Let $\{X\}_i~i = 1,2,\cdots$ be bounded and independent random variables. Form the sum $S_n = \sum_{k=1}^n X_k$. Let $S = S_{\infty} = \sum_{k=1}^{\infty} X_k$ i.e, $S_n$ converges to $S$ a.s [Ian's comment]
Suppose $S_n$ is Cauchy (in prob.).
Can we conclude that $S$ is bounded a.s?
Not true. Let $\{X_n\}_{n \geq 2}$ be independent random variables such that $P\{X_n=1\}=\frac 1 {n^{2}}$ and $P\{X_n=0\}=1-\frac 1 {n^{2}}$. Then $P\{X_n \neq 0\}=\frac 1 {n^{2}}$ so $\sum _n P\{X_n \neq 0\}< \infty$. By Borel Cantelli Lemma $P\{\lim \sup \{X_n \neq 0\}\}=0$ so the series $\sum X_n$ converges almost surely. This implies $S_n$ converges almost surely, hence in probability so $\{S_n\}$ is Cauchy in probability. However, $S=\sum_n X_n \geq k$ if $X_1=X_2=...=X_k=1$ and this event has positive probability [$(1-\frac 1 {2^{2}})(1-\frac 1 {3^{2}})\cdots (1-\frac 1 {k^{2}})$ to be precise]. It follows that $S$ is not a bounded random variable even though $|X_n|\leq 1$ for all $n$.