Explanation

(a)
Given series can be rewritten as
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sum_{r = 1}^{\infty}{\tan^{- 1}\left( \frac{1}{1 + r + r^{2}} \right)}$
Now, $\tan^{- 1}\left( \frac{1}{1 + r + r^{2}} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}\left( \frac{r + 1 - r}{1 + r(r + 1)} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}{(r + 1) - \tan^{- 1}{(r)}}$
$\therefore\ \sum_{r = 1}^{n}{\lbrack\tan^{- 1}{(r + 1) - \tan^{- 1}r}\rbrack}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}{(n + 1) - \tan^{- 1}{(1)}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}{(n + 1) - \frac{\pi}{4}}$
$\Rightarrow \ \ \sum_{r = 1}^{\infty}{\tan^{- 1}{\left( \frac{1}{1 + r + r^{2}} \right) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}}}$