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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.

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    [Possible useful question](http://math.stackexchange.com/questions/116089/evaluating-the-sum-after-reordering-an-infinite-series-1-dfrac-1-2-dfrac)2012-03-10

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If we write $z=a+ib$, we have $ \frac1{z+n}=\frac1{a+n+ib}=\frac{a+n-ib}{(a+n)^2+b^2}=\frac{a+n}{(a+n)^2+b^2} -i\frac{b}{(a+n)^2+b^2}. $ So the imaginary part converges absolutely and we can forget about it. The same for the part $a/((a+n)^2+b^2)$, i.e. the convergence/divergence of the series is decided by the terms of the form $ \frac{n}{(a+n)^2+b^2}. $ These terms are asymptotically $1/n$, so basically you have to test your assertion for the harmonic series.

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    My argument doesn't address the $p+q$ thing because I actually didn't spend time thinking about it. What my argument does is to show you that you only need to prove the $p+q$ argument for the sequence with absolute values $\{1/n\}$, because that argument will work for any other $z\ne0$ (it actually does work for $z=0$ but your series is not defined there): the first series converges for every $z$ because $(-1)^n/n$ does; and the second will diverge for every $z$ if the corresponding series $1,1/2,-1/3,-1/4,1/5,\ldots$ (say) diverges.2012-03-10