Question

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

• A. $2$
• B. $6$
• C. $1$
• D. $3$

Answer 1

The correct option is D $3$

$\underset{n\to \infty }{\text{lim}}6\tan \left\{\sum _{r=1}^n{\tan }^{-1}\left(\frac{1}{r^2+3r+3}\right)\right\}$

$=\underset{n\to \infty }{\text{lim}}6\tan \left\{\sum _{r=1}^n{\tan }^{-1}\left(\frac{(r+2)-(r+1)}{1+(r+2)(r+1)}\right)\right\}$

$=\underset{n\to \infty }{\text{lim}}6\tan \left\{\sum _{r=1}^n\right({\tan }^{-1}(r+2)-{\tan }^1(r+1)\left)\right\}$

$=\underset{n\to \infty }{\text{lim}}6\tan \left\{{\tan }^{-1}\right(n+2)-{\tan }^{-1}2\}$

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

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