Cesàro mean of $P(X_k = \pm k)=\frac{1}{2k\log(k+1)}$, $P(X_k = 0)=1-\frac{1}{k\log(k+1)}$ For the sequence $(X_n)_{n\geq 1}$ of independent random variables as described in the title, I would like to know if the sequence $Y_n:=\frac{1}{n}\sum_{k=1}^{n}X_k$ converges in probability (or even almost surely).
I feel like it does not converge in probability, althought I am having a hard time proving it.
I have been told to use the event {$ X_k=k \ \text{for infinitely many k} $}, and I found that it has probability $1$, but does it give any information on the convergence of the sequence?
Thank you very much in advance.

Assume $Y_n(\omega) \to Y(\omega)$. We will show that if $Y(\omega) < 1/8$ then only finitely many $X_k = k$, and if $Y(\omega) > -1/8$ then only finitely many $X_k = -k$. Together with the fact that a.s. infinitely many $X_k = k$ and infinitely many $X_k = -k$, this gives us that $P(Y_n \to Y) = 0$.
Assume $Y(\omega) < 1/8$.
As $Y_n(\omega) \to Y$, $\exists n_0 \forall n > n0: Y_n < 1/4 \wedge Y_{n + 1} - Y_n < 1/4$.
But $Y_{n + 1} - Y_n = \frac{X_{n + 1} - Y_n}{n + 1}$, so $X_{n + 1} < (n + 1) / 4 + Y_n < n / 4 + 1/2$. Thus, all $X_n$ for $n > n_0$ are not equal to $n$, or, in other words, only finitely many $X_k = k$.
Case $Y(\omega) > - 1/8$ is similar, we need to consider $Y_n - Y_{n + 1} < 1/4$ and $X_k = -k$.
There probably should be significantly more elegant way to prove it, but this also isn't to complicated I think.
