$ z \in \Bbb{C}: \sum_{n=1}^{\infty} \left(\frac{1}{(3n-2)^z} + \frac{1}{(3n-1)^z} - \frac{1}{(3n)^z}\right) $ is convergent? Is the series convergent ?
$$ \sum_{n=1}^{\infty} \left(\frac{1}{(3n-2)^z} + \frac{1}{(3n-1)^z} - \frac{1}{(3n)^z}\right) $$
$$ z \in \Bbb{C} $$

I had improved my knowledge a bit and I hope that this is a correct answer.

From Dirichlet Series theorem it is known that, if:
$$ f(z) = \sum_{n=1}^{\infty} \frac{a_n}{n^z} $$
converges for:
$$ \Re(z) = c $$
then it converges for:
$$ z = c + xi $$
$$ x \in \Bbb{R} $$
From this point we only have to check convergence for $\Re(z)$.

We known that:
$$ \frac{1}{(3n-2)^z} + \frac{1}{(3n-1)^z} - \frac{1}{(3n)^z} \ge \frac{1}{(3n)^z} + \frac{1}{(3n)^z} - \frac{1}{(3n)^z}$$
$$ \frac{1}{(3n-2)^z} + \frac{1}{(3n-1)^z} - \frac{1}{(3n)^z} \ge \frac{1}{(3n)^z}$$
Hence:
$$ \frac{1}{3^z} \sum_{n=1}^{\infty} \frac{1}{n^z} = \frac{1}{3^z} \zeta(z) $$
$$ \Re(z) > 1 $$
Finally the series $ \sum_{n=1}^{\infty} \left(\frac{1}{(3n-2)^z} + \frac{1}{(3n-1)^z} - \frac{1}{(3n)^z}\right) $ is divergent for $\Re(z) < 1$. The same method can be made to set some upperbound.
