# probability – Does \$x_k\$ identically distributed with finite mean implies \$x_k/k\$ converges to 0 \$a.s.\$?

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?