Question

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

• A. 1
• B. 2
• C. 3
• D. 6

Answer 1

Answer: (C)

$T_r={\tan }^{-1}\left[\frac{(r+2)-(r+1)}{1+(r+2)(r+1)}\right]$
$={\tan }^{-1}(r+2)-{\tan }^{-1}(r+1)$
$T_1={\tan }^{-1}3-{\tan }^{-1}2$
$T_2={\tan }^{-1}4-{\tan }^{-1}3$
$T_n={\tan }^{-1}(n+2)-{\tan }^{-1}(n+1)$
$\overline{S_n={\tan }^{-1}(n+2)-{\tan }^{-1}2}={\tan }^{-1}\left(\frac{n+2-2}{1+2(n+2)}\right)$
$={\tan }^{-1}\left(\frac{n}{2n+5}\right)$
$\underset{n\to \infty }{\lim }6\tan \left({\tan }^{-1}\left(\frac{n}{2n+5}\right)\right)$
$=\underset{n\to \infty }{\lim }\frac{6n}{2n+5}=\frac{6}{2}=3$