The value of _ n → ∞ 6tan \ ∑ _ r = 1 ^ n ^ - 1 ( 1 r ^ 2 + 3r + 3...

Let $S = \mathop{∑}_{r = 1}^{n}\tan^{- 1}\left(\dfrac{1}{{r}^{2} + 3r + 3}\right)$

So, ${T}_{r} = \tan^{- 1}\dfrac{1}{{r}^{2} + 3r + 2 + 1}$

$= \tan^{- 1}\left(\dfrac{( r + 2 ) - ( r + 1 )}{1 + ( r + 1 ) ( r + 2 )}\right)$

$= \tan^{- 1}\left(r + 2\right) - \tan^{- 1}\left(r + 1\right)$

${T}_{1} = \tan^{- 1}3 - \tan^{- 1}2$

${T}_{2} = \tan^{- 1}4 - \tan^{- 1}3 . .$
${T}_{n} = \tan^{- 1}\left(n + 2\right) - \tan^{- 1}\left(n + 1\right)$

$⇒ \mathop{∑}_{r = 1}^{n}\tan^{- 1}\dfrac{1}{{r}^{2} + 3r + 3} = \tan^{- 1}\left(n + 2\right) - \tan^{- 1}2$

i.e. $6\tan \left(\lim_{n \to \infty}\mathop{∑}_{r = 1}^{n}\tan^{- 1}\dfrac{1}{{r}^{2} + 3r + 3}\right) = 6\tan \left(\lim_{n \to \infty}\mathop{∑}_{r = 1}^{n}\left[\tan^{- 1}\left(n + 2\right) - \tan^{- 1}2\right]\right)$

$= 6\tan \left(\dfrac{\pi}{2} - \tan^{- 1}2\right)$

$= 6\tan \left(\tan^{- 1}\dfrac{1}{2}\right) = 3$