= Solution
For events $A_n$, their limsup is the event that infinitely many occur. The first <Borel-Cantelli lemma> states that $\sum_n\mathbb P(A_n)<\infty$ implies $\mathbb P(\limsup A_n)=0$, without independence. Indeed
$$
\mathbb P\!\left(\bigcup_{n\ge N}A_n\right)\le\sum_{n\ge N}\mathbb P(A_n)\longrightarrow0,
$$
and intersecting these decreasing tail-unions proves the claim.
The second <Borel-Cantelli lemma> states that independent events with $\sum_n\mathbb P(A_n)=\infty$ satisfy $\mathbb P(\limsup A_n)=1$. For each fixed $N$,
$$
\mathbb P\!\left(\bigcap_{n=N}^M A_n^c\right)=\prod_{n=N}^M(1-\mathbb P(A_n))\le\exp\!\left(-\sum_{n=N}^M\mathbb P(A_n)\right)\longrightarrow0.
$$
Thus the probability that no event occurs after $N$ is zero; take the countable union over $N$ to exclude finitely many occurrences.
\b[As printed, the requested real logarithm is undefined:] independent <Cauchy random variables> are negative with probability $1/2$, and the second <Borel-Cantelli lemma> shows that this happens infinitely often almost surely. There is therefore no real-valued sequence $\log X_n/\log n$ as stated. The natural corrected expression uses $\log|X_n|$ (or takes the limsup only over indices with $X_n>0$).
For the absolute-value interpretation,
$$
\mathbb P(|X_n|>t)=1-\frac2\pi\arctan t\sim\frac2{\pi t}\quad(t\to\infty).
$$
For every $q>1$ the events $|X_n|>n^q$ are summable and hence occur only finitely often; for $0<q\le1$ they are independent with divergent probability sum and occur infinitely often. Applying the two <Borel-Cantelli lemmas> to a countable sequence of $q$ approaching one from above and below gives
$$
\boxed{\limsup_{n\to\infty}\frac{\log|X_n|}{\log n}=1\quad\text{almost surely}.}
$$
The one-sided tail $\mathbb P(X_n>t)\sim1/(\pi t)$ gives the same constant $1$ for the positive-subsequence interpretation. It does not repair the undefined logarithms at the remaining indices unless that interpretation is explicitly adopted.
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