Applying the Borel-Cantelli lemma I'm working on the following two exercises that require the Borel-Cantelli lemma

*

*If the $X_n$ are identically distributed and $E|X_1|<\infty$, then $P(\lim_{n\to\infty}\frac{X_n}{n}=0)=1$.

*If the $X_n$ are i.i.d.  and $E|X_1|=\infty$, then $P(\limsup_{n\to\infty}\frac{|X_n|}{n}=\infty)=1$.

Following the basic framework of the lemma, in both cases I have to find appropriate sets $A_n$ that "appear infinitely often" and show in the first case that $\sum_nP(A_n)<\infty$ (from which follows $P(\limsup_nA_n)=0$) and in the second that $\sum_nP(A_n)=\infty$ (from which follows $P(\limsup_nA_n)=1$; here the $A_n$ should be independent). However, I'm having trouble finding the appropriate $A_n$.
In the first case the goal is $P(\lim_{n\to\infty}\frac{X_n}{n}=0)=1$, so I tried representing $\{\lim_{n\to\infty}\frac{X_n}{n}\neq0\}$ by the $\limsup$ of $A_n=\{|\frac{X_m}{m}|>\frac{1}{n}\text{ for infinitely many }m\}$, but I couldn't sum the probabilities. In the second case I tried $A_n=\{|\frac{X_n}{n}|>n\}$, but didn't get far either. Is there a general rule of thumb for finding the "right" $A_n$ to consider?
 A: EDIT 2: This answer is not correct. Please see my edit below and the comment from William.
I don't know if BC is relevant in the first case since we're interested in the limit, not the limsup.
For the first part it is sufficient to show that $P(\lim X_n/n > \epsilon)$ goes to $0$.
Apply a markov bound to the LHS:
$$P(\lim X_n/n > \epsilon) \leq \frac{\mathbb E[\lim X_n]/n}{\epsilon}$$
Then, since $E|X_i|$ is finite for all $i$, dominated convergence gives us (since $|X_i|$ dominates $X_i$ and |X_i| is integrable):
$$=\frac{\lim\mathbb E[ X_n]/n}{\epsilon}$$
which converges to $0$. Note that we have not used independence.
For the second part, we will use the second BC lemma (which requires independence).
It is sufficient to show that for any $M>0$: $P(\limsup |X_n|/n >M) = 1$. For a specific value of $n$:
$$P(|X_n|/n > M) = 1- P(|X_n|/n \leq M) \geq 1- \delta$$
for some constant $\delta \in (0,1)$. So that summing the terms above:
$$\sum_n P(|X_n|/n > M) \geq \sum_n 1- \delta$$
which diverges, so the second BC lemma applies (since the events are independent) and we have the desired conclusion.
EDIT: there is an error in the second part. Notice that $\delta$ actually depends on $n$.
A: Nice exercise!
We first show that $\mathbf{P}\left( \lim \dfrac{X_{n^2}}{n^2} = 0 \right) = 1,$ or equivalently, $\mathbf{P}\left( \limsup \dfrac{|X_{n^2}|}{n^2} > 0 \right) = 0.$ Notice that the event inside the last probability is the union over $j \in \mathbf{N}$ of the events where $\limsup \dfrac{|X_{n^2}|}{n^2} > \dfrac{1}{j}.$ Suffices to show that these events have probability zero (hence, their countable union will too). But this is precisely where Borel-Cantelli takes place for
$$
\left\{ \limsup \dfrac{|X_{n^2}|}{n^2} > \dfrac{1}{j} \right\} = \left\{ \dfrac{|X_{n^2}|}{n^2} > \dfrac{1}{j} \text{ i.o.} \right\}.
$$
Markov's inequality shows that the last event has probability $\leq \dfrac{\mu j}{n^2},$ where $\mu$ is common expectation. The $n^2$ divisor allows deducing the series associated to this sequence converges and Borel-Cantelli then concludes that the required probability is zero (as desired). Therefore, we have established that
$$
\mathbf{P}\left( \lim \dfrac{X_{n^2}}{n^2} = 0 \right) = 1.
$$
We now interpolate and show that a.s. $\dfrac{X_{k}}{k} - \dfrac{X_{n_k^2}}{n_k^2} \to 0$ where $n_k$ is the only integer such that $(n-1)^2 \leq k < n^2.$ For simplicity, write $n = n_k.$ Then,
$$
\dfrac{X_k}{k} - \dfrac{X_{n^2}}{n^2} = \dfrac{n^2 X_k - k X_{n^2}}{k n^2} = \dfrac{(n^2-k) X_k}{kn^2} + \dfrac{X_k - X_{n_2}}{n^2}.
$$
As before, Borel-Cantelli shows that each summand converges to zero a.s. (use that $n^2 - k \asymp n$ while $kn^2 \asymp n^4$).
As to how to solve your second item, we use the following
$$
\left\{ \limsup \dfrac{|X_n|}{n} = \infty \right\} = \bigcap_{k \in \mathbf{N}} \left\{ \limsup \dfrac{|X_n|}{n} > k \right\} = \bigcap_{k \in \mathbf{N}} \left\{ \dfrac{|X_n|}{n} > k \text{ i.o.} \right\}.
$$
Suffices to show each event in the intersection has probability one. By Borel-Cantelli, suffices to show that $\sum \mathbf{P}(|X_n| > nk) = \infty.$ Since the $(X_n)$ have the same distribution, so do the $Y_n = \frac{1}{k} |X_n|,$ and the $Y_n$ also have infinite expectation, we reach $\sum \mathbf{P}(Y_1 > n).$ We finish by proving a classical result, given a positive random variable $Z,$ $\mathbf{E}(Z)$ is finite if and only if $\sum \mathbf{P}(Z > n)$ converges: this follows from exchanging the two integrals
$$
\sum \mathbf{P}(Z > n) = \int\limits_{z > n} d(F \otimes \mu)(z,n) = \int [z] dF(z),
$$
where $[z]$ denotes the integer least or equal than $z,$ so that $z - 1 \leq [z] \leq z$ and $\mu$ is counting measure on the natural numebrs. Therefore the equivalence follows (since $\int 1 dF = 1$ is integrable). QED
