Asymptotics of $\max\limits_{1\leqslant k\leqslant n}X_k/n$ I found an assertion in this paper at the beginning of page 6, but i can't see how to justify it: Let $X_n \geq 0$ i.i.d. with finite expectation then:
$$ \frac1n\max\limits_{k \leq n}X_k \to 0 \quad\text{almost surely as  } \,n \to \infty$$ 
As Davide posted there is convergence in probability to zero. I was wondering if it is evident that we have almost sure convergence also.
 A: 
Let $M_n=n^{-1}\cdot\max\limits_{1\leqslant k\leqslant n}X_k$ for some nonnegative identically distributed integrable sequence $(X_n)_n$, then $M_n\to0$ almost surely. (The independence of the sequence $(X_n)_n$ is not needed.)

To prove this, let $x\gt0$. If $X_n\leqslant nx$ for every $n$ large enough, say for every $n\geqslant N$, then $nM_n\leqslant NM_N+nx$ for every $n\geqslant N$ and in particular $\limsup\limits_{n\to\infty}M_n\leqslant x$. Thus,
$$
[\limsup\limits_{n\to\infty}M_n\geqslant2x]\subseteq A_x,\qquad A_x=\limsup\limits_{n\to\infty}A^n_x,\qquad A^n_x=[X_n\geqslant nx].
$$
The sequence $(X_n)_n$ is identically distributed, hence
$$
\sum_{n\geqslant1}P[A_x^n]=\sum_{n\geqslant1}P[X_1\geqslant nx]\leqslant x^{-1}E[X_1],
$$
which is finite. By Borel-Cantelli lemma (easy part), $P[A_x]=0$. This holds for every $x\gt0$ hence $\limsup\limits_{n\to\infty}M_n=0$ almost surely. Since every $M_n\geqslant0$ almost surely, the conclusion follows.
Edit: To show the upper bound of the series, note that, for every nonnegative $\xi$,
$$
\sum_{n\geqslant1}\mathbf 1_{\xi\geqslant nx}=\lfloor x^{-1}\xi\rfloor\leqslant x^{-1}\xi,
$$
and integrate this pointwise inequality over $\xi$ with respect to the distribution of $X_1$.
