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

Question

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

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.

Answer 1

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, $$\dfrac{8n}{n^4-2n^2+5}=\dfrac{\dfrac{(n+1)^2}2-\dfrac{(n-1)^2}2}{\dfrac{(n+1)^2}2\cdot\dfrac{(n-1)^2}2+1}$$

Answer 2

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