Let $X_n$ be succession of independent exponentials of parameter $\lambda$.
Let $Z_n = max(X_1,...,X_n)$
I want to study the convergence of $Z_n$ and of $\frac{1}{\log{n}}Z_n$
First of all I find the CDF of $Z_n$: $$ F_{Z_n}(z) = (1-e^{-\lambda z})^n $$
The limit of n going to infinity is 0 but there doesn't exist a random function with CDF constant equal to 0, so It doesn't converge in law.
Now if I multiply for $\frac{1}{\log{n}}$, I get:
$$F_{\frac{1}{\log{n}}Z_n}(z) = (1- e^{-\lambda \log{n} z})^n$$
This should converge to the constant $\frac{1}{\lambda}$ but I am not sure why.
Note that \begin{align} P(Z_n\le z\log n) = \left(1-\frac{1}{n^{\lambda z}}\right)^n = \left(\left(1-\frac{1}{n^{\lambda z}}\right)^{n^{\lambda z}} \right)^{\frac{n}{n^{\lambda z}}} \\ \end{align} Therefore, \begin{align} \lim_n P(Z_n\le z\log n) &= \frac{1}{e}, \quad \mbox{ if } \lambda z=1\\ &= 1, \quad \mbox{ if } \lambda z>1\\ &= 0, \quad \mbox{ if } \lambda z<1 \end{align} which means that the sequence $Z_n/\log n$ converges in distribution to $\frac{1}{\lambda}$ (we do not care about the point $z=1/\lambda$ because it is not a point of continuity of the distribution function of the constant random variable $1/\lambda$). Since it converges to a constant, $Z_n/\log n$ also converges in probability.