Question

The value of
\lim_{{n \to \infty}} 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)

3

Answer 2

The correct answer is (C) : 3
=\lim_{{n \to \infty}} 6\tan\left\\{\sum_{{r=1}}^{n} \tan^{-1}\left(\frac{1}{{r^2+3r+3}}\right)\right\\}
=\lim_{{n \to \infty}} 6\tan\left\\{\sum_{{r=1}}^{n} \tan^{-1}\left(\frac{{(r+2) - (r+1)}}{{1 + (r+2)(r+1)}}\right)\right\\}
=\lim_{{n \to \infty}} 6\tan\left\\{\sum_{{r=1}}^{n} (\tan^{-1}(r+2) - \tan^{-1}(r+1))\right\\}
$\lim_{{n \to \infty}} 6\tan\left(\tan^{-1}(n+2) - \tan^{-1}(2)\right)$
=6\tan\left\\{\frac{\pi}{2} - \cot^{-1}\left(\frac{1}{2}\right)\right\\}
$=6\tan\left(\tan^{-1}\left(\frac{1}{2}\right)\right) = 3$