Calculate $\sum_{n=1}^{\infty} \arctan\bigl(\frac{2\sqrt2}{n^2+1}\bigr) $ $$ \lim_{n \to\infty} \sum_{k=1}^{n} \arctan\frac{2\sqrt2}{k^2+1}= \lim_{n \to\infty} \sum_{k=1}^{n} \arctan\frac{(\sqrt{k^2+2}+\sqrt2)-\sqrt{k^2+2}-\sqrt2)}{(\sqrt{k^2+2}+\sqrt2)(\sqrt{k^2+2}-\sqrt2)+1}= $$ 
$$\lim_{n \to\infty} \sum_{k=1}^{n} \arctan(\sqrt{k^2+2}+\sqrt2)-\arctan(\sqrt{k^2+2}-\sqrt2)  $$
But the elements do not reduce :( 
 A: Note that
$$\arctan\left(\frac{2\sqrt{2}}{k^2+1}\right)=\arctan\left(\frac{\left(\frac{k+1}{\sqrt{2}}\right)-\left(\frac{k-1}{\sqrt{2}}\right)}{\left(\frac{k+1}{\sqrt{2}}\right)\left(\frac{k-1}{\sqrt{2}}\right)+1}\right)=\arctan\left(\frac{k+1}{\sqrt{2}}\right)-\arctan\left(\frac{k-1}{\sqrt{2}}\right)\,.$$
Therefore,
$$\sum_{k=1}^n\,\arctan\left(\frac{2\sqrt{2}}{k^2+1}\right)=\arctan\left(\frac{n+1}{\sqrt{2}}\right)+\arctan\left(\frac{n}{\sqrt{2}}\right)-\arctan\left(\frac{1}{\sqrt{2}}\right)-\arctan\left(\frac{0}{\sqrt{2}}\right)\,.$$
Ergo,
$$\sum_{k=1}^\infty\,\arctan\left(\frac{2\sqrt{2}}{k^2+1}\right)=\pi-\arctan\left(\frac{1}{\sqrt{2}}\right)\approx 2.52611\,.$$
A: If $S_n = \sum_{k=1}^n \arctan(2 \sqrt{2}/(k^2+1))$, it appears that
$$ \tan(S_n) = - \frac{n(n+5)}{\sqrt{2}(n^2-n-1)}$$
You can prove this by induction.  The rest is just a matter of estimating the sum of the series to
check which interval $[(k-1/2)\pi, (k+1/2)\pi]$ it falls in.
A: Result
$$\pi -\frac{1}{2} \arctan \left(2 \sqrt{2}\right)\simeq 2.52611... $$
Derivation
More complicated than the elegant telescoping approach, but that was the way I found the result.
Observing that
$$\text{arctan}(t) = t \int_{0}^1 \frac{1}{1+t^2 x^2}\,dx$$
and interchanging sum and integral the sum can be written as
$$s =2 \sqrt{2} \int_{0}^1 \sum_{n=1}^\infty \frac{ \left(n^2+1\right)}{\left(n^2+1\right)^2+8 x^2}\,dx$$
The sum can be done:
$$\sum _{n=1}^{\infty } \frac{n^2+1}{\left(n^2+1\right)^2+8 x^2}=\frac{1}{2 \left(x^2+1\right)}\text{Re}\left(\pi  \sqrt{-1+i x} (1+i x) \cot \left(\pi  \sqrt{-1+i x}\right)-2\right)$$
and, suprisingly, also the final $x$-integral can be done with the result provided.
Corollary 1
With the same method we can easily find
$$s_{c}=\sum _{n=1}^{\infty } \arctan\left(\frac{1}{n^2+1}\right)\simeq 1.03729...\\= \frac{1}{8} \left(3 \pi -4 i \log \left(\sin \left(\sqrt{-1-i} \pi \right)\right)+4 i \log \left(\sin \left(\sqrt{-1+i} \pi \right)\right)\right)\\=\frac{3 \pi }{8}-\arg \left(\sin\pi \left(\sqrt{-1+i} \right)\right)\tag{c1}$$
This can be simplified to the real expression
$$s_{c}=\frac{3 \pi }{8}-\arctan \left(\frac{\cos \left(\sqrt[4]{2} \pi  \sin \left(\frac{\pi }{8}\right)\right) \sinh \left(\sqrt[4]{2} \pi  \cos \left(\frac{\pi }{8}\right)\right)}{\sin \left(\sqrt[4]{2} \pi  \sin \left(\frac{\pi }{8}\right)\right) \cosh \left(\sqrt[4]{2} \pi  \cos \left(\frac{\pi }{8}\right)\right)}\right)\tag{c2}$$
or
$$s_{c}=\frac{3 \pi }{8}-\arctan\left(\frac{\tanh \left(\pi  \sqrt{\frac{1}{2} \left(\sqrt{2}+1\right)}\right)}{\tan \left(\pi  \sqrt{\frac{1}{2} \left(\sqrt{2}-1\right)}\right)}\right)\tag{c3}$$
This is a rather trig-loaded expression.
I could not find a telescoping sum in this case. The factor $2^{\frac{3}{2}}$ in the OP was accurately tuned.
Corollary 2
Also
$$s_{c2} = \sum _{n=1}^{\infty } \arctan \left(\frac{1}{n^2}\right)\\=
\arctan \left(\frac{1-\cot \left(\frac{\pi }{\sqrt{2}}\right) \tanh \left(\frac{\pi }{\sqrt{2}}\right)}{1+\cot \left(\frac{\pi }{\sqrt{2}}\right) \tanh \left(\frac{\pi }{\sqrt{2}}\right)}\right)\\\simeq 1.42474...$$
