Prove that if a series $\sum U_n$ is absolutely convergent, then the series is convergent.
Definition: Given a series $\sum U_n$, $\sum U_n$ is absolutely convergent if $\sum|U_n|$ is convergent.
Definition: A sequence $S_n$ is convergent if and only if $S_n$ is a Cauchy sequence.
Definition: $S_n$ is a Cauchy Sequence if for any positive $\epsilon$, there exist an integer $N$ such that for any $p>N$ and $q>N$, $|S_p-S_q|<\epsilon$.
Attempt: $∑U_n$ is absolutely convergent => $∑|U_n|$ is convergent => $∑|U_n|$ is a Cauchy sequence,
Take $T_n=\sum U_n$.
For any $\varepsilon>0$, there exists an integer $N$ such that for any $p>N$, $q>N$, $|T_p-T_q|< \varepsilon$, that is to say, $\left|\sum\limits_{n=p+1}^q |U_n|\right| <\varepsilon$
Since $\left|\sum\limits_{n=p+1}^q U_n\right| \le \left|\sum\limits_{n=p+1}^q |U_n|\right|< \varepsilon$, then take $S_n=\sum\limits_{n=p+1}^q U_n$, we have $S_n$ as a Cauchy sequence. $S_n$ is convergent, the series is convergent
Easier solution (valid only for the real case): $$-|U_n|\le U_n\le|U_n|\implies 0\le U_n+|U_n|\le2|U_n|.$$ By comparison test, $\sum|U_n|$ converges $\implies$ $\sum U_n+|U_n|$ converges. Finally, $\sum U_n = \sum U_n+|U_n| - \sum |U_n|$ is the difference of two convergent series.