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

The Correct Option is C

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