Suppose $X_i$ are iid with $X_i \notin L^1$.
I would like to show that $P(|X_n| \geq n)$ i.0 $=1$.
I have shown the more general $P(|X_n| \geq nk \text{ i.o}) = P(|X_n|/n \geq k)$. $$\begin{aligned} \infty & = E[|X_1|]\\ & = \int_0^{\infty} P (|X_1| \geq t) dt\\ & = \sum_{n=0}^{\infty} \int_{nk}^{(n+1)k}P (|X_1| \geq t) dt\\ & \leq \sum_{n=0}^{\infty} \int_{nk}^{(n+1)k}P (|X_1| \geq nk) dt\\ & = \int_0^k P(|X_1| > 0) dt + \int_k^{2k} P(|X_1| \geq k) dt + ...\\ & \leq k + \sum_{n=1}^{\infty}kP(|X_1| \geq nk)\\ & = k + k \sum_{k=1}^{\infty}P(|X_n| \geq nk) \text{ (since the $X_i$ are identically distributed)}\\ \implies \infty & =\sum_{k=1}^{\infty}P(|X_n| \geq nk) \end{aligned}$$
So by BC2, $P(|X_n| \geq nk \text{ i.o}) =1 = P(|X_n|/n \geq k)$.
I think that if I can show $\sum_n P(|X_n|/n) = \infty$, then Borel-Cantelli will give the result, but this doesn't seem too easy to do. I think this is meant to be an easy question, any help would be much appreciated. Thanks!
Let me not exclude that there is a version that is neater than the one in your question.
But if there is one for $k=1$ then it can also be used in the more general case.
This because $X_i\notin L^1$ implies immediately that $X_i/k\notin L^1$.
Then with the neater version you can prove that $P(|X_n/k|\geq n\text { i.o})=1$ and conclude immediately that $P(|X_n|\geq nk\text { i.o})=1$.
So $k=1$ is not really a separate case here.