Question
The value of
$\mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {{r^2} + 3r + 3}}} \right)} } \right\}$ is equal to :
The value of
$\mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {{r^2} + 3r + 3}}} \right)} } \right\}$ is equal to :