@DanielFischer I didn't immediately understand what you were talking about yesterday. But then I realized you were simply saying
$$ \sum_{n=0}^{\infty} \text{Res} \left[ \frac{\tanh^{2m} z}{z^{2}}, i \pi (n + 1/2 ) \right] = \sum_{n=0}^{\infty} \text{Res} \left[ \frac{\tanh^{2m} (z+ i \pi n)}{(z+ i \pi n)^{2}}, i \pi /2 \right] = \text{Res} \left[ \tanh^{2m} (z) \sum_{n=0}^{\infty} \frac{1}{(z+ i \pi n)^{2}}, i \pi /2 \right].$$