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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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    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.2012-03-10
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    @Hardy Didn't you ask about reordering the H series a while ago? This is very similar to that question.2012-03-10
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    @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).2012-03-10
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    @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 $z$.2012-03-10
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    @PeterT.off, Yes I did ask a such a question I 'll go back and recheck and reanalyze.2012-03-10
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    @Hardy: you haven't read my answer. My argument easily shows that the first series converges (because it shows that one can apply Leibnitz criterion for the alternate series). It can also be used to show that the second series diverges.2012-03-10
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    @MartinArgerami, Firstly stop repeating that i have n't read your answer cause that is not true, i have read tried to understand your argument, just that your argument totally looks past the $(p+q)$ negative terms followed by p positive terms. Your argument states and i quote "basically you have to test your assertion for the harmonic series." All i understand from that line is that your series behaves like the harmonic series, which is well known to be divergent, hence u claim it is divergent. Cont2012-03-10
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    Cont: Based on identical reasoning, had u not brought up Leibnitz crterion, u could very easily argue the first series which is also rearranged harmonic series is also divergent, hence i think it may be possible to formulate a better argument.2012-03-10
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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