Proof:
let $s_n$ denote the $n^{th}$ partial sum of the series $\sum{a_n}$ and $t_n$ denote the $n^{th}$ partial sum of the series $\sum{|a_n|}$.
Since, the negative terms are finite
$\implies a_n \geq 0 $ $\forall n>K$
$\implies a_n = |a_n|$ $\forall n>K$
$\implies \forall m>n>K$: $s_m - s_n = t_m - t_n$
Now, since $\sum{a_n}$ is convergent, we can apply the cauchy criterion for series:
Let $\epsilon>0$ and $K \in \mathbb{N}$,
$|s_m - s_n| = |t_m - t_n| = |a_{n+1}+.....+a_m|<\epsilon$ $\forall m>n>K$
Can anyone verify this proof? I'm also unsure if I have used the cauchy criterion properly as it says: $\forall \epsilon>0$ , $ \exists K \in \mathbb{N}$ which I replaced with ''$\forall \epsilon>0$ and $K \in \mathbb{N}$" because I wanted to use the idea that the difference between partial sums of both series is equal after this K.
Thank you.
That looks fine. Using the Cauchy criterion in this case is a bit overkill. List the negative elements as $a_{n_1}, \ldots, a_{n_K}$ and let $M = \sum_{k=1}^K \lvert a_{n_i} \rvert$. Then for $n \geq n_K$ we have $$ t_n = s_n + 2M \leq \sum_{n=0}^{\infty} a_n + 2M < \infty$$ So as a monotonically increasing sequence bounded above, it must converge.