I am not sure whether the following statement is true. Let $x_k$ be identically distributed, and only assumed to have finite mean. Can we conclude that $x_k/k$ converges a.s.? Clearly it converges in probability. And if we compute

$$

P(|x_k|geq kepsilon)leq frac{E|x_k|}{kepsilon}$$

However, since we only assume finite mean, it is not sufficient to conclude the almost everywhere convergence. I can come up with counterexamples when $x_i$ are not identically distributed. But not for the identically distributed case. Does anyone have any idea about this?

Thanks in advance.