Here is the standard proof I got from Paul's math notes
First notice that $\left| {{a_n}} \right|$ is either ${a_n}$ or it is $- {a_n}$ depending on its sign. This means that we can then say,
$$0 \le {a_n} + \left| {{a_n}} \right| \le 2\left| {{a_n}} \right|$$ Now, since we are assuming that $\sum {\left| {{a_n}} \right|}$ is convergent then $\sum {2\left| {{a_n}} \right|}$ is also convergent since we can just factor the 2 out of the series and 2 times a finite value will still be finite. This however allows us to use the Comparison Test to say that $\sum \left({{a_n} + \left| {{a_n}}\right|}\right)$ is also a convergent series.
Finally, we can write,
$$\sum {{a_n}} = \sum \left({{a_n} + \left| {{a_n}} \right|}\right) -\sum {\left| {{a_n}} \right|}$$ and so $\sum {{a_n}}$ is the difference of two convergent series and so is also convergent.
So why are we doing this in a weird roundabout way with the difference of two convergent series?
Why can't we just do:
$${a_n} \le \left| {{a_n}} \right|$$
thus we can use the direct comparison test?
Is it the positive requirement?
Remembering what you have to prove and what you can just say hmmm that seems fine is the worst part of writing proofs :(
The proof I would use is even more complicated.
$\sum |a_i|$ is convergent
The Cauchy criteron says:
$\forall \epsilon>0, \exists N>0$ such that $n,m > N$ $\implies \sum_\limits{i=n}^m |a_i| < \epsilon$
$\sum_\limits{i=n}^m |a_n| \ge \left |\sum_\limits{i=n}^m a_n \right|$ by the triangle inequality.
i.e. $|a_n| + |a_{n+1}|+\cdots + |a_m| \ge |a_n + a_{n+1} + \cdots + a_m|$
$\epsilon > \sum_{i=n}^m |a_n|\ge \left|\sum_{i=n}^m a_n\right| \ge 0$
The Cauchy criterion holds for $\sum a_i$
$\sum a_i$ is convergent.