<s>[INST] Arccos telescopic sum: $\sum_{k=1}^{n} \arccos \frac{2k^2}{4k^4+1}$ Compute the finite sum:
$\arccos \frac{2 \cdot 1^2}{4\cdot 1^4+1}+...+\arccos \frac{2 \cdot n^2}{4\cdot n^4+1}$
Please help. I have no idea what to do. There should be a telescopation but I can’t find it.
 [/INST] This is not a complete answer, just a try. By partial fraction decomposition, you get
$$z_k:= \frac{2k^2}{4k^4+1} = \frac{k}{2(2k^2-2k+1)} - \frac{k}{2(2k^2+2k+1)} $$
Then you could use the fact that 
$$\arccos(z) = \frac \pi 2 - \arcsin(z) $$
and the identity
$$\arcsin\left(x\sqrt{1-y^2}-y\sqrt{1-x^2}\right)=\arcsin(x) - \arcsin(y), $$
which would lead to the system
$$\begin{split}
\frac{k}{2(2k^2-2k+1)} = x_k\sqrt{1-y_k^2} + c_k, \\
\frac{k}{2(2k^2+2k+1)} = y_k\sqrt{1-x_k^2} + c_k,
\end{split}$$
However, the solution $(x_k,y_k)$ even for $c_k =0$ is horrendous and does not lead to a telescoping sum, by which I mean
$$\arcsin(y_{k-1}) = \arcsin(x_k). $$
I doubt that you could find a $c_k$ such that the sum telescopes. If you managed to find such a $c_k$, then
$$\sum_{k=1}^n \arccos(z_k) = \frac{\pi n }{2} - \sum_{k=1}^n \arcsin(z_k) = \frac{\pi n}{2} - \left[\arcsin\left(\frac 2 5\right) + \arcsin\left(\frac{2n^2}{4n^4+1} \right) \right]. $$

EDIT. (Here I follow @bjorn93's suggestion in the comments, for which I am thankful.) If, on the other hand, you had 
$$z_k = \frac{2k^2}{\sqrt{4k^2+1}}, $$
then, by using the identity
$$\tan(\arccos(z)) = \frac{\sqrt{1-z^2}}{z} $$
you get
$$\begin{split}
\arccos(z_k) &= \arctan( \tan \arccos(z_k)) = \arctan\Bigg(\frac{\sqrt{1-z_k^2}}{z_k} \Bigg)\\ &= \arctan \left( \frac{\sqrt{4k^4 +1}}{2k^2} \sqrt{1-\frac{4k^4}{4k^4+1}}\right)  \\
&= \arctan\left( \frac{1}{2k^2} \right)
\end{split}$$
You can check that, by the arctangent identity
$$\arctan(x) - \arctan(y) = \arctan \left( \frac{x-y}{1+xy} \right),$$
we can split, for $k \geq 1$,
$$\arctan\left( \frac 1 {2k^2}\right) = \arctan\left( \frac{k}{k+1}\right) - \arctan\left( \frac{k-1}k\right). $$
(To find this decomposition, you may solve the recurrence relation $1+x_kx_{k-1} = 2k^2 (x_k - x_{k-1})$ with the initial condition $x_0 = 0$.) Hence the sum telescopes:
$$\sum_{k=1}^n \arccos\left( \frac{2k^2}{\sqrt{4k^4+1}} \right) = \sum_{k=1}^n \arctan\left( \frac{1}{2k^2} \right) = \arctan\left( \frac{n}{n+1} \right). $$
</s>