I'm trying to evaluate $$\sum_{n=1}^\infty \arctan \left(\frac{2a^2}{n^2}\right) \ , \ a >0. $$
The answer I get only seems to be correct for small values of $a$.
What accounts for this?
Using the principal branch of the logarithm, I get
$$ \begin{align} & \sum_{n=1}^\infty \arctan \left(\frac{2a^2}{n^2}\right) \\ & = \text{Im} \sum_{n=1}^\infty \log \left( 1 + \frac{2ia^2}{n^2} \right) \\ & = \text{Im} \log \prod_{n=1}^\infty \left(1 + \frac{2ia^2}{n^2} \right) \\ & = \text{Im} \log \prod_{n=1}^\infty \left(1 - \frac{(\sqrt{-2i}a)^2}{n^2} \right) \\ & =\text{Im} \log \left(\frac{\sin (\pi \sqrt{-2i}a)}{\pi \sqrt{-2i}a} \right) \\ & = \text{Im} \log \left(\frac{\sin \left(\pi (1-i)a\right)}{\pi (1-i)a} \right) \\ & = \text{Im} \log \left(\frac{\sin (\pi a) \cos (i\pi a) - \cos (\pi a) \sin (i \pi a)}{\pi (1-i)a} \right) \\ & = \text{Im} \log \left(\frac{\sin (\pi a) \cosh (\pi a) - i\cos (\pi a) \sinh (\pi a)}{\pi (1-i)a} \right) \\ & = \text{Im} \log \left(\frac{\sin (\pi a) \cosh (\pi a) + \cos(\pi a) \sinh (\pi a) + i \left( \sin (\pi a) \cosh (\pi a) - \cos (\pi a) \sinh ( \pi a) \right)}{2\pi a} \right) \\ & = \text{Arg} \ \Big(\sin (\pi a) \cosh (\pi a) + \cos(\pi a) \sinh (\pi a)+ i \left( \sin (\pi a) \cosh (\pi a) - \cos (\pi a) \sinh ( \pi a) \right) \Big) \end{align} $$