Convergence in probability of sequence of independent random variables implies almost sure convergence of their sum?

$\{X_n\}^{\infty}_{n=1}$ are independent random variables. Let $S_n = X_1 + \cdots + X_n$ be the random walk. Show that $\{S_n\}$ converges almost surely if and only if the random sequence converges in probability.

My approach (Only if part): Suppose $X_n \geq 0$. $|X_m - X_n| \leq |X_m|+|X_n| \leq |X_m + \cdots + X_n|$ for all $m,n \geq 1$.

$\therefore \sup_{m,n \geq N} |X_m - X_n| \leq \sup_{m,n \geq N} |X_m + \cdots + X_n|$ for all $N$.

$\therefore \mathbb{P}(\sup_{m,n \geq N} |X_m - X_n| > \epsilon) \leq \mathbb{P}(\sup_{m,n \geq N} |X_m + \cdots + X_n| > \epsilon)$ for all $N$.

Limit of RHS (over $N$) is $0$ (Cauchy criterion of convergence of series). Hence so is the LHS.

Now for general $\{X_n\}^{\infty}_{n=1}$, suppose $\sum X_n$ converges a.s. Therefore $\sum X^+_n$ and $\sum X^-_n$ also converge a.s. (Is this true?). By first part this means $X^+_n$ and $X^-_n$ converge in probability, to $X^+$ and $X^-$, say. Hence $(X^+_n - X^-_n)\xrightarrow[]{\mathbb{P}}(X^+ - X^-)$. (Algebraic property of probability convergence.)

However I'm not able to get the if part. Also, suppose $X_n$'s are non-zero iid. They converge in probability but $\sum X_n$ does not converge. So I'm not even sure I understand the if part of the question correctly.

• Are you assuming the $X_n$ are i.i.d. or not? – Math1000 May 2 '18 at 14:12
• @Math1000, nice joke. – zhoraster May 2 '18 at 15:21
• I think the question is supposed to be: show that $S_n$ converges almost surely if and only if $S_n$ converges in probability. As you say at the end, it's clearly not true that $S_n$ converges a.s. iff $X_n$ converges i.p. (Though "iid" is not a very good example, since in fact an iid sequence does not converge in probability, unless they are all constant. But having all the $X_n$ equal, say, 1, is a counterexample to "$S_n$ converges a.s. iff $X_n$ converges i.p.". – Nate Eldredge May 2 '18 at 16:11
• I agree with Nate. Also in general, if $S_n$ converges (I mean to a finite limit, do you use converge also for an infinite limit?), $X_n$ goes to 0. – Kolmo May 2 '18 at 21:19