$\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}}$ $$\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}}$$
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.

Hint:
$$\dfrac2{n^2+n+4}=\dfrac{\dfrac12}{1+\dfrac{n(n+1)}4}=\dfrac{\dfrac{n+1}2-\dfrac n2}{1+\dfrac n2\cdot\dfrac{n+1}2}$$
For the second, $$\dfrac{8n}{n^4-2n^2+5}=\dfrac{\dfrac{(n+1)^2}2-\dfrac{(n-1)^2}2}{\dfrac{(n+1)^2}2\cdot\dfrac{(n-1)^2}2+1}$$ 
