Strong LLN v.s. Weak LLN

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It is clear that SLLN implies WLLN since almost surely convergence implies converge in probability.  Here, we introduce a counterexample that satisfies WLLN but not SLLN.

$\bullet$ Counterexample. (WLLN $\not\Rightarrow$ SLLN)


Let $\{X_n\}$ be independent random variables with a common d.f. $F$ such that $$\mathscr{P}\{X_1=n\}=\mathscr{P}\{X_1=-n\}=\frac{c}{n^2\log{n}},\;n=3,4,\ldots,$$where $c$ is a constant $$c=\frac{1}{2}\left(\sum_{n=3}^\infty\frac{1}{n^2\log{n}}\right)^{-1}.$$According to the Extension of Weak Law of Large Number (1), let $b_n=n$, we have for large $n$, $$n\int_{|x|>n}\,dF(x)=n\sum_{k>n}\frac{c}{k^2\log{k}}\sim\frac{c}{\log{n}}=o(1),$$
$$\frac{1}{n^2}\cdot n\int_{|x|\leq n}x^2\,dF(x)=\frac{1}{n}\sum_{k=3}^n\frac{ck^2}{k^2\log{k}}\sim\frac{c}{\log{n}}=o(1).$$Thus the two conditions are satisfied and $$a_n=\sum_{j=1}^n\int_{|x|\leq n}x\,dF(x)=0.$$Hence, $S_n/n\rightarrow0$ in probability.

But, we have $$\mathscr{P}\{|X_1|>n\}=\int_{|x|>n}\,dF(x)=\sum_{k>n}\frac{c}{k^2\log{k}}\sim\frac{c}{n\log{n}}.$$By the r.v.'s sharing the same d.f., $$\sum_n\mathscr{P}\{|X_n|>n\}=\sum_n\mathscr{P}\{|X_1|>n\}=\infty.$$Thus, by Borel-Cantelli Lemma, we have $$\mathscr{P}\{|X_n|>n\mbox{ i.o.}\}=1.$$Now, let's discuss the convergence of $S_n$.  $|S_n-S_{n-1}|=|X_n|>n$ implies either $|S_n|>n/2$ or $|S_{n-1}|>n/2$.  It follows that $$\mathscr{P}\left\{|S_n|>\frac{n}{2}\mbox{ i.o.}\right\}=1,$$that is, $S_n/n\not\overset{a.s.}{\rightarrow}0.$

$\Box$