Let $b_n = (|a_n| + a_n)/2$ and $c_n = (|a_n|-a_n)/2.$
Then the partial sums satisfy
$$\sum_{n=1}^m a_n = \sum_{n=1}^m b_n - \sum_{n=1}^m c_n, \\ \sum_{n=1}^m |a_n| = \sum_{n=1}^m b_n + \sum_{n=1}^m c_n.$$
If $\sum a_n$ converges and $\sum |a_n|$ diverges, then both $\sum b_n$ and $\sum c_n$ diverge, since
$$2\sum_{n=1}^m b_n = \sum_{n=1}^m |a_n| + \sum_{n=1}^m a_n, \\ 2\sum_{n=1}^m c_n = \sum_{n=1}^m |a_n| - \sum_{n=1}^m a_n,$$
and the sum or difference of a divergent and convergent series is divergent.
Furthermore, we have divergence to $+\infty$ in each case, as the partial sums of $|a_n|$ form a non-negative, non-decreasing sequence.
Note that
$$\{b_n: n \in \mathbb{N}, b_n \neq 0\} = \{a_n: n \in A^+, a_n \neq 0\}, \\ \{c_n: n \in \mathbb{N}, c_n \neq 0\} = \{-a_n: n \in A^-\}, $$
and it easily shown that
$$ \sum_{n\in A_+} a_n=\sum_{n=1}^\infty b_n = +\infty\\ \sum_{n\in A_-} a_n = -\sum_{n=1}^\infty c_n = - \infty
$$
Best Answer
At first consider series $$\sum\limits_{n = 1}^\infty \frac{|c|^n}{n!}$$ It converges: apply ratio test for $c \ne 0$ (if $c = 0$ the convergence is obvious) $$\frac{|c|^{n+1}}{(n+1)!} \frac{n!}{|c|^n} = \frac{|c|}{n} \rightarrow 0$$
Then your series $$\sum\limits_{n = 1}^\infty \frac{c^n}{n! + n}$$ is absolutely convergent by comparison test: $$\frac{|c|^n}{n!} > \left|\frac{c^n}{n! + n}\right|$$