Question

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

• A. 1
• B. 2
• C. 3
• D. 6

Answer 1

Answer: (C)

$T _{ r }=\tan ^{-1}\left[\frac{( r +2)-( r +1)}{1+( r +2)( r +1)}\right]$
$=\tan ^{-1}( r +2)-\tan ^{-1}( r +1)$
$T _{1}=\tan ^{-1} 3-\tan ^{-1} 2$
$T _{2}=\tan ^{-1} 4-\tan ^{-1} 3$
$T_{n}=\tan ^{-1}(n+2)-\tan ^{-1}(n+1)$
$\overline{S_{n}=\tan ^{-1}(n+2)-\tan ^{-1} 2}=\tan ^{-1}\left(\frac{n+2-2}{1+2(n+2)}\right)$
$=\tan ^{-1}\left(\frac{n}{2 n+5}\right)$
$\displaystyle\lim _{n \rightarrow \infty} 6 \tan \left(\tan ^{-1}\left(\frac{n}{2 n+5}\right)\right)$
$=\displaystyle\lim _{n \rightarrow \infty} \frac{6 n}{2 n+5}=\frac{6}{2}=3$