Proving $\frac{1}{n} \sum_{k=1}^{n}X_{k}\to 0$ a.s. $\{X_{n}\}$ is a sequence of independent random variables, $EX_n=0$, and $\sum_{n=1}^{\infty}n^{-(r+1)}E(|X_n|^{2r})<\infty$. Proving $\frac{1}{n} \sum_{k=1}^{n}X_{k}\to 0$ a.s. and $r>1$
I think Borel-Cantelli lemma should be a very useful way to prove this kind of problem, but I don’t know how it should be applied to this one. Some approaches are welcome!
 A: It suffices to prove (in view of the Borel-Cantelli lemma and the fact that $a_n/n\to 0$ if and only if $\max_{1\leqslant n\leqslant 2^N}a_n/2^N\to 0$) that for each positive $\varepsilon$,
$$
\sum_{N\geqslant 1}\mathbb P\left(\max_{1\leqslant n\leqslant 2^N}\left\lvert \sum_{k=1}^nX_k\right\rvert >\varepsilon 2^N\right)<\infty.
$$
By Markov's inequality, we are reduced to establish the convergence of $\sum_{N\geqslant 1}b_N$, where
$$
b_N:=2^{-2Nr}\mathbb E\left[\max_{1\leqslant n\leqslant 2^N}\left\lvert \sum_{k=1}^nX_k\right\rvert^{2r}\right].
$$
Since $2r>1$, Doob's inequality gives
$$
b_N\leqslant \left(\frac{2r}{2r-1}\right)^{2r-1}2^{-2Nr}\mathbb E\left[ \left\lvert \sum_{k=1}^{2^N}X_k\right\rvert^{2r}\right].
$$
Using Marcinkiewicz–Zygmund inequality with $p=2r$, we get
$$
b_N\leqslant C_r2^{-Nr}\mathbb E\left[ \left(2^{-N} \sum_{k=1}^{2^N}X_k^2\right)^{r}\right],
$$
where $C_r$ depends only on $r$. Since $r>1$, we get by convexity of $t\mapsto t^r$ that
$$
b_N\leqslant C_r2^{-Nr}  2^{-N}\sum_{k=1}^{2^N}\mathbb E\left[\lvert X_k\rvert^{2r}\right] .
$$
As a consequence, $\sum_N b_N$ converges as long as
$$
\sum_N2^{-Nr}  2^{-N}\sum_{k=1}^{2^N}\mathbb E\left[\lvert X_k\rvert^{2r}\right]
$$
does, that is,
$$
\sum_{k\geqslant 1}\sum_{N: 2^N\geqslant k}2^{-N(r+1)}\mathbb E\left[\lvert X_k\rvert^{2r}\right]<\infty,
$$
which is exactly the mentioned condition because $\sum_{N: 2^N\geqslant k}2^{-N(r+1)}$ is of order $k^{-r-1}$.
As pointer out in the comments, it is not required that the $X_k$ share the same distribution.
