CA - Nice sum

The sum we have is: $$\sum_{n=1}^\infty \arctan\left({\frac 1 {8n^2}}\right)$$ This turns out to have the neat closed form of: $$\frac \pi 4 - \arctan \tanh \frac \pi 4$$ Remember that for a complex number $z$ with positive real and imaginary parts, the (principal) argument $\arg z = \arctan b/a$. With the principal argument we have the principal log: $$\log z = \log (|z|e^{i \arg z}) = \log |z| + i \arg z$$ We can then see for $x > 0$: $$\log (1 + ix) = \log\sqrt{1+x^2} + i \arctan \left(x\right)$$ So we have that $\operatorname{Im}\left\{\log(1+ix)\right\} = \arctan x$. So $\arctan \left({\frac 1 {8n^2}}\right) = \operatorname{Im} \left\{ \log \left({1 + \frac i {8n^2}}\right) \right\}$, so our sum becomes: $$\sum_{n=1}^\infty \operatorname{Im} \left\{ \log \left({1 + \frac i {8n^2}}\right) \right\}$$ Applying the addition formula for logs: $$\sum_{n=1}^\infty \operatorname{Im} \left\{ \log \left({1 + \frac i {8n^2}}\right) \right\} = \operatorname{Im} \left\{ \log \prod_{n=1}^\infty \left({1 + \frac i {8n^2}}\right) \right\}$$ We know that: $$\frac{\sin \pi \theta} {\pi \theta} = \prod_{n=1}^\infty \left({1 - \frac {\theta^2}{n^2}}\right)$$ So we want $-\theta^2 = \frac i 8$. Then $i\theta = \frac {\sqrt i} {2 \sqrt 2} = \frac{1 + i} {2 \sqrt 2 \sqrt 2} = \frac {1 + i} 4$. So in this case, dividing by $i$, $\theta = \frac {1 - i} 4$. So: $$\prod_{n=1}^\infty \left({1 + \frac i {8n^2}}\right) = \prod_{n=1}^\infty \left({1 - \frac{\left({\frac {1 - i} 4}\right)^2} {n^2}}\right) = \frac{\sin \left({\frac \pi 4 - \frac \pi 4 i}\right)} {\frac \pi 4 - \frac \pi 4 i}$$ Using $\sin(x+iy) = \sin x \cosh y + i \cos x \sinh y$: $$\sin \left({\frac \pi 4 - \frac \pi 4 i}\right) = \sin \frac \pi 4 \cosh \left({-\frac \pi 4}\right) + i \cos \frac \pi 4 \sinh \left({-\frac \pi 4}\right) = \frac {\sqrt 2} 2 \cosh \frac \pi 4 - i \frac{\sqrt 2} 2 \sinh \frac \pi 4$$ So now we have: $$\prod_{n=1}^\infty \left({1 + \frac i {8n^2}}\right) = \frac{\frac {\sqrt 2} 2 \left({ \cosh \frac \pi 4 - i \sinh \frac \pi 4 }\right)} {\frac \pi 4 - \frac \pi 4 i}$$ We just want to get out the real and imaginary parts, by multiplying through the conjugate of the denominator: $$\frac{\frac {\pi \sqrt 2} 8 \left({ \cosh \frac \pi 4 - i \sinh \frac \pi 4 }\right)(1+i)} {\left({\frac \pi 4 - \frac \pi 4 i}\right)\left({\frac \pi 4 + \frac \pi 4 i}\right)} = \frac {\frac {\pi \sqrt 2} 8} {\frac {\pi^2} 8} \left({\cosh \frac \pi 4 + i\cosh \frac \pi 4 - i \sinh \frac \pi 4 + \sinh \frac \pi 4}\right)$$ Though you can read off the argument from here, we can simplify it further for completeness to: $$\prod_{n=1}^\infty \left({1 + \frac i {8n^2}}\right) = \frac {\sqrt 2} \pi \left\{\left({\cosh \frac \pi 4 + \sinh \frac \pi 4}\right) + i\left({\cosh \frac \pi 4 - \sinh \frac \pi 4}\right)\right\}$$ So: $$\sum_{n=1}^\infty \arctan\left({\frac 1 {8n^2}}\right) = \operatorname{Im} \left\{\log \prod_{n=1}^\infty \left({1 + \frac i {8n^2}}\right)\right\} = \arctan \left({\frac{\cosh \frac \pi 4 - \sinh \frac \pi 4} {\cosh \frac \pi 4 + \sinh \frac \pi 4}}\right)$$ We're almost there, just got to do some simplification to get the form at the top of the post. First divide through $\cosh \frac \pi 4$ to get: $$\frac{\cosh \frac \pi 4 - \sinh \frac \pi 4} {\cosh \frac \pi 4 + \sinh \frac \pi 4} = \frac{\tan \frac \pi 4 - \tanh \frac \pi 4} {1 + \tan \frac \pi 4 \tanh \frac \pi 4}$$ Considering the addition formula for arctan, the arctan of this is equal to $$\arctan \tan \frac \pi 4 - \arctan \tanh \frac \pi 4 = \boxed{\frac \pi 4 - \arctan \tanh \frac \pi 4}$$ which is what we wanted to prove.