits? (At the moment, you consider F(log(n)x) which depends on n and is certainly not 1 for x>1 nor 0 for x<1.) Anyway, how do you prove this? – Did Sep 12 '11 at 14:34