= Solution
Write $S_n=\sum_{j=1}^nY_j$ for the <partial sum>. Fix an integer $M\geq1$. The <tail-sum formula for expectation>, applied to the nonnegative <random variable> $|Y_1|$, gives
$$
\sum_{n=1}^{\infty}\mathbb P(|Y_n|>Mn)
=\sum_{n=1}^{\infty}\mathbb P(|Y_1|>Mn)
=\infty,
$$
because $\mathbb E|Y_1|=\infty$. The events $\{|Y_n|>Mn\}$ are <independent random variables>[independent], so the second <Borel-Cantelli lemmas>[Borel--Cantelli lemma] shows that $|Y_n|/n>M$ infinitely often with probability one. Taking the countable intersection over $M\in\mathbb N$ yields
$$
\limsup_{n\to\infty}\frac{|Y_n|}{n}=\infty
$$
<almost sure convergence>[almost surely].
Now $Y_n=S_n-S_{n-1}$, and the <triangle inequality> gives
$$
\frac{|Y_n|}{n}
\leq \frac{|S_n|}{n}
+\frac{n-1}{n}\frac{|S_{n-1}|}{n-1}.
$$
If the <limit superior> of $|S_n|/n$ were finite, the right-hand side would have finite limit superior, contradicting the preceding conclusion. Therefore
$$
\limsup_{n\to\infty}\frac{|Y_1+\cdots+Y_n|}{n}=\infty
$$
almost surely.
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