Suppose $X_n\geq0$ are any sequence of random variables and $\dfrac{X_1+...+X_n}{n}\stackrel{p}{\to}1$. Then show that $\dfrac{\max(X_1,...,X_n)}{n}\stackrel{p}{\to}0$ and for any $r>1$, $\dfrac{1}{n^r}\sum_{k=1}^n X_k^r\stackrel{p}{\to}0$.
I could prove the almost sure version of this, that is, replacing convergence in probability by almost sure convergence everywhere in the statement. I thought of a subsequence argument (first showing for a subsequence that almost surely the results hold) but things don't seem to go through.
I have obtained that $X_n/n$ converges in probability to $0$. If this was an almost sure statement then I could immediately say that this implies $\max(X_1,...,X_n)/n$ also converges almost surely to $0$.
The convergence of the maximum $M_n:=n^{-1}\max_{1\leqslant k\leqslant n}X_k$ was treated here. Observe that for $1\leqslant k\leqslant n$, $X_k\leqslant nM_n$ hence $X_k^{r-1}\leqslant n^{r-1}M_n^{r-1}$; this leads to the bound $$ 0\leqslant\dfrac{1}{n^r}\sum_{k=1}^n X_k^r\leqslant \frac 1{n^r}\sum_{k=1}^n X_kn^{r-1}M_n^{r-1}=\left(\frac 1n\sum_{i=1}^nX_k\right)M_n^{r-1}. $$ The first factor converges in probability to one, the second one in probability to $0$ since $r-1>0$.