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.
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