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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 –  Pedro Tamaroff Mar 10 '12 at 4:59
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up vote 1 down vote accepted

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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Buddy although u made good argument but i am unconvinced that is the reason here. I think the solution to the problem has something to do with order of terms in the series, as it is a well known result that re-ordering the terms in a series can change it's sum. –  Hardy Mar 10 '12 at 3:35
    
@Hardy Didn't you ask about reordering the H series a while ago? This is very similar to that question. –  Pedro Tamaroff Mar 10 '12 at 4:58
    
@Hardy: it looks like you didn't read my answer. What I'm proving you is that your question is exactly about reordering the harmonic series, independently of $z$ (unless $z=0$, in which case your series is not defined). –  Martin Argerami Mar 10 '12 at 5:36
    
@MartinArgerami Both the series quoted in the question here are harmonic and the first converges and the second one does n't, according to your answer the first one should n't either which is not true as it definitively does for non exceptional values of $z$. –  Hardy Mar 10 '12 at 10:04
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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. –  Martin Argerami Mar 10 '12 at 15:52
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