The value of limlimits_{n rightarrow infty} 6 | vedclass

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(C) Let $T_r = 	an^{-1}\left(\frac{1}{r^2+3r+3}ight)$.We can rewrite the argument as $\frac{(r+2)-(r+1)}{1+(r+2)(r+1)}$.Thus,$T_r = 	an^{-1}(r+2)

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$3$

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(C) Let $T_r = 	an^{-1}\left(\frac{1}{r^2+3r+3}ight)$.We can rewrite the argument as $\frac{(r+2)-(r+1)}{1+(r+2)(r+1)}$.Thus,$T_r = 	an^{-1}(r+2) - 	an^{-1}(r+1)$.The sum $S_n = \sum_{r=1}^{n} T_r$ is a telescoping sum:$S_n = (	an^{-1}3 - 	an^{-1}2) + (	an^{-1}4 - 	an^{-1}3) + \dots + (	an^{-1}(n+2) - 	an^{-1}(n+1))$.$S_n = 	an^{-1}(n+2) - 	an^{-1}2$.Using the formula $	an^{-1}x - 	an^{-1}y = 	an^{-1}\left(\frac{x-y}{1+xy}ight)$:$S_n = 	an^{-1}\left(\frac{(n+2)-2}{1+(n+2)(2)}ight) = 	an^{-1}\left(\frac{n}{2n+5}ight)$.Now,we evaluate the limit:$\lim\limits_{n ightarrow \infty} 6 	an(S_n) = \lim\limits_{n ightarrow \infty} 6 	an\left(	an^{-1}\left(\frac{n}{2n+5}ight)ight)$.$= \lim\limits_{n ightarrow \infty} \frac{6n}{2n+5} = \frac{6}{2} = 3$.

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

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