Let $X_1,X_2,...$ be independent r.v.'s (not necessary to assume identically distributed), and $S_n = \sum^{n}_{i=1}X_i$ for any $n \ge 1$. Also,
$$P(\sup_{n\ge1}|S_n|>3\epsilon) \le 3\sup_{n\ge1}P(|S_n|>\epsilon)$$ is satisfied.
$\textbf{Prove that the series $\sum^{\infty}_{i=1}X_i$ converges in probability $\iff$ converges in almost surely.}$ (Hint: using the inequality above for $\Longrightarrow$)
To tackle this problem, I firstly to check the above is satisfied, I search some reference and online, I found the inequality is a little bit transformed by $\textbf{Etemadi's inequality}$ (which taking the limit in both side and using continuity). Some useful references can be found in https://en.wikipedia.org/wiki/Etemadi%27s_inequalityhte and PSQ Maximal inequality for a sequence of partial sums of independent random variables.
It is easy to check $\Longleftarrow$.
Unfortunately, for checking $\Longrightarrow$, I am struggle in how to use the inequality to prove the problem. As what I have learned, I only know a.s. convergence implies convergence in P. In addition, if the sequence $Y_n$ converges in P, then there must exists a subsequence ${Y_{n_k}}$ converges a.s. But the problem mentions that converges in P implies converge a.s. I am a little bit shocking.
My try:
Suppose we $\sum^{\infty}_{i=1}X_i$ converges in probability, then I took limit in BHS for the inequality, that is .
$$\lim_{n \rightarrow \infty}P(\sup_{n\ge1}|S_n|>3\epsilon) \le \lim_{n \rightarrow \infty}3\sup_{n\ge1}P(|S_n|>\epsilon)$$
$$\lim_{n \rightarrow \infty}P(\sup_{n\ge1}|S_n|>3\epsilon) \le 3\limsup\limits_{n \rightarrow \infty}P(|S_n|>\epsilon)$$
But how could I apply the condition $\sum^{\infty}_{i=1}X_i$ converges in probability. Do I need some additional steps to finish the proof? Could you give me some details.
Although I searched much information and got lots of helps from stackexchange mathematics perviously, but this is my first to post question here, I am very happy to join this amazing community :)
Thank U.