The first thing I need to do is to write the function $$f(n) = \begin{cases} 1, & \mbox{if } n \ne 0 \mbox{ mod 3} \\ -2, & \mbox{if } n\equiv 0 \mbox{ mod 3} \end{cases}$$ in the form of a Dirichlet series. My attempt produced the following
$$\sum_{n=1}^\infty f(n)n^{-s} = \sum_{n=1, 3 \nmid n}^\infty n^{-s} + (-2)\sum_{n=1, 3 \mid n}^\infty n^{-s}$$ ( not really sure if this is correct writing).
From here I understood there are some steps that need to be taken to prove the abscissa of convergence, but not quite got the actual ones with a given example. I read that it has some similarities with the radius of convergence from the alternate series for which you can use $$\frac{1}{R} = \limsup_{n \rightarrow \infty} |a_n|^{ \frac{1}{n}}$$ ( where $a_n$ I guess is the equivalent of f(n)). I would appreciate any correction or example on how to approach this exercise.
Your series is convergent for $\Re s>0$. This is a consequence of the common property of Dirichlet series (if it converges at $s=s_0$, then it does for any $s$ with $\Re s>\Re s_0$) and the fact that it converges if $s>0$ is real. The latter follows from $(3n-r)^{-s}-(3n)^{-s}\in\mathcal{O}(n^{-s-1})$ for $r\in\{1,2\}$ (and the comparison test).