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Show there exist independent random variables $\{X_n\}$ with $X_n\in\{-n, n, 0\}$, $\mathbb{E}(X_n)=0$, and $Y_n=\tfrac{1}{n}\sum_{k=1}^nX_k$ for all $n$, $\mathbb{P}(|Y_n|\ge\epsilon)\to 0$ for all $\epsilon>0$, and $\mathbb{P}(Y_n\to 0)<1$.

$\mathbb{E}(X_n)=0$, so $p_n=\mathbb{P}(X_n=-n)=\mathbb{P}(X_n=n)$. By Kolmogorov's criterion in http://mathworld.wolfram.com/StrongLawofLargeNumbers.html, if $S=\sum_n\tfrac{\mathbb{V}(X_n)}{n^2}$ converges, then $\mathbb{P}(Y_n\to 0)=1$, so we definitely need $S$ to diverge if we want $\{X_n\}$ to work. $$\mathbb{V}(X_n)=\mathbb{E}(X_n^2)=2p_nn^2,$$ so $S=\sum_n2p_n$ diverges.

If $\mathbb{V}(Y_n)\to 0$, we can use Chebyshev's inequality as in the proof of the weak LLN to get $\mathbb{P}(|Y_n|\ge\epsilon)\to 0$ for all $\epsilon>0$. If $\{p_nn^2\}$ is increasing, $$\mathbb{V}(Y_n)=\dfrac{1}{n^2}\sum_{k=1}^n\mathbb{V}(X_k)\le\dfrac{1}{n^2}\sum_{k=1}^n2p_nn^2=2np_n,$$ so the weak LLN will hold if $np_n\to 0$.

For all positive integers $k$, $p_n=\tfrac{1}{n\log^k(n)}$ satisfies $np_n\to 0$ and $\sum_np_n\to\infty$, where $\log^k(n)=\log(...\log(n)...)$, since $\sum_np_n$ grows like $\log^{k+1}(x)$, the integral of $p_x$. I know how to adapt the proof of the weak LLN to show that $\{X_n\}$ satisfies the weak LLN, but I don't know how to show that $\mathbb{P}(Y_n\to 0)< 1$. It might not even be true, since I don't know if the converse of Kolmogorov's criterion holds. At any rate, I can't think of a good way to show $\mathbb{P}(Y_n\to 0)<1$. You might be able to write a complicated expression for $\mathbb{P}(Y_n\to 0)$ and bound it, but that seems hard.

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If $p_n=\frac 1 {n log \, n}$ for $n >1$ then $\{Y_n\}$ does not tend to $0$ almost surely. This is because $\frac {X_n} n =\frac {nY_n-(n-1)Y_{n-1}} n$ so if $Y_n$ tends to $0$ then $ \frac {X_n} n \to 0$. By Borel Cantelli Lemma this implies $\sum P\{|\frac {X_n} n| >\frac 1 2\}<\infty$ which becomes $\sum 2p_n <\infty$ and this is not true.

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