The value of $\underset{\mathrm{n}\to \mathrm{\infty }}{lim} 6\tan \left\{\sum _{\mathrm{r}=1}^{\mathrm{n}} {\tan }^{-1}\left(\frac{1}{{\mathrm{r}}^2+3\mathrm{r}+3}\right)\right\}$ is equal to :

$\begin{array}{l}{\mathrm{T}}_{\mathrm{r}}={\tan }^{-1}\left[\frac{(\mathrm{r}+2)-(\mathrm{r}+1)}{1+(\mathrm{r}+2)(\mathrm{r}+1)}\right] \\ \left(\sin ce {\tan }^{-1}\left(\frac{x-y}{1+xy}\right)={\tan }^{-1}x-{\tan }^{-1}y\right)\\ ={\tan }^{-1}(\mathrm{r}+2)-{\tan }^{-1}(\mathrm{r}+1)\\ {\mathrm{T}}_1={\tan }^{-1}3-{\tan }^{-1}2\\ {\mathrm{T}}_2={\tan }^{-1}4-{\tan }^{-1}3\\ .......\\ {\mathrm{T}}_{\mathrm{n}}={\tan }^{-1}(\mathrm{n}+2)-{\tan }^{-1}(\mathrm{n}+1)\end{array}$

$\overline{{\mathrm{S}}_{\mathrm{n}}={\tan }^{-1}(\mathrm{n}+2)-{\tan }^{-1}2}={\tan }^{-1}\left(\frac{\mathrm{n}+2-2}{1+2(\mathrm{n}+2)}\right)={\tan }^{-1}\left(\frac{\mathrm{n}}{2\mathrm{n}+5}\right)$

$\begin{array}{l}\underset{\mathrm{n}\to \mathrm{\infty }}{lim} 6\tan \left({\tan }^{-1}\left(\frac{\mathrm{n}}{2\mathrm{n}+5}\right)\right)\\ =\underset{\mathrm{n}\to \mathrm{\infty }}{lim} \frac{6\mathrm{n}}{2\mathrm{n}+5}=\frac{6}{2}=3\end{array}$