We observe that the series $\dfrac {1} {z} -\dfrac {1} {z+1}+\dfrac {1} {z+2}- \dfrac {1} {z+3}+\ldots $ is conditionally convergent, except for certain exceptional values of $z$ ($z\in\mathbb{C}\setminus{{-\infty,\infty}}$ interpreted via ratio test), but the series $\dfrac {1} {z}+\dfrac {1} {z+1}+\ldots +\dfrac {1} {z+p-1}-\dfrac {1} {z+p}-\dfrac {1} {z+p+1}-.\ldots -\dfrac {1} {z+2p+q-1}+\dfrac {1} {z+2p+ q} +\ldots $ in which $(p + q)$ negative terms always follow $p$ positive terms, is divergent.
The second series i think can be rewritten as $\sum _{t=0}^{t=\infty }\left(\sum _{n=t\left( 2p+q\right)}^{n=t\left( 2p+q\right) + \left( p-1\right) }\dfrac {1} {z+n}-\sum _{n=t\left( 2p+q\right) + p}^{n=t\left( 2p+q\right) + \left(p+q-1\right) }\dfrac {1} {z+n}\right)$ but i am not sure how to proceed forward to prove this statement from here. Any help would be much appreciated.