I know that:
$$\sum_{k=1}^n\arctan(2k^2)=\frac{\pi n}{2}-\frac{1}{2}\arctan(\frac{2n(n+1)}{2n+1})$$
Can a similar closed form expressions be given for $\sum_{k=1}^n \arctan(k^2)$?
I was able to simplify it to:
$$\sum_{k=1}^n\arctan(k^2)=\frac{\pi n}{2}-\arctan(\frac{2n(n+1)}{2n+1})+\sum_{k=1}^n\arctan(\frac{1}{4k^6+3k^2})$$
But I can't simplify the sum on the right hand side.