Infinite sequence of exponentially distributed random variables Consider an infinite sequence of exponentially distributed random variables, $X_k$, where$ k \in \{1, \ldots, n\}$ with $\lambda = 1$. I am trying to evaluate:
$$\lim_{n\to\infty}  \frac{\max_{1 \leq k \leq n} (X_k)}{\ln n}.$$
So far, I have shown that $$\lim_{n \to \infty}E[[\dfrac{max_{1 \leq k \leq n} x_k}{log(n)}] = 1$$ 
If I show that the variance tends to zero in the limit, then I can assert that the limit is indeed 1. I am unable to do so. 

Edit: I wrote a computer simulation of this process. In fact, the monte carlo method agreed that the expected limit is 1; however, even with over 1 million k values, I am not getting exact convergence. For example, I am getting number like 1.02 and 0.98 . 
Here is the code for anyone interested:
#include <iostream>

#include <boost/math/distributions/exponential.hpp>
#include <boost/math/distributions.hpp> // For non-member functions of distributions
#include <boost/random.hpp> // Convenience header file
#include <math.h>       /* log */

using namespace std;

int main()
{
    // Don't forget to tell compiler which namespace
    using namespace boost::math;

    double scaleParameter = 1;

    boost::exponential_distribution<> exp(1);
    boost::lagged_fibonacci607 rng;
    rng.seed(static_cast<unsigned int> (std::time(0)));
    boost::variate_generator<boost::lagged_fibonacci607&, boost::exponential_distribution<> > Rng(rng, exp);

    std::vector<double> vec;

    int lim = 1000000;
    for (int i = 0; i < lim; ++i)
    {
        vec.push_back(Rng());
    }

    std::vector<double> xi;

    xi.reserve(vec.size());

    double max = vec[0];

    for (std::vector<double>::iterator it = vec.begin(); it != vec.end(); it++)
    {
        if (*it > max)
        {
            max = *it;
        }
        xi.push_back(max);
    }



    vector<double> Log;

    for (int i = 0; i < lim; ++i)
    {
        Log.push_back(log(i));
    }

    vector<double> eta;

    for (int i = 0; i < lim; ++i)
    {
        eta.push_back(xi[i] / Log[i]);
    }

    cout << eta[lim - 1] << endl;




    return 0;
}

 A: It is absolutely not correct that at least one should be infinite.  It is true that for every real number $a$, infinitely many of the random variables will be more than $a$, but every one of them will be finite.  But the fact that for every $a>0$, infinitely many of them are $>a$ is enough to imply that the limit is infinite.  I'm going to follow up on David Ulrich's comment.  We're trying to prove that $$\Pr\left( \lim_{n\to\infty} \max\{X_1,\ldots,X_n\} = \infty \right)=1.$$  Notice that $\max\{X_1,\ldots,X_n\}$ can only stay the same or get bigger every time $n$ is incremented by $1$.  It's monotone.  Every monotone non-decreasing sequence approaches a limit that is either some finite number or $\infty$.  Let $a>0$ be some finite number.  Consider
\begin{align}
& \Pr\left( \lim_{n\to\infty} \max\{X_1,\ldots,X_n\} \le a \right) \\[10pt]
\le {} & \Pr\left( X_1\le a\ \&\ \cdots\ \&\ X_n\le a \right) \\[10pt]
= {} & \left( \Pr(X_1\le a) \right)^n \to 0.
\end{align}
A: You are essentially asking for the distribution of the maximum order statistic of $n$ iid exponential random variables.  That is to say, what is $$\Pr\left[\max_{1\le k \le n} X_k \le x\right]?$$  Recall that $$\Pr[X_k \le x] = 1 - e^{-x}, \quad k = 1, 2, \ldots.$$  Since the $X_k$s are iid, it easily follows that $$\Pr\left[\max_{1\le k \le n} X_k \le x\right] = \Pr[(X_1 \le x) \cap (X_2 \le x) \cap \cdots \cap (X_n \le x)] \overset{\text{iid}}{=} \prod_{k=1}^n \Pr[X_k \le x].$$  Now substitute to obtain the desired CDF, and differentiate to obtain the PDF.  Now compute the expectation of the maximum order statistic, scaled by $1/\log n$; i.e., $$\operatorname{E}\left[X_{(n)}/\log n\right] = \frac{1}{\log n} \int_{x=0}^\infty x f_{X_{(n)}}(x) \, dx.$$  Is this finite?  If not, what does this suggest?  If so, what would you need to do next?

Consider the function $$I(n) = \int_{x=0}^\infty xe^{-x}(1-e^{-x})^n \, dx.$$ With the substitution $u = 1-e^{-x}$, $du = e^{-x} \, dx$, $x = -\log(1-u)$, we obtain the equivalent expression $$I(n) = -\int_{u=0}^1 u^n \log(1-u) \, du.$$ Now the series expansion $$-\log(1-u) = -\int \frac{du}{1-u} = \int \sum_{j=0}^\infty u^j \, du = \sum_{j=0}^\infty \frac{u^{j+1}}{j+1}$$ gives $$I(n) = \int_{u=0}^1 \sum_{j=1}^\infty \frac{u^{n+j}}{j} \, du = \sum_{j=1}^\infty \frac{1}{j(j+n+1)} = \frac{1}{n+1}\sum_{j=1}^\infty \left(\frac{1}{j} - \frac{1}{j+n+1}\right) = \frac{1}{n+1} \sum_{j=1}^{n+1} \frac{1}{j},$$ which one may recognize as $$I(n) = \frac{H_{n+1}}{n+1},$$ where $H_n$ is the $n^{\rm th}$ harmonic number.  Therefore the expectation is simply $$\operatorname{E}[X_{(n)}/\log n] = \frac{n I(n-1)}{\log n} = \frac{H_n}{\log n}.$$  Since we can establish the crude bounds $$H_n = \int_{x=1}^{n+1} \frac{dx}{\lfloor x \rfloor}  > \int_{x=1}^{n+1} \frac{dx}{x} = \log(n+1) > \log n,$$ and $$H_n = 1 + \int_{x=1}^n \frac{dx}{\lceil x \rceil} < 1 + \int_{x=1}^n \frac{dx}{x} = 1 + \log n,$$ we then have $$1 = \lim_{n \to \infty} \frac{\log n}{\log n} < \lim_{n \to \infty} \frac{H_n}{\log n} < \lim_{n \to \infty} \frac{1 + \log n}{\log n} = 1.$$  This establishes the asymptotic expectation is $1$.
One could then show that the variance tends to zero as $n \to \infty$, but this is more difficult if the method used is similar to that above.
