= Solution
{c}
Let $U$ have the <uniform distribution> on $(0,1)$, set $Y=1/U$, and take $X_n=Y/n$. The variable $Y$ is finite almost surely, so $X_n\to0$ almost surely. For any $\epsilon>0$ and all sufficiently large $n$,
$$
\mathbb P(|X_n|>\epsilon)=\mathbb P(U<1/(n\epsilon))=\frac1{n\epsilon}.
$$
The <harmonic series> diverges, so
$$
\boxed{\sum_n\mathbb P(|X_n|>\epsilon)=\infty.}
$$
This dependent sequence fails <complete convergence>. All its members are scalar multiples of the same nonconstant <random variable>, which explains why the independence hypothesis cannot be dropped.
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