How would you show that the following series is absolutely convergent $$ T\sum _{n=-\infty} ^\infty g(2\pi iTn) $$ where $$ g(z) = {1 \over z^2-a^2}, \space \space \space a>0,$$ and $2 \pi i Tn$ is simple pole of another function, h(z).
For a series of the form $$ \sum _{n=0} ^{\infty} f(z),$$ I would simply apply the ratio test, however I'm unsure how the fact that the summation is from $ -\infty$ to $ \infty$ changes things. Could you proceed as normal by splitting the summation into negative and positive parts and noting that the argument of g(z) is squared?
$$ \sum _{n=-\infty} ^\infty g(2\pi iTn) =\sum _{n=0} ^\infty g(2\pi iTn) + \sum _{n=1} ^\infty g(2\pi iT(-n)) $$
By doing this I get that $$ lim_{n \to\infty} \left \vert {{g(2\pi iT(n+1))} \over g(2\pi iT(n)) }\right \vert =1 $$ (and the same for the other summation).
Which means the result of the test is indetermined.