$\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}}$ $$\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}}$$
We know that : $\arctan{x} - \arctan{y} = \arctan{\frac{x-y}{1+xy}}$ for every $ xy > 1 $
I need to find two numbers which satisfy: $ab = n^2+2n+3 $ and $ a-b =2$ in order to telescope.
Edit: I am very sorry on the denominator it should have been $n^2+n+4$ not $n^2+2n+4$
Similarly here :
$$\sum\limits_{n=1}^{\infty}\arctan{\frac{8n}{n^4-2n^2+5}}$$
The result should be $ \arctan 2 $ on the first one and $ \pi/2 + \arctan2  $ on the second one.

$$\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}} = \sum\limits_{n=1}^{\infty}\arctan{\frac{2(n+1)-2n}{n(n+1)(1+\frac{4}{n(n+1)})}} =
\sum\limits_{n=1}^{\infty}\arctan{\frac{\frac{2}{n}-\frac{2}{n+1}}{1+\frac{2}
{n}\cdot\frac{2}{n+1}}} =\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n}-\arctan{\frac{2}{n+1}}} =\arctan2$$
