The value of $\displaystyle \lim_{n \to \infty} 6 \tan \left\{ \sum_{r=1}^{n} \tan^{-1} \left( \frac{1}{r^2 + 3r + 3} \right) \right\}$ is equal to:
\( 1 \)
\( 2 \)
\( 3 \)
\( 6 \)