A Borel-Cantelli problem showing $\sum P(E_k)<\infty$ Let $X_n, n ≥ 1$, be independent (but not necessarily identically distributed) random variables, $S_n =\sum_{k = 1}^n X_k$, and let $a_n$ be real numbers such that $a_n/a_{n+1} \leq C$ for all $n$ and $$P(\lim_{n\to\infty} S_n/a_n = 0) = 1$$
Show that $\sum_{k≥1} P(|X_k| ≥ a_k) < ∞$.
Here are my thoughts: I want to try and show that if instead $P(|X_k| ≥ a_k) = ∞$ then it would imply $P(|X_k| ≥ a_k \text{ i.o.}) = 1$ (since they are independent)
and then somehow use this to show that there is a subsequence for which $\lim S_k/k \neq 0$ this subsequence will be pulled from all indices where $|X_k|\geq a_k$.  But I'm not sure how to fill in the details( I would need to use the fact that $a_n/a_{n+1} \leq C$.  I'm not sure if this is the right way, even though it seems pretty plausible.  Any hints or solutions would be appreciated.  
source: 2011 Qualifying exam https://dornsife.usc.edu/assets/sites/990/docs/Spring_2011/507.pdf
 A: I believe that in addition to your assumptions you also need $a_n > 0$ (and thus implicitly $ C> 0$ since if you allow for negative $a_n$ you can find simple counterexamples to the statement.
The most direct way to approach this problem is not to use the common Borel Cantelli, but instead the converse result. In the setting of this question this result tells us that:
$$
\sum_{k≥1} P(|X_k| ≥ a_k) = ∞ \ \ \Longrightarrow \ \ \limsup_k \frac{|X_k|}{a_k} \ge 1 \ \text{a.s.}
$$
Now we see that a.s. also $S_{n-1}/a_n \to 0.$ Indeed
$$
\frac{|S_{n-1}|}{a_n} \le C \frac{|S_{n-1}|}{a_{n-1}} \to 0 \ \text{a.s}. 
$$
(we can even choose the same nullset for the a.s. convergence) Now if we look outside of a nullset of the above limits holde true. All that is left to do is show that this is impossible. In fact let us consider a subsequence $n_k$ such that for some $\alpha\ge 1$
$$
\lim_k \frac{|X_{n_k}|}{a_{n_k}} = \alpha
$$
Then it follows that
$$
\underbrace{\frac{S_{n_k}}{a_{n_k}}}_{\to 0} = \underbrace{\frac{S_{n_k-1}}{a_{n_k}}}_{\to 0} + \underbrace{\frac{X_{n_k}}{a_{n_k}}}_{\to \pm \alpha}
$$
which is impossible.
