This is Problem 3.6.11 from Rosenthal's 'First look at Rigorous Probability Theory'. I think I did it correctly, but suggestions and comments are very welcome.
For $X_n \sim Uniform\left[\frac1n\right], \ n \in \mathrm N$ find $P(\lim\sup X_n = 5)$.
By definition of $\limsup, \liminf$ and DeMorgan's Laws, $$ P(\limsup_n X_n) = 1-P((\lim\sup_n X_n)^c) = 1 - P(\lim\inf_n X^c_n) $$ where $X^c_n$ is event 'not 5' in $n^{th}$ rv. From the definition of $\liminf$ and subadditivity $$ 1 - P(\liminf_n X^c_n) = 1 - P\left(\bigcup_{n=4}^{\infty} \bigcap_{k=n}^\infty X^c_k\right) \geq 1 - \sum_{n=4}^{\infty}P\left(\bigcap_{k=n}^\infty X^c_k\right) $$ By independence, and definition of countable/discrete Uniform rv, $P(X^c_k) = \frac{k-1}k$ and the fact that the expression only makes sense for $n \geq 5$: $$ P\left(\bigcap_{k=4}^{\infty} X^c_k\right) = \lim_{M \to \infty} \prod_{k=4}^M(1-P(X_k)) = \lim_{M \to \infty}\frac45 \cdot\frac56\cdots\frac{M-1}M = \lim_{M \to \infty}\frac4M= 0 $$ Therefore, every term in the infinite sum converges to $0$, so the sum if $0$ $$ \sum_{n=4}^{\infty}P\left(\bigcap_{k=n}^\infty X^c_k\right) = 0 $$ Putting it all together, $$ P(\limsup_n X_n) = 1-0 = 1 $$ So the event 'get 5' happens infinitely often w.p. $1$.