Question: Let us have a series $\sum\_{n = 1}^{\infty} \frac{(-1)^n}{z + n}$, where $z \in \mathbb{C}$. I intended to use the Dirichlet test, where $\sum (-1)^n$ is a bounded complex series, and $\frac{1}{z + n}$ is actually going to zero. This would ensure the convergence of the series for all $z \in \mathbb{C} \cap z \notin \mathbb{Z}^+$. 

The problem is, the Dirichlet test requires the condition that $\frac{1}{z + n}$ is decreasing! However, since the term $z$ is complex, this assumption is violated. Thus, the test cannot be applied. 

How can this be solved in the most traditional way? Is it possible to apply Dirichlet test somehow here?

Answer: 
Note that

$${(-1)^n\over z+n}={(-1)^n(\overline z+n)\over(z+n)(\overline z+n)}={(-1)^n\overline z\over|z+n|^2}+{(-1)^nn\over|z+n|^2}$$

The sum $\sum{(-1)^n\overline z\over|z+n|^2}$ is absolutely convergent, while for each $z$, the sequence $n/|z+n|^2$ is eventually decreasing, so the alternating series $\sum{(-1)^nn\over|z+n|^2}$ converges conditionally.
The simplest way of seeing this is to combine two terms.

Note that $\frac{1}{z + 2n} - \frac{1}{z + 2n + 1} = \frac{1}{(z + 2n)(z + 2n + 1)}$.

Now you can use anything you know to show that the sum $\sum\_{n = 1}^\infty \frac{1}{(z + 2n)(z + 2n + 1)}$ converges (e.g. compare the absolute value with $\frac{c}{n^2}$ for some constant $c$). This, combined with the fact that $\frac{1}{z + n}$ tends to $0$, shows the convergence of the original series.