= Solution
Here
$$
\mathbb P(|X_n|>\varepsilon)=\frac1n\longrightarrow0
$$
for every $0<\varepsilon<1$, so $X_n\to0$ in probability. On the other hand, the events $A_n=\{X_n=1\}$ are independent and
$$
\sum_{n=1}^\infty\mathbb P(A_n)
=\sum_{n=1}^\infty\frac1n=\infty.
$$
The second <Borel-Cantelli lemmas> gives
$$
\mathbb P(X_n=1\text{ infinitely often})=1.
$$
\b[Thus $X_n$ cannot converge almost surely to zero. This <independent rare-event counterexample to almost-sure convergence> proves that convergence in probability need not imply almost-sure convergence of the full sequence.]
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