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There is one series and it seems pretty much easy to check either it is divergent or convergent but because of the complex denominator I am not able to get the solution by the certain convergent tests. Here is the question;

$\sum _{k=0}^{\infty} \left [ (-1)^{k}(2k)!\left(\frac {1} {(i+a)^{2k+1}}-\frac {1} {(-i+a)^{2k+1}}\right) \right ]$

Does anyone have an idea about this either it is convergent or divergent? a could be any number, for large k it seems it diverges but need a way to prove it? By comperition we could get rid of from $(2n)!$ but I do not know how to compare these complex donominators.

Thank you...

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    Defacing your questions is quite frowned upon; please don't do this.2013-03-27

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This might help:

$\frac {1} {(i+a)^{2k+1}}-\frac {1} {(-i+a)^{2k+1}}$ $= \frac {(a-i)^{2k+1}} {(a^2-i^2)^{2k+1}} - \frac {(a+i)^{2k+1}} {(a^2-i^2)^{2k+1}}$ $= \frac {(a-i)^{2k+1}-(a+i)^{2k+1}} {(a^2+1)^{2k+1}} $

which, for real $a$, looks imaginary to me. It also looks as if it tends to shrink more slowly than $(2k)!$ increases, but you might want to check that.

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    If $a$ is real then the difference is of the form $1/w - 1/\overline{w} = \frac{\overline{w}-w}{|w|^2}$ which is obviously purely imaginary.2012-04-09