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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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If we had either of those fractions multiplied by $(-1)^k(2k)!$, it would be easy to show by the ratio test that it diverges. With both of them together it's subtler. –  Michael Hardy Apr 9 '12 at 17:05
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Defacing your questions is quite frowned upon; please don't do this. –  Arthur Fischer Mar 27 '13 at 10:34

1 Answer 1

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. –  user20266 Apr 9 '12 at 17:12

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