$\{X_n\}^{\infty}_{n=1}$ 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.